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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 biholomorphi... | 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... | 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\... | 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... | 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 a... | 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 \... | 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 t... | 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}\lef... | 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 \... | 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}\... | 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 ... | 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}^{-... | 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 natur... | 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 dis... | 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} \... | 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 \... | 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 th... | 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 ... | 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 \nrightarr... | 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 \( \mat... | 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 \( X \), ... | 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 inte... | 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} ... | 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 ... | 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... | 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 \( A \), w... | 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... | 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 seco... | 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... | 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... | 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 \(... | 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\... | 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 ... | 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 \mathfr... | 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}... | 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 ... | 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... | 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 \(... | 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\ri... | 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 t... | 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 ab... | 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... | 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 ... | 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... | 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 \(... | 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 \( \mathfrak{p} \). T... | 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 un... | 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... | 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 \... | 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 \m... | 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 ... | 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( {\m... | 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 {\ma... | 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... | 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... | 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 \... | 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 th... | 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}... | 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 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 ... | 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} < \... | 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 \... | 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} \righta... | 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 ... | 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 forma... | 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) }... | 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 }\ri... | 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 ... | 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 \(... | 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 \( K\left( \alpha \right) = K\left( \beta \right) \) . Hen... | 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... | 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) ... | 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} ... | 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 \alph... | 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 o... | 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 res... | 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 \... | 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 \( > 1 \), t... | 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\... | 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... | 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... | 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... | 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 ... | 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 \( \subset \) . Co... | 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\nT... | 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... | 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}\;\mathfra... | 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... | 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 completio... | 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}... | 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 ... | 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 ... | 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 =... | 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 un... | 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}_... | 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 Pro... | 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}^{\p... | 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 hypot... | 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 \( \... | 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 \( {\mathfrak{D}}_{K/\mathbf{Q}} \) h... | 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 \... | 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 \( j > 1 \), t... | 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... | 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}... | 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}\r... | 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 } \... | 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}\righ... | 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 o... | 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 \( r \geqq 1 \) w... | 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 {... | 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... | 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 {\mathf... | 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}\parall... | 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 < \lambd... | Yes |
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