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Let \( k \) be a field and let \( k\left( \left( x\right) \right) \) be the set of all formal generalized power series in \( x \) with coefficients in \( k \) ; that is, the elements of \( k\left( \left( x\right) \right) \) are formal infinite sums \( \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \) with \( {n}_{0} \in \mathbb{Z} \) and each \( {a}_{n} \in k \) . We define addition and multiplication on \( k\left( \left( x\right) \right) \) by\n\n\[ \mathop{\sum }\limits_{n}^{\infty }{a}_{n}{x}^{n} + \mathop{\sum }\limits_{n}^{\infty }{b}_{n}{x}^{n} = \mathop{\sum }\limits_{n}^{\infty }\left( {{a}_{n} + {b}_{n}}\right) {x}^{n} \]\n\nand\n\n\[ \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \cdot \mathop{\sum }\limits_{{n = {n}_{1}}}^{\infty }{b}_{n}{x}^{n} = \mathop{\sum }\limits_{{n = {n}_{0} + {n}_{1}}}^{\infty }\left( {\mathop{\sum }\limits_{{k = {n}_{0}}}^{{n - {n}_{1}}}{a}_{n}{b}_{n - k}}\right) {x}^{n}. \] | A straightforward calculation shows that \( k\left( \left( x\right) \right) \) is a commutative ring with identity. Moreover, we can show that \( k\left( \left( x\right) \right) \) is a field. If \( f = \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \) is a nonzero element of \( k\left( \left( x\right) \right) \), we need to produce an inverse for \( f \) . Suppose that we have written the series so that \( {a}_{{n}_{0}} \) is the first nonzero coefficient. By multiplying by \( {a}_{{n}_{0}}^{-1}{x}^{-{n}_{0}} \), to find an inverse for \( f \) it suffices to assume that \( {n}_{0} = 0 \) and \( {a}_{{n}_{0}} = 1 \) . We can find the coefficients \( {b}_{n} \) of the inverse \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{b}_{n}{x}^{n} \) to \( f \) by recursion. To have \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{x}^{n} \cdot \mathop{\sum }\limits_{{n = 0}}^{\infty }{b}_{n}{x}^{n} = 1 \), we need \( {b}_{0} = 1 \) since \( {a}_{0} = 1 \) . For \( n > 0 \), the coefficient of \( {x}^{n} \) is\n\n\[ {b}_{n}{a}_{0} + {b}_{n - 1}{a}_{1} + \cdots + {b}_{0}{a}_{n} = 0 \]\n\nso if we have determined \( {b}_{0},\ldots ,{b}_{n - 1} \), then we determine \( {b}_{n} \) from the equation \( {b}_{n} = - \mathop{\sum }\limits_{{k = 1}}^{n}{b}_{n - k}{a}_{k} \) . By setting \( g \) to be the series with coefficients \( {b}_{n} \) determined by this information, | Yes |
Example 1.3 Let \( k \) be a field and let \( k\left( \left( x\right) \right) \) be the set of all formal generalized power series in \( x \) with coefficients in \( k \) ; that is, the elements of \( k\left( \left( x\right) \right) \) are formal infinite sums \( \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \) with \( {n}_{0} \in \mathbb{Z} \) and each \( {a}_{n} \in k \) . We define addition and multiplication on \( k\left( \left( x\right) \right) \) by\n\n\[ \mathop{\sum }\limits_{n}^{\infty }{a}_{n}{x}^{n} + \mathop{\sum }\limits_{n}^{\infty }{b}_{n}{x}^{n} = \mathop{\sum }\limits_{n}^{\infty }\left( {{a}_{n} + {b}_{n}}\right) {x}^{n} \]\n\nand\n\n\[ \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \cdot \mathop{\sum }\limits_{{n = {n}_{1}}}^{\infty }{b}_{n}{x}^{n} = \mathop{\sum }\limits_{{n = {n}_{0} + {n}_{1}}}^{\infty }\left( {\mathop{\sum }\limits_{{k = {n}_{0}}}^{{n - {n}_{1}}}{a}_{n}{b}_{n - k}}\right) {x}^{n}. \] | A straightforward calculation shows that \( k\left( \left( x\right) \right) \) is a commutative ring with identity. Moreover, we can show that \( k\left( \left( x\right) \right) \) is a field. If \( f = \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \) is a nonzero element of \( k\left( \left( x\right) \right) \), we need to produce an inverse for \( f \) . Suppose that we have written the series so that \( {a}_{{n}_{0}} \) is the first nonzero coefficient. By multiplying by \( {a}_{{n}_{0}}^{-1}{x}^{-{n}_{0}} \), to find an inverse for \( f \) it suffices to assume that \( {n}_{0} = 0 \) and \( {a}_{{n}_{0}} = 1 \) . We can find the coefficients \( {b}_{n} \) of the inverse \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{b}_{n}{x}^{n} \) to \( f \) by recursion. To have \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{x}^{n} \cdot \mathop{\sum }\limits_{{n = 0}}^{\infty }{b}_{n}{x}^{n} = 1 \), we need \( {b}_{0} = 1 \) since \( {a}_{0} = 1 \) . For \( n > 0 \), the coefficient of \( {x}^{n} \) is\n\n\[ {b}_{n}{a}_{0} + {b}_{n - 1}{a}_{1} + \cdots + {b}_{0}{a}_{n} = 0 \]\n\nso if we have determined \( {b}_{0},\ldots ,{b}_{n - 1} \), then we determine \( {b}_{n} \) from the equation \( {b}_{n} = - \mathop{\sum }\limits_{{k = 1}}^{n}{b}_{n - k}{a}_{k} \) . By setting \( g \) to be the series with coefficients \( {b}_{n} \) determined by this information, our computations yield \( {fg} = 1 \) . Thus, \( k\left( \left( x\right) \right) \) is a field. | Yes |
Example 1.4 The extension \( \mathbb{C}/\mathbb{R} \) is a finite extension since \( \left\lbrack {\mathbb{C} : \mathbb{R}}\right\rbrack = 2 \) . A basis for \( \mathbb{C} \) as an \( \mathbb{R} \) -vector space is \( \{ 1, i\} \) . As an extension of \( \mathbb{Q} \), both \( \mathbb{C} \) and \( \mathbb{R} \) are infinite extensions. If \( a \in \mathbb{C} \), let | \[ \mathbb{Q}\left( a\right) = \left\{ {\frac{\mathop{\sum }\limits_{i}{\alpha }_{i}{a}^{i}}{\mathop{\sum }\limits_{i}{\beta }_{i}{a}^{i}} : {\alpha }_{i},{\beta }_{i} \in \mathbb{Q},\mathop{\sum }\limits_{i}{\beta }_{i}{a}^{i} \neq 0}\right\} . \] We shall see in Proposition 1.8 that \( \mathbb{Q}\left( a\right) \) is a field extension of \( \mathbb{Q} \) . The degree of \( \mathbb{Q}\left( a\right) /\mathbb{Q} \) can be either finite or infinite depending on \( a \) . For instance, if \( a = \sqrt{-1} \) or \( a = \exp \left( {{2\pi i}/3}\right) \), then \( \left\lbrack {\mathbb{Q}\left( a\right) : \mathbb{Q}}\right\rbrack = 2 \) . These equalities are consequences of Proposition 1.15. On the other hand, we prove in Section 14 that \( \left\lbrack {\mathbb{Q}\left( \pi \right) : \mathbb{Q}}\right\rbrack = \infty \) . | Yes |
Example 1.5 If \( k \) is a field, let \( K = k\\left( t\\right) \) be the field of rational functions in \( t \) over \( k \) . If \( f \) is a nonzero element of \( K \), then we can use the construction of \( \\mathbb{Q}\\left( a\\right) \) in the previous example. Let \( F = k\\left( f\\right) \) be the set of all rational functions in \( f \) ; that is,\n\n\\[ F = \\left\\{ \\frac{\\mathop{\\sum }\\limits_{{i = 0}}^{n}{a}_{i}{f}^{i}}{\\mathop{\\sum }\\limits_{{j = 0}}^{m}{b}_{j}{f}^{j}} : {a}_{i},{b}_{j} \\in k\\text{ and }\\mathop{\\sum }\\limits_{{j = 0}}^{m}{b}_{j}{f}^{j} \\neq 0}\\right\\} .\n\\]\n\nIf \( f\\left( t\\right) = {t}^{2} \), then \( K/F \) is an extension of degree 2; a basis for \( K \) is \( \\{ 1, t\\} \) . | In Example 1.17, we shall see that \( K/F \) is a finite extension provided that \( f \) is not a constant, and in Chapter V we shall prove L\\\ | No |
Example 1.6 Let \( p\left( t\right) = {t}^{3} - 2 \in \mathbb{Q}\left\lbrack t\right\rbrack \) . Then \( p\left( t\right) \) is irreducible over \( \mathbb{Q} \) by the rational root test. Then the ideal \( \left( {p\left( t\right) }\right) \) generated by \( p\left( t\right) \) in \( \mathbb{Q}\left\lbrack t\right\rbrack \) is maximal; hence, \( K = \mathbb{Q}\left\lbrack t\right\rbrack /\left( {p\left( t\right) }\right) \) is a field. The set of cosets \( \{ a + \left( {p\left( t\right) }\right) : a \in \mathbb{Q}\} \) can be seen to be a field isomorphic to \( \mathbb{Q} \) under the map \( a \mapsto a + \left( {p\left( t\right) }\right) \) . We view the field \( \mathbb{Q}\left\lbrack t\right\rbrack /\left( {p\left( t\right) }\right) \) as an extension field of \( \mathbb{Q} \) by thinking of \( \mathbb{Q} \) as this isomorphic subfield. If \( f\left( t\right) \in \mathbb{Q}\left\lbrack t\right\rbrack \), then by the division algorithm, \( f\left( t\right) = q\left( t\right) p\left( t\right) + r\left( t\right) \) with \( r\left( t\right) = 0 \) or \( \deg \left( r\right) < \deg \left( p\right) = 3 \) . Moreover, \( f\left( t\right) \) and \( r\left( t\right) \) generate the same coset in \( \mathbb{Q}\left\lbrack t\right\rbrack /\left( {p\left( t\right) }\right) \) . What this means is that any element of \( K \) has a unique representation in the form \( a + {bt} + c{t}^{2} + \left( {p\left( t\right) }\right) \) for some \( a, b, c \in \mathbb{Q} \) . Therefore, the cosets \( 1 + \left( {p\left( t\right) }\right), t + \left( {p\left( t\right) }\right) \), and \( {t}^{2} + \left( {p\left( t\right) }\right) \) form a basis for \( K \) over \( \mathbb{Q} \), so \( \left\lbrack {K : \mathbb{Q}}\right\rbrack = 3 \) . Let \( a = t + \left( {p\left( t\right) }\right) \) . Then | \[ {a}^{3} - 2 = {t}^{3} + \left( {p\left( t\right) }\right) - (2 + \left( {p\left( t\right) }\right) = {t}^{3} - 2 + \left( {p\left( t\right) }\right) = 0. \] The element \( a \) is then a root of \( {x}^{3} - 2 \) in \( K \) . Note that we used the identification of \( \mathbb{Q} \) as a subfield in this calculation. | Yes |
Proposition 1.8 Let \( K \) be a field extension of \( F \) and let \( a \in K \) . Then\n\n\[ F\left\lbrack a\right\rbrack = \{ f\left( a\right) : f\left( x\right) \in F\left\lbrack x\right\rbrack \} \]\n\nand\n\n\[ F\left( a\right) = \{ f\left( a\right) /g\left( a\right) : f, g \in F\left\lbrack x\right\rbrack, g\left( a\right) \neq 0\} . \]\n\nMoreover, \( F\left( a\right) \) is the quotient field of \( F\left\lbrack a\right\rbrack \) . | Proof. The evaluation map \( {\operatorname{ev}}_{a} : F\left\lbrack x\right\rbrack \rightarrow K \) has image \( \{ f\left( a\right) : f \in F\left\lbrack x\right\rbrack \} \) , so this set is a subring of \( K \) . If \( R \) is a subring of \( K \) that contains \( F \) and \( a \), then \( f\left( a\right) \in R \) for any \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) by closure of addition and multiplication. Therefore, \( \{ f\left( a\right) : f\left( x\right) \in F\left\lbrack x\right\rbrack \} \) is contained in all subrings of \( K \) that contain \( F \) and \( a \) . Therefore, \( F\left\lbrack a\right\rbrack = \{ f\left( a\right) : f\left( x\right) \in F\left\lbrack x\right\rbrack \} \) . The quotient field of \( F\left\lbrack a\right\rbrack \) is then the set \( \{ f\left( a\right) /g\left( a\right) : f, g \in F\left\lbrack x\right\rbrack, g\left( a\right) \neq 0\} \) . It clearly is contained in any subfield of \( K \) that contains \( F\left\lbrack a\right\rbrack \) ; hence, it is equal to \( F\left( a\right) \) . | Yes |
Proposition 1.10 Let \( K \) be a field extension of \( F \) and let \( X \) be a subset of \( K \) . If \( \alpha \in F\left( X\right) \), then \( \alpha \in F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) for some \( {a}_{1},\ldots ,{a}_{n} \in X \) . Therefore,\n\n\[ F\left( X\right) = \bigcup \left\{ {F\left( {{a}_{1},\ldots ,{a}_{n}}\right) : {a}_{1},\ldots ,{a}_{n} \in X}\right\} \]\n\nwhere the union is over all finite subsets of \( X \) . | Proof. Each field \( F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) with the \( {a}_{i} \in X \) is contained in \( F\left( X\right) \) ; hence, \( \bigcup \left\{ {F\left( {{a}_{1},\ldots ,{a}_{n}}\right) : {a}_{i} \in X}\right\} \subseteq F\left( X\right) \) . This union contains \( F \) and \( X \), so if it is a field, then it is equal to \( F\left( X\right) \), since \( F\left( X\right) \) is the smallest subfield of \( K \) containing \( F \) and \( X \) . To show that this union is a field, let \( \alpha ,\beta \in \bigcup \left\{ {F\left( {{a}_{1},\ldots ,{a}_{n}}\right) : {a}_{i} \in X}\right\} \) . Then there are \( {a}_{i},{b}_{i} \in X \) with \( \alpha \in F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) and \( \beta \in F\left( {{b}_{1},\ldots ,{b}_{m}}\right) \) . Then both \( \alpha \) and \( \beta \) are contained in \( F\left( {{a}_{1},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{m}}\right) \), so \( \alpha \pm \beta ,{\alpha \beta } \), and \( \alpha /\beta \) (if \( \beta \neq 0) \) all lie in \( \bigcup \left\{ {F\left( {{a}_{1},\ldots ,{a}_{n}}\right) : {a}_{i} \in X}\right\} \) . This union is then a field, so \( F\left( X\right) = \bigcup \left\{ {F\left( {{a}_{1},\ldots ,{a}_{n}}\right) : {a}_{i} \in X}\right\} \) . | Yes |
Proposition 1.15 Let \( K \) be a field extension of \( F \) and let \( \alpha \in K \) be algebraic over \( F \) . 1. The polynomial \( \min \left( {F,\alpha }\right) \) is irreducible over \( F \) . 2. If \( g\left( x\right) \in F\left\lbrack x\right\rbrack \), then \( g\left( \alpha \right) = 0 \) if and only if \( \min \left( {F,\alpha }\right) \) divides \( g\left( x\right) \) . 3. If \( n = \deg \left( {\min \left( {F,\alpha }\right) }\right) \), then the elements \( 1,\alpha ,\ldots ,{\alpha }^{n - 1} \) form a basis for \( F\left( \alpha \right) \) over \( F \), so \( \left\lbrack {F\left( \alpha \right) : F}\right\rbrack = \deg \left( {\min \left( {F,\alpha }\right) }\right) < \infty \) . Moreover, \( F\left( \alpha \right) = F\left\lbrack \alpha \right\rbrack \) | Proof. If \( p\left( x\right) = \min \left( {F,\alpha }\right) \), then \( F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \cong F\left\lbrack \alpha \right\rbrack \) is an integral domain. Therefore, \( \left( {p\left( x\right) }\right) \) is a prime ideal, so \( p\left( x\right) \) is irreducible. To prove statement 2, if \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) with \( g\left( \alpha \right) = 0 \), then \( g\left( x\right) \in \ker \left( {\operatorname{ev}}_{\alpha }\right) \) . But this kernel is the ideal generated by \( p\left( x\right) \), so \( p\left( x\right) \) divides \( g\left( x\right) \) . For statement 3, we first prove that \( F\left\lbrack \alpha \right\rbrack = F\left( \alpha \right) \) . To see this, note that \( F\left\lbrack \alpha \right\rbrack \) is the image of the evaluation map \( {\mathrm{{ev}}}_{\alpha } \) . The kernel of \( {\mathrm{{ev}}}_{\alpha } \) is a prime ideal since \( {\mathrm{{ev}}}_{\alpha } \) maps \( F\left\lbrack x\right\rbrack \) into an integral domain. However, \( F\left\lbrack x\right\rbrack \) is a principal ideal domain, so every nonzero prime ideal of \( F\left\lbrack x\right\rbrack \) is maximal. Thus, \( \ker \left( {\mathrm{{ev}}}_{a}\right) \) is maximal, so \( F\left\lbrack \alpha \right\rbrack \cong F\left\lbrack x\right\rbrack /\ker \left( {\mathrm{{ev}}}_{\alpha }\right) \) is a field. Consequently, \( F\left\lbrack \alpha \right\rbrack = F\left( \alpha \right) \) . To finish the proof of statement 3, let \( n = \deg \left( {p\left( x\right) }\right) \) . If \( b \in F\left( \alpha \right) \), then \( b = g\left( \alpha \right) \) for some \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) . By the division algorithm, \( g\left( x\right) = q\left( x\right) p\left( x\right) + r\left( x\right) \), where \( r\left( x\right) = 0 \) or \( \deg \left( r\right) < n \) . Thus, \( b = g\left( \alpha \right) = r\left( \alpha \right) \) . Since \( r\left( \alpha \right) \) is an \( F \) -linear combination of \( 1,\alpha ,\ldots ,{\alpha }^{n - 1} \), we see that \( 1,\alpha ,\ldots ,{\alpha }^{n - 1} \) span \( F\left( \alpha \right) \) as an \( F \) -vector space. If \( \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}{a}_{i}{\alpha }^{i} = 0 \), then \( f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}{a}_{i}{x}^{i} \) is divisible by \( p\left( x\right) \) | Yes |
Proposition 1.15 Let \( K \) be a field extension of \( F \) and let \( \alpha \in K \) be algebraic over \( F \). 1. The polynomial \( \min \left( {F,\alpha }\right) \) is irreducible over \( F \). 2. If \( g\left( x\right) \in F\left\lbrack x\right\rbrack \), then \( g\left( \alpha \right) = 0 \) if and only if \( \min \left( {F,\alpha }\right) \) divides \( g\left( x\right) \). 3. If \( n = \deg \left( {\min \left( {F,\alpha }\right) }\right) \), then the elements \( 1,\alpha ,\ldots ,{\alpha }^{n - 1} \) form a basis for \( F\left( \alpha \right) \) over \( F \), so \( \left\lbrack {F\left( \alpha \right) : F}\right\rbrack = \deg \left( {\min \left( {F,\alpha }\right) }\right) < \infty \). Moreover, \( F\left( \alpha \right) = F\left\lbrack \alpha \right\rbrack \) | Proof. If \( p\left( x\right) = \min \left( {F,\alpha }\right) \), then \( F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \cong F\left\lbrack \alpha \right\rbrack \) is an integral domain. Therefore, \( \left( {p\left( x\right) }\right) \) is a prime ideal, so \( p\left( x\right) \) is irreducible. To prove statement 2, if \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) with \( g\left( \alpha \right) = 0 \), then \( g\left( x\right) \in \ker \left( {\operatorname{ev}}_{\alpha }\right) \). But this kernel is the ideal generated by \( p\left( x\right) \), so \( p\left( x\right) \) divides \( g\left( x\right) \). For statement 3, we first prove that \( F\left\lbrack \alpha \right\rbrack = F\left( \alpha \right) \). To see this, note that \( F\left\lbrack \alpha \right\rbrack \) is the image of the evaluation map \( {\mathrm{{ev}}}_{\alpha } \). The kernel of \( {\mathrm{{ev}}}_{\alpha } \) is a prime ideal since \( {\mathrm{{ev}}}_{\alpha } \) maps \( F\left\lbrack x\right\rbrack \) into an integral domain. However, \( F\left\lbrack x\right\rbrack \) is a principal ideal domain, so every nonzero prime ideal of \( F\left\lbrack x\right\rbrack \) is maximal. Thus, \( \ker \left( {\mathrm{{ev}}}_{a}\right) \) is maximal, so \( F\left\lbrack \alpha \right\rbrack \cong F\left\lbrack x\right\rbrack /\ker \left( {\mathrm{{ev}}}_{\alpha }\right) \) is a field. Consequently, \( F\left\lbrack \alpha \right\rbrack = F\left( \alpha \right) \). To finish the proof of statement 3, let \( n = \deg \left( {p\left( x\right) }\right) \). If \( b \in F\left( \alpha \right) \), then \( b = g\left( \alpha \right) \) for some \( g\left( x\right) \in F\left\lbrack x\right\rbrack \). By the division algorithm, \( g\left( x\right) = q\left( x\right) p\left( x\right) + r\left( x\right) \), where \( r\left( x\right) = 0 \) or \( \deg \left( r\right) < n \). Thus, \( b = g\left( \alpha \right) = r\left( \alpha \right) \). Since \( r\left( \alpha \right) \) is an \( F \) -linear combination of \( 1,\alpha ,\ldots ,{\alpha }^{n - 1} \), we see that \( 1,\alpha ,\ldots ,{\alpha }^{n - 1} \) span \( F\left( \alpha \right) \) as an \( F \) -vector space. If \( \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}{a}_{i}{\alpha }^{i} = 0 \), then \( f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}{a}_{i}{x}^{i} \) is divisible by \( p\left( x\right) \), so \( f\left( x\right) = 0 \), or else \( f \) is divisible by a polynomial of larger degree than itself. Thus, \( 1,\alpha ,\ldots ,{\alpha }^{n - 1} \) is a basis for \( F\left( \alpha \right) \) over \( F \). | Yes |
