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Theorem 6.6. Let \( S \) and \( {S}^{\prime } \) be tori and \( w : S \rightarrow {S}^{\prime } \) a quasiconformal mapping. Then the bijective isometry \( \left\lbrack \varphi \right\rbrack \rightarrow \left\lbrack {\varphi \circ {w}^{-1}}\right\rbrack \) of \( {T}_{S} \) onto \( {T}_{{S}^{\prime }} \) is biholomorphic. | Proof. The isometry depends only on the homotopy class of \( w \) . Therefore, by Theorem 6.3, there is no loss of generality in assuming that \( w \) has a lift \( \zeta \rightarrow \lambda \left( {\zeta + \mu \bar{\zeta }}\right) \)\n\nWe may assume that the lift of \( \varphi \) is \( \zeta \rightarrow \zeta + z\bar{\zeta } \) . Then \( \varphi \circ {w}^{-1} \) has the complex dilatation \( {e}^{i\theta }\left( {z - \mu }\right) /\left( {1 - \bar{\mu }z}\right) \), where \( \theta = 2\arg \lambda \) . It follows that \( \left\lbrack \varphi \right\rbrack \rightarrow \left\lbrack {\varphi \circ {w}^{-1}}\right\rbrack \) induces the conformal self-mapping \( z \rightarrow {e}^{i\theta }\left( {z - \mu }\right) /\left( {1 - \bar{\mu }z}\right) \) of the unit disc. | Yes |
Theorem 7.1. The mapping \( f \rightarrow {\Theta f} \) is a continuous, linear surjection of \( A\left( 1\right) \) onto \( A\left( G\right) \) of norm \( \leq 1 \) . | Proof. Let \( f \in A\left( 1\right) ,{z}_{0} \in D \), and \( r = \left( {1 - \left| {z}_{0}\right| }\right) /3 \) . There exist \( G \) -equivalent Dirichlet regions \( {N}_{1},\ldots ,{N}_{k} \) such that \( D\left( {{z}_{0},{2r}}\right) \subset {\bar{N}}_{1} \cup \cdots \cup {\bar{N}}_{k} \) . For every \( z \in D\left( {{z}_{0}, r}\right) \) we have\n\n\[ \pi {r}^{2}\mathop{\sum }\limits_{{g \in G}}\left| {f\left( {g\left( z\right) }\right) }\right| {\left| {g}^{\prime }\left( z\right) \right| }^{2} \leq \mathop{\sum }\limits_{{g \in G}}{\int }_{{D}_{r}\left( z\right) }\left| {f \circ g}\right| {\left| {g}^{\prime }\right| }^{2} \]\n\n\[ \leq \mathop{\sum }\limits_{{g \in G}}\mathop{\sum }\limits_{{j = 1}}^{k}{\int }_{{N}_{j}}\left| {f \circ g}\right| {\left| {g}^{\prime }\right| }^{2} = \mathop{\sum }\limits_{{j = 1}}^{k}\mathop{\sum }\limits_{{g \in G}}{\int }_{g\left( {N}_{j}\right) }\left| f\right| = k{\int }_{D}\left| f\right| < \infty , \]\n\nwhere (7.1) yields the first inequality. It follows that \( \mathop{\sum }\limits_{{g \in G}}\left| {f \circ g}\right| {\left| {g}^{\prime }\right| }^{2} \) is locally uniformly convergent in \( D \) .\n\nFrom this we conclude that \( {\Theta f} \in A\left( G\right) \) . Also,\n\n\[ \parallel {\Theta f}\parallel = {\int }_{N}\left| {\Theta f}\right| \leq \mathop{\sum }\limits_{{g \in G}}{\int }_{N}\left| {f \circ g}\right| {\left| {g}^{\prime }\right| }^{2} = \mathop{\sum }\limits_{{g \in G}}{\int }_{g\left( N\right) }\left| f\right| = \parallel f\parallel . \]\n\nThis implies continuity of the mapping \( f \rightarrow {\Theta f} \), since its linearity is clear. To prove surjectivity requires more analysis, and we refer to Lehner [2]. | No |
Theorem 7.2. A differential \( \mu \) is infinitesimally trivial if and only if\n\n\[ \mathop{\lim }\limits_{{w \rightarrow 0}}\frac{{s}_{w\mu }\left( z\right) }{w} = 0 \] \n\nfor every \( z \in E \) . | Proof. In I.4.4 and II.3.2 we established the representation formula\n\n\[ {f}_{w\mu }\left( z\right) = z + \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}\left( z\right) {w}^{n} \]\n\nwith\n\n\[ {a}_{1}\left( z\right) = {T\mu }\left( z\right) = - \frac{1}{\pi }{\iint }_{D}\frac{\mu \left( \zeta \right) }{\zeta - z}{d\xi d\eta }. \]\n\nFrom this we obtain by straightforward computation\n\n\[ \mathop{\lim }\limits_{{w \rightarrow 0}}\frac{{s}_{w\mu }\left( z\right) }{w} = {a}_{1}^{\prime \prime \prime }\left( z\right) = - \frac{6}{\pi }{\iint }_{D}\frac{\mu \left( \zeta \right) }{{\left( \zeta - z\right) }^{4}}{d\xi d\eta } \]\n\nbecause we are now dealing with points \( z \) for which \( \left| z\right| > 1 \) . Hence\n\n\[ \mathop{\lim }\limits_{{w \rightarrow 0}}\frac{{S}_{w\mu }\left( z\right) }{w} = \mathop{\sum }\limits_{{n = 0}}^{\infty }{b}_{n}{z}^{-\left( {n + 4}\right) } \]\n\nwhere\n\n\[ {b}_{n} = - \frac{\left( {n + 1}\right) \left( {n + 2}\right) \left( {n + 3}\right) }{\pi }{\iint }_{D}\mu \left( \zeta \right) {\zeta }^{n}{d\xi d\eta }. \]\n\nTherefore, the theorem follows if we prove that\n\n\[ {\iint }_{D}\mu \left( \zeta \right) {\zeta }^{n}{d\xi d\eta } = 0,\;n = 0,1,2,\ldots , \]\n\n(7.4)\n\nif and only if \( \mu \in N\left( G\right) \) .\n\nIf \( {\Theta }_{n} \) is the \( \Theta \) -series for \( f\left( z\right) = {z}^{n} \), then by formula (7.3),\n\n\[ {\iint }_{D}\mu \left( \zeta \right) {\zeta }^{n}{d\xi d\eta } = {\iint }_{N}\mu \left( \zeta \right) {\Theta }_{n}\left( \zeta \right) {d\xi d\eta } \]\n\nBy Theorem 7.1, the function \( {\Theta }_{n} \) is an element of \( A\left( G\right) \) . Hence, if \( \mu \in N\left( G\right) \) , then (7.4) follows.\n\nConversely, if (7.4) holds, it follows from an approximation theorem of Carleman [1] that \( \mu \) is orthogonal in \( D \) to all functions \( f \in A\left( 1\right) \) . Formula (7.3) then shows that \( \mu \) is orthogonal in \( N \) to all functions \( {\Theta f} \) with \( f \) in \( A\left( 1\right) \) . By Theorem 7.1, every element of \( A\left( G\right) \) is of this form, and so \( \mu \in N\left( G\right) \) . | Yes |
Theorem 7.3. If \( {f}_{\mu } \) has an infinitesimally trivial complex dilatation, then\n\n\[ \n{\begin{Vmatrix}{S}_{{f}_{\mu } \mid E}\end{Vmatrix}}_{q} \leq 6\parallel \mu {\parallel }_{\infty }^{2}.\n\] | Proof. Fix a point \( z \in E \) and consider the holomorphic function\n\n\[ \nw \rightarrow \psi \left( w\right) = {\left( {\left| z\right| }^{2} - 1\right) }^{2}{S}_{{f}_{{w\mu }/\parallel \mu {\parallel }_{x}}}\left( z\right) \n\]\n\nin the unit disc. It follows from Theorem 7.2 that \( \psi \) has a zero of order at least 2 at the origin. Application of Schwarz’s lemma to \( \psi \), which is bounded in absolute value by 6 , yields therefore\n\n\[ \n\left| {\psi \left( w\right) }\right| \leq 6{\left| w\right| }^{2} \n\]\n\nSetting \( w = \parallel \mu {\parallel }_{\infty } \), we get back our function \( {f}_{\mu } \), and the estimate (7.5) follows. | Yes |
Lemma 7.1. Let \( v \in N\left( G\right) \) and \( \parallel v{\parallel }_{\infty } < 2 \) . Then, for \( 0 \leq t \leq 1/4 \), there is a \( \sigma \left( t\right) \in \) \( \left\lbrack {tv}\right\rbrack \) such that \( \parallel \sigma \left( t\right) {\parallel }_{\infty } \leq {12}{t}^{2} \) . | Proof. By Theorem 7.3,\n\n\[ \n{\begin{Vmatrix}{S}_{{f}_{tv} \mid E}\end{Vmatrix}}_{q} \leq {24}{t}^{2} \n\]\n\nFor \( 0 \leq t \leq 1/4 \), we have \( {24}{t}^{2} < 2 \) . Hence by formula (4.6), there is a complex dilatation \( \sigma \left( t\right) \in \left\lbrack {tv}\right\rbrack \) for which\n\n\[ \n\parallel \sigma \left( t\right) {\parallel }_{\infty } = \frac{1}{2}{\begin{Vmatrix}{S}_{{f}_{tv} \mid E}\end{Vmatrix}}_{q} \leq {12}{t}^{2}. \n\] | Yes |
Theorem 7.4. If \( {f}^{\mu } \) is extremal in its equivalence class, then\n\n\[ \parallel \mu {\parallel }_{G}^{ * } = \parallel \mu {\parallel }_{\infty }.\n\] | Proof. By the Hahn-Banach extension theorem (Dunford-Schwartz [1], p. 63), there is a linear functional \( \lambda \) in \( {L}^{1}\left( G\right) \) with \( \lambda \mid A\left( G\right) = {l}_{\mu } \) and \( \parallel \lambda \parallel = \parallel \mu {\parallel }_{G}^{ * } \) . From what was said in 7.1 we deduce the existence of a \( \kappa \in {L}^{\infty }\left( G\right) \), such that \( \lambda \left( \varphi \right) = \int {\kappa \varphi } \) and \( \parallel \lambda \parallel = \parallel \kappa {\parallel }_{\infty } \) . Since \( \parallel \mu {\parallel }_{G}^{ * } \leq \parallel \mu {\parallel }_{\infty } \), we have to prove that \( \parallel \mu {\parallel }_{\infty } \leq \parallel \kappa {\parallel }_{\infty } \) .\n\nFrom \( \lambda \mid A\left( G\right) = {l}_{\mu } \) we conclude that\n\n\[ {\int }_{N}\left( {\kappa - \mu }\right) \varphi = 0 \]\n\nfor \( \varphi \in A\left( G\right) \) . Hence \( \mu - \kappa \in N\left( G\right) \), and by Lemma 7.2, \( \parallel \mu {\parallel }_{\infty } \leq \parallel \kappa {\parallel }_{\infty } \) . | Yes |
Theorem 7.6. On a compact Riemann surface, a quasiconformal mapping with the smallest maximal dilatation in its homotopy class is a Teichmüller mapping. | Proof. Let \( \left( {\varphi }_{n}\right) \) be a weakly convergent Hamilton sequence with limit \( \varphi \) . For a compact Riemann surface of genus \( > 1 \), the Dirichlet regions \( N \) are relatively compact in \( D \) (IV.5.1). Therefore\n\n\[ 1 = \lim {\int }_{N}\left| {\varphi }_{n}\right| = {\int }_{N}\left| \varphi \right| \]\n\nHence, \( \left( {\varphi }_{n}\right) \) is not degenerate and the result we wanted to prove follows from Theorem 7.5.\n\nSince we assumed in this section that the Riemann surface \( S \) has a disc as its universal covering surface, our reasoning does not apply as such to a torus. But this case was thoroughly handled in section 6. It follows from Theorem 6.3 that also in the case of a torus the extremals are Teichmüller mappings. | No |
Theorem 9.1. The mapping\n\n\\[ \n\\left\\lbrack {k\\bar{\\varphi }/\\left| \\varphi \\right| }\\right\\rbrack \\rightarrow {k\\varphi } \n\\]\n\n(9.1)\n\nwhere \\( \\varphi \\in {Q}_{S},\\parallel \\varphi \\parallel = 1 \\), and \\( 0 \\leq k < 1 \\), is a homeomorphism of \\( {T}_{S} \\) onto the open unit ball of \\( {Q}_{s} \\) . | Proof. By our previous remarks,(9.1) is a well defined injection of \\( {T}_{S} \\) into the open unit ball of \\( {Q}_{S} \\). It is clearly also a surjection onto this ball.\n\nIn order to prove that (9.1) is continuous, we assume that \\( {p}_{n} = \\left\\lbrack {{k}_{n}\\bar{\\varphi }_{n}/\\left| {\\varphi }_{n}\\right| }\\right\\rbrack \\) converges to \\( p = \\left\\lbrack {k\\bar{\\varphi }/\\left| \\varphi \\right| }\\right\\rbrack \\) in \\( {T}_{S} \\). We use in \\( {T}_{S} \\) its \\( \\beta \\) -metric (see III.2.2), which is topologically equivalent to the Teichmüller metric \\( {\\tau }_{S} \\) of \\( {T}_{S} \\). By Theorem 8.2, we have \\( {\\beta }_{S}\\left( {{p}_{n},0}\\right) = {k}_{n},{\\beta }_{S}\\left( {p,0}\\right) = k \\). Hence, by the triangle inequality \\( \\left| {{k}_{n} - k}\\right| \\leq {\\beta }_{S}\\left( {{p}_{n}, p}\\right) \\). It follows that \\( {k}_{n} \\rightarrow k \\). If \\( k = 0 \\), this implies that \\( \\parallel {k}_{n}{\\varphi }_{n} - {k\\varphi }\\parallel \\rightarrow 0 \\).\n\nAssuming that \\( k > 0 \\) we show that \\( {\\varphi }_{n} \\rightarrow \\varphi \\) in \\( {Q}_{S} \\). Let us lift the differentials \\( {\\varphi }_{n} \\) and \\( \\varphi \\) to the universal covering surface \\( H \\). In \\( H \\), the analytic functions \\( {\\varphi }_{n}, n = 1,2,\\ldots \\), constitute a normal family. If \\( \\begin{Vmatrix}{{\\varphi }_{n} - \\varphi }\\end{Vmatrix} \\) does not tend to 0 as \\( n \\rightarrow \\infty \\), there is a subsequence \\( \\left( {\\varphi }_{{n}_{i}}\\right) \\) and a differential \\( \\psi \\in {Q}_{S},\\psi \\neq \\varphi \\), \\( \\parallel \\psi \\parallel = 1 \\), such that \\( \\begin{Vmatrix}{{\\varphi }_{{n}_{i}} - \\psi }\\end{Vmatrix} \\rightarrow 0 \\). Then \\( {k}_{{n}_{i}}\\bar{\\varphi }_{{n}_{i}}/\\left| {\\varphi }_{{n}_{i}}\\right| \\rightarrow k\\bar{\\psi }/\\left| \\psi \\right| \\) a.e. By Lemma 3.1, \\( \\left\\lbrack {k\\bar{\\varphi }/\\left| \\varphi \\right| }\\right\\rbrack = \\left\\lbrack {k\\bar{\\psi }/\\left| \\psi \\right| }\\right\\rbrack \\). Hence \\( \\varphi = \\psi \\), which is a contradiction, and so \\( \\lim \\begin{Vmatrix}{{\\varphi }_{n} - \\varphi }\\end{Vmatrix} = 0 \\). From\n\n\\[ \n\\begin{Vmatrix}{{k}_{n}{\\varphi }_{n} - {k\\varphi }}\\end{Vmatrix} \\leq \\left| {{k}_{n} - k}\\right| + \\begin{Vmatrix}{{\\varphi }_{n} - \\varphi }\\end{Vmatrix} \n\\]\n\nwe finally see that \\( {k}_{n}{\\varphi }_{n} \\rightarrow {k\\varphi } \\), i.e., that (9.1) is continuous.\n\nThe continuity of the inverse of (9.1) follows from the invariance of domain. The space \\( {Q}_{S} \\) is finite dimensional and \\( {T}_{S} \\) is homeomorphic with an open subset of \\( {Q}_{S} \\) (Theorems 4.3 and 4.8). Hence (9.1), as a continuous injection of \\( {T}_{S} \\) into \\( {Q}_{S} \\), is a homeomorphism of \\( {T}_{S} \\) onto its image.\n\nWe call (9.1) the Teichmüller imbedding. | Yes |
Theorem 9.2. The mapping\n\n\\[ \n\\left\\lbrack {k\\bar{\\varphi }/\\left| \\varphi \\right| }\\right\\rbrack \\rightarrow \\left( {k{x}_{1},\\ldots, k{x}_{{3p} - 3}, k{y}_{1},\\ldots, k{y}_{{3p} - 3}}\\right) ,\n\\]\n\n(9.2)\n\nwhere \\( \\varphi \\in {Q}_{S},\\parallel \\varphi \\parallel = 1 \\), and \\( 0 \\leq k < 1 \\), is a homeomorphism of \\( {T}_{S} \\) onto the open unit ball of the euclidean space \\( {\\mathbb{R}}^{{6p} - 6} \\) . | The validity of this statement follows immediately from Theorem 9.1, by virtue of the fact that all norms in \\( {Q}_{S} \\) define the same topology. | No |
Lemma 9.1. Let \( {f}_{1} : S \rightarrow {S}_{1} \) and \( {f}_{2} : S \rightarrow {S}_{2} \) be Teichmüller mappings determined by \( \left( {{\varphi }_{1},{k}_{1}}\right) \) and \( \left( {{\varphi }_{2},{k}_{2}}\right) \) such that \( {\varphi }_{2}/{\varphi }_{1} \) is a constant. Then \( {f}_{2} \circ {f}_{1}^{-1} \) is a Teich-müller mapping. Up to constants, the initial differential of \( {f}_{2} \circ {f}_{1}^{-1} \) agrees with that of \( {f}_{1}^{-1} \), and the terminal differential with that of \( {f}_{2} \) . | Proof. Let \( p \in S \) be a regular point for \( {\varphi }_{1} \) and \( {\varphi }_{2} \) . We denote by \( \zeta \) and \( {\zeta }^{ * } \) natural parameters for \( {\varphi }_{1} \) and \( {\varphi }_{2} \) at \( p \) vanishing at \( p \), and by \( {\zeta }_{1} \) and \( {\zeta }_{2} \) the corresponding natural parameters for the terminal differentials of \( {f}_{1} \) and \( {f}_{2} \) at \( {f}_{1}\left( p\right) \) and \( {f}_{2}\left( p\right) \) .\n\nLet us consider the projection mapping \( {\zeta }_{1} \rightarrow {\zeta }_{2} \) of \( {f}_{2} \circ {f}_{1}^{-1} \) . It can be decomposed into mappings \( {\zeta }_{1} \rightarrow \zeta ,\zeta \rightarrow {\zeta }^{ * } \), and \( {\zeta }^{ * } \rightarrow {\zeta }_{2} \) (see Fig. 16). Here \( {\zeta }_{1} \rightarrow \zeta \) and \( {\zeta }^{ * } \rightarrow {\zeta }_{2} \) are stretchings. Furthermore, \( \zeta \rightarrow {\zeta }^{ * } \) is conformal and \( d{\zeta }^{ * }/{d\zeta } = \sqrt{{\varphi }_{2}/{\varphi }_{1}} \) . Since \( {\varphi }_{2}/{\varphi }_{1} \) is constant, we see that \( {\zeta }^{ * }/\zeta \) is constant. It follows that \( {\zeta }_{1} \rightarrow {\zeta }_{2} \) is an affine transformation. Explicitly, we write \( {\varphi }_{2}/{\varphi }_{1} = \) \( {a}^{2}{e}^{-{i\theta }}, a > 0,0 \leq \theta < {2\pi } \) . Then we can choose \( {\zeta }^{ * } = a{e}^{-{i\theta }/2}\zeta \) .\n\nThe stretchings \( {\zeta }_{1} \rightarrow \zeta \) and \( {\zeta }^{ * } \rightarrow {\zeta }_{2} \) can be determined from (9.3) and (8.2). It follows that\n\n\[{\zeta }_{2} = \frac{a{e}^{-{i\theta }/2}}{\left( {1 + {k}_{1}}\right) \left( {1 - {k}_{2}}\right) }\left\lbrack {\left( {1 - {k}_{1}{k}_{2}{e}^{i\theta }}\right) {\zeta }_{1} + \left( {{k}_{2}{e}^{i\theta } - {k}_{1}}\right) {\bar{\zeta }}_{1}}\right\rbrack .\](9.4)\n\nSince \( {\zeta }_{1} \) is the natural parameter of the terminal differential of \( {f}_{1} \), we see from (9.4) that \( {f}_{2} \circ {f}_{1}^{-1} \) is a Teichmüller mapping whose initial differential is a constant times the initial differential of \( {f}_{1}^{-1} \) . The result about the terminal differential of \( {f}_{2} \circ {f}_{1}^{-1} \) is obtained if we interchange the roles of \( {f}_{1} \) and \( {f}_{2} \) . | Yes |
