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Proposition 4.1. If \( f : X \rightarrow {X}^{\prime } \) and \( g : X \rightarrow {X}^{\prime } \) are homotopic chain maps in the good category \( \mathcal{C} \), then \( f \) and \( g \) induce the same maps \( {H}_{ * }\left( f\right) \) and \( {H}_{ * }\left( g\right) \) on homology, i.e., \( {H}_{n}\left( f\right... | Proof. Let \( h \) be a homotopy, and suppose that \( {\partial }_{n - 1}\left( {x}_{n}\right) = 0 \) . Then the computation \( {f}_{n}\left( {x}_{n}\right) - {g}_{n}\left( {x}_{n}\right) = {h}_{n - 1}{\partial }_{n - 1}\left( {x}_{n}\right) + {\partial }_{n}^{\prime }{h}_{n}\left( {x}_{n}\right) = 0 + {\partial }_{n}^... | Yes |
Proposition 4.2. Let \( X = {\left\{ \left( {X}_{n},{\partial }_{n}\right) \right\} }_{n = - \infty }^{\infty } \) be a chain complex in a good category with \( {\partial }_{n} \) in \( {\operatorname{Hom}}_{R}\left( {{X}_{n + 1},{X}_{n}}\right) \) for each \( n \) . Then the boundary operator \( {\partial }_{n - 1} \)... | Proof. To see that the map \( {\bar{\partial }}_{n - 1} \) carries coker \( {\partial }_{n} \) to \( \ker {\partial }_{n - 2} \), we write it as a composition\n\n\[ \operatorname{coker}{\partial }_{n} = {X}_{n}/\text{ image }{\partial }_{n} \rightarrow {X}_{n}/\ker {\partial }_{n - 1} \cong \text{ image }{\partial }_{n... | Yes |
Proposition 4.3. An additive functor \( F : \mathcal{C} \rightarrow {\mathcal{C}}^{\prime } \) between two good categories is exact if and only if it carries all short exact sequences into short exact sequences. | Proof. Necessity is obvious. For sufficiency, let\n\n\[ A\overset{\varphi }{ \rightarrow }B\overset{\psi }{ \rightarrow }C \]\n\nbe exact, and let the additive functor \( F \) be covariant, the contravariant case being completely analogous. Put \( {A}_{1} = \ker \varphi ,{B}_{1} = \ker \psi \), and \( {C}_{1} = \) imag... | Yes |
Proposition 4.4. Let \( F : \mathcal{C} \rightarrow {\mathcal{C}}^{\prime } \) be an additive functor between good categories, let \( X \) be a complex in \( \mathcal{C} \), and let \( F\left( X\right) \) be the corresponding complex in \( {\mathcal{C}}^{\prime } \) . If \( F \) is exact, then \( F \) carries the homol... | Proof IN THE COVARIANT CASE. Let\n\n\[ A\overset{\varphi }{ \rightarrow }B\overset{\psi }{ \rightarrow }C \]\n\nbe a given complex, thus having \( {\psi \varphi } = 0 \), and form the image complex\n\n\[ F\left( A\right) \overset{F\left( \varphi \right) }{ \rightarrow }F\left( B\right) \overset{F\left( \psi \right) }{ ... | Yes |
Proposition 4.5. If \( F \) is a covariant left exact functor, then \( F \) carries an exact sequence\n\n\[ 0 \rightarrow A\overset{\varphi }{ \rightarrow }B\overset{\psi }{ \rightarrow }C \]\n\ninto an exact sequence\n\n\[ 0 \rightarrow F\left( A\right) \overset{F\left( \varphi \right) }{ \rightarrow }F\left( B\right)... | Proof. Starting from the given exact sequence, let \( i \) : image \( \psi \rightarrow C \) be the inclusion, and let \( \bar{\psi } : B \rightarrow \) image \( \psi \) be \( \psi \) with its range space reduced. Then \( \psi = i\bar{\psi } \), and the sequences\n\n\[ 0 \rightarrow A\overset{\varphi }{ \rightarrow }B\o... | Yes |