The element \( \sqrt[3]{2} \) satisfies the polynomial \( {x}^{3} - 2 \) over \( \mathbb{Q} \) , which is irreducible by the Eisenstein criterion, so \( {x}^{3} - 2 \) is the minimal polynomial of \( \sqrt[3]{2} \) over \( \mathbb{Q} \) . Thus, \( \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rbrack = 3 \) . | If \( p \) is a prime, then \( {x}^{n} - p \) is irreducible over \( \mathbb{Q} \), again by Eisenstein, so \( \left\lbrack {\mathbb{Q}\left( \sqrt[n]{p}\right) : \mathbb{Q}}\right\rbrack = n \) . | Yes |
Example 1.17 Here is a very nice, nontrivial example of a finite field extension. Let \( k \) be a field and let \( K = k\\left( t\\right) \) be the field of rational functions in \( t \) over \( k \) . Let \( u \\in K \) with \( u \\notin k \) . Write \( u = f\\left( t\\right) /g\\left( t\\right) \) with \( f, g \\in k\\left\\lbrack t\\right\\rbrack \) and \( \\gcd \\left( {f\\left( t\\right), g\\left( t\\right) }\\right) = 1 \), and let \( F = k\\left( u\\right) \) . We claim that\n\n\[ \n\\left\\lbrack {K : F}\\right\\rbrack = \\max \\{ \\deg (f\\left( t\\right) ,\\deg \\left( {g\\left( t\\right) }\\right) \\} \n\]\n\nwhich will show that \( K/F \) is a finite extension. | To see this, first note that \( K = F\\left( t\\right) \) . By using Proposition 1.15, we need to determine the minimal polynomial of \( t \) over \( F \) to determine \( \\left\\lbrack {K : F}\\right\\rbrack \) . Consider the polynomial \( p\\left( x\\right) = {ug}\\left( x\\right) - f\\left( x\\right) \\in F\\left\\lbrack x\\right\\rbrack \) . Then \( t \) is a root of \( p\\left( x\\right) \) . Therefore, \( t \) is algebraic over \( F \), and so \( \\left\\lbrack {K : F}\\right\\rbrack < \\infty \) as \( K = F\\left( t\\right) \) . Say \( f\\left( t\\right) = \\mathop{\\sum }\\limits_{{i = 0}}^{n}{a}_{i}{t}^{i} \) and \( g\\left( t\\right) = \\mathop{\\sum }\\limits_{{i = 0}}^{m}{b}_{i}{t}^{i} \) . First note that \( \\deg \\left( {p\\left( x\\right) }\\right) = \\max \\{ \\deg \\left( {f\\left( t\\right) ,\\deg \\left( {g\\left( t\\right) }\\right) \\} \) . If this were false, then the only way this could happen would be if \( m = n \) and the coefficient of \( {x}^{n} \) in \( p\\left( x\\right) \) were zero. But this coefficient is \( u{b}_{n} - {a}_{n} \) , which is nonzero since \( u \\notin k \) . We now show that \( p\\left( x\\right) \) is irreducible over \( F \) , which will verify that \( \\left\\lbrack {K : F}\\right\\rbrack = \\max \\{ n, m\\} \) . We do this by viewing \( p\\left( x\\right) \) in two ways. The element \( u \) is not algebraic over \( k \), otherwise \( \\left\\lbrack {K : k}\\right\\rbrack = \\lbrack K \) : \( F\\rbrack \\cdot \\left\\lbrack {F : k}\\right\\rbrack < \\infty \), which is false. Therefore, \( u \) is transcendental over \( k \), so \( k\\left\\lbrack u\\right\\rbrack \\cong k\\left\\lbrack x\\right\\rbrack \) . Viewing \( p \) as a polynomial in \( u \), we have \( p \\in k\\left\\lbrack x\\right\\rbrack \\left\\lbrack u\\right\\rbrack \\subseteq k\\left( x\\right) \\left\\lbrack u\\right\\rbrack \) , and \( p \) has degree 1 in \( u \) . Therefore, \( p \) is irreducible over \( k\\left( x\\right) \) . Moreover, since \( \\gcd \\left( {f\\left( t\\right), g\\left( t\\right) }\\right) = 1 \), the polynomial \( p \) is primitive in \( k\\left\\lbrack x\\right\\rbrack \\left\\lbrack u\\right\\rbrack \) . Therefore, \( p \) is irreducible over \( k\\left\\lbrack x\\right\\rbrack \) . We have \( p \\in k\\left\\lbrack u\\right\\rbrack \\left\\lbrack x\\right\\rbrack = k\\left\\lbrack x\\right\\rbrack \\left\\lbrack u\\right\\rbrack \) (think about this!), so \( p \) is irreducible over \( k\\left\\lbrack u\\right\\rbrack \), as a polynomial in \( x \) . Therefore, \( p \) is irreducible over \( k\\left( u\\right) = F \), which shows that \( p \) is the minimal polynomial of \( u \) over \( F \), by Proposition 1.15. Therefore, we have \( \\left\\lbrack {K : F}\\right\\rbrack = \\max \\{ \\deg \\left( {f\\left( t\\right) ,\\deg \\left( {g\\left( t\\right) }\\right) \\} \), as desired. | Yes |
Lemma 1.19 If \( K \) is a finite extension of \( F \), then \( K \) is algebraic and finitely generated over \( F \) . | Proof. Suppose that \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) is a basis for \( K \) over \( F \) . Then every element of \( K \) is of the form \( \mathop{\sum }\limits_{i}{a}_{i}{\alpha }_{i} \) with \( {a}_{i} \in F \), so certainly we have \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) ; thus, \( K \) is finitely generated over \( F \) . If \( a \in K \), then \( \left\{ {1, a,\ldots ,{a}^{n}}\right\} \) is dependent over \( F \), since \( \left\lbrack {K : F}\right\rbrack = n \) . Thus, there are \( {\beta }_{i} \in F \), not all zero, with \( \mathop{\sum }\limits_{i}{\beta }_{i}{a}^{i} = 0 \) . If \( f\left( x\right) = \mathop{\sum }\limits_{i}{\beta }_{i}{x}^{i} \), then \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) and \( f\left( a\right) = 0 \) . Therefore, \( a \) is algebraic over \( F \), and so \( K \) is algebraic over \( F \) . | Yes |
Lemma 1.19 If \( K \) is a finite extension of \( F \), then \( K \) is algebraic and finitely generated over \( F \) . | Proof. Suppose that \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) is a basis for \( K \) over \( F \) . Then every element of \( K \) is of the form \( \mathop{\sum }\limits_{i}{a}_{i}{\alpha }_{i} \) with \( {a}_{i} \in F \), so certainly we have \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) ; thus, \( K \) is finitely generated over \( F \) . If \( a \in K \), then \( \left\{ {1, a,\ldots ,{a}^{n}}\right\} \) is dependent over \( F \), since \( \left\lbrack {K : F}\right\rbrack = n \) . Thus, there are \( {\beta }_{i} \in F \), not all zero, with \( \mathop{\sum }\limits_{i}{\beta }_{i}{a}^{i} = 0 \) . If \( f\left( x\right) = \mathop{\sum }\limits_{i}{\beta }_{i}{x}^{i} \), then \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) and \( f\left( a\right) = 0 \) . Therefore, \( a \) is algebraic over \( F \), and so \( K \) is algebraic over \( F \) . | Yes |
Proposition 1.20 Let \( F \subseteq L \subseteq K \) be fields. Then \[ \left\lbrack {K : F}\right\rbrack = \left\lbrack {K : L}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack \] | Proof. Let \( \left\{ {{a}_{i} : i \in I}\right\} \) be a basis for \( L/F \), and let \( \left\{ {{b}_{j} : j \in J}\right\} \) be a basis for \( K/L \) . Consider the set \( \left\{ {{a}_{i}{b}_{j} : i \in I, j \in J}\right\} \) . We will show that this set is a basis for \( K/F \) . If \( x \in K \), then \( x = \mathop{\sum }\limits_{j}{\alpha }_{j}{b}_{j} \) for some \( {\alpha }_{j} \in L \) , with only finitely many of the \( {b}_{j} \neq 0 \) . But \( {\alpha }_{j} = \mathop{\sum }\limits_{i}{\beta }_{ij}{a}_{j} \) for some \( {\beta }_{ij} \in F \) , with only finitely many \( {\beta }_{ij} \) nonzero for each \( j \) . Thus, \( x = \mathop{\sum }\limits_{{i, j}}{\beta }_{ij}{a}_{i}{b}_{j} \) , so the \( \left\{ {{a}_{i}{b}_{j}}\right\} \) span \( K \) as an \( F \) -vector space. For linear independence, if \( \mathop{\sum }\limits_{{i, j}}{\beta }_{ij}{a}_{i}{b}_{j} = 0 \) with \( {\beta }_{ij} \in F \), then the independence of the \( {b}_{j} \) over \( L \) shows that \( \mathop{\sum }\limits_{i}{\beta }_{ij}{a}_{i} = 0 \) for each \( j \) . But independence of the \( {a}_{i} \) over \( F \) gives \( {\beta }_{ij} = 0 \) for each \( i, j \) . Thus, the \( {a}_{i}{b}_{j} \) are independent over \( F \), so they form a basis for \( K/F \) . Therefore, \[ \left\lbrack {K : F}\right\rbrack = \left| \left\{ {{a}_{i}{b}_{j} : i \in I, j \in J}\right\} \right| \] \[ = \left| \left\{ {{a}_{i} : i \in I}\right\} \right| \cdot \left| \left\{ {{b}_{j} : j \in J}\right\} \right| = \left\lbrack {K : L}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack . \] | Yes |
Proposition 1.21 Let \( K \) be a field extension of \( F \) . If each \( {\alpha }_{i} \in K \) is algebraic over \( F \), then \( F\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\rbrack \) is a finite dimensional field extension of \( F \) with\n\n\[ \left\lbrack {F\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\rbrack : F}\right\rbrack \leq \mathop{\prod }\limits_{{i = 1}}^{n}\left\lbrack {F\left( {\alpha }_{i}\right) : F}\right\rbrack \] | Proof. We prove this by induction on \( n \) ; the case \( n = 1 \) follows from Proposition 1.15. If we set \( L = F\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n - 1}}\right\rbrack \), then by induction \( L \) is a field and \( \left\lbrack {L : F}\right\rbrack \leq \mathop{\prod }\limits_{{i = 1}}^{{n - 1}}\left\lbrack {F\left( {\alpha }_{i}\right) : F}\right\rbrack \) . Then \( F\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\rbrack = L\left\lbrack {\alpha }_{n}\right\rbrack \) is a field since \( {\alpha }_{n} \) is algebraic over \( L \), and since \( \min \left( {L,{\alpha }_{n}}\right) \) divides \( \min \left( {F,{\alpha }_{n}}\right) \) by Proposition 1.15, we have \( \left\lbrack {F\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\rbrack : L}\right\rbrack \leq \left\lbrack {F\left( {\alpha }_{n}\right) : F}\right\rbrack \) . Hence, by Proposition 1.20 and the induction hypothesis,\n\n\[ \left\lbrack {F\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\rbrack : F}\right\rbrack = \left\lbrack {F\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\rbrack : L}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack \leq \mathop{\prod }\limits_{{i = 1}}^{n}\left\lbrack {F\left( {\alpha }_{i}\right) : F}\right\rbrack . \]\n\nThis finishes the proof. | Yes |
Proposition 1.23 Let \( K \) be a field extension of \( F \), and let \( X \) be a subset of \( K \) such that each element of \( X \) is algebraic over \( F \) . Then \( F\\left( X\\right) \) is algebraic over \( F \) . If \( \\left| X\\right| < \\infty \), then \( \\left\\lbrack {F\\left( X\\right) : F}\\right\\rbrack < \\infty \) . | Proof. Let \( a \in F\\left( X\\right) \) . By Proposition 1.10, there are \( {\\alpha }_{1},\\ldots ,{\\alpha }_{n} \in X \) with \( a \in F\\left( {{\\alpha }_{1},\\ldots ,{\\alpha }_{n}}\\right) \) . By Proposition 1.21, \( F\\left( {{\\alpha }_{1},\\ldots ,{\\alpha }_{n}}\\right) \) is algebraic over \( F \) . Thus, \( a \) is algebraic over \( F \) and, hence, \( F\\left( X\\right) \) is algebraic over \( F \) . If \( \\left| X\\right| < \\infty \), then \( \\left\\lbrack {F\\left( X\\right) : F}\\right\\rbrack < \\infty \) by Proposition 1.21. | Yes |
Theorem 1.24 Let \( F \subseteq L \subseteq K \) be fields. If \( L/F \) and \( K/L \) are algebraic, then \( K/F \) is algebraic. | Proof. Let \( \alpha \in K \), and let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {x}^{n} \) be the minimal polynomial of \( \alpha \) over \( L \) . Since \( L/F \) is algebraic, the field \( {L}_{0} = F\left( {{a}_{0},\ldots ,{a}_{n - 1}}\right) \) is a finite extension of \( F \) by Corollary 1.22. Now \( f\left( x\right) \in {L}_{0}\left\lbrack x\right\rbrack \), so \( \alpha \) is algebraic over \( {L}_{0} \) . Thus,\n\n\[ \left\lbrack {{L}_{0}\left( \alpha \right) : F}\right\rbrack = \left\lbrack {{L}_{0}\left( \alpha \right) : {L}_{0}}\right\rbrack \cdot \left\lbrack {{L}_{0} : F}\right\rbrack < \infty . \]\n\nBecause \( F\left( \alpha \right) \subseteq {L}_{0}\left( \alpha \right) \), we see that \( \left\lbrack {F\left( \alpha \right) : F}\right\rbrack < \infty \), so \( \alpha \) is algebraic over \( F \) . Since this is true for all \( \alpha \in K \), we have shown that \( K/F \) is algebraic. | Yes |
Corollary 1.26 Let \( K \) be a field extension of \( F \), and let \( L \) be the algebraic closure of \( F \) in \( K \). Then \( L \) is a field, and therefore is the largest algebraic extension of \( F \) contained in \( K \). | Proof. Let \( a, b \in L \). Then \( F\left( {a, b}\right) \) is algebraic over \( F \) by Proposition 1.23, so \( F\left( {a, b}\right) \subseteq L \), and since \( a \pm b,{ab}, a/b \in F\left( {a, b}\right) \subseteq L \), the set \( L \) is closed under the field operations, so it is a subfield of \( K \). Each element of \( K \) that is algebraic over \( F \) lies in \( L \), which means that \( L \) is the largest algebraic extension of \( F \) contained in \( K \). | Yes |
Example 1.27 Let \( F = \mathbb{Q} \), and view all fields in this example as subfields of \( \mathbb{C} \) . Let \( \omega = {e}^{{2\pi i}/3} \), so \( {\omega }^{3} = 1 \) and \( \omega \neq 1 \) . The composite of \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) and \( \mathbb{Q}\left( {\omega \sqrt[3]{2}}\right) \) is \( \mathbb{Q}\left( {\omega ,\sqrt[3]{2}}\right) \) . | To see that this is the composite, note that both \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) and \( \mathbb{Q}\left( {\omega \sqrt[3]{2}}\right) \) are contained in \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \), so their composite is also contained in \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) . However, if a field \( L \) contains \( \sqrt[3]{2} \) and \( \omega \sqrt[3]{2} \) , then it also contains \( \omega = \omega \sqrt[3]{2}/\sqrt[3]{2} \) . Thus, \( L \) must contain \( \sqrt[3]{2} \) and \( \omega \), so it must contain \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) . Therefore, \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) is the smallest field containing both \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) and \( \mathbb{Q}\left( {\omega \sqrt[3]{2}}\right) \) . We can also show that \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) = \mathbb{Q}\left( {\sqrt[3]{2} + \omega }\right) \) , so \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) is generated by one element over \( \mathbb{Q} \) . If \( a = \omega + \sqrt[3]{2} \), then \( {\left( a - \omega \right) }^{3} = 2 \) . Expanding this and using the relation \( {\omega }^{2} = - 1 - \omega \), solving for \( \omega \) yields\n\n\[ \omega = \frac{{a}^{3} - {3a} - 3}{3{a}^{2} + {3a}} \]\n\nso \( \omega \in \mathbb{Q}\left( a\right) \) . Thus, \( \sqrt{2} = a - \omega \in \mathbb{Q}\left( a\right) \), so \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) = \mathbb{Q}\left( {\sqrt[3]{2} + \omega }\right) \) . | Yes |
Example 1.28 The composite of \( \mathbb{Q}\left( \sqrt{2}\right) \) and \( \mathbb{Q}\left( \sqrt{3}\right) \) is the field \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \) . This composite can be generated by a single element over \( \mathbb{Q} \) . In fact, \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) = \mathbb{Q}\left( {\sqrt{2} + \sqrt{3}}\right) \) . | To see this, the inclusion \( \supseteq \) is clear. For the reverse inclusion, let \( a = \sqrt{2} + \sqrt{3} \) . Then \( {\left( a - \sqrt{2}\right) }^{2} = 3 \) . Multiplying this and rearranging gives \( 2\sqrt{2}a = {a}^{2} - 1 \), so\n\n\[ \sqrt{2} = \frac{{a}^{2} - 1}{2a} \in \mathbb{Q}\left( a\right) \]\n\nSimilar calculations show that\n\n\[ \sqrt{3} = \frac{\left( {a}^{2} + 1\right) }{2a} \in \mathbb{Q}\left( a\right) \]\n\nTherefore, \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \subseteq \mathbb{Q}\left( a\right) \), which, together with the previous inclusion, gives \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) = \mathbb{Q}\left( a\right) \) . | Yes |
Lemma 2.2 Let \( K = F\left( X\right) \) be a field extension of \( F \) that is generated by a subset \( X \) of \( K \) . If \( \sigma ,\tau \in \operatorname{Gal}\left( {K/F}\right) \) with \( {\left. {\left. \sigma \right| }_{X} = \tau \right| }_{X} \), then \( \sigma = \tau \) . Therefore, \( F \) -automorphisms of \( K \) are determined by their action on a generating set. | Proof. Let \( a \in K \) . Then there is a finite subset \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \subseteq X \) with \( a \in F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) . This means there are polynomials \( f, g \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) with \( a = f\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) /g\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) ; say\n\n\[ f\left( {{x}_{1},\ldots ,{x}_{n}}\right) = \sum {b}_{{i}_{1}{i}_{2}\cdots {i}_{n}}{x}_{1}^{{i}_{1}}{x}_{2}^{{i}_{2}}\cdots {x}_{n}^{{i}_{n}}, \]\n\n\[ g\left( {{x}_{1},\ldots ,{x}_{n}}\right) = \sum {c}_{{i}_{1}{i}_{2}\cdots {i}_{n}}{x}_{1}^{{i}_{1}}{x}_{2}^{{i}_{2}}\cdots {x}_{n}^{{i}_{n}}, \]\n\nwhere each coefficient is in \( F \) . Since \( \sigma \) and \( \tau \) preserve addition and multiplication, and fix elements of \( F \), we have\n\n\[ \sigma \left( a\right) = \sum \frac{{b}_{{i}_{1}{i}_{2}\cdots {i}_{n}}\sigma {\left( {\alpha }_{1}\right) }^{{i}_{1}}\sigma {\left( {\alpha }_{2}\right) }^{{i}_{2}}\cdots \sigma {\left( {\alpha }_{n}\right) }^{{i}_{n}}}{{c}_{{i}_{1}{i}_{2}\cdots {i}_{n}}\sigma {\left( {\alpha }_{1}\right) }^{{i}_{1}}\sigma {\left( {\alpha }_{2}\right) }^{{i}_{2}}\cdots \sigma {\left( {\alpha }_{n}\right) }^{{i}_{n}}} \]\n\n\[ = \sum \frac{{b}_{{i}_{1}{i}_{2}\cdots {i}_{n}}\tau {\left( {\alpha }_{1}\right) }^{{i}_{1}}\tau {\left( {\alpha }_{2}\right) }^{{i}_{2}}\cdots \tau {\left( {\alpha }_{n}\right) }^{{i}_{n}}}{{c}_{{i}_{1}{i}_{2}\cdots {i}_{n}}\tau {\left( {\alpha }_{1}\right) }^{{i}_{1}}\tau {\left( {\alpha }_{2}\right) }^{{i}_{2}}\cdots \tau {\left( {\alpha }_{n}\right) }^{{i}_{n}}} \]\n\n\[ = \tau \left( a\right) \text{.} \]\n\nThus, \( \sigma = \tau \), so \( F \) -automorphisms are determined by their action on generators. | Yes |
Lemma 2.3 Let \( \tau : K \rightarrow L \) be an \( F \) -homomorphism and let \( \alpha \in K \) be algebraic over \( F \) . If \( f\left( x\right) \) is a polynomial over \( F \) with \( f\left( \alpha \right) = 0 \) , then \( f\left( {\tau \left( \alpha \right) }\right) = 0 \) . Therefore, \( \tau \) permutes the roots of \( \min \left( {F,\alpha }\right) \) . Also, \( \min \left( {F,\alpha }\right) = \min \left( {F,\tau \left( \alpha \right) }\right) \) . | Proof. Let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \) . Then\n\n\[ 0 = \tau \left( 0\right) = \tau \left( {f\left( \alpha \right) }\right) = \mathop{\sum }\limits_{i}\tau \left( {a}_{i}\right) \tau {\left( \alpha \right) }^{i}. \]\n\nBut, since each \( {a}_{i} \in F \), we have \( \tau \left( {a}_{i}\right) = {a}_{i} \) . Thus, \( 0 = \mathop{\sum }\limits_{i}{a}_{i}\tau {\left( \alpha \right) }^{i} \), so \( f\left( {\tau \left( \alpha \right) }\right) = 0 \) . In particular, if \( p\left( x\right) = \min \left( {F,\alpha }\right) \), then \( p\left( \widetilde{\tau \left( \alpha \right) }\right) = 0 \), so \( \min \left( {F,\tau \left( \alpha \right) }\right) \) divides \( p\left( x\right) \) . Since \( p\left( x\right) \) is irreducible, \( \min \left( {F,\tau \left( \alpha \right) }\right) = p\left( x\right) = \min \left( {F,\alpha }\right) \) . | Yes |
Corollary 2.4 If \( \left\lbrack {K : F}\right\rbrack < \infty \), then \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| < \infty \) . | Proof. We can write \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) for some \( {\alpha }_{i} \in K \) . Any \( F \) - automorphism of \( K \) is determined by what it does to the \( {\alpha }_{i} \) . By Lemma 2.3, there are only finitely many possibilities for the image of any \( {\alpha }_{i} \) ; hence, there are only finitely many automorphisms of \( K/F \) . | Yes |