Theorem 9.3. The mapping \( z \rightarrow \left\lbrack {z\bar{\varphi }/\left| \varphi \right| }\right\rbrack \) is an isometry of the hyperbolic unit disc onto the Teichmüller disc determined by \( \varphi \) . | Proof. For distances from the origin, isometry is clear:\n\n\[{\tau }_{S}\left( {0,\left\lbrack {z\bar{\varphi }/\left| \varphi \right| }\right\rbrack }\right) = \frac{1}{2}\log \frac{1 + \left| z\right| }{1 - \left| z\right| } = h\left( {0, z}\right) ,\]\n\nwhere \( h \) denotes the hyperbolic distance in the unit disc.\n\nIn the general case, let \( {f}_{1} \) and \( {f}_{2} \) be the Teichmüller mappings with complex dilatations \( {z}_{1}\bar{\varphi }/\left| \varphi \right| \) and \( {z}_{2}\bar{\varphi }/\left| \varphi \right| \) . By Lemma 9.1, the composition \( {f}_{2} \circ {f}_{1}^{-1} \) is also a Teichmüller mapping. From (9.5) (or from formula (I.4.4)) we see that its complex dilatation has the absolute value \( \left| {\left( {{z}_{1} - {z}_{2}}\right) /\left( {1 - {\bar{z}}_{1}{z}_{2}}\right) }\right| \) . Therefore, by Theorem 8.2,\n\n\[{\tau }_{S}\left( {\left\lbrack {{z}_{1}\bar{\varphi }/\left| \varphi \right| }\right\rbrack ,\left\lbrack {{z}_{2}\bar{\varphi }/\left| \varphi \right| }\right\rbrack }\right) = \frac{1}{2}\log \frac{\left| {1 - {\bar{z}}_{1}{z}_{2}}\right| + \left| {{z}_{1} - {z}_{2}}\right| }{\left| {1 - {\bar{z}}_{1}{z}_{2}}\right| - \left| {{z}_{1} - {z}_{2}}\right| } = h\left( {{z}_{1},{z}_{2}}\right) .\] | Yes |
Theorem 9.4. In the space \( {T}_{p}, p \geq 1 \), the hyperbolic metric and the Teichmüller metric are the same. | Proof. For \( p = 1 \), the theorem follows from (9.7) and the fact that \( {T}_{p} \) can then be identified with the hyperbolic unit disc (see 6.7).\n\nFor \( p > 1 \), let us consider a model \( {T}_{S} \) of \( {T}_{p} \) . In studying the distances between two points \( {q}_{1} \) and \( {q}_{2} \) of \( {T}_{S} \), we may assume, by replacing \( {T}_{S} \) by another model of \( {T}_{p} \), that \( {q}_{1} \) lies at the origin. Let \( {q}_{2} = \left\lbrack {k\bar{\varphi }/\left| \varphi \right| }\right\rbrack \) .\n\nLet us apply again the holomorphic mapping (9.6) of the unit disc into \( {T}_{S} \) . By Theorem 9.3,\n\n\[ {\tau }_{S}\left( {{q}_{1},{q}_{2}}\right) = h\left( {{z}_{1},{z}_{2}}\right) \]\n\n(9.8)\n\nwhere \( {z}_{1} \) and \( {z}_{2} \) are the preimages of \( {q}_{1} \) and \( {q}_{2} \) under (9.6). Here the holomorphic mapping of \( D \) into \( {T}_{S} \) depends on the given points. Therefore, we can only infer from (9.8) that the Teichmüller metric \( {\tau }_{S} \) is greater than or equal to the hyperbolic metric of \( {T}_{S} \) .\n\nEquality is in fact more difficult to establish; for the proof we refer to Royden [1]. | No |
Theorem 9.5. The Teichmüller space of a Riemann surface of type \( \left( {p, n}\right) \) , \( {2p} - 2 + n > 0 \), is homeomorphic to the euclidean space \( {\mathbb{R}}^{{6p} - 6 + {2n}} \) . | The proof can be reduced to the case of a compact surface. The idea is to construct suitable coverings of \( S \) branched at the points of \( S \smallsetminus {S}_{0} \) . The procedure is explained in Ahlfors [1], pp. 20-23. | No |
Proposition 2.1. Let \( \\left( {X,\\mathcal{T}}\\right) \) be a topological space, let \( C \\subseteq X \) be closed, and let \( x \\in X \) . If there is a sequence of points in \( C \) that converges to \( x \), then \( x \\in C \) . | Proof. Exercise. (This is essentially the same as the second result mentioned in the previous section.) | No |
Proposition 2.4. Let \( \left( {X,\mathcal{T}}\right) \) be a topological space and let \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \) be a sequence in \( X \) (not necessarily convergent). Let \( {\left\{ {x}_{{n}_{k}}\right\} }_{k = 1}^{\infty } \) be a subsequence of our sequence that converges to a point \( x \) . Then \( x \) is an accumulation point of the original sequence. | Proof. Let \( {\left\{ {x}_{{n}_{k}}\right\} }_{k = 1}^{\infty } \) be a subsequence as in the proposition, converging to \( x \) . Let \( U \) be an open set containing \( x \) . Then by definition of sequence convergence there is an \( N \in \mathbb{N} \) such that for all \( k \geq N,{x}_{{n}_{k}} \in U \) . Then the points \( {x}_{{n}_{N}},{x}_{{n}_{N + 1}},{x}_{{n}_{N + 2}},{x}_{{n}_{N + 3}},\ldots \) are all in \( U \), and so \( x \) is an accumulation point of the sequence. | Yes |
4. If \( \left( {{\mathcal{D}}_{1},{ \leq }_{1}}\right) \) and \( \left( {{\mathcal{D}}_{2},{ \leq }_{2}}\right) \) are directed sets, then \( \left( {{\mathcal{D}}_{1} \times {\mathcal{D}}_{2}, \leq }\right) \) is a directed set where \( \leq \) is defined by | \[ \left( {a, b}\right) \leq \left( {x, y}\right) \text{ if and only if }a{ \leq }_{1}x\text{ and }b{ \leq }_{2}y. \] (Show this!) | No |
Theorem 3.6. Let \( \left( {X,\mathcal{T}}\right) \) be a topological space, and let \( A \subseteq X \) . Then \( x \in \bar{A} \) if and only if there is a net \( w : \mathcal{D} \rightarrow A \) such that \( w \rightarrow x \) . | Proof. \( \left( \Leftarrow \right) \) . This proof is essentially the same as for sequences.\n\nSuppose \( w : \mathcal{D} \rightarrow A \) is a net such that \( w \rightarrow x \) . We want to show \( x \in \bar{A} \) . So fix an open set \( U \) containing \( x \) . By definition of net convergence, there is a \( d \in \mathcal{D} \) such that \( w\left( e\right) \in U \) for all \( e \geq d \) . Since \( w\left( e\right) \in A \) for all \( e \in \mathcal{D} \), in particular the intersection \( U \cap A \) is nonempty.\n\n\( \left( \Rightarrow \right) \) . Here is where nets help you.\n\nSuppose \( x \in \bar{A} \) . We need to show that there is a net \( w \) that converges to \( x \) . To do this we first need to define a directed set to be the domain of our net.\n\nLet \( {\mathcal{D}}_{x} = \{ U \in \mathcal{T} : x \in U\} \) . Equipped with the superset relation, this is a directed set. To be clear, we are saying \( U \leq V \) if and only if \( U \supseteq V \) ; going up in the order on \( {\mathcal{D}}_{x} \) is like \ | No |
Corollary 3.7. A subset \( A \) of a topological space \( \left( {X,\mathcal{T}}\right) \) is closed if and only if the limit points of all convergent nets in \( A \) are again in \( A \) . | Proof. By the previous result, we know that the closure \( \bar{A} \) of a set \( A \) is precisely the set of all limit points of nets in \( A \) . The result then follows from the fact that \( A \) is closed if and only if \( A = \bar{A} \) . | Yes |
Theorem 3.8. Let \( \left( {X,\mathcal{T}}\right) \) be a topological space. Then the space is Hausdorff if and only if every net in \( X \) converges to at most one point. | Proof. \( \left( \Rightarrow \right) \) . This proof is essentially the same as the proof for sequences.\n\nSuppose \( \left( {X,\mathcal{T}}\right) \) is Hausdorff. Let \( w : \mathcal{D} \rightarrow X \) be a net that converges to a point \( x \), and suppose that \( y \neq x \) . We want to show that \( w \nrightarrow y \) . Choose disjoint open sets \( U \) and \( V \) containing \( x \) and \( y \), respectively. By definition of net convergence, there is some tail \( {T}_{a} \) of the net in \( U \) , and therefore this tail is disjoint from \( V \) . This means the net cannot converge to \( y \), since if some other tail \( {T}_{b} \) was in \( V \), we could use directedness to find a \( c \in \mathcal{D} \) such that \( a \leq c \) and \( b \leq c \), from which it would follow that the tail \( {T}_{c} \) would have to be in both \( U \) and \( V \), which is impossible.\n\n\( \left( \Leftarrow \right) \) . We prove this by contrapositive. Suppose \( \left( {X,\mathcal{T}}\right) \) is not Hausdorff, and we will show that there is a net converging to more than one point.\n\nFind two points \( x, y \in X \) such that any pair of open sets \( U \) and \( V \) containing \( x \) and \( y \), respectively, intersect one another. Again using the idea of the directed set \( {\mathcal{D}}_{x} = \{ U \in \mathcal{T} : x \in U\} \) with the superset relation, define the directed set \( \mathcal{D} = {\mathcal{D}}_{x} \times {\mathcal{D}}_{y} \), with\n\n\[ \left( {{U}_{1},{V}_{1}}\right) \leq \left( {{U}_{2},{V}_{2}}\right) \text{if and only if}{U}_{1} \supseteq {U}_{2}\text{and}{V}_{1} \supseteq {V}_{2}\text{.} \]\n\nFor each \( U \in {\mathcal{D}}_{x} \) and \( V \in {\mathcal{D}}_{y} \), fix a point \( {x}_{U, V} \in U \cap V \) . Then the function \( w : \mathcal{D} \rightarrow X \) given by \( w\left( {U, V}\right) = {x}_{U, V} \) is a net that converges to both \( x \) and \( y \) . (Prove this!) | No |
Proposition 5.4. Any collection \( \mathcal{S} \subseteq \mathcal{P}\left( X\right) \) with the finite intersection property generates a unique, smallest filter that contains it. | Proof. Exercise. (Hint: First add finite intersections, then add supersets. Prove the resulting collection is a filter.) | No |
If \( \left( {X,\mathcal{T}}\right) \) is a topological space and \( x \in X \), the collection \( \{ U \in \mathcal{T} : x \in U\} \) has the finite intersection property (it's actually closed under finite intersections, by definition of a topology). | The filter it generates is the neighbourhood filter introduced in Example 5.2.2. | No |
Proposition 5.6. Let \( X \) be a set and \( \mathcal{U} \) a filter on \( X \). Then \( \mathcal{U} \) is an ultrafilter if and only if for any subset \( A \subseteq X \), either \( A \in \mathcal{U} \) or \( X \smallsetminus A \in \mathcal{U} \). | Proof. \( \left( \Rightarrow \right) \) Let \( \mathcal{U} \) be an ultrafilter on \( X \), and suppose \( A \) is a nonempty subset of \( X \) such that \( A \notin \mathcal{U} \). We want to show that \( X \smallsetminus A \in \mathcal{U} \).\n\nBy the maximality of \( \mathcal{U} \), it must be the case that \( \mathcal{U} \cup \{ A\} \) is not a filter, and moreover that it does not have the finite intersection property (if it did, it would generate a filter that contains \( \mathcal{U} \), again contradicting maximality). That means there is a \( B \in \mathcal{U} \) such that \( A \cap B = \varnothing \). But then \( B \subseteq X \smallsetminus A \), and therefore \( X \smallsetminus A \in \mathcal{U} \) since \( \mathcal{U} \) is closed upwards.\n\n\( \left( \Leftarrow \right) \) Exercise. (Hint: Suppose \( \mathcal{U} \) is properly contained in another filter, so in particular there is some set \( A \) such that \( \mathcal{U} \sqcup \{ A\} \) is contained in this larger filter. Use this set to contradict the property you assumed.) | No |
Corollary 5.7. Let \( \mathcal{U} \) be an ultrafilter on \( X \), and let \( A \in \mathcal{U} \). Given a subset \( B \subseteq A \), either \( B \in \mathcal{U} \) or \( A \smallsetminus B \in \mathcal{U} \). | ## Proof. Exercise. | No |
Proposition 5.10. Let \( \left( {X,\mathcal{T}}\right) \) be a topological space. If \( \mathcal{F} \) is a filter on \( X \) and \( \mathcal{F} \rightarrow x \), then \( x \) is an accumulation point of \( \mathcal{F} \) . Conversely, if \( x \) is an accumulation point of an ultrafilter \( \mathcal{U} \) on \( X \), then \( \mathcal{U} \rightarrow x \) . | Proof. First, suppose \( \mathcal{F} \) is a filter and \( \mathcal{F} \rightarrow x \) . Then \( {\mathcal{F}}_{x} \subseteq \mathcal{F} \), and so in particular for any open set \( U \in {\mathcal{F}}_{x} \) and any \( F \in \mathcal{F}, U \cap F \neq \varnothing \) since \( \mathcal{F} \) is closed under finite intersections and \( \varnothing \notin \mathcal{F} \) .\n\nSecond, suppose \( x \) is an accumulation point of an ultrafilter \( \mathcal{U} \) on \( X \) . We want to show that \( {\mathcal{F}}_{x} \subseteq \mathcal{U} \), so fix \( F \in {\mathcal{F}}_{x} \) . Since \( x \) is an accumulation point of \( \mathcal{U} \) and \( F \) contains an open set containing \( x \), we have that \( U \cap F \neq \varnothing \) for every \( U \in \mathcal{U} \) . But then the collection \( \mathcal{U} \cup \{ F\} \) has the finite intersection property, and so it must be contained in some filter. That filter must be \( \mathcal{U} \) itself, since \( \mathcal{U} \) is not properly contained in any other filter on \( X \) . | Yes |
Theorem 6.3. Let \( \left( {X,\mathcal{T}}\right) \) be a topological space, and let \( x \in X \). 1. If \( w : \mathcal{D} \rightarrow X \) is a net, then \( w \rightarrow x \) if and only if the derived filter \( {\mathcal{F}}_{w} \rightarrow x \). 2. If \( \mathcal{F} \) is a filter on \( X \), then \( \mathcal{F} \rightarrow x \) if and only if every derived net of \( \mathcal{F} \) converges to \( x \). | Proof. 1. Exercise. (There's almost nothing to do here. It's immediate from the definitions.) 2. ( \( \Rightarrow \) ). Assume \( \mathcal{F} \rightarrow x \), and let \( w : \mathcal{F} \rightarrow X \) be any derived net of \( \mathcal{F} \). Fix an open set \( U \) containing \( x \). We want to show that a tail of the net is in \( U \). By assumption, \( U \in \mathcal{F}\left( {\mathcal{F} \rightarrow x\text{implies that}\mathcal{F}\text{contains every open set containing}x}\right) \). We will show that the tail \( {T}_{U} \subseteq U \). Indeed, given \( V \in \mathcal{F} \) with \( U \leq V \) (ie. \( V \subseteq U \)), we immediately have \( w\left( V\right) \in V \subseteq U \). So \( {T}_{U} \subseteq U \), and therefore \( w \rightarrow x \). \( \left( \Leftarrow \right) \). Assume every derived net \( w : \mathcal{F} \rightarrow X \) converges to \( x \), and assume for a contradiction that \( \mathcal{F} \nrightarrow x \). This means that there is some \( A \subseteq X \) containing an open set \( U \) containing \( x \) such that \( A \notin \mathcal{F} \). In particular this means that \( U \notin \mathcal{F} \) or in other words \( F \neq U \) for all \( F \in \mathcal{F} \). Let \( w \) be any derived net of \( \mathcal{F} \) such that \( w\left( V\right) \in V \smallsetminus U \) for all \( V \in \mathcal{F} \). Then clearly this derived net does not converge to \( x \) (no point in \( \mathcal{F} \) gets mapped into \( U \) by \( w \), let alone a tail of \( w \) ), contradicting the assumption that all derived nets of \( \mathcal{F} \) do converge to \( x \). | No |
Proposition 1. Let \( A \) be a ring, \( K \) its quotient field, and \( x \) algebraic over \( K \) . Then there exists an element \( c \neq 0 \) of \( A \) such that \( {cx} \) is integral over \( A \) . | Proof. There exists an equation\n\n\[ \n{a}_{n}{x}^{n} + \cdots + {a}_{0} = 0 \n\]\n\nwith \( {a}_{i} \in A \) and \( {a}_{n} \neq 0 \) . Multiply it by \( {a}_{n}^{n - 1} \) . Then\n\n\[ \n{\left( {a}_{n}x\right) }^{n} + \cdots + {a}_{0}{a}_{n}^{n - 1} = 0 \n\]\n\nis an integral equation for \( {a}_{n}x \) over \( A \) . | Yes |
Proposition 2. If \( B \) is integral over \( A \) and finitely generated as an A-algebra, then \( B \) is a finitely generated \( A \) -module. | Proof. We may prove this by induction on the number of ring generators, and thus we may assume that \( B = A\left\lbrack x\right\rbrack \) for some element \( x \) integral over \( A \) . But we have already seen that our assertion is true in that case. | No |