Lemma 4.6 (Snake Lemma). In a good category \( \mathcal{C} \), a snake diagram as in Figure 4.1 induces a homomorphism \( \omega : \ker \gamma \rightarrow \operatorname{coker}\alpha \) with\n\n\[ \ker \omega = \psi \left( {\ker \beta }\right) \;\text{ and }\;\text{ image }\omega = {\varphi }^{\prime - 1}\left( {\text{ ... | Proof. First let us construct \( \omega \) and see that it is well defined. Let \( c \) be in \( \ker \gamma \) . Since \( \psi \) is onto \( C \), write \( c = \psi \left( b\right) \) for some \( b \in B \) . The commutativity of the second square in Figure 4.1 gives \( 0 = \gamma \left( c\right) = {\gamma \psi }\left... | Yes |
Theorem 4.7. Let \( A = \left\{ \left( {{A}_{n},{\alpha }_{n}}\right) \right\}, B = \left\{ \left( {{B}_{n},{\beta }_{n}}\right) \right\} \), and \( C = \left\{ \left( {{C}_{n},{\gamma }_{n}}\right) \right\} \) be chain complexes in a good category \( \mathcal{C} \), and suppose that \( \varphi = \left\{ {\varphi }_{n}... | Proof. We regard the top two displayed rows of the diagram in Remark 2 as a snake diagram. Applying the Snake Lemma (Lemma 4.6), we obtain a connecting homomorphism \( {\omega }_{n} \) and an exact sequence\n\n\[ \ker {\alpha }_{n}\overset{{\bar{\varphi }}_{n + 1}}{ \rightarrow }\ker {\beta }_{n}\overset{{\bar{\psi }}_... | Yes |
Corollary 4.8. If\n\n\\[ \n0 \rightarrow A\\overset{\\varphi }{ \\rightarrow }B\\overset{\\psi }{ \\rightarrow }C \rightarrow 0 \n\\]\n\nis an exact sequence of chain complexes in a good category and if \\( A \\) is exact, then \\( {H}_{n}\\left( B\\right) \\cong {H}_{n}\\left( C\\right) \\) for all \\( n \\) ; if inst... | Proof. Theorem 4.7 gives the long exact sequence\n\n\\[ \n\\cdots \rightarrow {H}_{n + 1}\\left( C\\right) \rightarrow {H}_{n}\\left( A\\right) \rightarrow {H}_{n}\\left( B\\right) \rightarrow {H}_{n}\\left( C\\right) \rightarrow {H}_{n - 1}\\left( A\\right) \rightarrow \\cdots . \n\\]\n\nIf \\( {H}_{n}\\left( A\\right... | Yes |
Lemma 4.9. In a good category \( \mathcal{C} \), the six-term exact sequence that is obtained from a snake diagram as in Figure 4.1 is functorial in the following sense: If there are two horizontal planar snake diagrams, one with tildes \( \left( \sim \right) \) over all modules and maps and the other as is, and if the... | Proof. For the first square from the left, the assumed commutativity shows that \( {f}_{{A}^{\prime }}\widetilde{\alpha } = \alpha {f}_{A} \), and thus \( x \in \ker \widetilde{\alpha } \) implies \( {f}_{A}\left( x\right) \in \ker \alpha \) ; similarly \( x \in \ker \widetilde{\beta } \) implies \( {f}_{B}\left( x\rig... | Yes |
Theorem 4.10. In a good category \( \mathcal{C} \), the long exact sequence that is obtained from a short exact sequence of chain complexes as in Theorem 4.7 is functorial in the following sense: if there are two short exact sequences of chain complexes as in the theorem, one with tildes \( \left( \sim \right) \) over ... | Proof. Theorem 4.7 was proved by three applications of Proposition 4.2, which includes its own assertion of functoriality, and two applications of Lemma 4.6, whose functoriality is addressed in Lemma 4.9. The argument involved only manipulations with diagrams, and functoriality is in place for every step. Hence functor... | Yes |