Consider the extension \( \mathbb{C}/\mathbb{R} \). We claim that \( \operatorname{Gal}\left( {\mathbb{C}/\mathbb{R}}\right) = \) \( \{ \mathrm{{id}},\sigma \} \), where \( \sigma \) is complex conjugation. | Both of these functions are \( \mathbb{R} \) - automorphisms of \( \mathbb{C} \), so they are contained in \( \operatorname{Gal}\left( {\mathbb{C}/\mathbb{R}}\right) \). To see that there is no other automorphism of \( \mathbb{C}/\mathbb{R} \), note that an element of \( \operatorname{Gal}\left( {\mathbb{C}/\mathbb{R}}\right) \) is determined by its action on \( i \), since \( \mathbb{C} = \mathbb{R}\left( i\right) \). Lemma 2.3 shows that if \( \tau \in \operatorname{Gal}\left( {\mathbb{C}/\mathbb{R}}\right) \), then \( \tau \left( i\right) \) is a root of \( {x}^{2} + 1 \), so \( \tau \left( i\right) \) must be either \( i \) or \( - i \). Therefore, \( \tau = \) id or \( \tau = \sigma \). | Yes |
The Galois group of \( \mathbb{Q}\left( \sqrt[3]{2}\right) /\mathbb{Q} \) is \( \langle \mathrm{{id}}\rangle \). | To see this, if \( \sigma \) is a \( \mathbb{Q} \) -automorphism of \( \mathbb{Q}\left( \sqrt[3]{2}\right) \), then \( \sigma \left( \sqrt[3]{2}\right) \) is a root of \( \min \left( {\mathbb{Q},\sqrt[3]{2}}\right) = {x}^{3} - 2 \). If \( \omega = {e}^{{2\pi i}/3} \), then the roots of this polynomial are \( \sqrt[3]{2},\omega \sqrt[3]{2} \), and \( {\omega }^{2}\sqrt[3]{2} \). The only root of \( {x}^{3} - 2 \) that lies in \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) is \( \sqrt[3]{2} \), since if another root lies in this field, then \( \omega \in \mathbb{Q}\left( \sqrt[3]{2}\right) \), which is false since \( \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rbrack = 3 \) and \( \left\lbrack {\mathbb{Q}\left( \omega \right) : \mathbb{Q}}\right\rbrack = 2 \). Therefore, \( \sigma \left( \sqrt[3]{2}\right) = \sqrt[3]{2} \), and since \( \sigma \) is determined by its action on the generator \( \sqrt[3]{2} \), we see that \( \sigma = \mathrm{{id}} \). | Yes |
Example 2.7 Let \( K = {\mathbb{F}}_{2}\left( t\right) \) be the rational function field in one variable over \( {\mathbb{F}}_{2} \), and let \( F = {\mathbb{F}}_{2}\left( {t}^{2}\right) \) . Then \( \left\lbrack {K : F}\right\rbrack = 2 \) . | The element \( t \) satisfies the polynomial \( {x}^{2} - {t}^{2} \in F\left\lbrack x\right\rbrack \), which has only \( t \) as a root, since \( {x}^{2} - {t}^{2} = {\left( x - t\right) }^{2} \) in \( K\left\lbrack x\right\rbrack \) . Consequently, if \( \sigma \) is an \( F \) -automorphism of \( K \), then \( \sigma \left( t\right) = t \), so \( \sigma = \mathrm{{id}} \) . This proves that \( \operatorname{Gal}\left( {K/F}\right) = \{ \mathrm{{id}}\} \) . | Yes |
Example 2.8 Let \( F = {\mathbb{F}}_{2} \) . The polynomial \( 1 + x + {x}^{2} \) is irreducible over \( F \) , since it has no roots in \( F \) . In fact, this is the only irreducible quadratic over \( F \) ; the three other quadratics factor over \( F \) . Let \( K = F\left\lbrack x\right\rbrack /\left( {1 + x + {x}^{2}}\right) \), a field that we can view as an extension field of \( F \) ; see Example 1.6 for details on this construction. To simplify notation, we write \( M = \left( {1 + x + {x}^{2}}\right) \) . Every element of \( K \) can be written in the form \( a + {bx} + M \) by the division algorithm. Let us write \( \alpha = x + M \) . The subfield \( \{ a + M : a \in F\} \) of \( K \) is isomorphic to \( F \) . By identifying \( F \) with this subfield of \( K \), we can write every element of \( K \) in the form \( a + {b\alpha } \) with \( a, b \in F \) . Then \( K = F\left( \alpha \right) \), so any \( F \) -automorphism of \( K \) is determined by its action on \( \alpha \) . By Lemma 2.3, if \( \sigma \) is an \( F \) -automorphism of \( K \), then \( \sigma \left( \alpha \right) \) is a root of \( 1 + x + {x}^{2} \) . By factoring \( 1 + x + {x}^{2} \) as \( \left( {x - \alpha }\right) \left( {x - \beta }\right) \) and expanding, we see that the other root of \( 1 + x + {x}^{2} \) is \( \alpha + 1 \) . Therefore, the only possibility for \( \sigma \left( \alpha \right) \) is \( \alpha \) or \( \alpha + 1 \), so \( \operatorname{Gal}\left( {K/F}\right) \) has at most two elements. | To see that \( \operatorname{Gal}\left( {K/F}\right) \) has exactly two elements, we need to check that there is indeed an automorphism \( \sigma \) with \( \sigma \left( \alpha \right) = \alpha + 1 \) . If \( \sigma \) does exist, then \( \sigma \left( {a + {b\alpha }}\right) = a + b\left( {\alpha + 1}\right) = \left( {a + b}\right) + {b\alpha } \) . We leave it as an exercise (Problem 7) to show that the function \( \sigma : K \rightarrow K \) defined by \( \sigma \left( {a + {b\alpha }}\right) = \left( {a + b}\right) + {b\alpha } \) is an \( F \) -automorphism of \( K \) . Therefore, \( \operatorname{Gal}\left( {K/F}\right) = \{ \mathrm{{id}},\sigma \} \) | No |
Lemma 2.9 Let \( K \) be a field.\n\n1. If \( {L}_{1} \subseteq {L}_{2} \) are subfields of \( K \), then \( \operatorname{Gal}\left( {K/{L}_{2}}\right) \subseteq \operatorname{Gal}\left( {K/{L}_{1}}\right) \) . | Proof. The first four parts are simple consequences of the definitions. We leave the proofs of parts \( 2,3 \), and 4 to the reader and prove part 1 for the sake of illustration. If \( \sigma \in \operatorname{Gal}\left( {K/{L}_{2}}\right) \), then \( \sigma \left( a\right) = a \) for all \( a \in {L}_{2} \) . Thus, \( \sigma \left( a\right) = a \) for all \( a \in {L}_{1} \), as \( {L}_{1} \subseteq {L}_{2} \), so \( \sigma \in \operatorname{Gal}\left( {K/{L}_{1}}\right) \) . | No |
Corollary 2.10 If \( K \) is a field extension of \( F \), then there is \( 1 - 1 \) inclusion reversing correspondence between the set of subgroups of \( \operatorname{Gal}\left( {K/F}\right) \) of the form \( \operatorname{Gal}\left( {K/L}\right) \) for some subfield \( L \) of \( K \) containing \( F \) and the set of subfields of \( K \) that contain \( F \) of the form \( \mathcal{F}\left( S\right) \) for some subset \( S \) of \( \operatorname{Aut}\left( K\right) \) . This correspondence is given by \( L \mapsto \operatorname{Gal}\left( {K/L}\right) \), and its inverse is given by \( H \mapsto \mathcal{F}\left( H\right) \) . | Proof. This follows immediately from the lemma. If \( \mathcal{G} \) and \( \mathcal{F} \) are respectively the set of groups and fields in question, then the map that sends a subfield \( L \) of \( K \) to the subgroup \( \operatorname{Gal}\left( {K/L}\right) \) of \( \operatorname{Aut}\left( K\right) \) sends \( \mathcal{F} \) to \( \mathcal{G} \) . This map is injective and surjective by part 5 of the lemma. Its inverse is given by sending \( H \) to \( \mathcal{F}\left( H\right) \) by part 6 . | Yes |
Lemma 2.12 (Dedekind’s Lemma) Let \( {\tau }_{1},\ldots ,{\tau }_{n} \) be distinct characters from \( G \) to \( {K}^{ * } \) . Then the \( {\tau }_{i} \) are linearly independent over \( K \) ; that is, if \( \mathop{\sum }\limits_{i}{c}_{i}{\tau }_{i}\left( g\right) = 0 \) for all \( g \in G \), where the \( {c}_{i} \in K \), then all \( {c}_{i} = 0 \) . | Proof. Suppose that the lemma is false. Choose \( k \) minimal (relabeling the \( {\tau }_{i} \) if necessary) so that there are \( {c}_{i} \in K \) with \( \mathop{\sum }\limits_{i}{c}_{i}{\tau }_{i}\left( g\right) = 0 \) for all \( g \in G \) . Then all \( {c}_{i} \neq 0 \) . Since \( {\tau }_{1} \neq {\tau }_{2} \), there is an \( h \in G \) with \( {\tau }_{1}\left( h\right) \neq {\tau }_{2}\left( h\right) \) . We have \( \mathop{\sum }\limits_{{i = 1}}^{k}\left( {{c}_{i}{\tau }_{1}\left( h\right) }\right) {\tau }_{i}\left( g\right) = 0 \) and \[ \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{\tau }_{i}\left( {hg}\right) = \mathop{\sum }\limits_{i}\left( {{c}_{i}{\tau }_{i}\left( h\right) }\right) {\tau }_{i}\left( g\right) = 0 \] for all \( g \) . Subtracting gives \( \mathop{\sum }\limits_{{i = 1}}^{k}\left( {{c}_{i}\left( {{\tau }_{1}\left( h\right) - {\tau }_{i}\left( h\right) }\right) }\right) {\tau }_{i}\left( g\right) = 0 \) for all \( g \) . This is an expression involving \( k - 1 \) of the \( {\tau }_{i} \) with not all of the coefficients zero. This contradicts the minimality of \( k \), so the lemma is proved. | Yes |
Proposition 2.13 If \( K \) is a finite field extension of \( F \), then \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| \leq \) \( \left\lbrack {K : F}\right\rbrack \) . | Proof. The group \( \operatorname{Gal}\left( {K/F}\right) \) is finite by Corollary 2.4. Let \( \operatorname{Gal}\left( {K/F}\right) = \) \( \left\{ {{\tau }_{1},\ldots ,{\tau }_{n}}\right\} \), and suppose that \( \left\lbrack {K : F}\right\rbrack < n \) . Let \( {\alpha }_{1},\ldots ,{\alpha }_{m} \) be a basis for \( K \) as an \( F \) -vector space. The matrix\n\n\[ A = \left( \begin{matrix} {\tau }_{1}\left( {\alpha }_{1}\right) & {\tau }_{1}\left( {\alpha }_{2}\right) & \cdots & {\tau }_{1}\left( {\alpha }_{m}\right) \\ {\tau }_{2}\left( {\alpha }_{1}\right) & {\tau }_{2}\left( {\alpha }_{2}\right) & \cdots & {\tau }_{2}\left( {\alpha }_{m}\right) \\ \vdots & \vdots & \ddots & \vdots \\ {\tau }_{n}\left( {\alpha }_{1}\right) & {\tau }_{n}\left( {\alpha }_{2}\right) & \cdots & {\tau }_{n}\left( {\alpha }_{m}\right) \end{matrix}\right) \]\n\nover \( K \) has \( \operatorname{rank}\left( A\right) \leq m < n \), so the rows of \( A \) are linearly dependent over \( K \) . Thus, there are \( {c}_{i} \in K \), not all zero, such that \( \mathop{\sum }\limits_{i}{c}_{i}{\tau }_{i}\left( {\alpha }_{j}\right) = 0 \) for all \( j \) . If we set \( G = {K}^{ * } \), then for \( g \in G \) there are \( {a}_{i} \in F \) with \( g = \mathop{\sum }\limits_{j}{a}_{j}{\alpha }_{j} \) . Thus,\n\n\[ \mathop{\sum }\limits_{i}{c}_{i}{\tau }_{i}\left( g\right) = \mathop{\sum }\limits_{i}{c}_{i}{\tau }_{i}\left( {\mathop{\sum }\limits_{j}{a}_{j}{\alpha }_{j}}\right) = \mathop{\sum }\limits_{i}{c}_{i}\left( {{a}_{j}\mathop{\sum }\limits_{j}{\tau }_{j}\left( {\alpha }_{j}\right) }\right) \]\n\n\[ = \mathop{\sum }\limits_{j}{a}_{j}\left( {\mathop{\sum }\limits_{i}{c}_{i}{\tau }_{i}\left( {\alpha }_{j}\right) }\right) = 0 \]\n\nAll the \( {c}_{i} \) are then 0 by Dedekind’s lemma. This contradiction proves that \( \operatorname{Gal}\left( {K/F}\right) \leq \left\lbrack {K : F}\right\rbrack \) | Yes |
Proposition 2.14 Let \( G \) be a finite group of automorphisms of \( K \) with \( F = \mathcal{F}\left( G\right) \) . Then \( \left| G\right| = \left\lbrack {K : F}\right\rbrack \), and so \( G = \operatorname{Gal}\left( {K/F}\right) \) . | Proof. By the previous proposition, \( \left| G\right| \leq \left\lbrack {K : F}\right\rbrack \) since \( G \subseteq \operatorname{Gal}\left( {K/F}\right) \) . Suppose that \( \left| G\right| < \left\lbrack {K : F}\right\rbrack \) . Let \( n = \left| G\right| \), and take \( {\alpha }_{1},\ldots ,{\alpha }_{n + 1} \in K \) linearly independent over \( F \) . If \( G = \left\{ {{\tau }_{1},\ldots ,{\tau }_{n}}\right\} \), let \( A \) be the matrix\n\n\[ A = \left( \begin{matrix} {\tau }_{1}\left( {\alpha }_{1}\right) & {\tau }_{1}\left( {\alpha }_{2}\right) & \cdots & {\tau }_{1}\left( {\alpha }_{n + 1}\right) \\ {\tau }_{2}\left( {\alpha }_{1}\right) & {\tau }_{2}\left( {\alpha }_{2}\right) & \cdots & {\tau }_{2}\left( {\alpha }_{n + 1}\right) \\ \vdots & \vdots & \ddots & \vdots \\ {\tau }_{n}\left( {\alpha }_{1}\right) & {\tau }_{n}\left( {\alpha }_{2}\right) & \cdots & {\tau }_{n}\left( {\alpha }_{n + 1}\right) \end{matrix}\right) \]\n\nThen the columns of \( A \) are linearly dependent over \( K \) . Choose \( k \) minimal so that the first \( k \) columns of \( A \) are linearly dependent over \( K \) (relabeling if necessary). Thus, there are \( {c}_{i} \in K \) not all zero with \( \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{\tau }_{j}\left( {\alpha }_{i}\right) = 0 \) for all \( j \) . Minimality of \( k \) shows all \( {c}_{i} \neq 0 \) . Thus, by dividing we may assume that \( {c}_{1} = 1 \) . If each \( {c}_{i} \in F \), then \( 0 = {\tau }_{j}\left( {\mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{\alpha }_{i}}\right) \) for each \( j \), so \( \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{\alpha }_{i} = 0 \) . This is false by the independence of the \( {\alpha }_{i} \) over \( F \) . Take \( \sigma \in G \) . Since \( \sigma \) permutes the elements of \( G \), we get \( \mathop{\sum }\limits_{{i = 1}}^{k}\sigma \left( {c}_{i}\right) {\tau }_{j}\left( {\alpha }_{i}\right) = 0 \) for all \( j \) . Subtracting this from the original equation and recalling that \( {c}_{1} = 1 \) gives \( \mathop{\sum }\limits_{{i = 2}}^{k}\left( {{c}_{i} - \sigma \left( {c}_{i}\right) }\right) {\tau }_{j}\left( {\alpha }_{i}\right) = 0 \) for all \( j \) . Minimality of \( k \) shows that \( {c}_{i} - \sigma \left( {c}_{i}\right) = 0 \) for each \( i \) . Since this is true for all \( \sigma \in G \), we get all \( {c}_{i} \in \mathcal{F}\left( G\right) = F \) . But we have seen that this leads to a contradiction. Thus \( \left| G\right| = \left\lbrack {K : F}\right\rbrack \) . In particular, \( G = \operatorname{Gal}\left( {K/F}\right) \), since \( G \subseteq \operatorname{Gal}\left( {K/F}\right) \) and \( \left| G\right| = \left\lbrack {K : F}\right\rbrack \geq \left| {\operatorname{Gal}\left( {K/F}\right) }\right| \) | Yes |
Corollary 2.16 Let \( K \) be a finite extension of \( F \) . Then \( K/F \) is Galois if and only if \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| = \left\lbrack {K : F}\right\rbrack \) . | Proof. If \( K/F \) is a Galois extension, then \( F = \mathcal{F}\left( {\operatorname{Gal}\left( {K/F}\right) }\right) \), so by Proposition 2.14, \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| = \left\lbrack {K : F}\right\rbrack \) . Conversely, if \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| = \left\lbrack {K : F}\right\rbrack \) , let \( L = \mathcal{F}\left( {\operatorname{Gal}\left( {K/F}\right) }\right) \) . Then \( \operatorname{Gal}\left( {K/L}\right) = \operatorname{Gal}\left( {K/F}\right) \) by Proposition 2.14, and so \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| = \left\lbrack {K : L}\right\rbrack \leq \left\lbrack {K : F}\right\rbrack \) . Since \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| = \left\lbrack {K : F}\right\rbrack \), this forces \( \left\lbrack {K : L}\right\rbrack = \left\lbrack {K : F}\right\rbrack \), so \( L = F \) . | Yes |
Corollary 2.17 Let \( K \) be a field extension of \( F \), and let \( a \in K \) be algebraic over \( F \) . Then \( \left| {\operatorname{Gal}\left( {F\left( a\right) /F}\right) }\right| \) is equal to the number of distinct roots of \( \min \left( {F, a}\right) \) in \( F\left( a\right) \) . Therefore, \( F\left( a\right) \) is Galois over \( F \) if and only if \( \min \left( {F, a}\right) \) has \( n \) distinct roots in \( F\left( a\right) \), where \( n = \deg \left( {\min \left( {F, a}\right) }\right) \) . | Proof. If \( \tau \in \operatorname{Gal}\left( {F\left( a\right) /F}\right) \), we have seen that \( \tau \left( a\right) \) is a root of \( \min \left( {F, a}\right) \) . Moreover, if \( \sigma ,\tau \in \operatorname{Gal}\left( {F\left( a\right) /F}\right) \) with \( \sigma \neq \tau \), then \( \sigma \left( a\right) \neq \tau \left( a\right) \), since \( F \) - automorphisms on \( F\left( a\right) \) are determined by their action on \( a \) . Therefore, \( \left| {\operatorname{Gal}\left( {F\left( a\right) /F}\right) }\right| \leq n \) . Conversely, let \( b \) be a root in \( F\left( a\right) \) of \( \min \left( {F, a}\right) \) . Define \( \tau : F\left( a\right) \rightarrow F\left( a\right) \) by \( \tau \left( {f\left( a\right) }\right) = f\left( b\right) \) for any \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) . This map is well defined precisely because \( b \) is a root of \( \min \left( {F, a}\right) \) . It is straightforward to show that \( \tau \) is an \( F \) -automorphism, and \( \tau \left( a\right) = b \) by the definition of \( \tau \) . Thus, \( \left| {\operatorname{Gal}\left( {F\left( a\right) /F}\right) }\right| \) is equal to the number of distinct roots of \( \min \left( {F, a}\right) \) in \( F\left( a\right) \) . Since \( \left\lbrack {F\left( a\right) : F}\right\rbrack = \deg \left( {\min \left( {F, a}\right) }\right) \), we see that \( F\left( a\right) \) is Galois over \( F \) if and only if \( \min \left( {F, a}\right) \) has \( n \) distinct roots in \( F\left( a\right) \) . | Yes |
Example 2.20 Let \( F \) be a field of characteristic not 2, and let \( a \in F \) be an element that is not the square of any element in \( F \) . Let \( K = F\left\lbrack x\right\rbrack /\left( {{x}^{2} - a}\right) \) , a field since \( {x}^{2} - a \) is irreducible over \( F \) . We view \( F \) as a subfield of \( K \) by identifying \( F \) with the subfield \( \left\{ {\alpha + \left( {{x}^{2} - a}\right) : \alpha \in F}\right\} \) of \( K \) . Under this identification, each coset is uniquely expressible in the form \( \alpha + {\beta x} + \left( {{x}^{2} - a}\right) \) and, hence, is an \( F \) -linear combination of \( 1 + \left( {{x}^{2} - a}\right) \) and \( x + \left( {{x}^{2} - a}\right) \) . Thus,1 and \( u = x + \left( {{x}^{2} - a}\right) \) form a basis for \( K \) as an \( F \) -vector space, so \( \left\lbrack {K : F}\right\rbrack = 2 \) . If \( \sigma \) is defined by | \[ \sigma \left( {\alpha + {\beta u}}\right) = \alpha - {\beta u} \] then \( \sigma \) is an automorphism of \( K \) since \( u \) and \( - u \) are roots in \( K \) of \( {x}^{2} - a \) . Thus, \( \mathrm{{id}},\sigma \in \mathrm{{Gal}}\left( {K/F}\right) \), so \( \left| {\mathrm{{Gal}}\left( {K/F}\right) }\right| = 2 = \left\lbrack {K : F}\right\rbrack \) . Consequently, \( K/F \) is a Galois extension. | Yes |
Example 2.21 The extension \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) /\mathbb{Q} \) is Galois, where \( \omega = {e}^{{2\pi i}/3} \). | In fact, the field \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) is the field generated over \( \mathbb{Q} \) by the three roots \( \sqrt[3]{2} \) , \( \omega \sqrt[3]{2} \), and \( {\omega }^{2}\sqrt[3]{2} \), of \( {x}^{3} - 2 \), and since \( \omega \) satisfies \( {x}^{2} + x + 1 \) over \( \mathbb{Q} \) and \( \omega \) is not in \( \mathbb{Q}\left( \sqrt[3]{2}\right) \), we see that \( \left\lbrack {\mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) : \mathbb{Q}}\right\rbrack = 6 \) . It can be shown (see Problem 3) that the six functions\n\n\[ \text{id} : \sqrt[3]{2} \rightarrow \sqrt[3]{2},\;\omega \rightarrow \omega \]\n\n\[ \sigma : \sqrt[3]{2} \rightarrow \omega \sqrt[3]{2},\;\omega \rightarrow \omega ,\]\n\n\[ \tau : \sqrt[3]{2} \rightarrow \sqrt[3]{2},\;\omega \rightarrow {\omega }^{2},\]\n\n\[ \rho : \sqrt[3]{2} \rightarrow \omega \sqrt[3]{2},\;\omega \rightarrow {\omega }^{2},\]\n\n\[ \mu : \sqrt[3]{2} \rightarrow {\omega }^{2}\sqrt[3]{2},\;\omega \rightarrow \omega ,\]\n\n\[ \xi : \sqrt[3]{2} \rightarrow {\omega }^{2}\sqrt[3]{2},\;\omega \rightarrow {\omega }^{2} \]\n\nextend to distinct automorphisms of \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) /\mathbb{Q} \) . Thus,\n\n\[ \left| {\operatorname{Gal}\left( {\mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) /\mathbb{Q}}\right) }\right| = \left\lbrack {\mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) : \mathbb{Q}}\right\rbrack \]\n\nand so \( \mathbb{Q}\left( {\omega ,\sqrt[3]{2}}\right) /\mathbb{Q} \) is Galois. | No |