Proposition 3. Let \( A \subset B \subset C \) be three rings. If \( B \) is integral over \( A \) and \( C \) is integral over \( B \), then \( C \) is integral over \( A \) . | Proof. Let \( x \in C \) . Then \( x \) satisfies an integral equation\n\n\[ {x}^{n} + {b}_{n - 1}{x}^{n - 1} + \cdots + {b}_{0} = 0 \]\n\nwith \( {b}_{i} \in B \) . Let \( {B}_{1} = A\left\lbrack {{b}_{0},\ldots ,{b}_{n - 1}}\right\rbrack \) . Then \( {B}_{1} \) is a finitely generated \( A \) -module by Proposition 2, and \( {B}_{1}\left\lbrack x\right\rbrack \) is a finitely generated \( {B}_{1} \) -module, whence a finitely generated \( A \) -module. Since multiplication by \( x \) maps \( {B}_{1}\left\lbrack x\right\rbrack \) into itself, it follows that \( x \) is integral over \( A \) . | Yes |
Proposition 4. Let \( A \subset B \) be two rings, and \( B \) integral over \( A \) . Let \( \sigma \) be a homomorphism of \( B \) . Then \( \sigma \left( B\right) \) is integral over \( \sigma \left( A\right) \) . | Proof. Apply \( \sigma \) to an integral equation satisfied by any element \( x \) of \( B \) . It will be an integral equation for \( \sigma \left( x\right) \) over \( \sigma \left( A\right) \) . | Yes |
Proposition 5. Let \( A \) be a ring contained in a field \( L \) . Let \( B \) be the set of elements of \( L \) which are integral over \( A \) . Then \( B \) is a ring, called the integral closure of \( A \) in \( L \) . | Proof. Let \( x, y \) lie in \( B \), and let \( M, N \) be two finitely generated \( A \) - modules such that \( {xM} \subset M \) and \( {yN} \subset N \) . Then \( {MN} \) is finitely generated, and is mapped into itself by multiplication with \( x \pm y \) and \( {xy} \) . | Yes |
Proposition 6. Let \( A \) be a Noetherian ring, integrally closed. Let \( L \) be a finite separable extension of its quotient field \( K \) . Then the integral closure of \( A \) in \( L \) is finitely generated over \( A \) . | Proof. It will suffice to show that the integral closure of \( A \) is contained in a finitely generated \( A \) -module, because \( A \) is assumed to be Noetherian.\n\nLet \( {w}_{1},\ldots ,{w}_{n} \) be a linear basis of \( L \) over \( K \) . After multiplying each \( {w}_{i} \) by a suitable element of \( A \), we may assume without loss of generality that the \( {w}_{i} \) are integral over \( A \) (Proposition 1). The trace \( \operatorname{Tr} \) from \( L \) to \( K \) is a \( K \) -linear map of \( L \) into \( K \), and is non-degenerate (i.e. there exists an element \( x \in L \) such that \( \operatorname{Tr}\left( x\right) \neq 0 \) ). If \( \alpha \) is a non-zero element of \( L \) , then the function \( \operatorname{Tr}\left( {\alpha x}\right) \) on \( L \) is an element of the dual space of \( L \) (as \( K \) -vector space), and induces a homomorphism of \( L \) into its dual space. Since the kernel is trivial, it follows that \( L \) is isomorphic to its dual under the bilinear form\n\n\[ \left( {x, y}\right) \mapsto \operatorname{Tr}\left( {xy}\right) \]\n\nLet \( {w}_{1}^{\prime },\ldots ,{w}_{n}^{\prime } \) be the dual basis of \( {w}_{1},\ldots ,{w}_{n} \), so that\n\n\[ \operatorname{Tr}\left( {{w}_{i}^{\prime }{w}_{j}}\right) = {\delta }_{ij} \]\n\nLet \( c \neq 0 \) be an element of \( A \) such that \( c{w}_{i}^{\prime } \) is integral over \( A \) . Let \( z \) be in \( L \), integral over \( A \) . Then \( {zc}{w}_{i}^{\prime } \) is integral over \( A \), and so is \( \operatorname{Tr}\left( {{cz}{w}_{i}^{\prime }}\right) \) for each \( i \) . If we write\n\n\[ z = {b}_{1}{w}_{1} + \cdots + {b}_{n}{w}_{n} \]\n\nwith coefficients \( {b}_{i} \in K \), then\n\n\[ \operatorname{Tr}\left( {{cz}{w}_{i}^{\prime }}\right) = c{b}_{i} \]\n\nand \( c{b}_{i} \in A \) because \( A \) is integrally closed. Hence \( z \) is contained in\n\n\[ A{c}^{-1}{w}_{1} + \cdots + A{c}^{-1}{w}_{n} \]\n\nSince \( z \) was selected arbitrarily in the integral closure of \( A \) in \( L \), it follows that this integral closure is contained in a finitely generated \( A \) -module, and our proof is finished. | Yes |
Proposition 7. If \( A \) is a unique factorization domain, then \( A \) is integrally closed. | Proof. Suppose that there exists a quotient \( a/b \) with \( a, b \in A \) which is integral over \( A \), and a prime element \( p \) in \( A \) which divides \( b \) but not \( a \) . We have, for some integer \( n \geqq 1 \) ,\n\n\[{\left( a/b\right) }^{n} + {a}_{n - 1}{\left( a/b\right) }^{n - 1} + \cdots + {a}_{0} = 0,\]\n\nwhence\n\n\[{a}^{n} + {a}_{n - 1}b{a}^{n - 1} + \cdots + {a}_{0}{b}^{n} = 0.\]\n\nSince \( p \) divides \( b \), it must divide \( {a}^{n} \), and hence must divide \( a \), contradiction. | Yes |
Theorem 1. Let \( A \) be a principal ideal ring, and \( L \) a finite separable extension of its quotient field, of degree \( n \) . Let \( B \) be the integral closure of \( A \) in \( L \) . Then \( B \) is a free module of rank \( n \) over \( A \) . | Proof. As a module over \( A \), the integral closure is torsion-free, and by the general theory of principal ideal rings, any torsion-free finitely generated module is in fact a free module. It is obvious that the rank is equal to the degree \( \left\lbrack {L : K}\right\rbrack \) . | Yes |
Proposition 8. Let \( A \) be a subring of a ring \( B \), integral over \( A \) . Let \( S \) be a multiplicative subset of \( A \) . Then \( {S}^{-1}B \) is integral over \( {S}^{-1}A \) . If \( A \) is integrally closed, then \( {S}^{-1}A \) is integrally closed. | Proof. If \( x \in B \) and \( s \in S \), and if \( M \) is a finitely generated \( A \) -module such that \( {xM} \subset M \), then \( {S}^{-1}M \) is a finitely generated \( {S}^{-1}A \) -module which is mapped into itself by \( {s}^{-1}x \), so that \( {s}^{-1}x \) is integral over \( {S}^{-1}A \) . As to the second assertion, let \( x \) be integral over \( {S}^{-1}A \), with \( x \) in the quotient field of \( A \) . We have an equation\n\n\[ \n{x}^{n} + \frac{{b}_{n - 1}}{{s}_{n - 1}}{x}^{n - 1} + \cdots + \frac{{b}_{0}}{{s}_{0}} = 0 \n\]\n\n\( {b}_{i} \in A \) and \( {s}_{i} \in S \) . Thus there exists an element \( s \in S \) such that \( {sx} \) is integral over \( A \), hence lies in \( A \) . This proves that \( x \) lies in \( {S}^{-1}A \) . | Yes |
Proposition 9. Let \( A \) be a ring, \( \mathfrak{p} \) a prime ideal, and \( B \) a ring containing \( A \) and integral over \( A \) . Then \( \mathfrak{p}B \neq B \), and there exists a prime ideal \( \mathfrak{P} \) of \( B \) lying above \( \mathfrak{p} \) . | Proof. We know that \( {B}_{\mathfrak{p}} \) is integral over \( {A}_{\mathfrak{p}} \), and that \( {A}_{\mathfrak{p}} \) is a local ring with maximal ideal \( {\mathfrak{m}}_{\mathfrak{p}} \) . Since we obviously have\n\n\[ \mathfrak{p}{B}_{\mathfrak{p}} = \mathfrak{p}{A}_{\mathfrak{p}}B = \mathfrak{p}{A}_{\mathfrak{p}}{B}_{\mathfrak{p}} = {\mathfrak{m}}_{\mathfrak{p}}{B}_{\mathfrak{p}}, \]\n\nit will suffice to prove our first assertion when \( A \) is a local ring. In that case, if \( \mathfrak{p}B = B \), then 1 has an expression as a finite linear combination of elements of \( B \) with coefficients in \( \mathfrak{p} \), \n\n\[ 1 = {a}_{1}{b}_{1} + \cdots + {a}_{n}{b}_{n} \]\n\nwith \( {a}_{i} \in \mathfrak{p} \) and \( {b}_{i} \in B \) . Let \( {B}_{0} = A\left\lbrack {{b}_{1},\ldots ,{b}_{n}}\right\rbrack \) . Then \( \mathfrak{p}{B}_{0} = {B}_{0} \) and \( {B}_{0} \) is a finite \( A \) -module by Proposition 2. Hence \( {B}_{0} = 0 \), contradiction.\n\nTo prove our second assertion, we go back to the original notation, and note the following commutative diagram:\n\n\[ B \rightarrow {B}_{\mathfrak{p}} \]\n\n\[ \uparrow \; \uparrow \;\text{(all arrows inclusions).} \]\n\n\[ A \rightarrow {A}_{\mathfrak{p}} \]\n\nWe have just proved that \( {m}_{\mathfrak{p}}{B}_{\mathfrak{p}} \neq {B}_{\mathfrak{p}} \) . Hence \( {m}_{\mathfrak{p}}{B}_{\mathfrak{p}} \) is contained in a maximal ideal \( \mathfrak{M} \) of \( {B}_{\mathfrak{p}} \), and \( \mathfrak{M} \cap {A}_{\mathfrak{p}} \) therefore contains \( {\mathfrak{m}}_{\mathfrak{p}} \) . Since \( {\mathfrak{m}}_{\mathfrak{p}} \) is maximal, it follows that\n\n\[ {\mathfrak{m}}_{\mathfrak{p}} = \mathfrak{M} \cap {A}_{\mathfrak{p}} \]\n\nLet \( \mathfrak{P} = \mathfrak{M} \cap B \) . Then \( \mathfrak{P} \) is a prime ideal of \( B \), and taking intersections with \( A \) going both ways around our diagram shows that \( \mathfrak{M} \cap A = \mathfrak{p} \) , so that\n\n\[ \mathfrak{P} \cap A = \mathfrak{p} \]\n\nas was to be shown. | Yes |
Proposition 10. Let \( A \) be a subring of \( B \), and assume \( B \) integral over \( A \) . Let \( \mathfrak{P} \) be a prime ideal of \( B \) lying over a prime ideal \( \mathfrak{p} \) of \( A \) . Then \( \mathfrak{P} \) is maximal if and only if \( \mathfrak{p} \) is maximal. | Proof. Assume \( \mathfrak{p} \) maximal in \( A \) . Then \( A/\mathfrak{p} \) is a field. We are reduced to proving that a ring which is integral over a field is a field. If \( k \) is a field and \( x \) is integral over \( k \), then it is standard from elementary field theory that the ring \( k\left\lbrack x\right\rbrack \) is itself a field, so \( x \) is invertible in the ring. Conversely, assume that \( \mathfrak{P} \) is maximal in \( B \) . Then \( B/\mathfrak{P} \) is a field, which is integral over the ring \( A/\mathfrak{p} \) . If \( A/\mathfrak{p} \) is not a field, it has a non-zero maximal ideal \( \mathfrak{m} \) . By Proposition 9, there exists a maximal ideal \( \mathfrak{M} \) of \( B/\mathfrak{P} \) lying above \( \mathfrak{m} \), contradiction. | Yes |
Proposition 11. Let \( A \) be a ring, integrally closed in its quotient field \( K \) . Let \( L \) be a finite Galois extension of \( K \) with group \( G \) . Let \( \mathfrak{p} \) be a maximal ideal of \( A \), and let \( \mathfrak{P} \) , \( \mathfrak{Q} \) be prime ideals of the integral closure of \( A \) in \( L \) lying above \( \mathfrak{p} \) . Then there exists \( \sigma \in G \) such that \( \sigma \mathfrak{P} = \mathfrak{Q} \) . | Proof. Suppose that \( \mathfrak{P} \neq \sigma \mathfrak{Q} \) for any \( \sigma \in G \) . There exists an element \( x \in B \) such that\n\n\[ x \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \]\n\n\[ x \equiv 1\left( {{\;\operatorname{mod}\;\sigma }\mathfrak{Q}}\right) ,\;\text{ all }\sigma \in G \]\n\n(use the Chinese remainder theorem). The norm\n\n\[ {N}_{K}^{L}\left( x\right) = \mathop{\prod }\limits_{{\sigma \in G}}{\sigma x} \]\n\nlies in \( B \cap K = A \) (because \( A \) is integrally closed), and lies in \( \mathfrak{P} \cap A = \mathfrak{p} \) .\n\nBut \( x \notin \sigma \mathfrak{Q} \) for all \( \sigma \in G \), so that \( {\sigma x} \notin \mathfrak{Q} \) for all \( \sigma \in G \) . This contradicts the fact that the norm of \( x \) lies in \( \mathfrak{p} = \mathfrak{Q} \cap A \) . | Yes |
Proposition 12. The field \( {L}^{d} \) is the smallest subfield \( E \) of \( L \) containing \( K \) such that \( \mathfrak{P} \) is the only prime of \( B \) lying above \( \mathfrak{P} \cap E \) (which is prime in \( B \cap E) \) . | Proof. Let \( E \) be as above, and let \( H \) be the Galois group of \( L \) over \( E \) . Let \( \mathfrak{q} = \mathfrak{P} \cap E \) . By Proposition 11, all primes of \( B \) lying above \( \mathfrak{q} \) are conjugate by elements of \( H \) . Since there is only one prime, namely \( \mathfrak{P} \) , it means that \( H \) leaves \( \mathfrak{P} \) invariant. Hence \( H \subset {G}_{\mathfrak{B}} \) and \( E \supset {L}^{d} \) . We have already observed that \( {L}^{d} \) has the required property. | Yes |
Proposition 13. Notation being as above, we have \( A/\mathfrak{p} = {B}^{d}/\mathfrak{Q} \) (under the canonical injection \( A/\mathfrak{p} \rightarrow {B}^{d}/\mathfrak{Q} \) ). | Proof. If \( \sigma \) is an element of \( G \), not in \( {G}_{\mathfrak{P}} \), then \( \sigma \mathfrak{P} \neq \mathfrak{P} \) and \( {\sigma }^{-1}\mathfrak{P} \neq \mathfrak{P} \) . Let\n\n\[ \n{\mathfrak{Q}}_{\sigma } = {\sigma }^{-1}\mathfrak{P} \cap {B}^{d} \n\]\n\nThen \( {\mathfrak{Q}}_{\sigma } \neq \mathfrak{Q} \) . Let \( x \) be an element of \( {B}^{d} \) . There exists an element \( y \) of \( {B}^{d} \) such that\n\n\[ \ny \equiv x\;\left( {\;\operatorname{mod}\;\mathfrak{Q}}\right) \n\]\n\n\[ \ny \equiv 1\;\left( {\;\operatorname{mod}\;{\mathfrak{O}}_{0}}\right) \n\]\n\nfor each \( \sigma \) in \( G \), but not in \( {G}_{\mathfrak{P}} \) . Hence in particular,\n\n\[ \ny \equiv x\;\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \n\]\n\n\[ \ny \equiv 1\;\left( {{\;\operatorname{mod}\;{\sigma }^{-1}}\mathfrak{P}}\right) \n\]\n\nfor each \( \sigma \) not in \( {G}_{\mathfrak{P}} \) . This second congruence yields\n\n\[ \n{\sigma y} \equiv 1\;\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \n\]\n\nfor all \( \sigma \notin {G}_{\mathfrak{P}} \) . The norm of \( y \) from \( {L}^{d} \) to \( K \) is a product of \( y \) and other factors \( {\sigma y} \) with \( \sigma \notin {G}_{\mathfrak{P}} \) . Thus we obtain\n\n\[ \n{N}_{K}^{{L}^{d}}\left( y\right) \equiv x\;\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) . \n\]\n\nBut the norm lies in \( K \), and even in \( A \), since it is a product of elements integral over \( A \) . This last congruence holds mod \( \mathfrak{Q} \), since both \( x \) and the norm lie in \( {B}^{d} \) . This is precisely the meaning of the assertion in our proposition. | Yes |
Proposition 14. Let \( A \) be integrally closed in its quotient field \( K \), and let \( B \) be its integral closure in a finite Galois extension \( L \) of \( K \), with group \( G \). Let \( \mathfrak{p} \) be a maximal ideal of \( A \), and \( \mathfrak{P} \) a maximal ideal of \( B \) lying above \( \mathfrak{p} \). Then \( B/\mathfrak{P} \) is a normal extension of \( A/\mathfrak{p} \), and the map \( \sigma \mapsto \bar{\sigma } \) induces a homomorphism of \( {G}_{\mathfrak{P}} \) onto the Galois group of \( B/\mathfrak{P} \) over \( A/\mathfrak{p} \). | Proof. Let \( \bar{B} = B/\mathfrak{P} \) and \( \bar{A} = A/\mathfrak{p} \). Any element of \( \bar{B} \) can be written as \( \bar{x} \) for some \( x \in B \). Let \( \bar{x} \) generate a separable subextension of \( \bar{B} \) over \( \bar{A} \), and let \( f \) be the irreducible polynomial for \( x \) over \( K \). The coefficients of \( f \) lie in \( A \) because \( x \) is integral over \( A \), and all the roots of \( f \) are integral over \( A \). Thus\n\n\[ f\left( X\right) = \mathop{\prod }\limits_{{i = 1}}^{m}\left( {X - {x}_{i}}\right) \]\n\nsplits into linear factors in \( B \). Since\n\n\[ \bar{f}\left( X\right) = \prod \left( {X - {\bar{x}}_{i}}\right) \]\n\nand all the \( {\bar{x}}_{i} \) lie in \( \bar{B} \), it follows that \( \bar{f} \) splits into linear factors in \( \bar{B} \). We observe that \( f\left( x\right) = 0 \) implies \( \bar{f}\left( \bar{x}\right) = 0 \). Hence \( \bar{B} \) is normal over \( \bar{A} \), and\n\n\[ \left\lbrack {\bar{A}\left( \bar{x}\right) : \bar{A}}\right\rbrack \leqq \left\lbrack {K\left( x\right) : K}\right\rbrack \leqq \left\lbrack {L : K}\right\rbrack .\n\nThis implies that the maximal separable subextension of \( \bar{A} \) in \( \bar{B} \) is of finite degree over \( \bar{A} \) (using the primitive element theorem of elementary field theory). This degree is in fact bounded by \( \left\lbrack {L : K}\right\rbrack \).\n\nThere remains to prove that the map \( \sigma \mapsto \bar{\sigma } \) gives a surjective homomorphism of \( {G}_{\mathfrak{P}} \) onto the Galois group of \( \bar{B} \) over \( \bar{A} \). To do this, we shall give an argument which reduces our problem to the case when \( \mathfrak{P} \) is the only prime ideal of \( B \) lying above \( \mathfrak{p} \). Indeed, by Proposition 13, the residue class fields of the ground ring and the ring \( {B}^{d} \) in the decomposition field are the same. This means that to prove our surjectivity, we may take \( {L}^{d} \) as ground field. This is the desired reduction, and we can assume \( K = {L}^{d} \), \( G = {G}_{\mathfrak{P}} \).\n\nThis being the case, take a generator of the maximal separable sub-extension of \( \bar{B} \) over \( \bar{A} \), and let it be \( \bar{x} \), for some element \( x \) in \( B \). Let \( f \) be the irreducible polynomial of \( x \) over \( K \). Any automorphism of \( \bar{B} \) is determined by its effect on \( \bar{x} \), and maps \( \bar{x} \) on some root of \( \bar{f} \). Suppose that \( x = {x}_{1} \). Given any root \( {x}_{i} \) of \( f \), there exists an element \( \sigma \) of \( G = {G}_{\mathfrak{P}} \) such that \( {\sigma x} = {x}_{i} \). Hence \( \bar{\sigma }\bar{x} = {\bar{x}}_{i} \). Hence the automorphism of \( \bar{B} \) over \( \bar{A} \) induced by elements of \( G \) operate transitively on the roots of \( \bar{f} \). Hence they give us all automorphisms of the residue class field, as was to be shown. | Yes |