Lemma 4.11. If \( P \) is projective in the good category \( \mathcal{C} \) and if the diagram\nin \( \mathcal{C} \) has \( \ker \varphi = \) image \( \psi \) and \( {\varphi \tau } = 0 \), then there exists a map \( \sigma : P \rightarrow {A}^{\prime \prime } \) in \( \mathcal{C} \) such that the diagram commutes. | Proof. The hypotheses force image \( \tau \subseteq \ker \varphi = \operatorname{image}\psi \) . Thus if we put \( B = \) image \( \psi \) and \( C = {A}^{\prime \prime } \), then the above diagram leads to the diagram in Figure 4.3. The hypothesis \ | No |
Theorem 4.12. Let \( X = {\left\{ \left( {X}_{n},{\partial }_{n}\right) \right\} }_{n = - \infty }^{\infty } \) and \( {X}^{\prime } = {\left\{ \left( {X}_{n}^{\prime },{\partial }_{n}^{\prime }\right) \right\} }_{n = - \infty }^{\infty } \) be chain complexes in the good category \( \mathcal{C} \), and let \( r \) be ... | Proof. For the existence of the chain map, it is enough by induction to construct \( {f}_{r + 1} \) . Matters are therefore as in the first of the above diagrams with \( n = r \) . Since \( {X}^{\prime } \) is exact at \( {X}_{r}^{\prime } \) and \( {X}_{r + 1} \) is projective, we are in the situation of Lemma 4.11 wi... | Yes |
Corollary 4.13. Let \( M \) be a module in a good category \( \mathcal{C} \) and let\n\n\[ X = \left( {{X}^{ + }\overset{\varepsilon }{ \rightarrow }M}\right) \;\text{ and }\;{X}^{\prime } = \left( {{X}^{\prime + }\overset{{\varepsilon }^{\prime }}{ \rightarrow }M}\right) \]\n\nbe two projective resolutions of \( M \) ... | Proof. The existence of \( f \) extending \( {f}_{-1} = {1}_{M} \) is immediate by applying the first part of Theorem 4.12 with \( r = - 1 \) . The hypotheses apply because \( {X}_{n} \) is projective for \( n > - 1 \) and \( {X}^{\prime } \) is exact at \( {X}_{n}^{\prime } \) for \( n \geq - 1 \) . A similar argument... | Yes |
Lemma 4.14. If \( I \) is injective in the good category \( \mathcal{C} \) and if the diagram\nin \( \mathcal{C} \) has \( \ker \varphi = \) image \( \psi \) and \( {\tau \psi } = 0 \), then there exists a map \( \sigma : {A}^{\prime \prime } \rightarrow I \) in \( \mathcal{C} \) such that the diagram commutes. | Proof. The hypotheses force \( \ker \tau \supseteq \operatorname{image}\psi = \ker \varphi \) . Thus \( \tau : A \rightarrow I \) and \( \varphi : A \rightarrow {A}^{\prime \prime } \) descend to maps \( \bar{\tau } : A/\ker \varphi \rightarrow I \) and \( \bar{\varphi } : A/\ker \varphi \rightarrow {A}^{\prime \prime ... | No |
Proposition 4.15. A unital left \( R \) module \( I \) is injective for the good category of all unital left \( R \) modules if and only if every \( R \) homomorphism of a left ideal \( J \) of \( R \) into \( I \) extends to an \( R \) homomorphism \( R \rightarrow I \) . | Proof. The necessity is immediate from Figure 4.4 and the definition of \ | No |
Theorem 4.16. Let \( X = {\left\{ \left( {X}_{n},{d}_{n}\right) \right\} }_{n = - \infty }^{\infty } \) and \( {X}^{\prime } = {\left\{ \left( {X}_{n}^{\prime },{d}_{n}^{\prime }\right) \right\} }_{n = - \infty }^{\infty } \) be cochain complexes in the good category \( \mathcal{C} \), and let \( r \) be an integer. Le... | Proof. For the existence of the cochain map, it is enough by induction to construct \( {f}_{r + 1} \) . Matters are therefore as in the first of the above diagrams with \( n = r \) . Since \( X \) is exact at \( {X}_{r} \) and \( {X}_{r + 1}^{\prime } \) is injective, we are in the situation of Lemma 4.14 with \( I = {... | Yes |