This example shows us that any finite group can occur as the Galois group of a Galois extension. We will use this example a number of times in later sections. Let \( k \) be a field and let \( K = k\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) be the field of rational functions in \( n \) variables over \( k \) . For each permutation \( \sigma \in {S}_{n} \), define \( \sigma \left( {x}_{i}\right) = {x}_{\sigma \left( i\right) } \) . Then \( \sigma \) has a natural extension to an automorphism of \( K \) by defining \[ \sigma \left( \frac{f\left( {{x}_{1},\ldots ,{x}_{n}}\right) }{g\left( {{x}_{1},\ldots ,{x}_{n}}\right) }\right) = \frac{f\left( {{x}_{\sigma \left( 1\right) },\ldots ,{x}_{\sigma \left( n\right) }}\right) }{g\left( {{x}_{\sigma \left( 1\right) },\ldots ,{x}_{\sigma \left( n\right) }}\right) }.\] | The straightforward but somewhat messy calculation that this does define a field automorphism on \( K \) is left to Problem 5. We can then view \( {S}_{n} \subseteq \) \( \operatorname{Aut}\left( K\right) \) . Let \( F = \mathcal{F}\left( {S}_{n}\right) \) . By Proposition 2.14, \( K/F \) is a Galois extension with \( \operatorname{Gal}\left( {K/F}\right) = {S}_{n} \) . The field \( F \) is called the field of symmetric functions in the \( {x}_{i} \) . The reason for this name is that if \( f\left( {{x}_{1},\ldots ,{x}_{n}}\right) /g\left( {{x}_{1},\ldots ,{x}_{n}}\right) \in \) \( F \), then \[ f\left( {{x}_{\sigma \left( 1\right) },\ldots ,{x}_{\sigma \left( n\right) }}\right) /g\left( {{x}_{\sigma \left( 1\right) },\ldots ,{x}_{\sigma \left( n\right) }}\right) = f\left( {{x}_{1},\ldots ,{x}_{n}}\right) /g\left( {{x}_{1},\ldots ,{x}_{n}}\right) \] for all \( \sigma \in {S}_{n} \) . | No |
Lemma 3.1 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) and \( \alpha \in F \) . Then \( \alpha \) is a root of \( f \) if and only if \( x - \alpha \) divides \( f \) . Furthermore, \( f \) has at most \( \deg \left( f\right) \) roots in any extension field of \( F \) . | Proof. By the division algorithm, \( f\left( x\right) = q\left( x\right) \cdot \left( {x - \alpha }\right) + r\left( x\right) \) for some \( q\left( x\right) \) and \( r\left( x\right) \) with \( r\left( x\right) = 0 \) or \( \deg \left( r\right) < \deg \left( {x - \alpha }\right) \) . In either case, we see that \( r\left( x\right) = r \) is a constant. But \( f\left( \alpha \right) = r \), so \( f\left( \alpha \right) = 0 \) if and only if \( x - \alpha \) divides \( f\left( x\right) \) .\n\nFor the second part, we argue by induction on \( n = \deg \left( f\right) \) . If \( n = 1 \), then \( f\left( x\right) = {ax} + b \) for some \( a, b \in F \) . The only root of \( f \) is \( - b/a \), so the result is true if \( n = 1 \) . Assume that any polynomial over an extension field of \( F \) of degree \( n - 1 \) has at most \( n - 1 \) roots in any extension field \( K \) of \( F \) . If \( f\left( x\right) \) has no roots in \( K \), then we are done. If instead \( \alpha \in K \) is a root of \( f \), then \( f\left( x\right) = \left( {x - \alpha }\right) \cdot g\left( x\right) \) for some \( g\left( x\right) \in K\left\lbrack x\right\rbrack \) by the first part of the lemma. Since \( g\left( x\right) \) has degree \( n - 1 \), by induction \( g \) has at most \( n - 1 \) roots in \( K \) . The roots of \( f \) consist of \( \alpha \) together with the roots of \( g \) . Thus, \( f \) has at most \( n \) roots. | Yes |
Theorem 3.3 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) have degree \( n \) . There is an extension field \( K \) of \( F \) with \( \left\lbrack {K : F}\right\rbrack \leq n \) such that \( K \) contains a root of \( f \) . In addition, there is a field \( L \) containing \( F \) with \( \left\lbrack {L : F}\right\rbrack \leq n \) ! such that \( f \) splits over \( L \) . | Proof. Let \( p\left( x\right) \) be an irreducible factor of \( f\left( x\right) \) in \( F\left\lbrack x\right\rbrack \), and let \( K \) be the field \( F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \) . Then \( F \) is isomorphic to a subfield of \( K \) ; namely the map \( \varphi : F \rightarrow K \) given by \( \varphi \left( a\right) = a + \left( {p\left( x\right) }\right) \) is an injection of fields. We will view \( F \subseteq K \) by replacing \( F \) with \( \varphi \left( F\right) \) . If \( \alpha = x + \left( {p\left( x\right) }\right) \in K \), then \( p\left( \alpha \right) = p\left( x\right) + \left( {p\left( x\right) }\right) = 0 + \left( {p\left( x\right) }\right) \) . Thus, \( \alpha \) is a root of \( p \) in \( K \) ; therefore, \( \alpha \) is a root of \( f \) . Since \( \left\lbrack {K : F}\right\rbrack = \deg \left( p\right) \leq n \), this proves the first part of the theorem.\n\nFor the second part, we use induction on \( n \) . By the first part, there is a field \( K \supseteq F \) with \( \left\lbrack {K : F}\right\rbrack \leq n \) such that \( K \) contains a root \( \alpha \) of \( f\left( x\right) \), say \( f\left( x\right) = \left( {x - \alpha }\right) \cdot g\left( x\right) \) with \( g\left( x\right) \in K\left\lbrack x\right\rbrack \) . By induction, there is a field \( L \supseteq K \) with \( \left\lbrack {L : K}\right\rbrack \leq \left( {n - 1}\right) \) ! such that \( g \) splits over \( L \) . But then \( f \) splits over \( L \) and \( \left\lbrack {L : F}\right\rbrack = \left\lbrack {L : K}\right\rbrack \cdot \left\lbrack {K : F}\right\rbrack \leq \left( {n - 1}\right) ! \cdot n = n! \) . | Yes |
Corollary 3.5 If \( {f}_{1}\left( x\right) ,\ldots ,{f}_{n}\left( x\right) \in F\left\lbrack x\right\rbrack \), then there is a splitting field for \( \left\{ {{f}_{1},\ldots ,{f}_{n}}\right\} \) over \( F \) . | Proof. Suppose that \( {f}_{1},\ldots ,{f}_{n} \in F\left\lbrack x\right\rbrack \) . Note that a splitting field of \( \left\{ {{f}_{1},\ldots ,{f}_{n}}\right\} \) is the same as a splitting field of the product \( {f}_{1}\cdots {f}_{n} \) . If \( f = {f}_{1}\cdots {f}_{n} \), then by Theorem 3.3, there is a field \( L \supseteq F \) such that \( f \) splits over \( L \) . Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in L \) be the roots of \( f \) . Then \( F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) is a splitting field for \( f \) over \( F \) . | Yes |
Corollary 3.8 Let \( F \) be a field and let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be a polynomial of degree \( n \) . If \( K \) is a splitting field of \( f \) over \( F \), then \( \left\lbrack {K : F}\right\rbrack \leq n! \) . | Proof. We prove this by induction on \( n = \deg \left( f\right) \) . If \( n = 1 \), then the result is clear. Suppose that \( n > 1 \) and that the result is true for polynomials of degree \( n - 1 \) . Let \( K \) be a splitting field of \( f \) over \( F \), and let \( a \) be a root of \( f \) in \( K \) . Then \( \left\lbrack {F\left( a\right) : F}\right\rbrack \leq n \), since \( \min \left( {F, a}\right) \) divides \( f \) . If \( f\left( x\right) = \left( {x - a}\right) g\left( x\right) \) , then \( \deg \left( g\right) = n - 1 \) and \( K \) is the splitting field of \( g \) over \( F\left( a\right) \) . By induction, \( \left\lbrack {K : F\left( a\right) }\right\rbrack \leq \left( {n - 1}\right) ! \) by Theorem 3.3, so\n\n\[ \left\lbrack {K : F}\right\rbrack = \left\lbrack {F\left( a\right) : F}\right\rbrack \cdot \left\lbrack {K : F\left( a\right) }\right\rbrack \]\n\n\[ \leq n \cdot \left( {n - 1}\right) ! = n!\text{.} \]\n\nThis proves the corollary. | Yes |
Corollary 3.8 Let \( F \) be a field and let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be a polynomial of degree \( n \) . If \( K \) is a splitting field of \( f \) over \( F \), then \( \left\lbrack {K : F}\right\rbrack \leq n! \) . | Proof. We prove this by induction on \( n = \deg \left( f\right) \) . If \( n = 1 \), then the result is clear. Suppose that \( n > 1 \) and that the result is true for polynomials of degree \( n - 1 \) . Let \( K \) be a splitting field of \( f \) over \( F \), and let \( a \) be a root of \( f \) in \( K \) . Then \( \left\lbrack {F\left( a\right) : F}\right\rbrack \leq n \), since \( \min \left( {F, a}\right) \) divides \( f \) . If \( f\left( x\right) = \left( {x - a}\right) g\left( x\right) \) , then \( \deg \left( g\right) = n - 1 \) and \( K \) is the splitting field of \( g \) over \( F\left( a\right) \) . By induction, \( \left\lbrack {K : F\left( a\right) }\right\rbrack \leq \left( {n - 1}\right) ! \) by Theorem 3.3, so\n\n\[ \left\lbrack {K : F}\right\rbrack = \left\lbrack {F\left( a\right) : F}\right\rbrack \cdot \left\lbrack {K : F\left( a\right) }\right\rbrack \]\n\n\[ \leq n \cdot \left( {n - 1}\right) ! = n!\text{.} \]\n\nThis proves the corollary. | Yes |
Example 3.9 Let \( k \) be a field, and let \( K = k\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) be the rational function field in \( n \) variables over \( k \) . We view the symmetric group \( {S}_{n} \) as a subgroup of \( \operatorname{Aut}\left( K\right) \) by defining\n\n\[ \sigma \left( \frac{f\left( {{x}_{1},\ldots ,{x}_{n}}\right) }{g\left( {{x}_{1},\ldots ,{x}_{n}}\right) }\right) = \frac{f\left( {{x}_{\sigma \left( 1\right) },\ldots ,{x}_{\sigma \left( n\right) }}\right) }{g\left( {{x}_{\sigma \left( 1\right) },\ldots ,{x}_{\sigma \left( n\right) }}\right) } \]\n\nfor \( \sigma \in {S}_{n} \), as in Example 2.22. Let \( F = \mathcal{F}\left( {S}_{n}\right) \), the field of symmetric functions in the \( {x}_{i} \) . Then \( {S}_{n} = \operatorname{Gal}\left( {K/F}\right) \) by Proposition 2.14, so \( \left\lbrack {K : F}\right\rbrack = \) \( \left| {S}_{n}\right| = n \) !. We wish to determine \( F \) . Let \( {s}_{1},{s}_{2},\ldots ,{s}_{n} \) be the elementary symmetric functions in the \( {x}_{i} \) ; that is,\n\n\[ {s}_{1} = {x}_{1} + {x}_{2} + \cdots + {x}_{n} \]\n\n\[ {s}_{2} = \mathop{\sum }\limits_{{i \neq j}}{x}_{i}{x}_{j} \]\n\n\[ {s}_{n} = {x}_{1}{x}_{2}\cdots {x}_{n} \]\n\nThen \( k\left( {{s}_{1},{s}_{2},\ldots ,{s}_{n}}\right) \subseteq F \) . We claim that \( F = k\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) . | To show this, we use the concept of splitting fields. Let\n\n\[ f\left( t\right) = {t}^{n} - {s}_{1}{t}^{n - 1} + \cdots + {\left( -1\right) }^{n}{s}_{n} \in k\left( {{s}_{1},{s}_{2},\ldots ,{s}_{n}}\right) \left\lbrack t\right\rbrack . \]\n\nThen \( f\left( t\right) = \left( {t - {x}_{1}}\right) \cdots \left( {t - {x}_{n}}\right) \) in \( K\left\lbrack x\right\rbrack \), which can be seen by expanding this product. Since \( K \) is generated over \( k \) by the \( {x}_{i} \), we see that \( K \) is a splitting field for \( f\left( t\right) \) over \( k\left( {{s}_{1},{s}_{2},\ldots ,{s}_{n}}\right) \) . We know that \( \left\lbrack {K : F}\right\rbrack = \left| {S}_{n}\right| = n! \) , and so \( \left\lbrack {K : k\left( {{s}_{1},{s}_{2},\ldots ,{s}_{n}}\right) }\right\rbrack \geq n \) !. However, \( \left\lbrack {K : k\left( {{s}_{1},{s}_{2},\ldots ,{s}_{n}}\right) }\right\rbrack \leq n \) ! by Corollary 3.8. Therefore, \( \left\lbrack {K : k\left( {{s}_{1},{s}_{2},\ldots ,{s}_{n}}\right) }\right\rbrack = \left\lbrack {K : F}\right\rbrack \) . This forces \( F = k\left( {{s}_{1},{s}_{2},\ldots ,{s}_{n}}\right) \) . Therefore, any symmetric function can be written in terms of the elementary symmetric functions. In fact, every symmetric polynomial can be written as a polynomial in the elementary symmetric functions (see Problem 17). | Yes |
Lemma 3.10 If \( K \) is a field, then the following statements are equivalent:\n\n1. There are no algebraic extensions of \( K \) other than \( K \) itself.\n\n2. There are no finite extensions of \( K \) other than \( K \) itself.\n\n3. If \( L \) is a field extension of \( K \), then \( K = \{ a \in L : a \) is algebraic over \( K\} \) .\n\n4. Every \( f\left( x\right) \in K\left\lbrack x\right\rbrack \) splits over \( K \) .\n\n5. Every \( f\left( x\right) \in K\left\lbrack x\right\rbrack \) has a root in \( K \) .\n\n6. Every irreducible polynomial over \( K \) has degree 1. | Proof. (1) \( \Rightarrow \) (2): This is clear, since any finite extension of \( F \) is an algebraic extension of \( F \) .\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) : Let \( a \in L \) be algebraic over \( K \) . Then \( K\left( a\right) \) is a finite extension of \( K \), so \( K\left( a\right) = K \) . Thus, \( a \in K \) .\n\n\( \left( 3\right) \Rightarrow \left( 4\right) \) : Let \( f\left( x\right) \in K\left\lbrack x\right\rbrack \), and let \( L \) be a splitting field of \( f \) over \( K \) . Since \( L \) is algebraic over \( K \), statement 3 shows that \( L = K \) ; that is, \( f \) splits over \( K \) .\n\n\( \left( 4\right) \Rightarrow \left( 5\right) \) : This is clear.\n\n\( \left( 5\right) \Rightarrow \left( 6\right) \) : Let \( f\left( x\right) \in K\left\lbrack x\right\rbrack \) be irreducible. By statement \( 5, f \) has a root in \( K \), so \( f \) has a linear factor. Since \( f \) is irreducible, this means \( f \) itself is linear, so \( \deg \left( f\right) = 1 \) .\n\n(6) \( \Rightarrow \) (1): Let \( L \) be an algebraic extension of \( K \) . Take \( a \in L \) and let \( p\left( x\right) = \min \left( {K, a}\right) \) . By statement 6, the degree of \( p \) is 1, so \( \left\lbrack {K\left( a\right) : K}\right\rbrack = 1 \) . Thus, \( a \in K \), so \( L = K \) . | Yes |
Lemma 3.13 If \( K/F \) is algebraic, then \( \left| K\right| \leq \max \{ \left| F\right| ,\left| \mathbb{N}\right| \} \) . | Proof. In this proof, we require some facts of cardinal arithmetic, facts that can be found in Proposition 2.1 in Appendix B. If \( a \in K \), pick a labeling \( {a}_{1},\ldots ,{a}_{n} \) of the roots of \( \min \left( {F, a}\right) \) in \( K \) . If \( \mathcal{M} \) is the set of all monic polynomials over \( F \), define \( f : K \rightarrow \mathcal{M} \times \mathbb{N} \) by \( f\left( a\right) = \left( {p\left( x\right), r}\right) \) if \( p\left( x\right) = \min \left( {F, a}\right) \) and \( a = {a}_{r} \) . This map is clearly injective, so\n\n\[ \left| K\right| \leq \left| {\mathcal{M} \times \mathbb{N}}\right| = \max \{ \left| \mathcal{M}\right| ,\left| \mathbb{N}\right| \}\]\n\nWe will be done by showing that \( \left| \mathcal{M}\right| \leq \max \{ \left| F\right| ,\left| \mathbb{N}\right| \} \) . For this, if \( {\mathcal{M}}_{n} \) is the set of monic polynomials over \( F \) of degree \( n \), then \( \left| {\mathcal{M}}_{n}\right| = \left| {F}^{n}\right| \), since the map \( \left( {{a}_{0},\ldots ,{a}_{n - 1}}\right) \mapsto {x}^{n} + \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}{a}_{i}{x}^{i} \) is a bijection between \( {F}^{n} \) and \( {\mathcal{M}}_{n} \) . If \( F \) is finite, then \( \left| {F}^{n}\right| = {\left| F\right| }^{n} \) is finite, and if \( F \) is infinite, then \( \left| {F}^{n}\right| = \left| F\right| \) . Therefore, since \( \mathcal{M} \) is the union of the disjoint sets \( {\mathcal{M}}_{n} \), we have \( \left| \mathcal{M}\right| = \left| {\mathop{\bigcup }\limits_{n}{\mathcal{M}}_{n}}\right| = \max \{ \left| F\right| ,\left| \mathbb{N}\right| \} \) . | Yes |
Theorem 3.14 Let \( F \) be a field. Then \( F \) has an algebraic closure. | Proof. Let \( S \) be a set containing \( F \) with \( \left| S\right| > \max \{ \left| F\right| ,\left| \mathbb{N}\right| \} \) . Let \( \mathcal{A} \) be the set of all algebraic extension fields of \( F \) inside \( S \) . Then \( \mathcal{A} \) is ordered by defining \( K \leq L \) if \( L \) is an extension field of \( K \) . By Zorn’s lemma, there is a maximal element \( M \) of \( \mathcal{A} \) . We claim that \( M \) is an algebraic closure of \( M \) . To show that \( M \) is algebraically closed, let \( L \) be an algebraic extension of M. By Lemma 3.13,\n\n\[ \left| L\right| \leq \max \{ \left| M\right| ,\left| \mathbb{N}\right| \} \leq \{ \left| F\right| ,\left| \mathbb{N}\right| \} < \left| S\right| .\n\]\n\nThus, there is a function \( f : L \rightarrow S \) with \( {\left. f\right| }_{M} = \) id. By defining + and \( \cdot \) on \( f\left( L\right) \) by \( f\left( a\right) + f\left( b\right) = f\left( {a + b}\right) \) and \( f\left( a\right) \cdot f\left( b\right) = f\left( {ab}\right) \), we see that \( f\left( L\right) \) is a field extension of \( M \) and \( f \) is a field homomorphism. Maximality of \( M \) shows that \( f\left( L\right) = M \), so \( L = M \) . Thus, \( M \) is algebraically closed. Since \( M \) is algebraic over \( F \), we see that \( M \) is an algebraic closure of \( F \) . | Yes |
Corollary 3.15 Let \( S \) be a set of nonconstant polynomials over \( F \) . Then \( S \) has a splitting field over \( F \) . | Proof. Let \( K \) be an algebraic closure of \( F \) . Then each \( f\left( x\right) \in S \) splits over \( K \) . Let \( X \) be the set of roots of all \( f \in S \) . Then \( F\left( X\right) \subseteq K \) is a splitting field for \( S \) over \( F \), since each \( f \) splits over \( F\left( X\right) \) and this field is generated by the roots of all the polynomials from \( S \) . | Yes |
Lemma 3.17 Let \( \sigma : F \rightarrow {F}^{\prime } \) be a field isomorphism. Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be irreducible, let \( \alpha \) be a root of \( f \) in some extension field \( K \) of \( F \), and let \( {\alpha }^{\prime } \) be a root of \( \sigma \left( f\right) \) in some extension \( {K}^{\prime } \) of \( {F}^{\prime } \) . Then there is an isomorphism \( \tau : F\left( \alpha \right) \rightarrow {F}^{\prime }\left( {\alpha }^{\prime }\right) \) with \( \tau \left( \alpha \right) = {\alpha }^{\prime } \) and \( {\left. \tau \right| }_{F} = \sigma \) . | Proof. Since \( f \) is irreducible and \( f\left( \alpha \right) = 0 \), the minimal polynomial of \( \alpha \) over \( F \) is a constant multiple of \( f \) . Thus, \( f \) and \( \min \left( {F,\alpha }\right) \) generate the same principal ideal in \( F\left\lbrack x\right\rbrack \) . We then have an \( F \) -isomorphism \( \varphi : F\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\right) \rightarrow F\left( \alpha \right) \) given by \( \varphi \left( {g\left( x\right) + \left( {f\left( x\right) }\right) }\right) = g\left( \alpha \right) \) and an \( {F}^{\prime } \) - isomorphism \( \psi : {F}^{\prime }\left\lbrack x\right\rbrack /\left( {{f}^{\prime }\left( x\right) }\right) \rightarrow {F}^{\prime }\left( {\alpha }^{\prime }\right) \) given by \( \psi \left( {g\left( x\right) + \left( {{f}^{\prime }\left( x\right) }\right) }\right) = \) \( g\left( {\alpha }^{\prime }\right) \) . Since \( \sigma \left( f\right) = {f}^{\prime } \), the map \( \nu \left( {g\left( x\right) + \left( {f\left( x\right) }\right) }\right) = \sigma \left( {g\left( x\right) }\right) + \left( {{f}^{\prime }\left( x\right) }\right) \) gives a well-defined isomorphism \( \nu : F\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\right) \rightarrow {F}^{\prime }\left\lbrack x\right\rbrack /\left( {{f}^{\prime }\left( x\right) }\right) \) which extends \( \sigma \) . We have the following sequence of field isomorphisms:\n\n\[ F\left( \alpha \right) \overset{{\varphi }^{-1}}{ \rightarrow }F\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\right) \overset{\nu }{ \rightarrow }{F}^{\prime }\left\lbrack x\right\rbrack /\left( {{f}^{\prime }\left( x\right) }\right) \overset{\psi }{ \rightarrow }{F}^{\prime }\left( {\alpha }^{\prime }\right) . \]\n\nTherefore, the composition \( {\varphi }^{-1} \circ \nu \circ \psi : F\left( \alpha \right) \rightarrow F\left( {\alpha }^{\prime }\right) \) is an isomorphism extending \( \sigma \) on \( F \) with \( \alpha \mapsto x + \left( {f\left( x\right) }\right) \mapsto x + \left( {{f}^{\prime }\left( x\right) }\right) \mapsto {\alpha }^{\prime } \) . | Yes |