Let \( A \) be a ring integrally closed in its quotient field \( K \) . Let \( L \) be a finite Galois extension of \( K \), and \( B \) the integral closure of \( A \) in \( L \) . Let \( \mathfrak{p} \) be a maximal ideal of \( A \) . Let \( \varphi : A \rightarrow A/\mathfrak{p} \) be the canonical homomorphism, and let \( {\psi }_{1},{\psi }_{2} \) be two homomorphisms of \( B \) extending \( \varphi \) in a given algebraic closure of \( A/\mathfrak{p} \) . Then there exists an automorphism \( \sigma \) of \( L \) over \( K \) such that\n\n\[{\psi }_{1} = {\psi }_{2} \circ \sigma\] | Proof. The kernels of \( {\psi }_{1},{\psi }_{2} \) are prime ideals of \( B \) which are conjugate by Proposition 11. Hence there exists an element \( \tau \) of the Galois group \( G \) such that \( {\psi }_{1},{\psi }_{2} \circ \tau \) have the same kernel. Without loss of generality, we may therefore assume that \( {\psi }_{1},{\psi }_{2} \) have the same kernel \( \mathfrak{P} \) . Hence there exists an automorphism \( \omega \) of \( {\psi }_{1}\left( B\right) \) onto \( {\psi }_{2}\left( B\right) \) such that \( \omega \circ {\psi }_{1} = {\psi }_{2} \) . There exists an element \( \sigma \) of \( {G}_{\mathfrak{P}} \) such that \( \omega \circ {\psi }_{1} = {\psi }_{1} \circ \sigma \), by the preceding proposition. This proves what we wanted. | Yes |
Corollary 2. Let the assumptions be as in Corollary 1, and assume that \( \mathfrak{P} \) is the only prime of \( B \) lying above \( \mathfrak{p} \) . Let \( f\left( X\right) \) be a polynomial in \( A\left\lbrack X\right\rbrack \) with leading coefficient 1. Assume that \( f \) is irreducible in \( K\left\lbrack X\right\rbrack \), and has a root \( \alpha \) in \( B \) . Then the reduced polynomial \( \bar{f} \) is a power of an irreducible polynomial in \( \bar{A}\left\lbrack X\right\rbrack \) . | Proof. By Corollary 1, we know that any two roots of \( \bar{f} \) are conjugate under some isomorphism of \( \bar{B} \) over \( \bar{A} \), and hence that \( \bar{f} \) cannot split into relative prime polynomials. Therefore, \( \bar{f} \) is a power of an irreducible polynomial. | Yes |
Corollary 3. Let \( K/k \) be abelian with group \( G \) . Let \( \mathfrak{p} \) be a prime of \( k \), let \( \mathfrak{P} \) be a prime of \( K \) lying above \( \mathfrak{p} \) and let \( {G}_{\mathfrak{P}} \) be its decomposition group. Let \( E \) be the fixed field of \( {G}_{\mathfrak{P}} \) . Then \( E \) is the maximal subfield of \( K \) containing \( k \) in which \( \mathfrak{p} \) splits completely | Proof. Let\n\n\[ G = \mathop{\bigcup }\limits_{{i = 1}}^{r}{\sigma }_{i}{G}_{\mathfrak{P}} \]\n\nbe a coset decomposition. Let \( \mathfrak{q} = \mathfrak{P} \cap E \) . Since a Galois group permutes the primes lying above a given prime transitively, we know that \( \mathfrak{P} \) is the only prime of \( K \) lying above \( \mathfrak{q} \) . For each \( i \), the prime \( {\sigma }_{i}\mathfrak{P} \) is the only prime lying above \( {\sigma }_{i}\mathfrak{q} \), and since \( {\sigma }_{1}\mathfrak{P},\ldots ,{\sigma }_{r}\mathfrak{P} \) are distinct, it follows that the primes \( {\sigma }_{1}\mathfrak{q},\ldots ,{\sigma }_{r}\mathfrak{q} \) are distinct. Since \( G \) is abelian, the primes \( {\sigma }_{i}\mathfrak{q} \) are primes of \( E \), and \( \left\lbrack {E : k}\right\rbrack = r \), so that \( \mathfrak{p} \) splits completely in \( E \) . Conversely, let \( F \) be an intermediate field between \( k \) and \( K \) in which \( \mathfrak{p} \) splits completely, and let \( H \) be the Galois group of \( K/F \) . If \( \sigma \in {G}_{\mathfrak{P}} \) and \( \mathfrak{P} \cap F = {\mathfrak{P}}_{F} \), then \( \sigma \) leaves \( {\mathfrak{P}}_{F} \) fixed. However, the decomposition group of \( {\mathfrak{P}}_{F} \) over \( \mathfrak{p} \) must be trivial since \( \mathfrak{p} \) splits completely in \( F \) . Hence the restriction of \( \sigma \) to \( F \) is the identity, and therefore \( {G}_{\mathfrak{P}} \subset H \) . This proves that \( F \subset E \), and concludes the proof of our corollary. | Yes |
Proposition 15. Let \( \mathfrak{o} \) be a Dedekind ring with only a finite number of prime ideals. Then \( \mathfrak{o} \) is a principal ideal ring. | Proof. Let \( {\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{s} \) be the prime ideals. Given any ideal\n\n\[ \mathfrak{a} = {\mathfrak{p}}_{1}^{{r}_{1}}\cdots {\mathfrak{p}}_{s}^{{r}_{s}} \neq 0 \]\n\nselect an element \( {\pi }_{i} \) in \( {\mathfrak{p}}_{i} \) but not in \( {\mathfrak{p}}_{i}^{2} \) and find an element \( \alpha \) of \( \mathfrak{o} \) such that\n\n\[ \alpha \equiv {\pi }_{i}^{{r}_{i}}\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}_{i}^{{r}_{i} + 1}}\right) . \]\n\nIf\n\n\[ \left( \alpha \right) = {\mathfrak{p}}_{1}^{{e}_{1}}\cdots {\mathfrak{p}}_{s}^{{e}_{s}} \]\n\nis a factorization of the ideal generated by \( \alpha \), then one sees immediately that \( {e}_{i} = {r}_{i} \) for all \( i \), and hence that \( \mathfrak{a} = \left( \alpha \right) \) . | Yes |
Proposition 16. Let \( A \) be a Dedekind ring and \( S \) a multiplicative subset of \( A \) . Then \( {S}^{-1}A \) is a Dedekind ring. The map \[ \mathfrak{a} \mapsto {S}^{-1}\mathfrak{a} \] is a homomorphism of the group of fractional ideals of \( A \) onto the group of fractional ideals of \( {S}^{-1}A \), and the kernel consists of those fractional ideals of \( A \) which meet \( S \) . | Proof. If \( \mathfrak{p} \) meets \( S \), then \[ {S}^{-1}\mathfrak{p} = {S}^{-1}A \] because 1 lies in \( {S}^{-1}\mathfrak{p} \) . If \( \mathfrak{a},\mathfrak{b} \) are two ideals of \( A \), then \[ {S}^{-1}\left( {\mathfrak{a}\mathfrak{b}}\right) = \left( {{S}^{-1}\mathfrak{a}}\right) \left( {{S}^{-1}\mathfrak{b}}\right) \] so multiplication by \( {S}^{-1} \) induces a homomorphism of the group of (fractional) ideals. If \( {S}^{-1}\mathfrak{a} = {S}^{-1}A \), then we can write \( 1 = \alpha /s \) for some \( \alpha \in \mathfrak{a} \) and \( s \in S \) . Thus \( \alpha = s \) and \( \mathfrak{a} \) meets \( S \) . This proves that the kernel of our homomorphism is what we said it is. Our mapping is surjective since we saw in \( §1 \) that every ideal of \( {S}^{-1}A \) is of type \( {S}^{-1}\mathfrak{a} \) for some ideal \( \mathfrak{a} \) of \( A \) . The same applies of course to fractional ideals. This proves our proposition. | Yes |
Proposition 17. Let \( A \) be a Dedekind ring, and assume that its group of ideal classes is finite. Let \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{r} \) be representative fractional ideals of the ideal classes, and let \( b \) be a non-zero element of \( A \) which lies in all the \( {\mathfrak{a}}_{i} \) . Let \( S \) be the multiplicative subset of \( A \) generated by the powers of \( b \) . Then every ideal of \( {S}^{-1}A \) is principal. | Proof. All the ideals \( {S}^{-1}{\mathfrak{a}}_{1},\ldots ,{S}^{-1}{\mathfrak{a}}_{r} \) map on the unit ideal in the homomorphism of Proposition 16. Since every ideal of \( A \) is equal to some \( {\mathfrak{a}}_{i} \) times a principal ideal, our proposition follows from the surjectivity of Proposition 16. | No |
Proposition 18. Let \( A \) be a Dedekind ring and \( M, N \) two modules over \( A \). If \( \mathfrak{p} \) is a prime of \( A \), denote by \( {S}_{\mathfrak{p}} \) the multiplicative set \( A - \mathfrak{p} \). Assume that \( {S}_{\mathfrak{p}}^{-1}M \subset {S}_{\mathfrak{p}}^{-1}N \) for all \( \mathfrak{p} \). Then \( M \subset N \). | Proof. Let \( a \in M \). For each \( \mathfrak{p} \) we can find \( {x}_{\mathfrak{p}} \in N \) and \( {s}_{\mathfrak{p}} \in {S}_{\mathfrak{p}} \) such that \( a = {x}_{\mathfrak{p}}/{s}_{\mathfrak{p}} \). Let \( \mathfrak{b} \) be the ideal generated by the \( {s}_{\mathfrak{p}} \). Then \( \mathfrak{b} \) is the unit ideal, and we can write \[ 1 = \sum {y}_{\mathfrak{p}}{s}_{\mathfrak{p}} \] with elements \( {y}_{\mathfrak{p}} \in A \) all but a finite number of which are 0. This yields \[ a = \sum {y}_{\mathfrak{p}}{s}_{\mathfrak{p}}a = \sum {y}_{\mathfrak{p}}{x}_{\mathfrak{p}} \] and shows that \( a \) lies in \( N \), as desired. | Yes |
Proposition 19. Let \( A \) be a local ring and \( M \) a free module of rank \( n \) over \( A \). Let \( \mathfrak{p} \) be the maximal ideal of \( A \). Then \( M/\mathfrak{p}M \) is a vector space of dimension \( n \) over \( A/\mathfrak{p \) . | Proof. This is obvious, because if \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) is a basis for \( M \) over \( A \), so\n\n\[ M = \sum A{x}_{i}\;\text{ (direct sum),} \]\n\n then\n\n\[ M/\mathfrak{p}M \approx \sum \left( {A/\mathfrak{p}}\right) {\bar{x}}_{i}\;\text{ (direct sum),} \]\n\n where \( {\bar{x}}_{i} \) is the residue class of \( {x}_{i}{\;\operatorname{mod}\;\mathfrak{p}} \) . | No |
Proposition 20. Let \( A \) be a Dedekind ring, \( K \) its quotient field, \( K \subset E \subset L \) two finite separable extensions, and \( A \subset B \subset C \) the corresponding tower of integral closures of \( A \) in \( E \) and \( L \) . Let \( \mathfrak{p} \) be a prime of \( A,\mathfrak{q} \) a prime of \( B \) lying above \( \mathfrak{p} \), and \( \mathfrak{P} \) a prime of \( C \) lying above \( \mathfrak{q} \) . Then\n\n\[ e\left( {\mathfrak{P}/\mathfrak{p}}\right) = e\left( {\mathfrak{P}/\mathfrak{q}}\right) e\left( {\mathfrak{q}/\mathfrak{p}}\right) \]\n\n\[ f\left( {\mathfrak{P}/\mathfrak{p}}\right) = f\left( {\mathfrak{P}/\mathfrak{q}}\right) f\left( {\mathfrak{q}/\mathfrak{p}}\right) . \] | Proof. Obvious. | No |
Proposition 21. Let \( A \) be a Dedekind ring, \( K \) its quotient field, \( L \) a finite separable extension of \( K \), and \( B \) the integral closure of \( A \) in \( L \) . Let \( \mathfrak{p} \) be a prime of \( A \) . Then\n\n\[ \left\lbrack {L : K}\right\rbrack = \mathop{\sum }\limits_{{\mathfrak{P} \mid \mathfrak{p}}}{e}_{\mathfrak{P}}{f}_{\mathfrak{P}} \] | Proof. We can localize at \( \mathfrak{p} \) (multiplying \( A \) and \( B \) by \( {S}_{\mathfrak{p}}^{-1} \) ), and thus may assume that \( A \) is a discrete valuation ring. In that case, \( B \) is a free module of rank \( n = \left\lbrack {L : K}\right\rbrack \) over \( A \), and \( B/\mathfrak{p}B \) is a vector space of dimension \( n \) over \( A/\mathfrak{p} \) .\n\nLet \( \mathfrak{p}B = {\mathfrak{P}}_{1}^{{e}_{1}}\cdots {\mathfrak{P}}_{r}^{{e}_{r}} \) be the factorization of \( \mathfrak{p} \) in \( B \) . Since \( {\mathfrak{P}}_{i}^{{e}_{i}} \supset \mathfrak{p}B \) for each \( i \), we have a well-defined homomorphism\n\n\[ B \rightarrow B/\mathfrak{p}B \rightarrow B/{\mathfrak{P}}_{i}^{{e}_{i}} \]\n\nand therefore a homomorphism into the direct sum\n\n\[ B \rightarrow B/\mathfrak{p}B \rightarrow \mathop{\prod }\limits_{{i = 1}}^{r}B/{\mathfrak{P}}_{i}^{{e}_{i}} \]\n\nEach \( B/{\mathfrak{P}}_{i}^{{e}_{i}} \) can be viewed as an \( A/\mathfrak{p} \) -vector space, and hence so can the direct sum. The kernel of our homomorphism consists of those elements of \( B \) lying in all the \( {\mathfrak{P}}_{i}^{{e}_{i}} \), and is therefore \( \mathfrak{p}B \) . Furthermore, our map is surjective by the Chinese remainder theorem. It is obviously an \( A/\mathfrak{p} \) - homomorphism, and thus \( B/\mathfrak{p}B \) is \( A/\mathfrak{p} \) -isomorphic to the above direct sum.\n\nWe shall now determine the dimension of \( B/{\mathfrak{P}}^{e} \) (if \( \mathfrak{P} \) is some \( {\mathfrak{P}}_{i} \) and \( e = {e}_{i} \) ).\n\nLet \( \Pi \) be a generator of \( \mathfrak{P} \) in \( B \) . (We know from Proposition 15 that \( \mathfrak{P} \) is principal.) Let \( j \) be an integer \( \geqq 1 \) . We can view \( {\mathfrak{P}}^{j}/{\mathfrak{P}}^{j + 1} \) as an \( A/\mathfrak{p} \) - vector space, since \( \mathfrak{p}{\mathfrak{P}}^{j} \subset {\mathfrak{P}}^{j + 1} \) . We consider the map\n\n\[ B/\mathfrak{P} \rightarrow {\mathfrak{P}}^{j}/{\mathfrak{P}}^{j + 1} \]\n\ninduced by multiplying an element of \( B \) by \( {\Pi }^{j} \) . This map is an \( A/\mathfrak{p} \) - homomorphism, which is clearly injective and surjective. Hence \( B/\mathfrak{P} \) and \( {\mathfrak{P}}^{j}/{\mathfrak{P}}^{j + 1} \) are \( A/\mathfrak{p} \) -isomorphic.\n\nThe \( A/\mathfrak{p} \) -vector space \( B/{\mathfrak{P}}^{e} \) has a composition series induced by the inclusions\n\n\[ B \supset \mathfrak{P} \supset {\mathfrak{P}}^{2} \supset \cdots \supset {\mathfrak{P}}^{e} \]\n\nThe dimension of \( B/\mathfrak{P} \) over \( A/\mathfrak{p} \) is \( {f}_{\mathfrak{P}} \), by definition. From this it follows that the dimension of \( B/{\mathfrak{P}}^{e} \) over \( A/\mathfrak{p} \) is \( {e}_{\mathfrak{P}}{f}_{\mathfrak{P}} \), thereby proving our proposition. | Yes |
Corollary 1. Let \( \mathfrak{a} \) be a fractional ideal of \( A \) . Then\n\n\[ {N}_{K}^{L}\left( {\mathfrak{a}B}\right) = {\mathfrak{a}}^{\left\lbrack L : K\right\rbrack }.\] | Proof. Immediate. | No |
Assume that \( L \) is Galois over \( K \) . Then all the \( {e}_{\mathfrak{P}} \) are equal to the same number \( e \) (for \( \mathfrak{P} \mid \mathfrak{p} \) ), all the \( {f}_{\mathfrak{P}} \) are equal to the same number \( f \) (for \( \mathfrak{P} \mid \mathfrak{p} \) ), and if\n\n\[ \mathfrak{p}B = {\left( {\mathfrak{P}}_{1}\cdots {\mathfrak{P}}_{r}\right) }^{e} \]\n\nthen\n\n\[ \text{efr} = \left\lbrack {L : K}\right\rbrack \text{.} \] | Proof. All the \( \mathfrak{P} \) lying above \( \mathfrak{p} \) are conjugate to each other, and hence all the ramification indices and residue class degrees are equal. The last formula is clear. | No |