Corollary 4.17. Let \( M \) be a module in a good category \( \mathcal{C} \) and let\n\n\[ X = \left( {{X}^{ + }\overset{\varepsilon }{ \leftarrow }M}\right) \;\text{ and }\;{X}^{\prime } = \left( {{X}^{\prime + }\overset{{\varepsilon }^{\prime }}{ \leftarrow }M}\right) \]\n\nbe two injective resolutions of \( M \) . T... | Proof. The existence of \( f \) extending \( {f}_{-1} = {1}_{M} \) is immediate by applying the first part of Theorem 4.16 with \( r = - 1 \) . The hypotheses apply because \( X \) is exact at \( {X}_{n} \) for \( n \geq - 1 \) and \( {X}_{n}^{\prime } \) is injective for \( n > - 1 \) . A similar argument shows the ex... | Yes |
Lemma 4.18. Let \( \mathcal{C} \) be a good category of unital left \( R \) modules, and let\n\n\[ 0 \rightarrow A\overset{\varphi }{ \rightarrow }B\overset{\psi }{ \rightarrow }C \rightarrow 0 \]\n\nbe an exact sequence in \( \mathcal{C} \). Then the following conditions are equivalent:\n\n(a) \( B \) is a direct sum ... | Proof. If (a) holds, then \( {\left. \psi \right| }_{{B}^{\prime }} \) is one-one from \( {B}^{\prime } \) onto \( C \). Let \( \sigma \) be its inverse. Then \( \sigma : C \rightarrow {B}^{\prime } \) is one-one with \( {\psi \sigma } = {1}_{C} \). So (b) holds.\n\nIf (b) holds, then any \( b \) in \( B \) has the pro... | Yes |
Proposition 4.20. If \( R \) is any ring with identity, then the category of all unital left \( R \) modules has enough injectives. | Proof. We treat first the case that \( R = \mathbb{Z} \) . In view of Example 2 of injectives, we are to exhibit an arbitrary abelian group \( A \) as isomorphic to a subgroup of a divisible group. We know that \( A \) is isomorphic to a quotient of some free abelian group. Write \( A \cong F/S \) with \( F \) a direct... | Yes |
Proposition 4.21. In the four situations of derived functors in Figure 4.5, under the assumption that the domain category for \( F \) has enough projectives or enough injectives as appropriate, the \( {0}^{\text{th }} \) derived functor of \( F \) is naturally isomorphic to \( F \) . | Proof IF \( F \) IS COVARIANT AND RIGHT EXACT. Let\n\n\[ \n{X}_{1}\overset{{\partial }_{0}}{ \rightarrow }{X}_{0}\overset{\varepsilon }{ \rightarrow }M \rightarrow 0 \n\] \n\nbe the terms in degree \( 1,0, - 1, - 2 \) of a projective resolution of \( M \) . By Proposition 4.5 and its remark, the right exactness and cov... | Yes |
Theorem 4.24. Let \( F : \mathcal{C} \rightarrow {\mathcal{C}}^{\prime } \) be an additive functor between two good categories. Suppose that \( F \) either is contravariant right exact or is covariant left exact, and suppose that \( \mathcal{C} \) has enough injectives. Whenever there are three modules and two maps in ... | Proof. The necessary modifications to the proof of Theorem 4.22 are fairly straightforward, but some comments are in order concerning how Lemma 4.23 is to be modified. In the diagram in the statement of Lemma 4.23, all the horizontal arrows are to be reversed, the projectives \( {P}_{A} \) and \( {P}_{C} \) are to be r... | Yes |
Theorem 4.25. Let \( F : \mathcal{C} \rightarrow {\mathcal{C}}^{\prime } \) be an additive functor between two good categories. Suppose that \( F \) either is covariant right exact or is contravariant left exact, and suppose that \( \mathcal{C} \) has enough projectives. Then the passage as in Theorem 4.22 from short e... | PROOF. The proof of Theorem 4.22 involved constructing a diagram\n\n\n\nwith exact rows and commuting squares in which each \( {X}_{n} \) and \( {Z}_{n} \) is projective, and a similar diagram corresponds to the give... | Yes |