Lemma 3.18 Let \( \sigma : F \rightarrow {F}^{\prime } \) be a field isomorphism, let \( K \) be a field extension of \( F \), and let \( {K}^{\prime } \) be a field extension of \( {F}^{\prime } \) . Suppose that \( K \) is a splitting field of \( \left\{ {f}_{i}\right\} \) over \( F \) and that \( \tau : K \rightarrow {K}^{\prime } \) is a homomorphism with \( {\left. \tau \right| }_{F} = \sigma \) . If \( {f}_{i}^{\prime } = \sigma \left( {f}_{i}\right) \), then \( \tau \left( K\right) \) is a splitting field of \( \left\{ {f}_{i}^{\prime }\right\} \) over \( {F}^{\prime } \) . | Proof. Because \( K \) is a splitting field of a set \( \left\{ {f}_{i}\right\} \) of polynomials over \( F \) , given \( {f}_{i} \) there are \( a,{\alpha }_{1},\ldots ,{\alpha }_{n} \in K \) with \( {f}_{i}\left( x\right) = a\mathop{\prod }\limits_{j}\left( {x - {\alpha }_{j}}\right) \) . Therefore, \( \tau \left( {{f}_{i}\left( x\right) }\right) = \tau \left( a\right) \mathop{\prod }\limits_{j}\left( {x - \tau \left( {\alpha }_{j}\right) }\right) \) . Hence, each \( {f}_{i}^{\prime } = \sigma \left( {f}_{i}\right) = \tau \left( {f}_{i}\right) \) splits over \( \tau \left( K\right) \) . Since \( K \) is generated over \( F \) by the roots of the \( {f}_{i} \), the field \( \tau \left( K\right) \) is generated over \( {F}^{\prime } \) by the images of the roots of the \( {f}_{i} \) ; that is, \( \tau \left( K\right) \) is generated over \( {F}^{\prime } \) by the roots of the \( {f}_{i}^{\prime } \) . Thus, \( \tau \left( K\right) \) is a splitting field over \( {F}^{\prime } \) for \( \left\{ {f}_{i}^{\prime }\right\} \) . | Yes |
Theorem 3.19 Let \( \sigma : F \rightarrow {F}^{\prime } \) be a field isomorphism, let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) , and let \( \sigma \left( f\right) \in {F}^{\prime }\left\lbrack x\right\rbrack \) be the corresponding polynomial over \( {F}^{\prime } \) . Let \( K \) be the splitting field of \( f \) over \( F \), and let \( {K}^{\prime } \) be the splitting field of \( \sigma \left( f\right) \) over \( {F}^{\prime } \) . Then there is an isomorphism \( \tau : K \rightarrow {K}^{\prime } \) with \( {\left. \tau \right| }_{F} = \sigma \) . Furthermore, if \( \alpha \in K \) and if \( {\alpha }^{\prime } \) is any root of \( \sigma \left( {\min \left( {F,\alpha }\right) }\right) \) in \( {K}^{\prime } \), then \( \tau \) can be chosen so that \( \tau \left( \alpha \right) = {\alpha }^{\prime } \) . | Proof. We prove this by induction on \( n = \left\lbrack {K : F}\right\rbrack \) . If \( n = 1 \), then \( f \) splits over \( F \), and the result is trivial in this case. So, suppose that \( n > 1 \) and that the result is true for splitting fields of degree less than \( n \) . If \( f \) splits over \( F \), then the result is clear. If not, let \( p\left( x\right) \) be a nonlinear irreducible factor of \( f\left( x\right) \), let \( \alpha \) be a root of \( p \), and let \( {\alpha }^{\prime } \) be a root of \( \sigma \left( p\right) \) . Set \( L = F\left( \alpha \right) \) and \( {L}^{\prime } = F\left( {\alpha }^{\prime }\right) \) . Then \( \left\lbrack {L : F}\right\rbrack > 1 \), so \( \left\lbrack {K : L}\right\rbrack < n \) . By Lemma 3.17, there is a field isomorphism \( \rho : L \rightarrow {L}^{\prime } \) with \( \rho \left( \alpha \right) = {\alpha }^{\prime } \) . Since \( K \) is the splitting field over \( L \) for \( f\left( x\right) \) and \( {K}^{\prime } \) is the splitting field over \( {L}^{\prime } \) for \( \sigma \left( f\right) \), by induction the isomorphism \( \tau \) extends to an isomorphism \( \tau : K \rightarrow {K}^{\prime } \) . The isomorphism \( \tau \) is then an extension of \( \sigma \) (and \( \rho \) ), and \( \tau \left( \alpha \right) = \rho \left( \alpha \right) = {\alpha }^{\prime } . | Yes |
Theorem 3.20 (Isomorphism Extension Theorem) Let \( \sigma : F \rightarrow {F}^{\prime } \) be a field isomorphism. Let \( S = \left\{ {{f}_{i}\left( x\right) }\right\} \) be a set of polynomials over \( F \), and let \( {S}^{\prime } = \left\{ {\sigma \left( {f}_{i}\right) }\right\} \) be the corresponding set over \( {F}^{\prime } \) . Let \( K \) be a splitting field for \( S \) over \( F \), and let \( {K}^{\prime } \) be a splitting field for \( {S}^{\prime } \) over \( {F}^{\prime } \) . Then there is an isomorphism \( \tau : K \rightarrow {K}^{\prime } \) with \( {\left. \tau \right| }_{F} = \sigma \) . Furthermore, if \( \alpha \in K \) and \( {\alpha }^{\prime } \) is any root of \( \sigma \left( {\min \left( {F,\alpha }\right) }\right) \) in \( {K}^{\prime } \), then \( \tau \) can be chosen so that \( \tau \left( \alpha \right) = {\alpha }^{\prime } \) . | Proof. We prove this with a Zorn’s lemma argument. Let \( \mathcal{S} \) be the set of all pairs \( \left( {L,\varphi }\right) \) such that \( L \) is a subfield of \( K \) and \( \varphi : L \rightarrow {K}^{\prime } \) is a homomorphism extending \( \sigma \) . This set is nonempty since \( \left( {F,\sigma }\right) \in \mathcal{S} \) . Furthermore, \( \mathcal{S} \) is partially ordered by defining \( \left( {L,\varphi }\right) \leq \left( {{L}^{\prime },{\varphi }^{\prime }}\right) \) if \( L \subseteq {L}^{\prime } \) and \( {\left. {\varphi }^{\prime }\right| }_{L} = \varphi \) . Let \( \left\{ \left( {{L}_{i},{\varphi }_{i}}\right) \right\} \) be a chain in \( \mathcal{S} \) . If \( L = \mathop{\bigcup }\limits_{i}{L}_{i} \) and \( \varphi : L \rightarrow {K}^{\prime } \) is defined by \( \varphi \left( a\right) = {\varphi }_{i}\left( a\right) \) if \( a \in {L}_{i} \), then it is not hard to see that \( L \) is a field extension of all the \( {L}_{i} \) and \( \varphi \) is a homomorphism extending \( \sigma \) . Thus, \( \left( {L,\varphi }\right) \) is an upper bound in \( \mathcal{S} \) for this chain. Therefore, by Zorn’s lemma there is a maximal element \( \left( {M,\tau }\right) \) in \( \mathcal{S} \) . We claim that \( M = K \) and \( \tau \left( M\right) = {K}^{\prime } \) . If \( M \neq K \), then there is an \( f \in S \) that does not split over \( M \) . Let \( \alpha \in K \) be a root of \( f \) that is not in \( M \), and let \( p\left( x\right) = \min \left( {F, a}\right) \) . Set \( {p}^{\prime } = \sigma \left( p\right) \in {F}^{\prime }\left\lbrack x\right\rbrack \) and let \( {\alpha }^{\prime } \in {K}^{\prime } \) be a root of \( {p}^{\prime } \) . Such an \( {\alpha }^{\prime } \) exists since \( {p}^{\prime } \) divides \( {f}^{\prime } \) and \( {f}^{\prime } \) splits over \( {K}^{\prime } \) . By Lemma 3.17, there is a \( \rho : M\left( \alpha \right) \rightarrow \tau \left( M\right) \left( {\alpha }^{\prime }\right) \) that extends \( \tau \) . Then \( \left( {M\left( \alpha \right) ,\rho }\right) \in S \) is larger than \( \left( {M,\tau }\right) \), a contradiction to the maximality of \( \left( {M,\tau }\right) \) . This proves that \( M = K \) . The equality \( \tau \left( K\right) = {K}^{\prime } \) follows immediately from Lemma 3.18, since \( \tau \left( K\right) \subseteq {K}^{\prime } \) is a splitting field for \( {S}^{\prime } \) over \( {F}^{\prime } \) . | Yes |
Corollary 3.21 Let \( F \) be a field, and let \( S \) be a subset of \( F\left\lbrack x\right\rbrack \) . Any two splitting fields of \( S \) over \( F \) are \( F \) -isomorphic. In particular, any two algebraic closures of \( F \) are \( F \) -isomorphic. | Proof. For the proof of the first statement, the isomorphism extension theorem gives an isomorphism extending id on \( F \) between any two splitting fields of \( S \) . The second statement follows from the first, since any algebraic closure of \( F \) is a splitting field of the set of all nonconstant polynomials in \( F\left\lbrack x\right\rbrack \) . | Yes |
Corollary 3.22 Let \( F \) be a field, and let \( N \) be an algebraic closure of \( F \). If \( K \) is an algebraic extension of \( F \), then \( K \) is isomorphic to a subfield of \( N \). | Proof. Let \( M \) be an algebraic closure of \( K \). By Theorem 1.24, \( M \) is algebraic over \( F \); hence, \( M \) is also an algebraic closure of \( F \). Therefore, by the previous corollary, \( M \cong N \). If \( f : M \rightarrow N \) is an \( F \)-isomorphism, then \( f\left( K\right) \) is a subfield of \( N \) isomorphic to \( K \). | Yes |
Example 3.24 If \( \left\lbrack {K : F}\right\rbrack = 2 \), then \( K \) is normal over \( F \) . | For, if \( a \in K - F \) , then \( K = F\left( a\right) \), since \( \left\lbrack {K : F}\right\rbrack = 2 \) . If \( p\left( x\right) = \min \left( {F, a}\right) \), then \( p \) has one root in \( K \) ; hence, since \( \deg \left( p\right) = 2 \), this polynomial factors over \( K \) . Because \( K \) is generated over \( F \) by the roots of \( p\left( x\right) \), we see that \( K \) is a splitting field for \( p\left( x\right) \) over \( F \) . | Yes |
Example 3.25 If \( F \subseteq L \subseteq K \) are fields such that \( K/F \) is normal, then \( K/L \) is normal. | This is true because if \( K \) is the splitting field over \( F \) of a set of polynomials \( S \subseteq F\left\lbrack x\right\rbrack \), then \( K \) is generated over \( F \) by the roots of the polynomials in \( S \) . Consequently, \( K \) is generated by the roots as an extension of \( L \), so \( K \) is a splitting field of \( S \) over \( L \), and so \( K \) is normal over \( L \) . | Yes |
Example 3.27 Let \( F \) be a field of characteristic \( p > 0 \), and suppose that \( K = F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) with \( {a}_{i}^{p} \in F \) for each \( i \) . Then we show that \( K \) is normal over \( F \) . | The minimal polynomial of \( {a}_{i} \) divides \( {x}^{p} - {a}_{i}^{p} \), which factors completely over \( K \) as \( {x}^{p} - {a}_{i}^{p} = {\left( x - {a}_{i}\right) }^{p} \) ; hence, \( \min \left( {F,{a}_{i}}\right) \) splits over \( K \) . Thus, \( K \) is the splitting field of \( \left\{ {\min \left( {F,{a}_{i}}\right) : 1 \leq i \leq n}\right\} \) over \( F \) . Note that each \( \min \left( {F,{a}_{i}}\right) \) has only one distinct root, and any \( F \) -automorphism of \( K \) is determined by its action on the generators \( {a}_{1},\ldots ,{a}_{n} \), so \( \operatorname{Gal}\left( {K/F}\right) = \{ \mathrm{{id}}\} \) . For instance, if \( k\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) is the rational function field in \( n \) variables over a field \( k \) of characteristic \( p \), then \( k\left( {{x}_{1},\ldots ,{x}_{n}}\right) /k\left( {{x}_{1}^{p},\ldots ,{x}_{n}^{p}}\right) \) is a normal extension. | Yes |
Proposition 3.28 If \( K \) is algebraic over \( F \), then the following statements are equivalent:\n\n1. The field \( K \) is normal over \( F \) .\n\n2. If \( M \) is an algebraic closure of \( K \) and if \( \tau : K \rightarrow M \) is an \( F \) - homomorphism, then \( \tau \left( K\right) = K \) .\n\n3. If \( F \subseteq L \subseteq K \subseteq N \) are fields and if \( \sigma : L \rightarrow N \) is an \( F \) - homomorphism, then \( \sigma \left( L\right) \subseteq K \), and there is a \( \tau \in \operatorname{Gal}\left( {K/F}\right) \) with \( {\left. \tau \right| }_{L} = \sigma \) .\n\n4. For any irreducible \( f\left( x\right) \in F\left\lbrack x\right\rbrack \), if \( f \) has a root in \( K \), then \( f \) splits over \( K \) . | Proof. (1) \( \Rightarrow \) (2): Let \( M \) be an algebraic closure of \( K \), and let \( \tau : K \rightarrow M \) be an \( F \) -homomorphism. If \( K \) is the splitting field for \( S \subseteq F\left\lbrack x\right\rbrack \) over \( F \) , then so is \( \tau \left( K\right) \subseteq M \) by Lemma 3.17. Since \( K \) and \( \tau \left( K\right) \) are generated over \( F \) by the same set of roots, \( K = \tau \left( K\right) \) .\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) : Suppose that \( F \subseteq L \subseteq K \subseteq N \) are fields and that \( \sigma : L \rightarrow N \) is an \( F \) -homomorphism. Since \( L \subseteq K \), the extension \( L/F \) is algebraic, and so \( \sigma \left( L\right) \subseteq N \) is algebraic over \( F \) . Let \( {M}^{\prime } \) be the algebraic closure of \( F \) in \( N \) and let \( M \) be an algebraic closure of \( {M}^{\prime } \) . Then \( M \) is also an algebraic closure of \( K \) . By the isomorphism extension theorem, there is an extension \( \rho : M \rightarrow M \) with \( {\left. \rho \right| }_{L} = \sigma \) . Let \( \tau = {\left. \rho \right| }_{K} \) . By condition 2 we have \( \tau \left( K\right) = K \) , so \( \sigma \left( L\right) = \tau \left( L\right) \subseteq \tau \left( K\right) = K \) . Thus, \( \tau \in \operatorname{Gal}\left( {K/F}\right) \) .\n\n(3) \( \Rightarrow \left( 4\right) \) : Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be irreducible over \( F \), and let \( \alpha \in K \) be a root of \( f \) . Let \( L = F\left( \alpha \right) \subseteq K \) and let \( N \) be an algebraic closure of \( K \) . If \( \beta \in M \) is any root of \( f \), then there is an \( F \) -homomorphism \( \sigma : L \rightarrow M \) given by \( g\left( \alpha \right) \mapsto g\left( \beta \right) \) . By condition \( 3,\sigma \left( L\right) \subseteq K \), so \( \beta \in K \) . Hence, all roots of \( f \) lie in \( K \), so \( f \) splits over \( K \) .\n\n\( \left( 4\right) \Rightarrow \left( 1\right) \) : Condition 4 shows that \( \min \left( {F,\alpha }\right) \) splits over \( K \) for each \( \alpha \in K \) . Thus, \( K \) is the splitting field over \( F \) of \( \{ \min \left( {F, a}\right) : a \in K\} \), so \( K \) is normal over \( F \) . | Yes |
Lemma 4.3 Let \( f\left( x\right) \) and \( g\left( x\right) \) be polynomials over a field \( F \) . 1. If \( f \) has no repeated roots in any splitting field, then \( f \) is separable over \( F \) . 2. If \( g \) divides \( f \) and if \( f \) is separable over \( F \), then \( g \) is separable over \( F \) . 3. If \( {f}_{1},\ldots ,{f}_{n} \) are separable polynomials over \( F \), then the product \( {f}_{1}\cdots {f}_{n} \) is separable over \( F \) . 4. If \( f \) is separable over \( F \), then \( f \) is separable over any extension field of \( F \) . | Proof. For property 1, if \( f \) has no repeated roots in any splitting field, then neither does any irreducible factor of \( f \) . Thus, \( f \) is separable over \( F \) . To show property 2, if \( g \) divides \( f \) with \( f \) separable over \( F \), then no irreducible factor of \( f \) has a repeated root. However, the irreducible factors of \( g \) are also irreducible factors of \( f \) . Thus, \( g \) is separable over \( F \) . To prove property 3, we see that the set of irreducible factors of the \( {f}_{i} \) is precisely the set of irreducible factors of the polynomial \( {f}_{1}\cdots {f}_{n} \) . Each of these irreducible factors have no repeated roots, so \( {f}_{1}\cdots {f}_{n} \) is separable over \( F \) . Finally, for property 4, let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be separable over \( F \), and let \( K \) be an extension of \( F \) . If \( p\left( x\right) \) is an irreducible factor of \( f\left( x\right) \) in \( K\left\lbrack x\right\rbrack \), let \( \alpha \) be a root of \( p \) in some algebraic closure of \( K \), and set \( q\left( x\right) = \min \left( {F,\alpha }\right) \) . Then \( q\left( x\right) \in K\left\lbrack x\right\rbrack \) , so \( p \) divides \( q \) . But \( q \) has no repeated roots, since \( q \) is an irreducible factor of \( f \) . Thus, \( p \) has no repeated roots, so \( f \) is separable over \( K \) . | Yes |
If \( f \) has no repeated roots in any splitting field, then \( f \) is separable over \( F \) . | For property 1, if \( f \) has no repeated roots in any splitting field, then neither does any irreducible factor of \( f \) . Thus, \( f \) is separable over \( F \) . | Yes |
Proposition 4.5 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be a nonconstant polynomial. Then \( f \) has no repeated roots in a splitting field if and only if \( \gcd \left( {f,{f}^{\prime }}\right) = 1 \) in \( F\left\lbrack x\right\rbrack \) . | Proof. We first point out that \( f \) and \( {f}^{\prime } \) are relatively prime in \( F\left\lbrack x\right\rbrack \) if and only if they are relatively prime in \( K\left\lbrack x\right\rbrack \) . To prove this, suppose that \( \gcd \left( {f,{f}^{\prime }}\right) = 1 \) in \( F\left\lbrack x\right\rbrack \) . Then there are polynomials \( g, h \in F\left\lbrack x\right\rbrack \) with \( 1 = \) \( {fg} + {f}^{\prime }h \) . This also is an equation in \( K\left\lbrack x\right\rbrack \), so the gcd in \( K\left\lbrack x\right\rbrack \) of \( f \) and \( {f}^{\prime } \) must divide 1. Thus, \( \gcd \left( {f,{f}^{\prime }}\right) = 1 \) in \( K\left\lbrack x\right\rbrack \) . Conversely, suppose that \( \gcd \left( {f,{f}^{\prime }}\right) = 1 \) in \( K\left\lbrack x\right\rbrack \) . If \( d \) is the gcd of \( f \) and \( {f}^{\prime } \) in \( F\left\lbrack x\right\rbrack \), then \( d \in K\left\lbrack x\right\rbrack \) , so \( d \) divides 1 ; thus, \( f \) and \( {f}^{\prime } \) are relatively prime in \( F\left\lbrack x\right\rbrack \) .\n\nSuppose that \( f \) and \( {f}^{\prime } \) are relatively prime in \( F\left\lbrack x\right\rbrack \) . In particular, let \( K \) be a splitting field of \( \left\{ {f,{f}^{\prime }}\right\} \) over \( F \) . If \( f \) and \( {f}^{\prime } \) have a common root \( \alpha \in K \) , then \( x - \alpha \) divides both \( f \) and \( {f}^{\prime } \) in \( K\left\lbrack x\right\rbrack \) . This would contradict the fact that \( f \) and \( {f}^{\prime } \) are relatively prime in \( K\left\lbrack x\right\rbrack \) . Therefore, \( f \) and \( {f}^{\prime } \) have no common roots.\n\nConversely, if \( f \) and \( {f}^{\prime } \) have no common roots in a splitting field \( K \) of \( \left\{ {f,{f}^{\prime }}\right\} \), let \( d\left( x\right) \) be the greatest common divisor in \( K\left\lbrack x\right\rbrack \) of \( f\left( x\right) \) and \( {f}^{\prime }\left( x\right) \) . Then \( d \) splits over \( K \) since \( f \) splits over \( K \) and \( d \) divides \( f \) . Any root of \( d \) is then a common root of \( f \) and \( {f}^{\prime } \) since \( d \) also divides \( {f}^{\prime } \) . Thus, \( d\left( x\right) \) has no roots, so \( d = 1 \) . Therefore, \( f \) and \( {f}^{\prime } \) are relatively prime over \( K \) ; hence, they are also relatively prime over \( F \) . | Yes |