Corollary 3. Assume again that \( L \) is Galois over \( K \) with group \( G \), and let \( \mathfrak{P} \) be a prime of \( B \) lying above \( \mathfrak{p} \) in \( A \) . Then\n\n\[ \n{N}_{K}^{L}\mathfrak{P} \cdot B = \mathop{\prod }\limits_{{\sigma \in G}}\sigma \mathfrak{P} = {\left( {\mathfrak{P}}_{1}\cdots {\mathfrak{P}}_{r}\right) }^{ef} \]\n\n(with \( e, f, r \) as in Corollary 2, and the ideal on the left is viewed as embedded in \( I\left( B\right) \) ). The number ef is the order of the decomposition group of \( \mathfrak{P} \), and \( e \) is the order of the inertia group. | Proof. The group \( G \) operates transitively on the primes of \( B \) lying above \( \mathfrak{p} \), and the order of \( {G}_{\mathfrak{P}} \) is the order of the isotropy group. Our assertions are therefore obvious, taking into account Proposition 14 of \( §5 \) . | No |
Proposition 22. Let \( A \) be a Dedekind ring, \( K \) its quotient field, \( E \) a finite separable extension of \( K \), and \( B \) the integral closure of \( A \) in \( E \) . Let \( \mathfrak{b} \) be a fractional ideal of \( B \), and assume \( \mathfrak{b} \) is principal, \( \mathfrak{b} = \left( \beta \right) ,\beta \neq 0 \) . Then:\n\n\[ \n{N}_{K}^{E}\mathfrak{b} = \left( {{N}_{K}^{E}\left( \beta \right) }\right) \n\]\n\nthe norm on the left being the norm of a fractional ideal as defined above, and the norm on the right being the usual norm of elements of \( E \) . | Proof. Let \( L \) be the smallest Galois extension of \( K \) containing \( E \) . The norm from \( L \) to \( E \) of \( \mathfrak{b} \) and of \( \beta \) simply raises these to the power \( \left\lbrack {L : E}\right\rbrack \) . Since our proposition asserts an equality between fractional ideals, it will suffice to prove it when the extension is Galois over \( K \) . In that case, it follows at once from Corollary 3 above. | Yes |
Proposition 23. Let \( A \) be a discrete valuation ring, \( K \) its quotient field, \( L \) a finite separable extension of \( K \), and \( B \) the integral closure of \( A \) in \( L \) . Assume that there exists only one prime \( \mathfrak{P} \) of \( B \) lying above the maximal ideal \( \mathfrak{p} \) of \( A \) . Let \( \beta \) be an element of \( B \) such that its residue class \( {\;\operatorname{mod}\;\mathfrak{P}} \) generates \( B/\mathfrak{P} \) over \( A/\mathfrak{p} \) and \( \Pi \) an element of \( B \) which is of order \( 1 \) at \( \mathfrak{P} \) . Then \( A\left\lbrack {\beta ,\Pi }\right\rbrack = B \) . | Proof. Let \( C \) be the ring \( A\left\lbrack {\beta ,\Pi }\right\rbrack \) . It can be viewed as a submodule of \( B \) over \( A \), and by Nakayama’s lemma, applied to the factor module \( B/C \) , it will suffice to prove that\n\n\[ \mathfrak{p}B + C = B. \]\n\nBut \( \mathfrak{p}B = {\mathfrak{P}}^{e} \), and the products \( {\beta }^{i}{\Pi }^{j} \) generate \( B/{\mathfrak{P}}^{e} \) over \( A/\mathfrak{p} \), as in Proposition 21. Hence every element \( x \in B \) is such that\n\n\[ x \equiv \sum {c}_{ij}{\beta }^{i}{\Pi }^{j}\;\left( {{\;\operatorname{mod}\;\mathfrak{p}}B}\right) \]\n\nfor some \( {c}_{ij} \in A \) . This proves our proposition. | Yes |
Proposition 24. Let \( A \) be a Dedekind ring, and a a non-zero ideal. Let \( {n}_{\mathfrak{p}} = {\operatorname{ord}}_{\mathfrak{p}}\mathfrak{a} \) . Then the canonical map\n\n\[ A \rightarrow \mathop{\prod }\limits_{\mathfrak{p}}A/{\mathfrak{p}}^{{n}_{\mathfrak{p}}} \]\n\ninduces an isomorphism of \( A/\mathfrak{a} \) onto the product. | Proof. The map is surjective according to the Chinese remainder theorem, and it is clear that its kernel is exactly \( \mathfrak{a} \) . | Yes |
Proposition 25. Let \( A \) be a Dedekind ring with quotient field \( K \) . Let \( E \) be a finite separable extension of \( K \) . Let \( B \) be the integral closure of \( A \) in \( E \) and assume that \( B = A\left\lbrack \alpha \right\rbrack \) for some element \( \alpha \) . Let \( f\left( X\right) \) be the irreducible polynomial of \( \alpha \) over \( K \) . Let \( \mathfrak{p} \) be a prime of \( A \) . Let \( \bar{f} \) be the reduction of \( f{\;\operatorname{mod}\;\mathfrak{p}} \), and let\n\n\[ \bar{f}\left( X\right) = {\bar{P}}_{1}{\left( X\right) }^{{c}_{1}}\cdots {\bar{P}}_{r}{\left( X\right) }^{{c}_{r}} \]\n\nbe the factorization of \( \bar{f} \) into powers of irreducible factors over \( \bar{A} = A/\mathfrak{p} \) , with leading coefficients 1. Then\n\n\[ \mathfrak{p}B = {\mathfrak{P}}_{1}^{{e}_{1}}\cdots {\mathfrak{P}}_{r}^{{e}_{r}} \]\n\nis the factorization of \( \mathfrak{p} \) in \( B \), so that \( {e}_{i} \) is the ramification index of \( {\mathfrak{P}}_{i} \) over \( \mathfrak{p} \) , and we have\n\n\[ {\mathfrak{P}}_{i} = \mathfrak{p}B + {P}_{i}\left( \alpha \right) B \]\n\nif \( {P}_{i}\left( X\right) \in A\left\lbrack X\right\rbrack \) is a polynomial with leading coefficient 1 whose reduction \( {\;\operatorname{mod}\;\mathfrak{p}}i \) ’s \( {\bar{P}}_{i} \) . | Proof. Let \( \bar{P} \) be an irreducible factor of \( \bar{f} \), let \( \bar{\alpha } \) be a root of \( \bar{P} \), and let \( \mathfrak{P} \) be the prime of \( B \) which is the kernel of the map\n\n\[ A\left\lbrack \alpha \right\rbrack \rightarrow \bar{A}\left\lbrack \bar{\alpha }\right\rbrack \]\n\nIt is clear that \( \mathfrak{p}B + P\left( \alpha \right) B \) is contained in \( \mathfrak{P} \) . Conversely, let \( g\left( \alpha \right) \in \mathfrak{P} \) for some \( g\left( X\right) \in A\left\lbrack X\right\rbrack \) . Then \( \bar{g} = \overline{Ph} \) with some \( \bar{h} \in \bar{A}\left\lbrack X\right\rbrack \), and hence \( g - {Ph} \), which is a polynomial with coefficients in \( A \), in fact has coefficients in \( \mathfrak{p} \) . This proves the reverse inclusion, and proves the last formula of our proposition.\n\nFinally, let \( {e}_{i}^{\prime } \) be the ramification index of \( {\mathfrak{P}}_{i} \), so that\n\n\[ \mathfrak{p}B = {\mathfrak{P}}_{1}^{{e}_{1}^{\prime }}\cdots {\mathfrak{P}}_{r}^{{e}_{r}^{\prime }} \]\n\nand let \( {d}_{i} \) be the residue class degree \( \left\lbrack {B/{\mathfrak{P}}_{i} : A/\mathfrak{p}}\right\rbrack \) . It is clear that \( {d}_{i} \) is the degree of \( {\bar{P}}_{i} \) . Since \( f\left( \alpha \right) = 0 \), and since\n\n\[ f\left( X\right) - {P}_{1}{\left( X\right) }^{{e}_{1}}\cdots {P}_{r}{\left( X\right) }^{{e}_{r}} \in \mathfrak{p}A\left\lbrack X\right\rbrack ,\]\n\nit follows that\n\n(*)\n\n\[ {P}_{1}{\left( \alpha \right) }^{{e}_{1}}\cdots {P}_{r}{\left( \alpha \right) }^{{e}_{r}} \in \mathfrak{p}B. \]\n\nOn the other hand, we see that\n\n\[ {\mathfrak{P}}_{i}^{{e}_{i}} \subset \mathfrak{p}B + {P}_{i}{\left( \alpha \right) }^{{e}_{i}}B \]\n\nwhence using \( \left( *\right) \) we find\n\n\[ {\mathfrak{P}}_{1}^{{e}_{1}}\cdots {\mathfrak{P}}_{r}^{{e}_{r}} \subset \mathfrak{p}B + {P}_{1}{\left( \alpha \right) }^{{e}_{1}}\cdots {\mathfrak{P}}_{r}{\left( \alpha \right) }^{{e}_{r}}B \subset \mathfrak{p}B = {\mathfrak{P}}_{1}^{{e}_{1}^{\prime }}\cdots {\mathfrak{P}}_{r}^{{e}_{r}^{\prime }}. \]\n\nThis proves that \( {e}_{i} \geqq {e}_{i}^{\prime } \) for all \( i \) . But we know that\n\n\[ \sum {e}_{i}{d}_{i} = \deg f = \left\lbrack {E : F}\right\rbrack = \sum {e}_{i}^{\prime }{d}_{i}. \]\n\nIt follows that \( {e}_{i} = {e}_{i}^{\prime } \) for all \( i \), thus proving our theorem. | Yes |
Theorem 1. Let \( K \) be a field and \( {\left. \left| {}_{1},\ldots ,\right| \right| }_{s} \) non-trivial pairwise independent absolute values on \( K \). Let \( {x}_{1},\ldots ,{x}_{s} \) be elements of \( K \), and \( \epsilon > 0 \). Then there exists \( x \in K \) such that \[ {\left| x - {x}_{i}\right| }_{i} < \epsilon \] for all \( i \). | Proof. Consider first two of our absolute values, say \( {v}_{1} \) and \( {v}_{s} \). By hypothesis we can find \( \alpha \in K \) such that \( {\left| \alpha \right| }_{1} < 1 \) and \( {\left| \alpha \right| }_{s} \geqq 1 \). Similarly, we can find \( \beta \in K \) such that \( {\left| \beta \right| }_{1} \geqq 1 \) and \( {\left| \beta \right| }_{s} < 1 \). Let \( y = \beta /\alpha \). Then \( {\left| y\right| }_{1} > 1 \) and \( {\left| y\right| }_{s} < 1 \). We shall prove that there exists \( z \in K \) such that \( {\left| z\right| }_{1} > 1 \) and \( {\left| z\right| }_{j} < 1 \) for \( j = 2,\ldots, s \). We prove this by induction, the case \( s = 2 \) having just been proved. Suppose that we have found \( z \in K \) satisfying \[ {\left| z\right| }_{1} > 1\;\text{ and }\;{\left| z\right| }_{j} < 1\;\text{ for }j = 2,\ldots, s - 1. \] If \( {\left| z\right| }_{s} \leqq 1 \), then the element \( {z}^{n}y \) for large \( n \) will satisfy our requirements. If \( {\left| z\right| }_{s} > 1 \), then the sequence \[ {t}_{n} = {z}^{n}/\left( {1 + {z}^{n}}\right) \] tends to 1 at \( {v}_{1} \) and \( {v}_{s} \), but tends to 0 at \( {v}_{j}\left( {j = 2,\ldots, s - 1}\right) \). For large \( n \), it is then clear that \( {t}_{n}y \) satisfies our requirements. Using the element \( z \) that we have just constructed, we see that the sequence \[ \frac{{z}^{n}}{1 + {z}^{n}} \] tends to 1 at \( {v}_{1} \) and to 0 at \( {v}_{j} \) for \( j = 2,\ldots, s \). For each \( i = 1,\ldots, s \) we can therefore construct an element \( {z}_{i} \) which is very close to 1 at \( {v}_{i} \) and very close to 0 at \( {v}_{j} \) for \( j \neq i \). The element \[ x = {z}_{1}{x}_{1} + \cdots + {z}_{s}{x}_{s} \] then satisfies the requirement of the theorem. | Yes |
Theorem 2. Let \( K \) be a number field, \( v \) one of its canonical absolute values, \( E \) a finite extension of \( K \) . Two embeddings \( \sigma ,\tau : E \rightarrow {\bar{K}}_{v} \) over \( K \) give rise to the same absolute value on \( E \) if and only if they are conjugate over \( {K}_{v} \) . | Proof. Suppose that the two embeddings are conjugate over \( {K}_{v} \) . Then the uniqueness of the extension of the absolute value from \( {K}_{v} \) to \( {\bar{K}}_{v} \) guarantees that the induced absolute values on \( E \) are equal. Conversely, suppose that this is the case. Let\n\n\[ \lambda : {\tau E} \rightarrow {\sigma E} \]\n\nbe an isomorphism over \( K \) . We shall prove that \( \lambda \) extends to an isomorphism of \( {\tau E} \cdot {K}_{v} \) onto \( {\sigma E} \cdot {K}_{v} \) over \( {K}_{v} \) . Since \( {\tau E} \) is dense in \( {\tau E} \cdot {K}_{v} \) , an element \( x \in {\tau E} \cdot {K}_{v} \) can be written\n\n\[ x = \lim \tau {x}_{n} \]\n\nwith \( {x}_{n} \in E \) . Since the absolute values induced by \( \sigma \) and \( \tau \) on \( E \) coincide, it follows that the sequence\n\n\[ \left\{ {{\lambda \tau }{x}_{n}}\right\} = \left\{ {\sigma {x}_{n}}\right\} \]\n\nconverges to an element of \( {\sigma E} \cdot {K}_{v} \) which we denote by \( {\lambda x} \) . One then verifies immediately that \( {\lambda x} \) is independent of the particular sequence \( \tau {x}_{n} \) used, and that the map\n\n\[ \lambda : {\tau E} \cdot {K}_{v} \rightarrow {\sigma E} \cdot {K}_{v} \]\n\nis an isomorphism, which clearly leaves \( {K}_{v} \) fixed. This proves our assertion. | Yes |
Corollary 1. Let \( K \) be a number field and \( E \) a finite extension, of degree \( n \) . Let \( v \in {M}_{K} \) and for each absolute value \( w \) on \( E \) extending \( v \), let \( {n}_{w} \) be the local degree, \[ {n}_{w} = \left\lbrack {{E}_{w} : {K}_{v}}\right\rbrack \] Then \[ \mathop{\sum }\limits_{{w \mid v}}{n}_{w} = n \] | Proof. Immediate from Theorem 2 and the fact that for a finite separable extension, the degree is equal to the number of conjugates. | No |
Proposition 1. Let \( m \) be a positive integer such that\n\n\[ m ≢ 0\;\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) .\n\]\n\nThen for any \( x \in \mathfrak{p} \) the binomial series of \( {\left( 1 + x\right) }^{1/m} \) converges to an \( m \) -th root of \( 1 + x \) in \( {\mathfrak{o}}^{ * } \) . | Proof. Obvious, because the binomial coefficients have no \( p \) in the denominators. | No |
Proposition 2. Let \( f\left( X\right) \) be a polynomial with coefficients in 0 . Let \( {\alpha }_{0} \) be an element of \( \mathfrak{o} \) such that\n\n\[ \left| {f\left( {\alpha }_{0}\right) }\right| < \left| {{f}^{\prime }{\left( {\alpha }_{0}\right) }^{2}}\right| \]\n\n(here \( {f}^{\prime } \) denotes the formal derivative of \( {f}^{\prime } \) ). Then the sequence\n\n\[ {\alpha }_{i + 1} = {\alpha }_{i} - \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }\left( {\alpha }_{i}\right) } \]\n\nconverges to a root \( \alpha \) of \( f\left( X\right) \) in \( \mathfrak{o} \) . Furthermore,\n\n\[ \left| {\alpha - {\alpha }_{0}}\right| \leqq \left| \frac{f\left( {\alpha }_{0}\right) }{{f}^{\prime }{\left( {\alpha }_{0}\right) }^{2}}\right| < 1 \] | Proof. Let \( c = \left| {f\left( {\alpha }_{0}\right) /{f}^{\prime }{\left( {\alpha }_{0}\right) }^{2}}\right| < 1 \) . We show inductively that\n\n(i) \( \left| {\alpha }_{i}\right| \leqq 1 \),\n\n(ii) \( \left| {{\alpha }_{i} - {\alpha }_{0}}\right| \leqq c \),\n\n(iii) \( \left| \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }{\left( {\alpha }_{i}\right) }^{2}}\right| \leqq {c}^{{2}^{i}} \).\n\nThese three conditions obviously imply our proposition. If \( i = 0 \), they are hypotheses. By induction, assume them for \( i \) . Then:\n\n(i) \( \left| \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }{\left( {\alpha }_{i}\right) }^{2}}\right| \leqq {c}^{{2}^{i}} \) gives \( \left| {{\alpha }_{i + 1} - {\alpha }_{i}}\right| \leqq {c}^{{2}^{i}} < 1 \),\n\nwhence \( \left| {\alpha }_{i + 1}\right| \leqq 1 \).\n\n(ii) \( \left| {{\alpha }_{i + 1} - {\alpha }_{0}}\right| \leqq \max \left\{ {\left| {{\alpha }_{i + 1} - {\alpha }_{i}}\right| ,\left| {{\alpha }_{i} - {\alpha }_{0}}\right| }\right\} = c \).\n\n(iii) By Taylor's expansion, we have\n\n\[ f\left( {\alpha }_{i + 1}\right) = f\left( {\alpha }_{i}\right) - {f}^{\prime }\left( {\alpha }_{i}\right) \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }\left( {\alpha }_{i}\right) } + \beta {\left( \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }\left( {\alpha }_{i}\right) }\right) }^{2} \]\n\nfor some \( \beta \in \mathfrak{o} \), and this is less than or equal to\n\n\[ {\left| \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }\left( {\alpha }_{i}\right) }\right| }^{2} \]\n\nin absolute value.\n\nUsing Taylor’s expansion on \( {f}^{\prime }\left( {\alpha }_{i + 1}\right) \) we conclude that\n\n\[ \left| {{f}^{\prime }\left( {\alpha }_{i + 1}\right) }\right| = \left| {{f}^{\prime }\left( {\alpha }_{i}\right) }\right| \]\n\nFrom this we get\n\n\[ \left| \frac{f\left( {\alpha }_{i + 1}\right) }{f{\left( {}^{\prime }{\alpha }_{i + 1}\right) }^{2}}\right| \leqq {c}^{{2}^{i + 1}} \]\n\nas desired. | Yes |
Proposition 3. Let \( \alpha ,\beta \) be two elements of the algebraic closure of \( K \) , and assume that \( \alpha \) is separable over \( K\left( \beta \right) \) . Assume that for all isomorphisms \( \sigma \) of \( K\left( \alpha \right) \) over \( K,\sigma \neq {id} \), we have\n\n\[ \left| {\beta - \alpha }\right| < \left| {{\sigma \alpha } - \alpha }\right| \]\n\nThen \( K\left( \alpha \right) \subset K\left( \beta \right) \) . | Proof. It suffices to show that for all isomorphisms of \( K\left( {\beta ,\alpha }\right) \) over \( K\left( \beta \right) \) the element \( \alpha \) remains fixed. Let \( \tau \) be such an isomorphism. By the uniqueness of extensions of absolute values over complete fields, applying \( \tau \) to \( \beta - \alpha \) yields for all \( \sigma \neq {id} \) :\n\n\[ \left| {\beta - {\tau \alpha }}\right| < \left| {{\sigma \alpha } - \alpha }\right| \]\n\nUsing the hypothesis, we obtain\n\n\[ \left| {{\tau \alpha } - \alpha }\right| = \left| {{\tau \alpha } - \beta + \beta - \alpha }\right| < \left| {{\sigma \alpha } - \alpha }\right| .\n\nThis implies that \( \tau \) is the identity, hence \( K\left( {\beta ,\alpha }\right) = K\left( \beta \right) \), as desired. | Yes |
Proposition 4. If \( f \) is irreducible and separable, then any polynomial \( g \) sufficiently close to \( f \) is also irreducible. (Both \( f \) and \( g \) are still assumed to have leading coefficient 1, and the same degree.) Furthermore, given a root \( \alpha \) of \( f \), there exists a root \( \beta \) of \( g \) belonging to \( \alpha \), and \( K\left( \alpha \right) = K\left( \beta \right) \) . | Proof. If \( g \) is sufficiently close to \( f \), then its roots have multiplicity 1, and belong to the distinct roots of \( f \) . If \( \beta \) is a root of \( g \) very close to the root \( \alpha \) of \( f \), then Krasner’s lemma immediately shows that \( K\left( \alpha \right) = K\left( \beta \right) \) . Hence \( g \) is irreducible, since it has the same degree as \( f \) . | Yes |
Proposition 5. If \( \mathfrak{o}/\mathfrak{p} \) is finite, then \( \mathfrak{o} \) and \( U \) are compact. | Proof. We observe that \( \mathfrak{o} \) is the projective limit of the finite groups \( \mathfrak{o}/{\mathfrak{p}}^{i} \) and hence is compact. (It can be viewed as a closed subgroup of the Cartesian product of the \( \mathfrak{o}/{\mathfrak{p}}^{i} \) .) The same argument applies to \( U \) as a projective limit of \( U/{U}_{i} \) . | Yes |
Proposition 6. Let \( K \) be a \( \mathfrak{p} \)-adic field and \( U \) the units of its ring of integers. Let \( m \) be a positive integer. Then\n\n\[ \left( {U : {U}^{m}}\right) = \frac{1}{\parallel m{\parallel }_{\mathfrak{p}}}\left( {{K}_{m}^{ * } : 1}\right) \]\n\nand\n\n\[ \left( {{K}^{ * } : {K}^{*m}}\right) = \frac{m}{\parallel m{\parallel }_{\mathfrak{p}}}\left( {{K}_{m}^{ * } : 1}\right) \]\n\n(where \( {K}_{m}^{ * } \) is the group of \( m \)-th roots of unity contained in \( K \) ). | Proof. The second formula follows from the first by recalling that \( {K}^{ * } \approx \mathbf{Z} \times U. \)\n\nWe now consider the unit index, and the proof is taken from Artin [1].\n\nTake \( r \) so large that \( \left| {m{\pi }^{r + 1}}\right| \geqq \left| {\pi }^{2r}\right| \) and consider the group \( {U}_{r} \). Then for any integral \( x \),\n\n\[ {\left( 1 + x{\pi }^{r}\right) }^{m} \equiv 1 + {mx}{\pi }^{r}\;\left( {{\;\operatorname{mod}\;m}{\pi }^{r + 1}}\right) .\n\nThus if \( {\operatorname{ord}}_{\mathfrak{p}}m = s \), we have\n\n\[ {U}_{r}^{m} = {U}_{r + s} \]\n\nTake \( r \) sufficiently large that no \( m \)-th root of unity except 1 lies in \( {U}_{r} \). We apply the lemma to the homomorphism \( f\left( a\right) = {a}^{m} \), applied to the units. We obtain\n\n\[ \left( {U : {U}_{r}}\right) = \left( {{U}^{m} : {U}_{r + s}}\right) \left( {{K}_{m}^{ * } : 1}\right) \]\n\n\[ = \frac{\left( U : {U}_{r + s}\right) }{\left( U : {U}^{m}\right) }\left( {{K}_{m}^{ * } : 1}\right) \]\n\nHence\n\n\[ \left( {U : {U}^{m}}\right) = \frac{\left( U : {U}_{r + s}\right) }{\left( U : {U}_{r}\right) }\left( {{K}_{m}^{ * } : 1}\right) = \left( {{U}_{r} : {U}_{r + s}}\right) \left( {{K}_{m}^{ * } : 1}\right) .\n\nBut \( \left( {{U}_{r} : {U}_{r + s}}\right) = {\left( \mathbf{N}\mathfrak{p}\right) }^{s} \) and our assertion follows. [II, \( §4 \) ] | Yes |
Proposition 7. Let \( E \) be finite over \( K \), and assume that \( \mathfrak{P} \) is unramified over \( \mathfrak{p} \) . Let \( \bar{\alpha } \in {B}^{\varphi } \) be such that \( {B}^{\varphi } = {A}^{\varphi }\left( \bar{\alpha }\right) \) and let \( \alpha \) be an element of \( B \) such that \( {\varphi \alpha } = \bar{\alpha } \) . Then \( E = K\left( \alpha \right) \), and the irreducible polynomial \( g\left( X\right) \) of \( \alpha \) over \( K \) is such that \( {g}^{\varphi } \) is irreducible. Conversely, if \( E = K\left( \alpha \right) \) for some \( \alpha \in B \) satisfying a polynomial \( g\left( X\right) \) in \( A\left( X\right) \) having leading coefficient \( 1 \) and such that \( {g}^{\varphi } \) has no multiple root, then \( \mathfrak{P} \) is unramified over \( \mathfrak{p} \) and \( {B}^{\varphi } = {A}^{\varphi }\left( {\varphi \alpha }\right) \) | Proof. First assume \( \mathfrak{P} \) unramified. Let \( \bar{g}\left( X\right) \) be the irreducible polynomial of \( \bar{\alpha } \) over \( {A}^{\varphi } \) . Let \( \alpha \) be an element of \( B \) such that \( {\varphi \alpha } = \bar{\alpha } \), and let \( g\left( X\right) \) be its irreducible polynomial over \( K \) . Then \( \alpha \) is integral over \( A \) , and \( \bar{\alpha } \) is a root of \( {g}^{\varphi } \), whence \( \bar{g} \) divides \( {g}^{\varphi } \) . On the other hand\n\n\[ \deg \bar{g} = \left\lbrack {{B}^{\varphi } : {A}^{\varphi }}\right\rbrack = \left\lbrack {E : K}\right\rbrack \geqq \deg g \]\n\nand so \( \bar{g} = {g}^{\varphi } \) . This proves the first statement.\n\nConversely, if \( \alpha \) satisfies the stated condition, then we may assume without loss of generality that its irreducible polynomial \( g\left( X\right) \) is such that \( {g}^{\varphi } \) has no multiple roots. We can now apply Corollary 2 of Proposition 14, Chapter I, \( §5 \) (to the smallest Galois extension of \( K \) containing \( E \) ) to conclude that \( {g}^{\varphi } \) is a power of an irreducible polynomial, and hence is irreducible. Using the inequalities\n\n\[ \left\lbrack {{A}^{\varphi }\left( {\varphi \alpha }\right) : {A}^{\varphi }}\right\rbrack \leqq \left\lbrack {{B}^{\varphi } : {A}^{\varphi }}\right\rbrack \leqq \left\lbrack {E : K}\right\rbrack \]\n\nwe now conclude that we must have an equality everywhere, and that\n\n\[ {B}^{\varphi } = {A}^{\varphi }\left( {\varphi \alpha }\right) \]\n\nThis proves our proposition. | Yes |
Proposition 8. Let \( E \) be a finite extension of \( K \). (i) If \( E \supset F \supset K \), then \( E \) is unramified over \( K \) if and only if \( E \) is unramified over \( F \) and \( F \) is unramified over \( K \). | Proof. The first assertion comes from the fact that the degrees of residue class field extensions are bounded by the degrees of the field extensions, and their multiplicativity property in towers. One must also use the fact that assertion (i) holds when \ | No |
For each finite extension \( E \) of \( K \) in a given algebraic closure, let \( {B}_{E} \) be the integral closure of \( A \) in \( E \) . Let \( \bar{A} \) be the integral closure of \( A \) in the algebraic closure \( \overline{K} \) of \( K.{Let\varphi } \) be a homomorphism of \( \overline{A} \) such that its restriction to \( {B}_{E} \) has the maximal ideal \( {\mathfrak{P}}_{E} \) as kernel. Then the map \[ {B}_{E} \mapsto {B}_{E}^{\varphi } \] induces a bijection between unramified extensions \( E \) of \( K \) and separable extensions of \( {A}^{\varphi } . | Proof. We have shown in Proposition 7 that every finite separable extension of \( {A}^{\varphi } \) is obtainable as an image \( {B}_{E}^{\varphi } \) for some finite extension \( E \) of \( K \), unramified over \( K \) . We now must prove the uniqueness. If \( {E}_{1} \subset {E}_{2} \) are unramified, then clearly \( \varphi {B}_{{E}_{1}} \subset \varphi {B}_{{E}_{2}} \) . Let \( {E}_{1} = K\left( {\alpha }_{1}\right) \) and \( {E}_{2} = K\left( {\alpha }_{2}\right) \) be unramified extensions, generated by elements \( {\alpha }_{1},{\alpha }_{2} \) respectively satisfying polynomials over \( A \) having leading coefficient 1, and whose reductions \( {\;\operatorname{mod}\;\mathfrak{p}} \) have no multiple roots. Then \( {E}_{1}{E}_{2} = {E}_{2}\left( {\alpha }_{1}\right) \) , and \( {\alpha }_{1} \) satisfies with respect to \( {E}_{2} \) a similar condition (with the same polynomial as over \( K \) ). Let \( E = {E}_{1}{E}_{2} \) . Using Proposition 7 once more, we conclude that \[ \varphi {B}_{E} = \varphi {B}_{{E}_{2}}\left( {\varphi {\alpha }_{1}}\right) = {A}^{\varphi }\left( {\varphi {\alpha }_{1},\varphi {\alpha }_{2}}\right) = \left( {\varphi {B}_{{E}_{1}}}\right) \left( {\varphi {B}_{{E}_{2}}}\right) . \] If \( \varphi {B}_{{E}_{1}} = \varphi {B}_{{E}_{2}} \), we conclude that \( {E}_{1} = {E}_{2} \), thus proving our proposition. | Yes |
Proposition 10. Let \( E \) be a finite extension of \( K \). Let \( {E}_{u} \) be the com-positum of all unramified subfields over \( K \). Then \( {E}_{u} \) is unramified over \( K \), and \( E \) is totally ramified over \( {E}_{u} \). | Proof. The first statement comes from Proposition 8 of the preceding section. As to the second, we consider the towers\n\n\n\nIf the residue class field extension in the upper level of the tower had degree \( > 1 \), then it could be lifted back to an unramified subfield of \( E \) over \( {E}_{u} \), of the same degree, contradicting the maximality of \( {E}_{u} \). Hence the degree must be equal to 1, and therefore \( E \) is totally ramified over \( {E}_{u} \). | Yes |
Proposition 11. Assume that \( E \) is totally ramified over \( K \) . Let \( \Pi \) be an element of order 1 at \( \mathfrak{P} \) . Then \( \Pi \) satisfies an Eisenstein equation\n\n\[{X}^{e} + {a}_{e - 1}{X}^{e - 1} + \cdots + {a}_{0} = 0\]\n\nwhere \( {a}_{i} \in \mathfrak{p} \) for all \( i \) and \( {a}_{0} ≢ 0\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{2}}\right) \) . Conversely, such an equation is irreducible, and a root generates a totally ramified extension of degree e. | Proof. All conjugates of \( \Pi \) over \( K \) have the same absolute value (by the uniqueness of the extension of \( \mathfrak{p} \) to any finite extension), and hence the coefficients of its irreducible equation, which are polynomial functions of the roots, lie in \( \mathfrak{P} \cap A = \mathfrak{p} \) . The last coefficient \( {a}_{0} \) is the product of \( \Pi \) and its conjugates, and there are \( e \) of those. Hence\n\n\[\\left| {a}_{0}\\right| = {\\left| \\Pi \\right| }^{e}\]\n\nso \( {a}_{0} = \\pi \) is an element of order 1 at \( \\mathfrak{p} \) . As to the converse, an Eisenstein equation is irreducible. If \( \\beta \) is a root, then the same argument we applied to II before now applies to \( \\beta \) and shows that \( {\\left| \\beta \\right| }^{e} = \\left| \\pi \\right| \) . Hence \( e = \\left\\lbrack {K\\left( \\beta \\right) : K}\\right\\rbrack \) | Yes |
Proposition 12. Let \( E \) be totally and tamely ramified over \( K \) . Then there exists an element \( \Pi \) of order 1 at \( \mathfrak{P} \) in \( E \) satisfying an equation\n\n\[ {X}^{e} - \pi = 0 \]\n\nwith \( \pi \) of order 1 at \( \mathfrak{p} \) in \( K \) . Conversely, let a be an element of \( A \), and e a positive integer not divisible by \( p \) . Then any root of an equation\n\n\[ {X}^{e} - a = 0 \]\n\ngenerates a tamely ramified extension of \( K \), and this extension is totally ramified if the order at \( \mathfrak{p} \) of \( a \) is relatively prime to \( e \) . | Proof. Let \( f\left( X\right) = {X}^{e} - a \) with \( a \in A \) and \( e \) not divisible by \( p \) . Let \( \alpha \) be any root of \( f \) . Write \( a = {\pi }^{r}u \) with some integer \( r \) and a unit \( u \) of \( A \) .\n\nThen \( K\left( \alpha \right) \) is contained in \( K\left( {\zeta ,{u}^{1/e},{\pi }^{1/e}}\right) \), where \( \zeta \) is a primitive \( e \) -th root of unity. The extension \( F = K\left( {\zeta ,{u}^{1/e}}\right) \) is unramified over \( K \), and hence \( \pi \) is still a prime element in \( {\mathfrak{P}}_{F} \) . The extension \( F\left( {\pi }^{1/e}\right) \) is totally and tamely ramified, and hence the ramification index of \( K\left( \alpha \right) \) over \( K \) divides that of \( K\left( {\zeta ,{u}^{1/e},{\pi }^{1/e}}\right) \) over \( K \) . This proves that \( K\left( \alpha \right) \) is tamely ramified over \( K \) . If the order of \( a \) at \( \mathfrak{p} \) is relatively prime to \( e \), then we can find two integers \( s, t \) such that\n\n\[ {se} + {tr} = 1\text{.} \]\n\nLet \( \beta = {\alpha }^{t}{\pi }^{s} \) . Then \( {\beta }^{e} \) and \( \pi \) have the same order at \( \mathfrak{P} \), whence the ramification index is at least equal to \( e \) . It must therefore be equal to \( e \) (because \( \left\lbrack {K\left( \alpha \right) : K}\right\rbrack \leqq e \) ), and our extension is totally tamely ramified. | Yes |
Proposition 14. Let \( K \) be a \( \mathfrak{p} \) -adic field (finite extension of \( {\mathbf{Q}}_{p} \) ). Given an integer \( n \), there exists only a finite number of extensions of degree \( \leqq n \) . | Proof. Since there is exactly one unramified extension of a given degree, corresponding to an extension of the residue class field, and since every extension is a tower of an unramified and totally ramified extension, it will suffice to prove that there is only a finite number of totally ramified extensions of a given degree \( e \) . But such extensions are obtained by Eisenstein equations\n\n\[ \n{X}^{e} + {a}_{e - 1}{X}^{e - 1} + \cdots + {u}_{0}\pi = 0,\n\]\n\nwhere the coefficients \( {a}_{i} \) belong to \( \mathfrak{p} \) and \( {u}_{0} \) is a unit \( (\pi \) being a fixed prime element of \( \mathfrak{p} \) ). The Cartesian product\n\n\[ \n\mathfrak{p} \times \cdots \times \mathfrak{p} \times U\n\]\n\nof the units and of \( \mathfrak{p} \) taken \( e - 1 \) times is compact. Any point in it can be viewed as determining a finite number of extensions of degree \( e \) (corresponding to the distinct roots of the equation). By Krasner's lemma, it follows that a neighborhood of such a point determines the same extensions (Proposition 4 of §2), and by compactness the finiteness follows. | Yes |
Proposition 1. If \( {w}_{1},\ldots ,{w}_{n} \) is a basis of \( E \) over \( K \) and\n\n\[ L = A{w}_{1} + \cdots + A{w}_{n} \]\n\nthen\n\n\[ {L}^{\prime } = A{w}_{1}^{\prime } + \cdots + A{w}_{n}^{\prime } \]\n\nwhere \( \left\{ {w}_{i}^{\prime }\right\} \) is the dual basis relative to the trace. | Proof. Let \( \alpha \in {L}^{\prime } \) and write\n\n\[ \alpha = {a}_{1}{w}_{1}^{\prime } + \cdots + {a}_{n}{w}_{n}^{\prime } \]\n\nwith \( {a}_{i} \in K \) . Then \( \operatorname{Tr}\left( {\alpha {w}_{i}}\right) = {a}_{i} \), whence \( {a}_{i} \in A \) for all \( i \) . This proves the inclusion \( \subset \) . Conversely,\n\n\[ \operatorname{Tr}\left( {A{w}_{i}^{\prime }L}\right) = A \cdot \operatorname{Tr}\left( {{w}_{i}^{\prime }L}\right) \subset A \]\n\nso the inclusion \( \supset \) is equally trivial. | Yes |
Proposition 2. Let \( E = K\left( \alpha \right) \) be a finite separable extension, of degree \( n \) . Let \( f \) be the irreducible polynomial of \( \alpha \) over \( K,{f}^{\prime } \) its derivative, and\n\n\[ \n\frac{f\left( X\right) }{X - \alpha } = {b}_{0} + {b}_{1}X + \cdots + {b}_{n - 1}{X}^{n - 1}.\n\]\n\nThen the dual basis of \( 1,\alpha ,\ldots ,{\alpha }^{n - 1} \) is\n\n\[ \n\frac{{b}_{0}}{{f}^{\prime }\left( \alpha \right) },\ldots ,\frac{{b}_{n - 1}}{{f}^{\prime }\left( \alpha \right) }.\n\] | Proof. Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be the distinct roots of \( f \) . Then\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{n}\frac{f\left( X\right) }{\left( X - {\alpha }_{i}\right) }\frac{{\alpha }_{i}^{r}}{{f}^{\prime }\left( {\alpha }_{i}\right) } = {X}^{r},\;0 \leqq r \leqq n - 1.\n\]\n\nTo see this, let \( g\left( X\right) \) be the difference of the left- and right-hand side of this equality. Then \( g \) has degree \( \leqq n - 1 \), and has \( n \) roots \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) ; hence \( g \) is identically 0 .\n\nThe polynomials\n\n\[ \n\frac{f\left( X\right) }{X - {\alpha }_{i}}\frac{{\alpha }_{i}^{r}}{{f}^{\prime }\left( {\alpha }_{i}\right) }\n\]\n\nare all conjugate to each other. If we define the trace of a polynomial with coefficients in \( E \) to be the polynomial obtained by applying the trace to the coefficients, then\n\n\[ \n\operatorname{Tr}\left\lbrack {\frac{f\left( X\right) }{X - \alpha }\frac{{\alpha }^{r}}{{f}^{\prime }\left( \alpha \right) }}\right\rbrack = {X}^{r}\n\]\n\nLooking at the coefficient of each power of \( X \) in this equation, we see that\n\n\[ \n\operatorname{Tr}\left( {{\alpha }^{i}\frac{{b}_{j}}{{f}^{\prime }\left( \alpha \right) }}\right) = {\delta }_{ij}\n\]\n\nthereby proving our assertion. | Yes |