Proposition 4.6 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial.\n\n1. If \( \operatorname{char}\left( F\right) = 0 \), then \( f \) is separable over \( F \) . If \( \operatorname{char}\left( F\right) = p > 0 \), then \( f \) is separable over \( F \) if and only if \( {f}^{\prime }\left( x\right) \neq 0 \), and this occurs if and only if \( f\left( x\right) \notin F\left\lbrack {x}^{p}\right\rbrack \) .\n\n2. If \( \operatorname{char}\left( F\right) = p \), then \( f\left( x\right) = g\left( {x}^{{p}^{m}}\right) \) for some integer \( m \geq 0 \) and some \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) that is irreducible and separable over \( F \) . | Proof. If \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) is irreducible over \( F \), then the only possibility for \( \gcd \left( {f,{f}^{\prime }}\right) \) is 1 or \( f \) . If \( \operatorname{char}\left( F\right) = 0 \), then \( \deg \left( {f}^{\prime }\right) = \deg \left( f\right) - 1 \) ; thus, \( f \) does not divide \( {f}^{\prime } \), and so \( \gcd \left( {f,{f}^{\prime }}\right) = 1 \) . Therefore, by Proposition 4.5, \( f \) has no repeated roots, so \( f \) is separable over \( F \) . If \( \operatorname{char}\left( F\right) = p > 0 \), the same reasoning shows \( \gcd \left( {f,{f}^{\prime }}\right) = f \) if and only if \( f \) divides \( {f}^{\prime } \), if and only if \( {f}^{\prime }\left( x\right) = 0 \), if and only if \( f\left( x\right) \in F\left\lbrack {x}^{p}\right\rbrack \) .\n\nFor statement 2, suppose that \( \operatorname{char}\left( F\right) = p \), and let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) . Let \( m \) be maximal such that \( f\left( x\right) \in F\left\lbrack {x}^{{p}^{m}}\right\rbrack \) . Such an \( m \) exists, since \( f \in F\left\lbrack {x}^{{p}^{0}}\right\rbrack \) and \( f \) lies in \( F\left\lbrack {x}^{{p}^{r}}\right\rbrack \) for only finitely many \( r \) because any nonconstant polynomial in \( F\left\lbrack {x}^{{p}^{r}}\right\rbrack \) has degree at least \( {p}^{r} \) . Say \( f\left( x\right) = g\left( {x}^{{p}^{m}}\right) \) . Then \( g\left( x\right) \notin F\left\lbrack {x}^{p}\right\rbrack \) by maximality of \( m \) . Moreover, \( g\left( x\right) \) is irreducible over \( F \) , since if \( g\left( x\right) = h\left( x\right) \cdot k\left( x\right) \), then \( f\left( x\right) = h\left( {x}^{{p}^{m}}\right) \cdot k\left( {x}^{{p}^{m}}\right) \) is reducible over \( F \) . By statement \( 2, g \) is separable over \( F \) . | Yes |
Corollary 4.10 Let \( L \) be a finite extension of \( F \). 1. \( L \) is separable over \( F \) if and only if \( L \) is contained in a Galois extension of \( F \). | Proof. If \( L \subseteq K \) with \( K/F \) Galois, then \( K/F \) is separable by Theorem 4.9. Hence, \( L/F \) is separable. Conversely, suppose that \( L/F \) is separable. Since \( \left\lbrack {L : F}\right\rbrack < \infty \), we may write \( L = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \), and each \( {\alpha }_{i} \) is separable over \( F \) . If \( K \) is the splitting field of \( \left\{ {\min \left( {F,{\alpha }_{i}}\right) : 1 \leq i \leq n}\right\} \), then \( L \subseteq K \) , and \( K/F \) is Galois by Theorem 4.9. | Yes |
Theorem 4.13 Let \( F \) be a field of characteristic \( p \) . Then \( F \) is perfect if and only if \( {F}^{p} = F \) . | Proof. Suppose that \( F \) is perfect. Let \( a \in F \), and consider the field \( K = \) \( F\left( \alpha \right) \), where \( \alpha \) is a root of \( {x}^{p} - a \) . The minimal polynomial of \( \alpha \) divides \( {x}^{p} - a = {\left( x - \alpha \right) }^{p} \) . However, \( K \) is separable over \( F \) since \( F \) is perfect; thus, this minimal polynomial has no repeated roots. This means \( \alpha \in F \) , so \( a \in {F}^{p} \) . Conversely, suppose that \( {F}^{p} = F \) . Let \( K \) be an algebraic extension of \( F \) , and let \( \alpha \in K \) . If \( p\left( x\right) = \min \left( {F,\alpha }\right) \), then by Proposition 4.6 there is an \( m \) with \( p\left( x\right) = g\left( {x}^{{p}^{m}}\right) \) for some \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) with \( g \) irreducible and separable over \( F \) . If \( g\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {x}^{r} \), then there are \( {b}_{i} \in F \) with \( {b}_{i}^{p} = {a}_{i} \) for all \( i \) . Then \( p\left( x\right) = \mathop{\sum }\limits_{i}{b}_{i}{x}^{{p}^{m}i} = {\left( \mathop{\sum }\limits_{i}{b}_{i}{x}^{{p}^{m - 1}i}\right) }^{p} \) . This contradicts the irreducibility of \( p \) unless \( m = 1 \) . Thus, \( p = g \) is separable over \( F \), so \( \alpha \) is separable over \( F \) . Therefore, any algebraic extension of \( F \) is separable, so \( F \) is perfect. | Yes |
Theorem 4.13 Let \( F \) be a field of characteristic \( p \) . Then \( F \) is perfect if and only if \( {F}^{p} = F \) . | Proof. Suppose that \( F \) is perfect. Let \( a \in F \), and consider the field \( K = \) \( F\left( \alpha \right) \), where \( \alpha \) is a root of \( {x}^{p} - a \) . The minimal polynomial of \( \alpha \) divides \( {x}^{p} - a = {\left( x - \alpha \right) }^{p} \) . However, \( K \) is separable over \( F \) since \( F \) is perfect; thus, this minimal polynomial has no repeated roots. This means \( \alpha \in F \) , so \( a \in {F}^{p} \) .\n\nConversely, suppose that \( {F}^{p} = F \) . Let \( K \) be an algebraic extension of \( F \) , and let \( \alpha \in K \) . If \( p\left( x\right) = \min \left( {F,\alpha }\right) \), then by Proposition 4.6 there is an \( m \) with \( p\left( x\right) = g\left( {x}^{{p}^{m}}\right) \) for some \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) with \( g \) irreducible and separable over \( F \) . If \( g\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {x}^{r} \), then there are \( {b}_{i} \in F \) with \( {b}_{i}^{p} = {a}_{i} \) for all \( i \) . Then \( p\left( x\right) = \mathop{\sum }\limits_{i}{b}_{i}{x}^{{p}^{m}i} = {\left( \mathop{\sum }\limits_{i}{b}_{i}{x}^{{p}^{m - 1}i}\right) }^{p} \) . This contradicts the irreducibility of \( p \) unless \( m = 1 \) . Thus, \( p = g \) is separable over \( F \), so \( \alpha \) is separable over \( F \) . Therefore, any algebraic extension of \( F \) is separable, so \( F \) is perfect. | Yes |
Any finite field is perfect | The map \( \varphi : F \rightarrow F \) given by \( \varphi \left( a\right) = {a}^{p} \) is a nonzero field homomorphism, so \( \varphi \) is injective. Since \( F \) is finite, \( \varphi \) is also surjective. Thus, \( {F}^{p} = \operatorname{im}\left( \varphi \right) = F \), so \( F \) is perfect by Theorem 4.13. | Yes |
Lemma 4.16 Let \( F \) be a field of characteristic \( p > 0 \) . If \( \alpha \) is algebraic over \( F \), then \( \alpha \) is purely inseparable over \( F \) if and only if \( {\alpha }^{{p}^{n}} \in F \) for some \( n \) . When this happens, \( \min \left( {F,\alpha }\right) = {\left( x - \alpha \right) }^{{p}^{n}} \) for some \( n \) . | Proof. If \( {\alpha }^{{p}^{n}} = a \in F \), then \( \alpha \) is a root of the polynomial \( {x}^{{p}^{n}} - a \) . This polynomial factors over \( F\left( \alpha \right) \) as \( {\left( x - \alpha \right) }^{{p}^{n}} \), and \( \min \left( {F,\alpha }\right) \) divides this polynomial, so \( \min \left( {F,\alpha }\right) \) has only \( \alpha \) as a root. Conversely, suppose that \( \alpha \) is purely inseparable over \( F \), and let \( f\left( x\right) = \min \left( {F,\alpha }\right) \) . There is a separable irreducible polynomial \( g\left( x\right) \) over \( F \) with \( f\left( x\right) = g\left( {x}^{{p}^{m}}\right) \) by Proposition 4.6. If \( g \) factors over a splitting field as \( g\left( x\right) = \left( {x - {b}_{1}}\right) \cdots \left( {x - {b}_{r}}\right) \), then \( f\left( x\right) = \left( {{x}^{{p}^{m}} - {b}_{i}}\right) \cdots \left( {{x}^{{p}^{m}} - {b}_{r}}\right) \) . If \( r > 1 \), then separability of \( g \) says that the \( {b}_{i} \) are distinct. By assumption, the only root of \( f \) is \( \alpha \) . Thus, \( {b}_{i} = {\alpha }^{{p}^{m}} \) for each \( i \) . Hence, \( r = 1 \), so \( f\left( x\right) = {x}^{{p}^{m}} - {b}_{1} \) . Therefore, \( {\alpha }^{{p}^{m}} \in F \), and \( \min \left( {F,\alpha }\right) = {x}^{{p}^{m}} - {b}_{1} = {\left( x - \alpha \right) }^{{p}^{m}}. \ | Yes |
1. If \( \alpha \in K \) is separable and purely inseparable over \( F \), then \( \alpha \in F \) . | Proof. Suppose that \( \alpha \in K \) is both separable and purely inseparable over \( F \) . Then \( \min \left( {F,\alpha }\right) \) has only one distinct root, and it also has no repeated roots. Therefore, \( p\left( x\right) = x - \alpha \), so \( \alpha \in F \) . | Yes |
Proposition 4.20 Let \( K \) be a field extension of \( F \). If \( S \) and \( I \) are the separable and purely inseparable closures of \( F \) in \( K \), respectively, then \( S \) and \( I \) are field extensions of \( F \) with \( S/F \) separable, \( I/F \) purely inseparable, and \( S \cap I = F \). If \( K/F \) is algebraic, then \( K/S \) is purely inseparable. | Proof. Let \( a, b \in S \) . Then \( F\left( {a, b}\right) \) is a separable extension of \( F \) by Lemma 4.10. Hence, \( a \pm b,{ab} \), and \( a/b \) are separable over \( F \), so they all lie in \( S \) . Thus, \( S \) is a field. For \( I \), if \( c, d \in I \), then there are \( n, m \) with \( {c}^{{p}^{n}} \in F \) and \( {d}^{{p}^{m}} \in F \) . Setting \( N = {nm} \), we have \( {\left( c \pm d\right) }^{{p}^{N}},{\left( cd\right) }^{{p}^{N}} \), and \( {\left( c/d\right) }^{{p}^{N}} \in F \) . Thus, \( c \pm d,{cd} \), and \( c/d \) belong to \( I \), so \( I \) is a field. The equality \( S \cap I = F \) holds, since \( S \cap I \) is both separable and purely inseparable over \( F \) . Finally, suppose that \( K/F \) is algebraic. If \( \alpha \in K \), then \( \min \left( {F,\alpha }\right) = g\left( {x}^{{p}^{n}}\right) \) for some separable, irreducible polynomial \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) by Proposition 4.6. If \( a = {\alpha }^{{p}^{n}} \), then \( g\left( a\right) = 0 \), so \( g\left( x\right) = \min \left( {F, a}\right) \) . Therefore, \( a \) is separable over \( F \), so \( {\alpha }^{{p}^{n}} = a \in S \) . Thus, \( K/S \) is purely inseparable. | Yes |
Proposition 4.20 Let \( K \) be a field extension of \( F \) . If \( S \) and \( I \) are the separable and purely inseparable closures of \( F \) in \( K \), respectively, then \( S \) and \( I \) are field extensions of \( F \) with \( S/F \) separable, \( I/F \) purely inseparable, and \( S \cap I = F \) . If \( K/F \) is algebraic, then \( K/S \) is purely inseparable. | Proof. Let \( a, b \in S \) . Then \( F\left( {a, b}\right) \) is a separable extension of \( F \) by Lemma 4.10. Hence, \( a \pm b,{ab} \), and \( a/b \) are separable over \( F \), so they all lie in \( S \) . Thus, \( S \) is a field. For \( I \), if \( c, d \in I \), then there are \( n, m \) with \( {c}^{{p}^{n}} \in F \) and \( {d}^{{p}^{m}} \in F \) . Setting \( N = {nm} \), we have \( {\left( c \pm d\right) }^{{p}^{N}},{\left( cd\right) }^{{p}^{N}} \), and \( {\left( c/d\right) }^{{p}^{N}} \in F \) . Thus, \( c \pm d,{cd} \), and \( c/d \) belong to \( I \), so \( I \) is a field. The equality \( S \cap I = F \) holds, since \( S \cap I \) is both separable and purely inseparable over \( F \) . Finally, suppose that \( K/F \) is algebraic. If \( \alpha \in K \), then \( \min \left( {F,\alpha }\right) = g\left( {x}^{{p}^{n}}\right) \) for some separable, irreducible polynomial \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) by Proposition 4.6. If \( a = {\alpha }^{{p}^{n}} \), then \( g\left( a\right) = 0 \), so \( g\left( x\right) = \min \left( {F, a}\right) \) . Therefore, \( a \) is separable over \( F \), so \( {\alpha }^{{p}^{n}} = a \in S \) . Thus, \( K/S \) is purely inseparable. | Yes |
Proposition 4.21 If \( F \subseteq L \subseteq K \) are fields such that \( L/F \) and \( K/L \) are separable, then \( K/F \) is separable. | Proof. Let \( S \) be the separable closure of \( F \) in \( K \) . Then \( L \subseteq S \), as \( L/F \) is separable. Also, since \( K/L \) is separable, \( K/S \) is separable. But \( K/S \) is purely inseparable, so \( K = S \) . Thus, \( K \) is separable over \( F \) . | Yes |
Example 4.22 Let \( K \) be a finite extension of \( F \), and suppose that \( \operatorname{char}\left( F\right) \) does not divide \( \left\lbrack {K : F}\right\rbrack \) . We show that \( K/F \) is separable. | If \( \operatorname{char}\left( F\right) = 0 \) , then this is clear, so suppose that \( \operatorname{char}\left( F\right) = p > 0 \) . Let \( S \) be the separable closure of \( F \) in \( K \) . Then \( K/S \) is purely inseparable, so \( \left\lbrack {K : S}\right\rbrack = {p}^{n} \) for some \( n \) by Lemma 4.17. However, since \( p \) does not divide \( \left\lbrack {K : F}\right\rbrack \), this forces \( \left\lbrack {K : S}\right\rbrack = 1 \) . Thus, \( K = S \), so \( K \) is separable over \( F \) . | Yes |
Theorem 4.23 Let \( K \) be a normal extension of \( F \), and let \( S \) and \( I \) be the separable and purely inseparable closures of \( F \) in \( K \) , respectively. Then \( S/F \) is Galois, \( I = \mathcal{F}\left( {\operatorname{Gal}\left( {K/F}\right) }\right) \), and \( \operatorname{Gal}\left( {S/F}\right) \cong \operatorname{Gal}\left( {K/I}\right) \) . Thus, \( K/I \) is Galois. Moreover, \( K = {SI} \) . | Proof. Let \( a \in S \), and set \( f\left( x\right) = \min \left( {F, a}\right) \) . Since \( K \) is normal over \( F \) , the polynomial \( f \) splits over \( K \) . Since \( a \) is separable over \( F \), the polynomial \( f \) has no repeated roots, so all its roots are separable over \( S \) . Thus, \( f \) splits over \( S \) . Hence, \( S \) is normal over \( F \) by Proposition 3.28, and since \( S \) is separable over \( F \), we see by Theorem 4.9 that \( S \) is Galois over \( F \) . The map \( \theta : \operatorname{Gal}\left( {K/F}\right) \rightarrow \operatorname{Gal}\left( {S/F}\right) \) given by \( {\left. \theta \left( \sigma \right) = \sigma \right| }_{S} \) is a well-defined group homomorphism. The kernel of \( \theta \) is \( \operatorname{Gal}\left( {K/S}\right) \), and this group is trivial by Lemma 4.17 since \( K \) is purely inseparable over \( S \) . By the isomorphism extension theorem, if \( \tau \in \operatorname{Gal}\left( {S/F}\right) \), there is a \( \sigma \in \operatorname{Gal}\left( {K/F}\right) \) with \( {\left. \sigma \right| }_{S} = \tau \) . Thus, \( \theta \) is an isomorphism.\n\nTo show that \( I = \mathcal{F}\left( {\operatorname{Gal}\left( {K/F}\right) }\right) \), if \( a \in I \), then \( {a}^{{p}^{n}} \in F \) for some \( n \) . For \( \sigma \in \operatorname{Gal}\left( {K/F}\right) \), we have \( {a}^{{p}^{n}} = \sigma \left( {a}^{{p}^{n}}\right) = \sigma {\left( a\right) }^{{p}^{n}} \), so \( \sigma \left( a\right) = a \) . Thus, \( I \subseteq \mathcal{F}\left( {\operatorname{Gal}\left( {K/F}\right) }\right) \) . Conversely, take \( b \in \mathcal{F}\left( {\operatorname{Gal}\left( {K/F}\right) }\right) \) . There is an \( n \) with \( {b}^{{p}^{n}} \in S \) because \( K/S \) is purely inseparable. Let \( \tau \in \operatorname{Gal}\left( {S/F}\right) \) . Since \( \theta \) is surjective, there is a \( \sigma \in \operatorname{Gal}\left( {K/F}\right) \) with \( \tau = {\left. \theta \left( \sigma \right) = \sigma \right| }_{S} \) . Then \( \tau \left( {b}^{{p}^{n}}\right) = \) \( \sigma \left( {b}^{{p}^{n}}\right) = {b}^{{p}^{n}} \) . This is true for each \( \tau \) ; hence, \( {b}^{{p}^{n}} \in \mathcal{F}\left( {\operatorname{Gal}\left( {S/F}\right) }\right) = F \) . This equality holds since \( S \) is Galois over \( F \) . Thus, \( b \) is purely inseparable over \( F \) . This proves \( I = \mathcal{F}\left( {\operatorname{Gal}\left( {K/F}\right) }\right) \), so \( \operatorname{Gal}\left( {K/F}\right) = \operatorname{Gal}\left( {K/I}\right) \) . Therefore, \( K \) is Galois over \( I \) ; hence, \( K/I \) is separable. Finally, \( K \) is separable over \( {SI} \) since \( I \subseteq {SI} \), and \( K \) is purely inseparable over \( {SI} \) since \( S \subseteq {SI} \) . Therefore, \( K = {SI} \). | Yes |
We give an example of a field extension \( K/F \) in which \( K \) is not separable over the purely inseparable closure \( I \) of \( F \) in \( K \) . This is also an example of a nonseparable field extension \( K/F \) in which the purely inseparable closure is \( F \) . Let \( k \) be a field of characteristic 2, let \( F \) be the rational function field \( F = k\left( {x, y}\right) \), let \( S = F\left( u\right) \), where \( u \) is a root of \( {t}^{2} + t + x \), and let \( K = S\left( \sqrt{uy}\right) \) . Then \( K/S \) is purely inseparable and \( S/F \) is separable, so \( S \) is the separable closure of \( F \) in \( K \) . We will show that \( I = F \) , which will prove that \( K/I \) is not separable since \( K/S \) is not separable. | To do this, we show that if \( a \in K \) with \( {a}^{2} \in F \), then \( a \in F \) . A basis for \( K/F \) is \( 1, u,\sqrt{uy} \), and \( u\sqrt{uy} \) . Say \( {a}^{2} \in F \) and write \( a = \alpha + {\beta u} + \gamma \sqrt{uy} + {\delta u}\sqrt{uy} \) with \( \alpha ,\beta ,\gamma ,\delta \in F \) . Then\n\n\[ \n{a}^{2} = {\alpha }^{2} + {\beta }^{2}\left( {u + x}\right) + {\gamma }^{2}\left( {uy}\right) + {\delta }^{2}\left( {u + x}\right) {uy}.\n\]\n\nThe coefficient of \( u \) is zero since \( {a}^{2} \in F \), so\n\n\[ \n{\beta }^{2} + {\left( \gamma + \delta \right) }^{2}y + {\delta }^{2}{xy} = 0.\n\]\n\nIf \( \delta = 0 \), then \( {\beta }^{2} + {\gamma }^{2}y = 0 \), so \( \gamma = 0 \) since \( y \) is not a square in \( F \) . But then \( \beta = 0 \), so \( a \in F \) . If \( \delta \neq 0 \), then\n\n\[ \nx = \frac{{\beta }^{2} + {\left( \gamma + \delta \right) }^{2}y}{{\delta }^{2}y} = {\left( \frac{\gamma }{\delta } + 1\right) }^{2} + {\left( \frac{\beta }{\delta }\right) }^{2}y\n\]\n\nwhich means that \( x \in {F}^{2}\left( y\right) \) . But this is impossible. Thus, \( \delta = 0 \), and so we conclude that \( a \in F \) . Thus, \( I = F \), so \( K/I \) is not separable. Note that \( K \neq {SI} \) also. | Yes |
Theorem 5.1 (Fundamental Theorem of Galois Theory) Let \( K \) be a finite Galois extension of \( F \), and let \( G = \operatorname{Gal}\left( {K/F}\right) \). Then there is a \( 1 - 1 \) inclusion reversing correspondence between intermediate fields of \( K/F \) and subgroups of \( G \), given by \( L \mapsto \operatorname{Gal}\left( {K/L}\right) \) and \( H \mapsto \mathcal{F}\left( H\right) \). Furthermore, if \( L \leftrightarrow H \), then \( \left\lbrack {K : L}\right\rbrack = \left| H\right| \) and \( \left\lbrack {L : F}\right\rbrack = \left\lbrack {G : H}\right\rbrack \). Moreover, \( H \) is normal in \( G \) if and only if \( L \) is Galois over \( F \). When this occurs, \( \operatorname{Gal}\left( {L/F}\right) \cong G/H \). | Proof. We have seen in Lemma 2.9 that the maps \( L \mapsto \operatorname{Gal}\left( {K/L}\right) \) and \( H \mapsto \mathcal{F}\left( H\right) \) give injective inclusion reversing correspondences between the set of fixed fields \( L \) with \( F \subseteq L \subseteq K \) and the set of subgroups of \( G \) of the form \( \operatorname{Gal}\left( {K/L}\right) \) for some \( L \) with \( F \subseteq L \subseteq K \). Let \( L \) be a subfield of \( K \) containing \( F \). Since \( K \) is Galois over \( F \), the extension \( K \) is normal and separable over \( F \). Thus, \( K \) is also normal and separable over \( L \), so \( K \) is Galois over \( L \). Hence, \( L = \mathcal{F}\left( {\operatorname{Gal}\left( {K/L}\right) }\right) \), so any intermediate field is a fixed field. Also, if \( H \) is a subgroup of \( G \), then \( H \) is a finite group, so \( H = \operatorname{Gal}\left( {K/\mathcal{F}\left( H\right) }\right) \) by Proposition 2.14. Every subgroup of \( G \) is therefore such a Galois group. The maps above then yield the desired correspondences. Recall that \( \left| {\operatorname{Gal}\left( {K/F}\right) }\right| = \left\lbrack {K : F}\right\rbrack \) if \( K \) is Galois over \( F \) by Proposition 2.14. Thus, if \( L \leftrightarrow H \), we have \( \left| H\right| = \left\lbrack {K : L}\right\rbrack \), since \( K \) is Galois over \( L \) and \( H = \operatorname{Gal}\left( {K/L}\right) \). Therefore, | Yes |
Which is the Galois group of the field \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) over \( \mathbb{Q} \)? | The subfield \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) is not Galois over \( \mathbb{Q} \), since the minimal polynomial of \( \sqrt[3]{2} \) does not split over \( \mathbb{Q}\left( \sqrt[3]{2}\right) \). Therefore, the corresponding subgroup is not normal in \( G \). However, every subgroup of an Abelian group is normal, so our Galois group is non-Abelian. Thus, \( G = \operatorname{Gal}\left( {\mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) /\mathbb{Q}}\right) \cong {S}_{3} \). We can also explicitly demonstrate this isomorphism. By the isomorphism extension theorem, there are \( \mathbb{Q} \)-automorphisms \( \sigma ,\tau \) of \( \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) \) with\n\n\[ \sigma : \sqrt[3]{2} \rightarrow \omega \sqrt[3]{2},\;\omega \rightarrow \omega ,\]\n\n\[ \tau : \sqrt[3]{2} \rightarrow \sqrt[3]{2},\;\omega \rightarrow {\omega }^{2}. \]\n\nIt is easy to check that \( \sigma \) has order \( 3,\tau \) has order 2, and \( {\sigma \tau } \neq {\tau \sigma } \). The subgroups of the Galois group are then\n\n\[ \text{(id)},\langle \sigma \rangle ,\langle \tau \rangle ,\langle {\sigma \tau }\rangle ,\left\langle {{\sigma }^{2}\tau }\right\rangle, G.\]\n\nThe corresponding fixed fields are\n\n\[ \mathbb{Q}\left( {\sqrt[3]{2},\omega }\right) ,\mathbb{Q}\left( \omega \right) ,\mathbb{Q}\left( \sqrt[3]{2}\right) ,\mathbb{Q}\left( {{\omega }^{2}\sqrt[3]{2}}\right) ,\mathbb{Q}\left( {\omega \sqrt[3]{2}}\right) ,\mathbb{Q}. \] | Yes |
Example 5.3 Let \( K = \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \). Then \( K \) is the splitting field of \( \left\{ {{x}^{2} - 2,{x}^{2} - 3}\right\} \) over \( \mathbb{Q} \) or, alternatively, the splitting field of \( \left( {{x}^{2} - 2}\right) \left( {{x}^{2} - 3}\right) \) over \( \mathbb{Q} \). The dimension of \( K/\mathbb{Q} \) is 4. The four automorphisms of \( K/\mathbb{Q} \) are given by | \[ \text{id} : \sqrt{2} \rightarrow \sqrt{2},\;\sqrt{3} \rightarrow \sqrt{3}\text{,} \] \[ \sigma : \sqrt{2} \rightarrow - \sqrt{2},\;\sqrt{3} \rightarrow \sqrt{3} \] \[ \tau : \sqrt{2} \rightarrow \sqrt{2},\;\sqrt{3} \rightarrow - \sqrt{3} \] \[ {\sigma \tau } : \sqrt{2} \rightarrow - \sqrt{2},\;\sqrt{3} \rightarrow - \sqrt{3}. \] This Galois group is Abelian and is isomorphic to \( \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \). The subgroups of \( G = \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) are \[ \text{(id)},\langle \sigma \rangle ,\langle \tau \rangle ,\langle {\sigma \tau }\rangle, G\text{.} \] The corresponding intermediate fields are \[ K,\mathbb{Q}\left( \sqrt{3}\right) ,\mathbb{Q}\left( \sqrt{2}\right) ,\mathbb{Q}\left( \sqrt{6}\right) ,\mathbb{Q}\text{.} \] | Yes |
Example 5.4 Let \( F = \mathbb{C}\left( t\right) \) be the rational function field in one variable over \( \mathbb{C} \), and let \( f\left( x\right) = {x}^{n} - t \in F\left\lbrack x\right\rbrack \) . The polynomial \( f \) is irreducible over \( F \) by the Eisenstein criterion, since \( F \) is the quotient field of the unique factorization domain \( \mathbb{C}\left\lbrack t\right\rbrack \) and \( t \) is an irreducible element of \( \mathbb{C}\left\lbrack t\right\rbrack \) . Let \( K \) be the splitting field of \( f \) over \( F \) . Then \( K = F\left( \alpha \right) \), where \( \alpha \) is any root of \( f\left( x\right) \) . | To see this, if \( \omega = \exp \left( {{2\pi i}/n}\right) \), then \( {\omega }^{n} = 1 \), so \( {\omega }^{i}\alpha \) is a root of \( f\left( x\right) \) for each \( i \) . There are exactly \( n \) distinct powers of \( \omega \), so the \( n \) distinct elements \( \alpha ,{\omega \alpha },\ldots ,{\omega }^{n - 1}\alpha \) are precisely the roots of \( f \) . All of these lie in \( F\left( \alpha \right) \) and generate \( F\left( \alpha \right) \), so \( K = F\left( \alpha \right) \) . The extension \( K/F \) is then Galois since \( f \) has no repeated roots. We see that \( \left\lbrack {K : F}\right\rbrack = \deg \left( f\right) = n \) . The isomorphism extension theorem tells us that there is an automorphism \( \sigma \) of \( K \) defined by \( \sigma \left( \alpha \right) = {\omega \alpha } \) . This formula yields that \( {\sigma }^{i}\left( \alpha \right) = {\omega }^{i}\alpha \) for each \( i \), so \( {\sigma }^{i}\left( \alpha \right) = \alpha \) if and only if \( n \) divides \( i \) . Thus, \( \sigma \) has order \( n \) in \( \operatorname{Gal}\left( {K/F}\right) \) . This forces \( \operatorname{Gal}\left( {K/F}\right) \) to be the cyclic group generated by \( \sigma \) . | Yes |
Theorem 5.5 (Natural Irrationalities) Let \( K \) be a finite Galois extension of \( F \), and let \( L \) be an arbitrary extension of \( F \) . Then \( {KL}/L \) is Galois and \( \operatorname{Gal}\left( {{KL}/L}\right) \cong \operatorname{Gal}\left( {K/K \cap L}\right) \) . Moreover, \( \left\lbrack {{KL} : L}\right\rbrack = \left\lbrack {K : K \cap L}\right\rbrack \) . | Proof. Define \( \theta : \operatorname{Gal}\left( {{KL}/L}\right) \rightarrow \operatorname{Gal}\left( {K/F}\right) \) by \( \theta \left( \sigma \right) = {\left. \sigma \right| }_{K} \) . This map is well defined since \( K \) is normal over \( F \), and \( \theta \) is a group homomorphism. The kernel of \( \theta \) is \( \left\{ {\sigma \in \operatorname{Gal}\left( {{KL}/L}\right) : {\left. \sigma \right| }_{K} = \mathrm{{id}}}\right\} \) . However, if \( \sigma \in \ker \left( \theta \right) \) , then \( {\left. \sigma \right| }_{L} = \) id and \( {\left. \sigma \right| }_{K} = \) id. Thus, the fixed field of \( \sigma \) contains both \( K \) and \( L \), so it contains \( {KL} \) . That means \( \sigma = \mathrm{{id}} \), so \( \theta \) is injective. Since the image of \( \theta \) is a subgroup of \( \operatorname{Gal}\left( {K/F}\right) \), this image is equal to \( \operatorname{Gal}\left( {K/E}\right) \), where \( E \) is the fixed field of this image. We show that \( E = K \cap L \) . If \( a \in K \cap L \), then \( a \) is fixed by \( {\left. \sigma \right| }_{K} \) for each \( \sigma \in \operatorname{Gal}\left( {{KL}/L}\right) \) . Therefore, \( a \in E \), so \( K \cap L \subseteq E \) . For the reverse inclusion, let \( a \in E \) . Then \( a \in K \) and \( {\left. \sigma \right| }_{K}\left( a\right) = a \) for all \( \sigma \in \operatorname{Gal}\left( {{KL}/L}\right) \) . Thus, \( \sigma \left( a\right) = a \) for all such \( \sigma \), so \( a \in L \) . This shows \( E \subseteq K \cap L \), and so \( E = K \cap L \) . We have thus proved that\n\n\[ \operatorname{Gal}\left( {{KL}/L}\right) \cong \operatorname{im}\left( \theta \right) = \operatorname{Gal}\left( {K/K \cap L}\right) \]\n\nThe degree formula follows immediately from this isomorphism. | Yes |
Theorem 5.6 (Primitive Element Theorem) A finite extension \( K/F \) is simple if and only if there are only finitely many fields \( L \) with \( F \subseteq L \subseteq K \) . | Proof. We prove this with the assumption that \( \left| F\right| = \infty \) . The case for finite fields requires a different proof, which we will handle in Section 6. Suppose that there are only finitely many intermediate fields of \( K/F \) . Since \( \left\lbrack {K : F}\right\rbrack < \infty \), we can write \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) for some \( {\alpha }_{i} \in K \) . We use induction on \( n \) ; the case \( n = 1 \) is trivial. If \( L = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n - 1}}\right) \) , then since any field between \( F \) and \( L \) is an intermediate field of \( K/F \) , by induction \( L = F\left( \beta \right) \) for some \( \beta \) . Then \( K = F\left( {{\alpha }_{n},\beta }\right) \) . For \( a \in F \), set \( {M}_{a} = F\left( {{\alpha }_{n} + {a\beta }}\right) \), an intermediate field of \( K/F \) . Since there are only finitely many intermediate fields of \( K/F \) but infinitely many elements of \( F \), there are \( a, b \in F \) with \( a \neq b \) and \( {M}_{a} = {M}_{b} \) . Therefore,\n\n\[ \beta = \frac{\left( {{\alpha }_{n} + {b\beta }}\right) - \left( {{\alpha }_{n} + {a\beta }}\right) }{b - a} \in {M}_{b}. \]\n\nHence, \( {\alpha }_{n} = \left( {{\alpha }_{n} + {b\beta }}\right) - {b\beta } \in {M}_{b} \), so \( K = F\left( {{\alpha }_{n},\beta }\right) = {M}_{b} \) . Thus, \( K \) is a simple extension of \( F \) .\n\nConversely, suppose that \( K = F\left( \alpha \right) \) for some \( \alpha \in F \) . Let \( M \) be a field with \( F \subseteq M \subseteq K \) . Then \( K = M\left( \alpha \right) \) . Let \( p\left( x\right) = \min \left( {F,\alpha }\right) \) and \( q\left( x\right) = \) \( \min \left( {M,\alpha }\right) \in M\left\lbrack x\right\rbrack \) Then \( q \) divides \( p \) in \( M\left\lbrack x\right\rbrack \) . Suppose that \( q\left( x\right) = {a}_{0} + \) \( {a}_{1}x + \cdots + {x}^{r} \), and set \( {M}_{0} = F\left( {{a}_{0},\ldots ,{a}_{r - 1}}\right) \subseteq M \) . Then \( q \in {M}_{0}\left\lbrack x\right\rbrack \), so \( \min \left( {{M}_{0},\alpha }\right) \) divides \( q \) . Thus,\n\n\[ \left\lbrack {K : M}\right\rbrack = \deg \left( q\right) \geq \deg \left( {\min \left( {{M}_{0},\alpha }\right) }\right) = \left\lbrack {K : {M}_{0}}\right\rbrack \]\n\n\[ = \left\lbrack {K : M}\right\rbrack \cdot \left\lbrack {M : {M}_{0}}\right\rbrack \]\n\nThis implies that \( \left\lbrack {M : {M}_{0}}\right\rbrack = 1 \), so \( M = {M}_{0} \) . Therefore, \( M \) is determined by \( q \) . However, there are only finitely many monic divisors of \( p \) in \( K\left\lbrack x\right\rbrack \), so there are only finitely many such \( M \) . | Yes |
Corollary 5.7 If \( K/F \) is finite and separable, then \( K = F\left( \alpha \right) \) for some \( \alpha \in K \) . | Proof. If \( K \) is finite and separable over \( F \), then \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) for some \( {\alpha }_{i} \) . Let \( N \) be the splitting field over \( F \) of \( \left\{ {\min \left( {F,{\alpha }_{i}}\right) : 1 \leq i \leq n}\right\} \) . Then \( N/F \) is Galois by Theorem 4.9 since each \( \min \left( {F,{\alpha }_{i}}\right) \) is separable over \( F \) . Moreover, \( K \subseteq N \) . By the fundamental theorem, the intermediate fields of \( N/F \) are in 1-1 correspondence with the subgroups of the finite group \( \operatorname{Gal}\left( {N/F}\right) \) . Any finite group has only finitely many subgroups, so \( N/F \) has only finitely many intermediate fields. In particular, \( K/F \) has only finitely many intermediate fields. Therefore, \( K = F\left( \alpha \right) \) for some \( \alpha \) by the primitive element theorem. | Yes |
Corollary 5.8 If \( K/F \) is finite and \( F \) has characteristic 0, then \( K = F\left( \alpha \right) \) for some \( \alpha \) . | Proof. This corollary follows immediately from the preceding corollary since any finite extension of a field of characteristic 0 is separable. | No |
Proposition 5.9 Let \( K \) be an algebraic extension of \( F \), and let \( N \) be the normal closure of \( K/F \). 1. The field \( N \) is a normal extension of \( F \) containing \( K \). Moreover, if \( M \) is a normal extension of \( F \) with \( K \subseteq M \subseteq N \), then \( M = N \). 2. If \( K = F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \), then \( N \) is the splitting field of the polynomials \( \min \left( {F,{a}_{1}}\right) ,\ldots ,\min \left( {F,{a}_{n}}\right) \) over \( F \). 3. If \( K/F \) is a finite extension, then so is \( N/F \). 4. If \( K/F \) is separable, then \( N/F \) is Galois. | Proof. Since \( N \) is a splitting field over \( F \) of a set of polynomials, \( N \) is normal over \( F \). It is clear that \( N \) contains \( K \). Suppose that \( M \) is a normal extension of \( F \) with \( K \subseteq M \subseteq N \). If \( a \in K \), then \( a \in M \), so by normality \( \min \left( {F, a}\right) \) splits over \( M \). However, if \( X \) is the set of roots of the polynomials \( \{ \min \left( {F, a}\right) : a \in K\} \), we have \( N = F\left( X\right) \). But since these polynomials split over \( M \), all of the roots of these polynomials lie in \( M \). Thus, \( X \subseteq M \), and so \( N = F\left( X\right) \subseteq M \). Therefore, \( M = N \). | Yes |
Corollary 5.10 Let \( K \) be an algebraic extension of \( F \), and let \( N \) be the normal closure of \( K/F \). If \( {N}^{\prime } \) is any normal extension of \( F \) containing \( K \), then there is an \( F \)-homomorphism from \( N \) to \( {N}^{\prime } \). Consequently, if \( {N}^{\prime } \) does not contain any proper subfield normal over \( F \) that contains \( K \), then \( N \) and \( {N}^{\prime } \) are \( F \)-isomorphic. | Proof. Suppose that \( {N}^{\prime } \) is normal over \( F \) and contains \( K \). Then \( \min \left( {F, a}\right) \) splits over \( {N}^{\prime } \) for each \( a \in K \). By the isomorphism extension theorem, the identity map on \( F \) extends to a homomorphism \( \sigma : N \rightarrow {N}^{\prime } \). Then \( \sigma \left( N\right) \) is a splitting field of \( \{ \min \left( {F, a}\right) : a \in K\} \) in \( {N}^{\prime } \), so \( \sigma \left( N\right) \) is normal over \( F \) and contains \( K \). Therefore, if \( {N}^{\prime } \) does not contain any proper subfield normal over \( F \) that contains \( K \), then \( \sigma \left( N\right) = {N}^{\prime } \), so \( N \) and \( {N}^{\prime } \) are \( F \)-isomorphic. | Yes |
Lemma 5.13 Let \( f\left( x\right) \in \mathbb{R}\left\lbrack x\right\rbrack \) .\n\n1. If \( f\left( x\right) = {x}^{2} - a \) for some \( a > 0 \), then \( f \) has a root in \( \mathbb{R} \) . Therefore, every nonnegative real number has a real square root.\n\n2. If \( \deg \left( f\right) \) is odd, then \( f \) has a root in \( \mathbb{R} \) . Consequently, the only odd degree extension of \( \mathbb{R} \) is \( \mathbb{R} \) itself. | Proof. Suppose that \( f\left( x\right) = {x}^{2} - a \) with \( a > 0 \) . Then \( f\left( 0\right) < 0 \) and \( f\left( u\right) > 0 \) for \( u \) sufficiently large. Therefore, there is a \( c \in \left\lbrack {0, u}\right\rbrack \) with \( f\left( c\right) = 0 \) by the intermediate value theorem. In other words, \( \sqrt{a} = c \in \mathbb{R} \) .\n\nFor part 2, suppose that the leading coefficient of \( f \) is positive. Then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}f\left( x\right) = \infty \text{ and }\mathop{\lim }\limits_{{x \rightarrow - \infty }}f\left( x\right) = - \infty .\n\]\n\nBy another use of the intermediate value theorem, there is a \( c \in \mathbb{R} \) with \( f\left( c\right) = 0 \) . If \( L/\mathbb{R} \) is an odd degree extension, take \( a \in L - \mathbb{R} \) . Then \( \mathbb{R}\left( a\right) /\mathbb{R} \) is also of odd degree, so \( \deg \left( {\min \left( {\mathbb{R}, a}\right) }\right) \) is odd. However, this polynomial has a root in \( \mathbb{R} \) by what we have just shown. Since this polynomial is irreducible, this forces \( \min \left( {\mathbb{R}, a}\right) \) to be linear, so \( a \in \mathbb{R} \) . Therefore, \( L = \mathbb{R} \) . | Yes |
Lemma 5.14 Every complex number has a complex square root. Therefore, there is no field extension \( N \) of \( \mathbb{C} \) with \( \left\lbrack {N : \mathbb{C}}\right\rbrack = 2 \) . | Proof. To prove this, we use the polar coordinate representation of complex numbers. Let \( a \in \mathbb{C} \), and set \( a = r{e}^{i\theta } \) with \( r \geq 0 \) . Then \( \sqrt{r} \in \mathbb{R} \) by Lemma 5.13, so \( b = \sqrt{r}{e}^{{i\theta }/2} \in \mathbb{C} \) . We have \( {b}^{2} = r{\left( {e}^{{i\theta }/2}\right) }^{2} = r{e}^{i\theta } = a \) . If \( N \) is an extension of \( \mathbb{C} \) with \( \left\lbrack {N : \mathbb{C}}\right\rbrack = 2 \), then there is an \( a \in \mathbb{C} \) with \( N = \mathbb{C}\left( \sqrt{a}\right) \) . But, the first part of the lemma shows that \( \mathbb{C}\left( \sqrt{a}\right) = \mathbb{C} \), so there are no quadratic extensions of \( \mathbb{C} \). | Yes |
Lemma 6.1 If \( K \) is a field and \( G \) is a finite subgroup of \( {K}^{ * } \), then \( G \) is cyclic. | Proof. Let \( n = \left| G\right| \) and \( m = \exp \left( G\right) \) . Then \( m \) divides \( n \) by Lagrange’s theorem. If \( g \in G \), then \( {g}^{m} = 1 \), so each element of \( G \) is a root of the polynomial \( {x}^{m} - 1 \) . This polynomial has at most \( m \) roots in the field \( K \) . However, \( {x}^{m} - 1 \) has at least the elements of \( G \) as roots, so \( n \leq m \) . Therefore, \( \exp \left( G\right) = \left| G\right| \), so \( G \) is cyclic. | Yes |