Proposition 3. Assume that \( A \) is a discrete valuation ring, that there is only one prime \( \mathfrak{P} \) of \( B \) above \( A \), and that \( B/\mathfrak{P} \) is separable over \( A/\mathfrak{p} \) . Then there exists \( \alpha \in B \) such that \( B = A\left\lbrack \alpha \right\rbrack \) . | Proof. Let \( \beta \) be an element of \( B \) whose residue class \( {\;\operatorname{mod}\;\mathfrak{P}} \) generates \( B/\mathfrak{P} \) over \( A/\mathfrak{p} \) . Let \( f \) be a polynomial with leading coefficient 1 and coefficients in \( A \) such that its reduced polynomial \( {\;\operatorname{mod}\;\mathfrak{p}} \) is an irreducible polynomial for \( \beta \) mod \( \mathfrak{P} \) . Let \( \Pi \) be an element of order one at \( \mathfrak{P} \) in \( B \) . Then\n\n\[ f\left( {\beta + \Pi }\right) \equiv f\left( \beta \right) + {f}^{\prime }\left( \beta \right) \Pi \;\left( {\;\operatorname{mod}\;{\mathfrak{P}}^{2}}\right) ,\]\n\nand \( {f}^{\prime }\left( \beta \right) ≢ 0\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \) . Hence taking either \( \beta \) or \( \beta + \Pi \) yields an element \( \alpha \) such that its residue class generates \( B/\mathfrak{P} \) over \( A/\mathfrak{p} \) and such that there exists an element of order 1 at \( \mathfrak{P} \) in the ring \( A\left\lbrack \alpha \right\rbrack \) . We conclude by Proposition 23 of Chapter I, \( §7 \) that \( B = A\left\lbrack \alpha \right\rbrack \) . | Yes |
Proposition 4. Let \( \\mathfrak{b} \) be a fractional ideal of \( B \) . Then\n\n\[ \n{\\mathfrak{b}}^{\\prime } = {B}^{\\prime }{\\mathfrak{b}}^{-1}.\n\] | Proof. We have\n\n\[ \n\\operatorname{Tr}\\left( {{B}^{\\prime }{\\mathfrak{b}}^{-1}\\mathfrak{b}}\\right) = \\operatorname{Tr}\\left( {{B}^{\\prime }B}\\right) \\subset A\n\]\n\nwhence \( {B}^{\\prime }{\\mathfrak{b}}^{-1} \\subset {\\mathfrak{b}}^{\\prime } \) . The converse is equally clear. | Yes |
Proposition 5. Let \( E \supset F \supset K \) be two separable extensions, \( C \) the integral closure of \( A \) in \( F \), and \( B \) the integral closure of \( A \) in \( E \). Then\n\n\[ \n{B}^{\prime }{}_{E/K} = {B}^{\prime }{}_{E/F}{C}^{\prime }{}_{F/K} \n\] | Proof. We prove first the inclusion \( \supset \). We have\n\n\[ \n{\operatorname{Tr}}_{K}^{E}\left( {{B}_{E/F}^{\prime }{C}_{F/K}^{\prime }B}\right) = {\operatorname{Tr}}_{K}^{F}{\operatorname{Tr}}_{F}^{E}\left( {{B}_{E/F}^{\prime }{C}^{\prime }{}_{F/K}B}\right) \n\]\n\n\[ \n= {\operatorname{Tr}}_{K}^{F}\left( {{C}_{F/K}^{\prime }{\operatorname{Tr}}_{F}^{E}\left( {{B}_{E/F}^{\prime }B}\right) }\right) \n\]\n\n\[ \n\subset A\text{.} \n\]\n\nThis proves the desired inclusion.\n\nConversely, let \( \beta \in {B}_{E/K}^{\prime } \). Then\n\n\[ \n{\operatorname{Tr}}_{K}^{E}\left( {\beta B}\right) = {\operatorname{Tr}}_{K}^{F}\left( {C{\operatorname{Tr}}_{F}^{E}\left( {\beta B}\right) }\right) \subset A \n\]\n\n(we can insert \( C \) since \( {CB} = B \) ). Thus\n\n\[ \n{\operatorname{Tr}}_{F}^{E}\left( {\beta B}\right) \subset {C}_{F/K}^{\prime } \n\]\n\nand\n\n\[ \n{C}_{F/K}^{\prime - 1}{\operatorname{Tr}}_{F}^{E}\left( {\beta B}\right) \subset C. \n\]\n\nThe \( C \) -fractional ideal \( {C}_{F/K}^{\prime - 1} \) can be taken inside the trace \( {\operatorname{Tr}}_{F}^{E} \) because it is contained in \( F \). Hence\n\n\[ \n\beta {C}_{F/K}^{\prime - 1} \subset {B}_{E/F}^{\prime } \n\]\n\nMultiplying by \( {C}_{F/K}^{\prime } \) shows that \( \beta \in {C}_{F/K}^{\prime }{B}_{E/F}^{\prime } \) and concludes the proof of the reverse inclusion. | Yes |
Proposition 6. Let \( S \) be a multiplicative subset of \( A \) . Then\n\n\[{\mathfrak{D}}_{{S}^{-1}B/{S}^{-1}A} = {S}^{-1}{\mathfrak{D}}_{B/A}\] | Proof. Obvious. | No |
Proposition 7. Let \( A \) be a discrete valuation ring, \( v \) its valuation, and \( \mathfrak{P} \) a prime of \( B \) lying above the prime \( \mathfrak{p} \) of \( A \) . Let \( {w}_{\mathfrak{P}} \) be the valuation corresponding to \( \mathfrak{P} \) and \( {A}_{v},{B}_{w\mathfrak{P}} \) the respective completions. Then:\n\n\[ \n{\mathfrak{D}}_{B/A}{B}_{{w}_{\mathfrak{P}}} = {\mathfrak{D}}_{{B}_{{w}_{\mathfrak{P}}}/{A}_{v}} \n\] | Proof. Since the differents are ideals, it suffices to prove that\n\n\[ \n{\operatorname{ord}}_{\mathfrak{P}}{\mathfrak{D}}_{B/A} = {\operatorname{ord}}_{\mathfrak{P}}{\mathfrak{D}}_{{B}_{{w}_{\mathfrak{P}}}/{A}_{v}}. \n\]\n\nLet \( \operatorname{Tr} \) denote the trace from \( E \) to \( K \) and \( {\operatorname{Tr}}_{w} \) the local trace from \( {E}_{w} \) to \( {K}_{v} \) for any \( w \) extending \( v \) in \( E \) . Then\n\n\[ \n\operatorname{Tr} = \mathop{\sum }\limits_{{w \mid v}}{\operatorname{Tr}}_{w} \n\]\n\n(as an operator on \( E \) ).\n\nLet \( x \in {E}_{w\mathfrak{p}} \) and assume that \( {\operatorname{Tr}}_{w\mathfrak{p}}\left( {x{B}_{w\mathfrak{p}}}\right) \subset {A}_{v} \) . Select an element \( \xi \) of \( E \) which is very close to \( x \) at \( {w}_{\mathfrak{P}} \) and very close to 0 at all other \( w \mid v \) . Let \( y \in B \) . Then \( {\operatorname{Tr}}_{w}\left( {\xi y}\right) \) is close to 0 if \( w \neq {w}_{\mathfrak{P}} \) and \( {\operatorname{Tr}}_{w}\left( {\xi y}\right) \) lies in \( {A}_{v} \) if \( w = {w}_{\mathfrak{P}} \), by assumption and the fact that the local trace is continuous. This implies that \( \operatorname{Tr}\left( {\xi y}\right) \) lies in \( A \) and hence that \( \xi \) lies in the complementary module \( {B}^{\prime } \) .\n\nConversely, let \( x \) be an element in \( {B}^{\prime } \) and let \( y \in {B}_{w\mathfrak{y}} \) . Find an element \( \xi \) of \( E \) which is close to \( x \) at \( {w}_{\mathfrak{P}} \) and close to 0 at the other \( w \mid v \) . Find an element \( \eta \) of \( B \) close to \( y \) at \( {w}_{\mathfrak{P}} \) and close to 0 at the other \( w \mid v \) . Then\n\n\[ \n\operatorname{Tr}\left( {\xi \eta }\right) = {\operatorname{Tr}}_{w\mathfrak{P}}\left( {\xi \eta }\right) + \mathop{\sum }\limits_{{w \neq w\mathfrak{P}}}{\operatorname{Tr}}_{w}\left( {\xi \eta }\right) \n\]\n\nThe global trace on the left lies in \( A \) . Each term in the sum on the right lies in \( {A}_{v} \) . Hence \( {\operatorname{Tr}}_{w\mathfrak{P}}\left( {\xi \eta }\right) \) lies in \( {A}_{v} \) . Since \( \xi \) and \( \eta \) are close to \( x, y \) respectively, it follows that \( {\operatorname{Tr}}_{w\mathfrak{P}}\left( {xy}\right) \) also lies in \( {A}_{v} \).\n\nThe above arguments show that \( {B}^{\prime } \) is dense in \( {B}_{w\mathfrak{B}}^{\prime }( = \) local complementary module with respect to \( {\operatorname{Tr}}_{w\mathfrak{P}} \) ) and the proposition follows. | Yes |
Proposition 8. Let \( \mathfrak{P} \) be a prime of \( B \) lying above \( \mathfrak{p} \), and let \( e \) be its ramification index. Then \( {\mathfrak{P}}^{e - 1} \) divides \( {\mathfrak{D}}_{B/A} \) . If \( \mathfrak{P} \) is strongly ramified, then \( {\mathfrak{P}}^{e} \) divides \( {\mathfrak{D}}_{B/A} \) . If \( \mathfrak{P} \) is unramified, then \( \mathfrak{P} \) does not divide \( {\mathfrak{D}}_{B/A} \) . There is only a finite number of ramified primes. Finally, \( {\mathfrak{D}}_{B/A} \) is the greatest common divisor of all ideals \( \left( {{f}^{\prime }\left( \alpha \right) }\right) \), where \( \alpha \) is an integral generator of \( E \) over \( K \), and \( f \) the irreducible polynomial for \( \alpha \) over \( K \) . | Proof. In view of the fact that ramification theory and the theory of the different localize to the completion, we may prove the first assertions under the assumption that \( K \) is complete.\n\nSince we work over a complete field, we can apply Proposition 3 of \( §1 \) , the Corollary of Proposition 2, \( §1 \), and Proposition 23 of Chapter I, \( §7 \) . If \( \mathfrak{P} \) is unramified, this yields \( {\mathfrak{D}}_{B/A} = \left( 1\right) \) . Using Proposition 5 of \( §1 \) (multiplicativity in towers), we may also assume that \( \mathfrak{P} \) is totally ramified. In that case, we can write \( B = A\left\lbrack \Pi \right\rbrack \) for some element \( \Pi \) of order 1 at \( \mathfrak{P} \) , and \( \Pi \) satisfies an Eisenstein equation\n\n\[ f\left( \Pi \right) = {\Pi }^{e} + {a}_{n - 1}{\Pi }^{e - 1} + \cdots + \pi = 0, \]\n\nfor \( {a}_{i} \in \mathfrak{p} \) and \( \pi \in A \) of order 1 at \( \mathfrak{p} \) . Then\n\n\[ {f}^{\prime }\left( \Pi \right) \equiv e{\Pi }^{e - 1}\;\left( {\;\operatorname{mod}\;{\mathfrak{P}}^{e}}\right) , \]\n\nand the second assertion of the proposition follows from the definitions. | Yes |
Proposition 9. Notation as above, the discriminant \( {D}_{E/K}\left( W\right) \) lies in \( K \) , and lies in \( A \) if the components of \( W \) lie in \( B \) . The discriminant is \( \neq 0 \) if and only if \( W \) is a basis of \( E \) over \( K \) . | Proof. Applying any isomorphism \( \sigma \) of \( E \) over \( K \) to the determinant \( \det \left( {{\sigma }_{i}{w}_{j}}\right) \) interchanges the rows, hence multiplies the determinant by \( \pm 1 \) . Taking the square gets rid of \( \pm 1 \) . If \( \alpha \) is a generator of \( E \) over \( K \), i.e. \( E = K\left( \alpha \right) \), then the discriminant \( {D}_{E/K}\left( {1,\alpha ,\ldots ,{\alpha }^{n - 1}}\right) \) is the Vander-monde determinant, and hence is \( \neq 0 \) . The same holds therefore for any basis \( V \) of \( E \) over \( K \) by a preceding remark concerning the change of the discriminant under linear transformations. If the coordinates of \( W \) are linearly dependent over \( K \), it is clear that the discriminant is 0 . If they are all integral over \( A \), it is also clear that the discriminant lies in \( A \) (because the integral closure of \( A \) in a Galois extension containing \( E \) is a ring). This proves our proposition. | Yes |
Proposition 10. Let \( {M}_{1} \subset {M}_{2} \) be two free modules of rank \( n \) over \( A \) , contained in \( E \) . Then \( {D}_{E/K}\left( {M}_{1}\right) \) divides \( {D}_{E/K}\left( {M}_{2}\right) \) (as principal ideals). If \( {D}_{E/K}\left( {M}_{1}\right) = {D}_{E/K}\left( {M}_{2}\right) u \) for some unit \( u \) of \( A \), then \( {M}_{1} = {M}_{2} \) . | Proof. The first statement is obvious. The second statement asserts that the matrix going from a basis of \( {M}_{1} \) to a basis of \( {M}_{2} \) is invertible in \( A \), and hence that \( {M}_{1} = {M}_{2} \) . | No |
Proposition 11. Let \( \mathfrak{b} \) be a fractional ideal of \( B \) and \( S \) a multiplicative subset of \( A \) . Then\n\n\[ {S}^{-1}{D}_{E/K}\left( \mathfrak{b}\right) = {D}_{E/K}\left( {{S}^{-1}\mathfrak{b}}\right) . \] | Proof. Trivial from the definitions. | No |
Proposition 12. Assume in addition that \( A \) is a discrete valuation ring.\n\nLet \( \mathfrak{b} \) be a fractional ideal of \( B,\mathfrak{b} = \left( \beta \right) \) for some \( \beta \neq 0 \) in \( E \) . Then\n\n\[ {D}_{E/K}\left( \mathfrak{b}\right) = {\left( {N}_{K}^{E}\left( \beta \right) \right) }^{2}{D}_{E/K}\left( B\right) . \] | Proof. Let \( W \) be a basis of \( B \) over \( A \) . Then \( {\beta W} \) is a basis of \( \mathfrak{b} \) over \( A \) , and the assertion is obvious from the definitions. | No |
Proposition 13. Let \( A \) be arbitrary again, and \( \mathfrak{b} \) a fractional ideal of \( B \) . Then \[ {D}_{E/K}\left( \mathfrak{b}\right) = {\left( {N}_{K}^{E}\left( \mathfrak{b}\right) \right) }^{2}{D}_{E/K}\left( B\right) , \] the norm being the norm of ideals as in Chapter I, §7. | Proof. It suffices to verify this relation for each \( \mathfrak{p} \) -component, \( \mathfrak{p} \) a prime of \( A \) . Thus we may assume that \( A \) is a discrete valuation ring by Proposition 11. In that case \( \mathfrak{b} = \left( \beta \right) \) for some \( \beta \in E \), and our assertion follows from Proposition 23 of Chapter I, §7. | Yes |
Proposition 14. The discriminant and different are related by the formula\n\n\[ \n{N}_{K}^{E}{\mathfrak{D}}_{B/A} = {D}_{E/K}\left( B\right) \n\] | Proof. Using Proposition 6 of \( \$ 1 \) and Proposition 11, we may assume that \( A \) is a discrete valuation ring, and hence that \( B \) is a free module over \( A \) . If \( W \) is a basis for \( B \) over \( A \), then \( {D}_{E/K}\left( B\right) \) is generated by \( {D}_{E/K}\left( W\right) \) . Let \( {W}^{\prime } \) be the complementary basis to \( W \) under the trace. Then the complementary module \( {B}^{\prime } \) is generated by \( {W}^{\prime } \) over \( A \) . Thus\n\n\[ \n{D}_{E/K}\left( {B}^{\prime }\right) = {D}_{E/K}\left( {W}^{\prime }\right) A. \n\]\n\nBut we see directly from the definition of the discriminant of a basis that\n\n\[ \n{D}_{E/K}\left( W\right) {D}_{E/K}\left( {W}^{\prime }\right) = 1 \n\]\n\nHence \( {D}_{E/K}\left( B\right) {D}_{E/K}\left( {B}^{\prime }\right) = A \) . Using Proposition 4 of \( §1 \) and Proposition 13 yields what we want. | Yes |
Proposition 15. We have\n\n\[ \n{D}_{E/K}\left( \beta \right) = {\left( -1\right) }^{n\left( {n - 1}\right) /2}{N}_{K}^{E}{\mathfrak{D}}_{E/K}\left( \beta \right) .\n\] | Proof. Exercise in permuting the rows of a determinant. | No |
Proposition 16. Let \( \alpha \in B \) and let \( \mathfrak{p} \) be a prime of \( A \) . If \( \mathfrak{p} \) does not divide \( {D}_{E/K}\left( \alpha \right) /{D}_{E/K}\left( B\right) \) then \( {B}_{\mathfrak{p}} = {A}_{\mathfrak{p}}\left\lbrack \alpha \right\rbrack \) . | Proof. By Theorem 1 of Chapter I, \( §2 \) we know that \( {B}_{\mathfrak{p}} \) is a free module over \( {A}_{\mathfrak{p}} \) . Furthermore\n\n\[ \n{D}_{E/K}\left( {1,\alpha ,\ldots ,{\alpha }^{n - 1}}\right) = {D}_{E/K}\left( B\right) {c}^{2} \n\]\n\nwhere \( c \) is an element of \( {A}_{\mathfrak{p}} \) . By hypothesis, this element \( c \) is a unit in \( {A}_{\mathfrak{p}} \) , and hence our proposition follows from Proposition 10. | No |