Corollary 6.4 If \( K/F \) is an extension of finite fields, then \( K \) is a simple extension of \( F \) . | Proof. By the previous corollary, the group \( {K}^{ * } \) is cyclic. Let \( \alpha \) be a generator of the cyclic group \( {K}^{ * } \) . Every nonzero element of \( K \) is a power of \( \alpha \) , so \( K = F\left( \alpha \right) \) . Therefore, \( K \) is a simple extension of \( F \) . | Yes |
Theorem 6.5 Let \( F \) be a finite field with \( \operatorname{char}\left( F\right) = p \), and set \( \left| F\right| = {p}^{n} \) . Then \( F \) is the splitting field of the separable polynomial \( {x}^{{p}^{n}} - x \) over \( {\mathbb{F}}_{p} \) . Thus, \( F/{\mathbb{F}}_{p} \) is Galois. Furthermore, if \( \sigma \) is defined on \( F \) by \( \sigma \left( a\right) = {a}^{p} \), then \( \sigma \) generates the Galois group \( \operatorname{Gal}\left( {F/{\mathbb{F}}_{p}}\right) \), so this Galois group is cyclic. | Proof. Let \( \left| F\right| = {p}^{n} \), so \( \left| {F}^{ * }\right| = {p}^{n} - 1 \) . By Lagrange’s theorem, if \( a \in {F}^{ * } \) , then \( {a}^{{p}^{n} - 1} = 1 \) . Multiplying by \( a \) gives \( {a}^{{p}^{n}} = a \) . This equation also holds for \( a = 0 \) . Therefore, the elements of \( F \) are roots of the polynomial \( {x}^{{p}^{n}} - x \) . However, this polynomial has at most \( {p}^{n} \) roots, so the elements of \( F \) are precisely the roots of \( {x}^{{p}^{n}} - x \) . This proves that \( F \) is the splitting field over \( {\mathbb{F}}_{p} \) of \( {x}^{{p}^{n}} - x \), and so \( F \) is normal over \( {\mathbb{F}}_{p} \) . Moreover, the derivative test shows that \( {x}^{{p}^{n}} - x \) has no repeated roots, so \( {x}^{{p}^{n}} - x \) is separable over \( {\mathbb{F}}_{p} \) . Thus, \( F \) is Galois over \( {\mathbb{F}}_{p} \) .\n\nDefine \( \sigma : F \rightarrow F \) by \( \sigma \left( a\right) = {a}^{p} \) . An easy computation shows that \( \sigma \) is an \( {\mathbb{F}}_{p} \) -homomorphism, and \( \sigma \) is surjective since \( F \) is finite. Hence, \( \sigma \) is an \( {\mathbb{F}}_{p} \) -automorphism of \( F \) . The fixed field of \( \sigma \) is \( \left\{ {a \in F : {a}^{p} = a}\right\} \supseteq {\mathbb{F}}_{p} \) . Each element in \( \mathcal{F}\left( \sigma \right) \) is a root of \( {x}^{p} - x \), so there are at most \( p \) elements in \( \mathcal{F}\left( \sigma \right) \) . This proves that \( {\mathbb{F}}_{p} = \mathcal{F}\left( \sigma \right) \), so \( \operatorname{Gal}\left( {F/{\mathbb{F}}_{p}}\right) \) is the cyclic group generated by \( \sigma \) . | Yes |
Corollary 6.6 Any two finite fields of the same size are isomorphic. | Proof. The proof of Theorem 6.5 shows that any two fields of order \( {p}^{n} \) are splitting fields over \( {\mathbb{F}}_{p} \) of \( {x}^{{p}^{n}} - x \), so the corollary follows from the isomorphic extension theorem. | No |
Corollary 6.7 If \( K/F \) is an extension of finite fields, then \( K/F \) is Galois with a cyclic Galois group. Moreover, if \( \operatorname{char}\left( F\right) = p \) and \( \left| F\right| = {p}^{n} \), then \( \operatorname{Gal}\left( {K/F}\right) \) is generated by the automorphism \( \tau \) defined by \( \tau \left( a\right) = {a}^{{p}^{n}} \) . | Proof. Say \( \left\lbrack {K : {\mathbb{F}}_{p}}\right\rbrack = m \) . Then \( \operatorname{Gal}\left( {K/{\mathbb{F}}_{p}}\right) \) is a cyclic group of order \( m \) by Theorem 6.5, so the order of the Frobenius automorphism \( \sigma \) of \( K \) is \( m \) . The group \( \operatorname{Gal}\left( {K/F}\right) \) is a subgroup of \( \operatorname{Gal}\left( {K/{\mathbb{F}}_{p}}\right) \), so it is also cyclic. If \( s = \left| {\operatorname{Gal}\left( {K/F}\right) }\right| \) and \( m = {ns} \), then a generator of \( \operatorname{Gal}\left( {K/F}\right) \) is \( {\sigma }^{n} \) . By induction, we see that the function \( {\sigma }^{n} \) is given by \( {\sigma }^{n}\left( a\right) = {a}^{{p}^{n}} \) . Also, since \( s = \left\lbrack {K : F}\right\rbrack \), we have that \( n = \left\lbrack {F : {\mathbb{F}}_{p}}\right\rbrack \), so \( \left| F\right| = {p}^{n} \) . | Yes |
Theorem 6.8 Let \( N \) be an algebraic closure of \( {\mathbb{F}}_{p} \) . For any positive integer \( n \), there is a unique subfield of \( N \) of order \( {p}^{n} \) . If \( K \) and \( L \) are subfields of \( N \) of orders \( {p}^{m} \) and \( {p}^{n} \), respectively, then \( K \subseteq L \) if and only if \( m \) divides \( n \) . When this occurs, \( L \) is Galois over \( K \) with Galois group generated by \( \tau \) , where \( \tau \left( a\right) = {a}^{{p}^{n}} \) . | Proof. Let \( n \) be a positive integer. The set of roots in \( N \) of the polynomial \( {x}^{{p}^{n}} - x \) has \( {p}^{n} \) elements and is a field. Thus, there is a subfield of \( N \) of order \( {p}^{n} \) . Since any two fields of order \( {p}^{n} \) in \( N \) are splitting fields of \( {x}^{{p}^{n}} - x \) over \( {\mathbb{F}}_{p} \) by Theorem 6.5, any subfield of \( N \) of order \( {p}^{n} \) consists exactly of the roots of \( {x}^{{p}^{n}} - x \) . Therefore, there is a unique subfield of \( N \) of order \( {p}^{n} \) . Let \( K \) and \( L \) be subfields of \( N \), of orders \( {p}^{m} \) and \( {p}^{n} \), respectively. First, suppose that \( K \subseteq L \) . Then \[ n = \left\lbrack {L : {\mathbb{F}}_{p}}\right\rbrack = \left\lbrack {L : K}\right\rbrack \cdot \left\lbrack {K : {\mathbb{F}}_{p}}\right\rbrack = m\left\lbrack {L : K}\right\rbrack \] so \( m \) divides \( n \) . Conversely, suppose that \( m \) divides \( n \) . Each element \( a \) of \( K \) satisfies \( {a}^{{p}^{m}} = a \) . Since \( m \) divides \( n \), each \( a \) also satisfies \( {a}^{{p}^{n}} = a \), so \( a \in L \) . This proves that \( K \subseteq L \) . When this happens \( L \) is Galois over \( K \) by Corollary 6.7. That corollary also shows that \( \operatorname{Gal}\left( {L/K}\right) \) is generated by \( \tau \) , where \( \tau \) is defined by \( \tau \left( a\right) = {a}^{\left| K\right| } \) . | Yes |
Corollary 6.9 Let \( F \) be a finite field, and let \( f\left( x\right) \) be a monic irreducible polynomial over \( F \) of degree \( n \) . 1. If \( a \) is a root of \( f \) in some extension field of \( F \), then \( F\left( a\right) \) is a splitting field for \( f \) over \( F \) . Consequently, if \( K \) is a splitting field for \( f \) over \( F \), then \( \left\lbrack {K : F}\right\rbrack = n \) . 2. If \( \left| F\right| = q \), then the set of roots of \( f \) is \( \left\{ {{a}^{{q}^{r}} : r \geq 1}\right\} \) . | Proof. Let \( K \) be a splitting field of \( f \) over \( F \) . If \( a \in K \) is a root of \( f\left( x\right) \) , then \( F\left( a\right) \) is an \( n \) -dimensional extension of \( F \) inside \( K \) . By Theorem 6.5, \( F\left( a\right) \) is a Galois extension of \( F \) ; hence, \( f\left( x\right) = \min \left( {F, a}\right) \) splits over \( F\left( a\right) \) . Therefore, \( F\left( a\right) \) is a splitting field of \( f \) over \( F \), so \( K = F\left( a\right) \) . This proves the first statement. For the second, we note that \( \operatorname{Gal}\left( {K/F}\right) = \langle \sigma \rangle \), where \( \sigma \left( c\right) = {c}^{q} \) for any \( c \in K \), by Theorem 6.8. Each root of \( f \) is then of the form \( {\sigma }^{r}\left( a\right) = {a}^{{q}^{r}} \) by the isomorphism extension theorem, which shows that the set of roots of \( f \) is \( \left\{ {{a}^{{q}^{r}} : r \geq 1}\right\} \) . | Yes |
Example 6.12 Let \( f\left( x\right) = {x}^{2} + 1 \) . If \( p \) is an odd prime, then we show that \( f \) is reducible over \( F = {\mathbb{F}}_{p} \) if and only if \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . | To prove this, if \( a \in F \) is a root of \( {x}^{2} + 1 \), then \( {a}^{2} = - 1 \), so \( a \) has order 4 in \( {F}^{ * } \) . By Lagrange’s theorem,4 divides \( \left| {F}^{ * }\right| = p - 1 \), so \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . Conversely, if \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), then 4 divides \( p - 1 \), so there is an element \( a \in {F}^{ * } \) of order 4, since \( {F}^{ * } \) is a cyclic group of order \( p - 1 \) . Thus, \( {a}^{4} = 1 \) and \( {a}^{2} \neq 1 \) . This forces \( {a}^{2} = - 1 \), so \( a \) is a root of \( f \) . | Yes |
Proposition 6.14 Let \( n \) be a positive integer. Then \( {x}^{{p}^{n}} - x \) factors over \( {\mathbb{F}}_{p} \) into the product of all monic irreducible polynomials over \( {\mathbb{F}}_{p} \) of degree a divisor of \( n \) . | Proof. Let \( F \) be a field of order \( {p}^{n} \) . Then \( F \) is the splitting field of \( {x}^{{p}^{n}} - x \) over \( {\mathbb{F}}_{p} \) by Theorem 6.5. Recall that \( F \) is exactly the set of roots of \( {x}^{{p}^{n}} - x \) . Let \( a \in F \), and set \( m = \left\lbrack {{\mathbb{F}}_{p}\left( a\right) : {\mathbb{F}}_{p}}\right\rbrack \), a divisor of \( \left\lbrack {F : {\mathbb{F}}_{p}}\right\rbrack \) . The polynomial \( \min \left( {{\mathbb{F}}_{p}, a}\right) \) divides \( {x}^{{p}^{n}} - x \), since \( a \) is a root of \( {x}^{{p}^{n}} - x \) . Conversely, if \( f\left( x\right) \) is a monic irreducible polynomial over \( {\mathbb{F}}_{p} \) of degree \( m \), where \( m \) divides \( n \) , let \( K \) be the splitting field of \( f \) over \( {\mathbb{F}}_{p} \) inside some algebraic closure of \( F \) . If \( a \) is a root of \( f \) in \( K \), then \( K = {\dot{\mathbb{F}}}_{p}\left( a\right) \) by Corollary 6.9. Therefore, \( \left\lbrack {K : {\mathbb{F}}_{p}}\right\rbrack = m \), so \( K \subseteq F \) by Theorem 6.8. Thus, \( a \in F \), so \( a \) is a root of \( {x}^{{p}^{n}} - x \) . Since \( f \) is irreducible over \( {\mathbb{F}}_{p} \), we have \( f = \min \left( {{\mathbb{F}}_{p}, a}\right) \), so \( f \) divides \( {x}^{{p}^{n}} - x \) . Since \( {x}^{{p}^{n}} - x \) has no repeated roots, \( {x}^{{p}^{n}} - x \) factors into distinct irreducible factors over \( {\mathbb{F}}_{p} \) . We have shown that the irreducible factors of \( {x}^{{p}^{n}} - x \) are exactly the irreducible polynomials of degree a divisor of \( n \) ; hence, the proposition is proven. | Yes |
The monic irreducible polynomials of degree 5 over \( {\mathbb{F}}_{2} \) can be determined by factoring \( {x}^{{2}^{5}} - x \), which we see factors as | \[ {x}^{{2}^{5}} - x = x\left( {x + 1}\right) \left( {{x}^{5} + {x}^{3} + 1}\right) \left( {{x}^{5} + {x}^{2} + 1}\right) \times \left( {{x}^{5} + {x}^{4} + {x}^{3} + x + 1}\right) \left( {{x}^{5} + {x}^{4} + {x}^{2} + x + 1}\right) \times \left( {{x}^{5} + {x}^{4} + {x}^{3} + {x}^{2} + 1}\right) \left( {{x}^{5} + {x}^{3} + {x}^{2} + x + 1}\right) . \] This factorization produces the six monic irreducible polynomials of degree 5 over \( {\mathbb{F}}_{2} \) . | Yes |
Proposition 7.2 Suppose that \( \operatorname{char}\left( F\right) \) does not divide \( n \), and let \( K \) be a splitting field of \( {x}^{n} - 1 \) over \( F \) . Then \( K/F \) is Galois, \( K = F\left( \omega \right) \) is generated by any primitive nth root of unity \( \omega \) , and \( \mathrm{{Gal}}\left( {K/F}\right) \) is isomorphic to a subgroup of \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) . Thus, \( \operatorname{Gal}\left( {K/F}\right) \) is Abelian and \( \left\lbrack {K : F}\right\rbrack \) divides \( \phi \left( n\right) \) . | Proof. Since \( \operatorname{char}\left( F\right) \) does not divide \( n \), the derivative test shows that \( {x}^{n} - 1 \) is a separable polynomial over \( F \) . Therefore, \( K \) is both normal and separable over \( F \) ; hence, \( K \) is Galois over \( F \) . Let \( \omega \in K \) be a primitive \( n \) th root of unity. Then all \( n \) th roots of unity are powers of \( \omega \), so \( {x}^{n} - 1 \) splits over \( F\left( \omega \right) \) . This proves that \( K = F\left( \omega \right) \) . Any automorphism of \( K \) that fixes \( F \) is determined by what it does to \( \omega \) . However, any automorphism restricts to a group automorphism of the set of roots of unity, so it maps the set of primitive \( n \) th roots of unity to itself. Any primitive \( n \) th root of unity in \( K \) is of the form \( {\omega }^{t} \) for some \( t \) relatively prime to \( n \) . Therefore, the map \( \theta : \operatorname{Gal}\left( {K/F}\right) \rightarrow {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) given by \( \sigma \mapsto t + n\mathbb{Z} \), where \( \sigma \left( \omega \right) = {\omega }^{t} \) , is well defined. If \( \sigma ,\tau \in \operatorname{Gal}\left( {K/F}\right) \) with \( \sigma \left( \omega \right) = {\omega }^{t} \) and \( \tau \left( \omega \right) = {\omega }^{s} \), then \( \left( {\sigma \tau }\right) \left( \omega \right) = \sigma \left( {\omega }^{s}\right) = {\omega }^{st} \), so \( \theta \) is a group homomorphism. The kernel of \( \theta \) is the set of all \( \sigma \) with \( \sigma \left( \omega \right) = \omega \) ; that is, \( \ker \left( \theta \right) = \langle \mathrm{{id}}\rangle \) . Thus, \( \theta \) is injective, so \( \operatorname{Gal}\left( {K/F}\right) \) is isomorphic to a subgroup of the Abelian group \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \), a group of order \( \phi \left( n\right) \) . This finishes the proof. | Yes |
Lemma 7.6 Let \( n \) be any positive integer. Then \( {x}^{n} - 1 = \mathop{\prod }\limits_{{d \mid n}}{\Psi }_{d}\left( x\right) \) . Moreover, \( {\Psi }_{n}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) . | Proof. We know that \( {x}^{n} - 1 = \prod \left( {x - \omega }\right) \), where \( \omega \) ranges over the set of all \( n \) th roots of unity. If \( d \) is the order of \( \omega \) in \( {\mathbb{C}}^{ * } \), then \( d \) divides \( n \), and \( \omega \) is a primitive \( d \) th root of unity. Gathering all the \( d \) th root of unity terms together in this factorization proves the first statement. For the second, we use induction on \( n \) ; the case \( n = 1 \) is clear since \( {\Psi }_{1}\left( x\right) = x - 1 \) . Suppose that \( {\Psi }_{d}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) for all \( d < n \) . Then from the first part, we have\n\n\[ \n{x}^{n} - 1 = \left( {\mathop{\prod }\limits_{{d \mid n, d < n}}{\Psi }_{d}\left( x\right) }\right) \cdot {\Psi }_{n}\left( x\right)\n\]\n\nSince \( {x}^{n} - 1 \) and \( \mathop{\prod }\limits_{{d \mid n}}{\Psi }_{d}\left( x\right) \) are monic polynomials in \( \mathbb{Z}\left\lbrack x\right\rbrack \), the division algorithm, Theorem 3.2 of Appendix A, shows that \( {\Psi }_{n}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) . | Yes |
Corollary 7.8 If \( K \) is a splitting field of \( {x}^{n} - 1 \) over \( \mathbb{Q} \), then \( \left\lbrack {K : \mathbb{Q}}\right\rbrack = \) \( \phi \left( n\right) \) and \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \cong {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) . Moreover, if \( \omega \) is a primitive nth root of unity in \( K \), then \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) = \left\{ {{\sigma }_{i} : \gcd \left( {i, n}\right) = 1}\right\} \), where \( {\sigma }_{i} \) is determined by \( {\sigma }_{i}\left( \omega \right) = {\omega }^{i} \) . | Proof. The first part of the corollary follows immediately from Proposition 7.2 and Theorem 7.7. The description of \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) is a consequence of the proof of Proposition 7.2. | Yes |
Example 7.9 Let \( K = {\mathbb{Q}}_{7} \), and let \( \omega \) be a primitive seventh root of unity in \( \mathbb{C} \). By Corollary 7.8, \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \cong {\left( \mathbb{Z}/7\mathbb{Z}\right) }^{ * } \), which is a cyclic group of order 6. The Galois group of \( K/\mathbb{Q} \) is \( \left\{ {{\sigma }_{1},{\sigma }_{2},{\sigma }_{3},{\sigma }_{4},{\sigma }_{5},{\sigma }_{6}}\right\} \), where \( {\sigma }_{i}\left( \omega \right) = \) \( {\omega }^{i} \). Thus, \( {\sigma }_{1} = \mathrm{{id}} \), and it is easy to check that \( {\sigma }_{3} \) generates this group. Moreover, \( {\sigma }_{i} \circ {\sigma }_{j} = {\sigma }_{ij} \), where the subscripts are multiplied modulo 7. The subgroups of \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) are then | \[ \text{(id),}\left\langle {\sigma }_{3}^{3}\right\rangle ,\left\langle {\sigma }_{3}^{2}\right\rangle ,\left\langle {\sigma }_{3}\right\rangle \] whose orders are \( 1,2,3 \), and 6, respectively. Let us find the corresponding intermediate fields. If \( L = \mathcal{F}\left( {\sigma }_{3}^{3}\right) = \mathcal{F}\left( {\sigma }_{6}\right) \), then \( \left\lbrack {K : L}\right\rbrack = \left| \left\langle {\sigma }_{6}\right\rangle \right| = 2 \) by the fundamental theorem. To find \( L \), we note that \( \omega \) must satisfy a quadratic over \( L \) and that this quadratic is \[ \left( {x - \omega }\right) \left( {x - {\sigma }_{6}\left( \omega \right) }\right) = \left( {x - \omega }\right) \left( {x - {\omega }^{6}}\right) . \] Expanding, this polynomial is \[ {x}^{2} - \left( {\omega + {\omega }^{6}}\right) x + \omega {\omega }^{6} = {x}^{2} - \left( {\omega + {\omega }^{6}}\right) x + 1. \] Therefore, \( \omega + {\omega }^{6} \in L \). If we let \( \omega = \exp \left( {{2\pi i}/7}\right) = \cos \left( {{2\pi }/7}\right) + i\sin \left( {{2\pi }/7}\right) \), then \( \omega + {\omega }^{6} = 2\cos \left( {{2\pi }/7}\right) \). Therefore, \( \omega \) satisfies a quadratic over \( \mathbb{Q}\left( {\cos \left( {{2\pi }/7}\right) }\right) \); hence, \( L \) has degree at most 2 over this field. This forces \( L = \mathbb{Q}\left( {\cos \left( {{2\pi }/7}\right) }\right) \). | Yes |
Example 8.2 Let \( F \) be any field, and let \( K = F\left( \sqrt{d}\right) \) for some \( d \in F - {F}^{2} \) . A convenient basis for \( K \) is \( \{ 1,\sqrt{d}\} \) . If \( \alpha = a + b\sqrt{d} \) with \( a, b \in F \), we determine the norm and trace of \( \alpha \) . | The linear transformation \( {L}_{\alpha } \) is equal to \( a{L}_{1} + b{L}_{\sqrt{d}} \), so we first need to find the matrix representations for \( {L}_{1} \) and \( {L}_{\sqrt{d}} \) . The identity transformation \( {L}_{1} \) has matrix \( \left( \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \) . For \( {L}_{\sqrt{d}} \) , we have\n\n\[ \n{L}_{\sqrt{d}}\left( 1\right) = \sqrt{d} = 0 \cdot 1 + 1 \cdot \sqrt{d} \n\]\n\n\[ \n{L}_{\sqrt{d}}\left( \sqrt{d}\right) = d = d \cdot 1 + 0 \cdot \sqrt{d} \n\]\n\nTherefore, the matrix for \( {L}_{\sqrt{d}} \) is \( \left( \begin{array}{ll} 0 & d \\ 1 & 0 \end{array}\right) \) . The matrix for \( {L}_{\alpha } \) is then\n\n\[ \na\left( \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) + b\left( \begin{array}{ll} 0 & d \\ 1 & 0 \end{array}\right) = \left( \begin{matrix} a & {bd} \\ b & a \end{matrix}\right) .\n\]\n\nFrom this we obtain \( {N}_{K/F}\left( {a + b\sqrt{d}}\right) = {a}^{2} - {b}^{2}d \) and \( {T}_{K/F}\left( {a + b\sqrt{d}}\right) = {2a} \) . In particular, \( {N}_{K/F}\left( \sqrt{d}\right) = - d \) and \( {T}_{K/F}\left( \sqrt{d}\right) = 0 \) . | Yes |
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