Proposition 17. Let \( K, E \) be two number fields. Assume that their discriminants are relatively prime and that the fields are linearly disjoint (i.e. if \( {w}_{1},\ldots ,{w}_{n} \) is a basis of \( K \) over \( \mathbf{Q} \) and \( {v}_{1},\ldots ,{v}_{m} \) is a basis of \( E \) over \( \mathbf{Q} \) , then \( \left\{ {{w}_{i}{v}_{j}}\right\} \) is a basis of \( {KE} \) over \( \mathbf{Q} \) ). Then\n\n\[{\mathfrak{o}}_{KE} = {\mathfrak{o}}_{K}{\mathfrak{o}}_{E}\]\n\nand\n\n\[{D}_{KE} = {D}_{K}^{m}{D}_{E}^{n}\] | Proof. From the fundamental properties of the different, we know that \( {\mathfrak{D}}_{{KE}/\mathbf{Q}} \) is equal to\n\n\[{\mathfrak{D}}_{{KE}/K}{\mathfrak{D}}_{K/\mathbf{Q}} = {\mathfrak{D}}_{{KE}/E}{\mathfrak{D}}_{E/\mathbf{Q}}\]\n\nBut \( {\mathfrak{D}}_{E/\mathbf{Q}} \) and \( {\mathfrak{D}}_{K/\mathbf{Q}} \) have no factor in common (viewed as ideals of \( {\mathfrak{o}}_{KE} \) ). The same holds for the other two factors. Hence\n\n\[{\mathfrak{D}}_{{KE}/E} = {\mathfrak{D}}_{K/\mathbf{Q}}\;\text{ and }\;{\mathfrak{D}}_{{KE}/K} = {\mathfrak{D}}_{E/\mathbf{Q}}.\]\n\nLet \( W \) be a basis for \( {\mathfrak{o}}_{K} \) over \( \mathbf{Z} \) and \( V \) a basis for \( {\mathfrak{o}}_{E} \) over \( \mathbf{Z} \) . Then the above remark implies that the complementary basis \( {W}^{\prime } \) of \( W \), which generates \( {\mathfrak{D}}_{K/\mathbf{Q}}^{-1} \), also generates \( {\mathfrak{D}}_{{KE}/E}^{-1} \) . This is the complementary module of \( {\mathfrak{o}}_{KE} \) relative to \( {\mathfrak{o}}_{E} \) . Dualizing again shows that \( W \) generates \( {\mathfrak{o}}_{KE} \) over \( {\mathfrak{o}}_{E} \) and proves the assertion concerning the rings of integers. We leave the assertion on the discriminants as an exercise. | No |
Theorem 2. Let \( m \) be a positive integer and \( \omega \) a primitive \( m \) -th root of unity. Then \( \left\lbrack {\mathbf{Q}\left( \omega \right) : \mathbf{Q}}\right\rbrack = \varphi \left( m\right) \) . The only ramified primes \( p \) in \( \mathbf{Q}\left( \omega \right) \) are those dividing \( m \) . If \[ m = {p}_{1}^{{r}_{1}}\cdots {p}_{s}^{{r}_{s}} \] is the prime power decomposition of \( m,{\omega }_{j} \) is a primitive \( {p}^{{r}_{j}} \) -th root of unity, then \[ \mathbf{Q}\left( \omega \right) = \mathbf{Q}\left( {{\omega }_{1},\ldots ,{\omega }_{s}}\right) = \mathbf{Q}\left( {\omega }_{1}\right) \ldots \mathbf{Q}\left( {\omega }_{s}\right) \] is the compositum of the \( \mathbf{Q}\left( {\omega }_{j}\right) \) . | Proof. Let \( g\left( X\right) = {X}^{m} - 1 \) . Then \( \omega \) satisfies \( g\left( X\right) = 0 \), and \[ {g}^{\prime }\left( \omega \right) = m{\omega }^{m - 1} \] is divisible only by primes dividing \( m \) . Hence any other prime is unramified in \( \mathbf{Q}\left( \omega \right) \) . For any \( j > 1 \), the field \( \mathbf{Q}\left( {\omega }_{j}\right) \) is an abelian extension of \( \mathbf{Q} \) whose intersection with \( \mathbf{Q}\left( {{\omega }_{1},\ldots ,{\omega }_{j - 1}}\right) \) is \( \mathbf{Q} \), because \( {p}_{j} \) is totally ramified in \( \mathbf{Q}\left( {\omega }_{j}\right) \) and unramified in the other field. Hence \( \mathbf{Q}\left( {{\omega }_{1},\ldots ,{\omega }_{j}}\right) \) has degree \( \varphi \left( {p}^{{r}_{j}}\right) \) over \( \mathbf{Q}\left( {{\omega }_{1},\ldots ,{\omega }_{j - 1}}\right) \) . This proves our theorem. | Yes |
Theorem 3. Let \( \omega \) be a primitive \( {p}^{r} \) -th root of unity, and \( K = \mathbf{Q}\left( \omega \right) \) . Then \( {\mathfrak{o}}_{K} = \mathbf{Z}\left\lbrack \omega \right\rbrack \) . The discriminant is given by \[ {D}_{K} = \pm {p}^{{p}^{r - 1}\left( {{pr} - r - 1}\right) } \] where the - sign holds when \( {p}^{r} = 4 \) or \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \), and the + sign holds otherwise. | Proof. We shall give the proof only when \( r = 1 \) . The principle is the same in general. Thus we deal with the \( p \) -th roots of unity. Let \( B = \mathbf{Z}\left\lbrack \omega \right\rbrack \) . To prove that \( B = {\mathfrak{o}}_{K} \) it suffices to prove that the discriminant of \( B \) and \( {\mathfrak{o}}_{K} \) as modules over \( \mathbf{Z} \) coincide as \( \mathbf{Z} \) -ideals by Proposition 10 of Chapter III, §3. To do this, it suffices to prove it locally for each prime. All primes except \( p \) are unramified, and consequently such primes do not contribute either to the discriminant of \( {\mathfrak{o}}_{K} \) or of \( B \) . As for \( p \), it is totally ramified, and using Proposition 23 of Chapter I, \( §7 \), we conclude that \( {S}_{p}^{-1}B = {S}_{p}^{-1}{\mathfrak{o}}_{K} \) if \( {S}_{p} \) is the complement of the principal ideal \( \left( p\right) \) in \( \mathbf{Z} \) . Hence the \( p \) -component of the discriminants is the same in both cases. This proves that \( B = {\mathfrak{o}}_{K} \) . The assertion concerning the exact value of the discriminant comes from taking the discriminant of the element \( \omega \) itself, and paying attention to the sign. There is no difficulty in this (use Proposition 15 of Chapter III, §3). | No |
Theorem 4. Let \( m \) be a positive integer, and \( \omega \) a primitive \( m \) -th root of unity. Then \( \mathbf{Z}\left\lbrack \omega \right\rbrack \) is the integral closure of \( \mathbf{Z} \) in \( \mathbf{Q}\left( \omega \right) \) . | Proof. It is clearly the compositum of the rings of integers of various prime power cyclotomic fields which satisfy the conditions of Proposition 17, Chapter III, §3. | No |
Theorem 5. Let \( m \) be a square-free integer \( \neq 0 \), and let \( K = \mathbf{Q}\left( \sqrt{m}\right) \) . If \( m \equiv 2 \) or 3 (mod 4), then \( \left\lbrack {1,\sqrt{m}}\right\rbrack \) is a basis for \( {\mathfrak{o}}_{K} \) over \( \mathbf{Z} \) . If \( m \equiv 1 \) \( \left( {\;\operatorname{mod}\;4}\right) \), then \[ \left\lbrack {1,\frac{1 + \sqrt{m}}{2}}\right\rbrack \] is a basis for \( {\mathfrak{o}}_{K} \) over \( \mathbf{Z} \) . | Proof. Exercise. To verify that an element \( x + y\sqrt{m} \) with \( x, y \in \mathbf{Q} \) is integral over \( \mathbf{Z} \), it is necessary and sufficient that its norm and trace lie in \( \mathbf{Z} \) . From this, there is no difficulty in verifying the assertion of the theorem. | No |
Theorem 6. Let \( \zeta \) be a primitive \( p \) -th root of unity for \( p \) odd, and\n\n\[ S = \mathop{\sum }\limits_{\nu }\left( \frac{\nu }{p}\right) {\zeta }^{\nu } \]\n\nthe sum being taken over non-zero residue classes \( {\;\operatorname{mod}\;p} \) . Then\n\n\[ {S}^{2} = \left( \frac{-1}{p}\right) p \] | Proof. The last statement follows at once from the explicit expression of \( \pm p \) as a square in \( \mathbf{Q}\left( \zeta \right) \) and also \( {\left( 1 + i\right) }^{2} = {2i} \) . As for the sum, we have\n\n\[ {S}^{2} = \mathop{\sum }\limits_{{\nu ,\mu }}\left( \frac{\nu \mu }{p}\right) {\zeta }^{\nu + \mu } \]\n\nAs \( \nu \) ranges over non-zero residue classes, so does \( {\nu \mu } \) for any fixed \( \mu \), and hence replacing \( \nu \) by \( {\nu \mu } \) yields\n\n\[ {S}^{2} = \mathop{\sum }\limits_{{\nu ,\mu }}\left( \frac{\nu {\mu }^{2}}{p}\right) {\zeta }^{\mu \left( {\nu + 1}\right) } = \mathop{\sum }\limits_{{\nu ,\mu }}\left( \frac{\nu }{p}\right) {\zeta }^{\mu \left( {\nu + 1}\right) } \]\n\n\[ = \mathop{\sum }\limits_{\mu }\left( \frac{-1}{p}\right) {\zeta }^{0} + \mathop{\sum }\limits_{{\nu \neq - 1}}\left( \frac{\nu }{p}\right) \mathop{\sum }\limits_{\mu }{\zeta }^{\mu \left( {\nu + 1}\right) }. \]\n\nBut \( 1 + \zeta + \cdots + {\zeta }^{p - 1} = 0 \), and the sum on the right over \( \mu \) consequently yields -1 . Hence\n\n\[ {S}^{2} = \left( \frac{-1}{p}\right) \left( {p - 1}\right) + \left( {-1}\right) \mathop{\sum }\limits_{{\nu \neq - 1}}\left( \frac{\nu }{p}\right) \]\n\n\[ = p\left( \frac{-1}{p}\right) - \mathop{\sum }\limits_{\nu }\left( \frac{\nu }{p}\right) \]\n\n\[ = \left( \frac{-1}{p}\right) p \]\n\nas desired. | Yes |
Theorem 7. We have \( {T}^{2}f = q{f}^{ - } \), i.e. \( {T}^{2}f\left( z\right) = {qf}\left( {-z}\right) \) . | Proof. We have\n\n\[ \n{T}^{2}f\left( z\right) = \mathop{\sum }\limits_{y}\mathop{\sum }\limits_{x}f\left( x\right) \lambda \left( {yx}\right) \lambda \left( {zy}\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{x}f\left( {x - z}\right) \mathop{\sum }\limits_{y}\lambda \left( {yx}\right) \n\]\n\n\[ \n= {qf}\left( {-z}\right) , \n\]\n\nas desired. | Yes |
Theorem 8. For complex functions \( f, g \) on \( F \), we have\n\n\[ T\left( {f * g}\right) = \left( {Tf}\right) \left( {Tg}\right) \]\n\n\[ T\left( {fg}\right) = \frac{1}{q}{Tf} * {Tg} \] | Proof. For the first formula, we have\n\n\[ T\left( {f * g}\right) \left( z\right) = \mathop{\sum }\limits_{y}\left( {f * g}\right) \left( y\right) \lambda \left( {zy}\right) = \mathop{\sum }\limits_{y}\mathop{\sum }\limits_{x}f\left( x\right) g\left( {y - x}\right) \lambda \left( {zy}\right) .\n\nWe change the order of summation, let \( y - x = t, y = x + t \), and find\n\n\[ \mathop{\sum }\limits_{x}f\left( x\right) \lambda \left( {zx}\right) \mathop{\sum }\limits_{t}g\left( t\right) \lambda \left( {zt}\right) \]\n\nwhich is precisely \( \left( {Tf}\right) \left( {Tg}\right) \left( z\right) \), thus proving the first formula. The second formula follows from the first because \( T \) is an isomorphism on \( \mathfrak{F} \), so that we can write \( f = T{f}_{1}, g = T{g}_{1} \) for some functions \( {f}_{1},{g}_{1} \) . We then combine the first formula with Theorem 7 to get the second. | Yes |
Theorem 10. Let \( f \) be the order of \( p{\;\operatorname{mod}\;m} \), and \( q = {p}^{f} \) . Let \( x \) be a character of \( F = {F}_{q} \) such that\n\n\[ \chi \left( a\right) \equiv {a}^{-\left( {q - 1}\right) /m}\;\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) . \]\n\nThen for any integer \( r \geqq 1 \) we have the factorization\n\n\[ \tau \left( {\chi }^{r}\right) \sim {\mathfrak{P}}^{a\left( r\right) } \]\n\nwhere \( \alpha \left( r\right) \) is the element of the group ring given by\n\n\[ \alpha \left( r\right) = \frac{1}{f}\mathop{\sum }\limits_{\mu }s\left( \frac{\left( {q - 1}\right) {\mu r}}{m}\right) {\sigma }_{\mu }^{-1}, \]\n\nand the sum is taken over all \( \mu {\;\operatorname{mod}\;m} \), prime to \( m \) . | Proof. This is essentially a reformulation of Theorem 9. Let \( K = \mathbf{Q}\left( {\zeta ,\omega }\right) \) and let \( \varphi \) be a homomorphism of \( {\mathfrak{o}}_{K} \) into \( {\bar{F}}_{p} \) corresponding to \( \mathfrak{P} \), i.e. inducing an injection\n\n\[ {\mathfrak{o}}_{K}/\mathfrak{P} \rightarrow {\bar{F}}_{p} \]\n\nThen we may assume that \( {\chi }^{r} = {\chi }_{\varphi }^{\left( {q - 1}\right) r/m} \) . We know from \( §1 \) that \( \omega - 1 \) is a prime element in \( \mathbf{Q}\left( \omega \right) \), and remains unramified in \( K \) . Hence by Theorem 9,\n\n\[ {\operatorname{ord}}_{\mathfrak{P}}\tau \left( {x}^{r}\right) = s\left( \frac{\left( {q - 1}\right) r}{m}\right) \]\n\nWe also have\n\n\[ {\sigma }_{\mu }\tau \left( {x}^{r}\right) = \tau \left( {x}^{r\mu }\right) \]\n\nso that\n\n\[ {\operatorname{ord}}_{{\sigma }_{\mu }^{-1}\mathfrak{B}}\tau \left( {x}^{r}\right) = {\operatorname{ord}}_{\mathfrak{B}}{\sigma }_{\mu }\tau \left( {x}^{r}\right) = s\left( \frac{\left( {q - 1}\right) {r\mu }}{m}\right) . \]\n\nAs \( \mu \) ranges over \( {\left( \mathbf{Z}/m\mathbf{Z}\right) }^{ * } \), each conjugate of \( \mathfrak{P} \) appears \( f \) times. This proves Theorem 10. | Yes |
Theorem 11. Let \( k = \mathbf{Q}\left( {\zeta }_{m}\right) \) where \( {\zeta }_{m} \) is a primitive \( m \) -th root of unity. Let \( p \) be a prime number, \( p \nmid m \), and let \( \mathfrak{p} \mid p \) in \( k \) . For positive integers \( a, b \) such that \( {ab}\left( {a + b}\right) ≢ 0{\;\operatorname{mod}\;m} \), let\n\n\[ \n{\theta }_{a, b} = \mathop{\sum }\limits_{\mu }\left( {\left\lbrack \frac{\left( {a + b}\right) \mu }{m}\right\rbrack - \left\lbrack \frac{a\mu }{m}\right\rbrack - \left\lbrack \frac{b\mu }{m}\right\rbrack }\right) {\sigma }_{\mu }^{-1}.\n\]\n\nThen \( {\mathfrak{p}}^{{\theta }_{a, b}} \) is principal, and in fact\n\n\[ \n{\mathfrak{p}}^{{\theta }_{a, b}} \sim \psi \left( {{\chi }^{a},{\chi }^{b}}\right)\n\]\n\nwhere \( x \) is the character described in Theorem 10. | Proof. We just transform the expression of Theorem 10 and use GS 4. We have\n\n\[ \n\psi \left( {{\chi }^{a},{\chi }^{b}}\right) = \frac{\tau \left( {\chi }^{a}\right) \tau \left( {\chi }^{b}\right) }{\tau \left( {\chi }^{a + b}\right) }\n\]\n\nand hence\n\n\[ \n\psi \left( {{\chi }^{a},{\chi }^{b}}\right) \sim {\mathfrak{P}}^{\beta \left( {a, b}\right) }\n\]\n\nwhere\n\n\[ \n\beta \left( {a, b}\right) =\n\]\n\n\[ \n\frac{1}{f}\mathop{\sum }\limits_{\mu }\left( {s\left( \frac{\left( {q - 1}\right) {a\mu }}{m}\right) + s\left( \frac{\left( {q - 1}\right) {b\mu }}{m}\right) - s\left( \frac{\left( {q - 1}\right) \left( {a + b}\right) \mu }{m}\right) }\right) {\sigma }_{\mu }^{-1}\n\]\n\n\[ \n= \frac{p - 1}{f}\mathop{\sum }\limits_{\mu }\left( {\mathop{\sum }\limits_{{j = 0}}^{{f - 1}}\left\lbrack \frac{\left( {a + b}\right) {p}^{j}\mu }{m}\right\rbrack - \left\lbrack \frac{a{p}^{j}\mu }{m}\right\rbrack - \left\lbrack \frac{b{p}^{j}\mu }{m}\right\rbrack }\right) {\sigma }_{\mu }^{-1},\n\]\n\nusing the lemma at the end of \( §3 \) . The decomposition group \( {G}_{\mathfrak{p}} \) of \( \mathfrak{p} \) in \( k \) is \( \left\{ {1,{\sigma }_{p},{\sigma }_{{p}^{2}},\ldots }\right\} \) . Hence we can replace \( {\sigma }_{\mu } \) by \( {\sigma }_{{p}^{j}\mu } \), and since \( \mu \) ranges over \( {\left( \mathbf{Z}/m\mathbf{Z}\right) }^{ * } \), so does \( {p}^{j}\mu \) . Consequently we find\n\n\[ \n\beta \left( {a, b}\right) = \left( {p - 1}\right) {\theta }_{a, b}\n\]\n\nSince \( \mathfrak{p} = {\mathfrak{P}}^{p - 1} \), we see that Theorem 11 is proved. | Yes |
Theorem 0. Let \( k \) be a number field. There exist two numbers \( {c}_{1},{c}_{2} > 0 \) depending only on \( k \), such that for any \( {M}_{k} \) -divisor \( \mathfrak{c} \), we have\n\n\[ \n{c}_{1}\parallel \mathfrak{c}{\parallel }_{k} < \lambda \left( \mathfrak{c}\right) \leqq \sup \left\lbrack {1,{c}_{2}\parallel \mathfrak{c}{\parallel }_{k}}\right\rbrack \n\] | Proof. Suppose that there is at least one complex absolute value \( {v}_{0} \) in \( {M}_{k} \) . We identify \( {k}_{{v}_{0}} \) with the complex plane, and consider the square centered at the origin, with sides of length \( 2\mathfrak{c}\left( {v}_{0}\right) \) . Let \( m \) be an integer such that\n\n\[ \nm < \lambda {\left( \mathfrak{c}\right) }^{1/2} \leqq m + 1. \n\]\n\nWithout loss of generality, we may assume that \( m \neq 0 \), and so \( m \geqq 1 \) . Cut up each side of the square into \( m \) equal parts, thus giving rise to \( {m}^{2} \) small squares inside the big one. Our set \( L\left( \mathrm{c}\right) \) is embedded inside the big square at \( {k}_{{v}_{0}} \) . Since it contains more than \( {m}^{2} \) elements, there exist two distinct elements \( x, y \in L\left( \mathrm{c}\right) \) lying in the same small square. Hence we can estimate their difference by\n\n\[ \n{\left| x - y\right| }_{{v}_{0}} \leqq \frac{2\sqrt{2}\mathfrak{c}\left( {v}_{0}\right) }{m}. \n\]\n\nIf \( v \) is any other archimedean absolute value of \( {M}_{k} \), then\n\n\[ \n{\left| x - y\right| }_{v} \leqq 2\mathrm{c}\left( v\right) \n\]\n\nand if \( v \) is non-archimedean, then\n\n\[ \n{\left| x - y\right| }_{v} \leqq \mathfrak{c}\left( v\right) \n\]\n\nTaking the product, we obtain\n\n\[ \n1 = \mathop{\prod }\limits_{{v \in {M}_{k}}}{\left| x - y\right| }_{v}^{{N}_{v}} \leqq \frac{{c}_{3}\parallel \mathrm{c}{\parallel }_{k}}{{m}^{2}} \n\]\n\nwith a suitable constant \( {c}_{3} \) . Since \( {\left( m + 1\right) }^{2} \leqq 4{m}^{2} \), the inequality on the right in Theorem 0 follows immediately.\n\nIf there is no complex absolute value in \( {M}_{k} \), then we proceed in a similar manner, using a real one \( {v}_{0} \), and cut up the interval centered at the origin of length \( 2\mathfrak{c}\left( {v}_{0}\right) \) into \( m \) equal parts, giving rise to \( m \) small intervals, with\n\n\[ \nm < \lambda \left( \mathfrak{c}\right) \leqq m + 1 \n\]\n\nThe arguments then proceed in the same way. | Yes |
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