Theorem 4.27. Let \( \mathcal{C} \) and \( {\mathcal{C}}^{\prime } \) be two good categories, let \( F \) be an additive functor from \( \mathcal{C} \) to \( {\mathcal{C}}^{\prime } \) that is covariant and left exact, and suppose that \( \mathcal{C} \) has enough injectives. If a module \( A \) in \( \mathcal{C} \) ha... | Proof. The injective resolution is at our disposal, according to Corollary 4.17. Using the hypothesis that \( \mathcal{C} \) has enough injectives, choose for each \( n \) an injective \( {J}_{n} \) containing \( {X}_{n} \), let \( {g}_{n} : {X}_{n} \rightarrow {J}_{n} \) be the inclusion, and make \( \left\{ {J}_{n}\r... | Yes |
Proposition 4.28. Let \( \mathcal{C} \) and \( {\mathcal{C}}^{\prime } \) be two good categories, let \( F, G, H \) be three additive functors from \( \mathcal{C} \) to \( {\mathcal{C}}^{\prime } \), suppose that \( F, G, H \) are covariant and right exact, and suppose that \( \mathcal{C} \) has enough projectives. If ... | Proof. If \( P = \left( {{P}^{ + } \rightarrow A}\right) \) is a projective resolution of \( A \), then the natural transformations \( S \) and \( T \) give us a planar diagram\n\n\n\nin which the columns are complex... | Yes |
Theorem 4.31. Let \( \mathcal{C} \) and \( {\mathcal{C}}^{\prime } \) be good categories of unital left \( R \) modules, and suppose that \( \mathcal{C} \) has enough projectives and \( {\mathcal{C}}^{\prime } \) has enough injectives. Then \( {\operatorname{Ext}}_{R}^{n}\left( {\cdot , \cdot }\right) \) and \( {\opera... | Proof. We induct on \( n \) for \( n \geq 0 \) . Several steps are involved in the proof, and we complete all of them for a particular \( n \) before going on to \( n + 1 \) . The steps for a particular \( n \) are\n\n(i) to define \( {T}_{\left( n, A, B\right) } \) in the presence of an injective \( I \) and a one-one... | Yes |
In an additive category, let \( \left( {C,{p}_{A},{p}_{B}}\right) \) be a product of two objects \( A \) and \( B \) . Then there exist unique \( {i}_{A} \in \operatorname{Hom}\left( {A, C}\right) \) and \( {i}_{B} \in \operatorname{Hom}\left( {B, C}\right) \) such that\n\n\[ \n{p}_{A}{i}_{A} = {1}_{A},\;{p}_{B}{i}_{B}... | Proof. To the pair \( {1}_{A} \in \operatorname{Hom}\left( {A, A}\right) \) and \( 0 \in \operatorname{Hom}\left( {A, B}\right) \), the product \( C \) associates a unique \( {i}_{A} \in \operatorname{Hom}\left( {A, C}\right) \) with \( {p}_{A}{i}_{A} = {1}_{A} \) and \( {p}_{B}{i}_{A} = 0 \) . Similarly the coproduct ... | Yes |
Proposition 4.33. Let \( \mathcal{C} \) be an additive category. If an element \( u \) of \( \operatorname{Hom}\left( {A, B}\right) \) has a kernel \( \left( {K, i}\right) \) and if \( m \in \operatorname{Hom}\left( {B,{B}^{\prime }}\right) \) is a monomorphism, then \( \left( {K, i}\right) \) is also a kernel of \( {m... | Proof. We test whether \( i = \ker u \) is a kernel of \( {mu} \) . We know that \( \left( {mu}\right) i = \) \( m\left( {ui}\right) = 0 \) . Suppose that \( {mu}{i}^{\prime } = 0 \) with \( {i}^{\prime } \in \operatorname{Morph}\left( {{K}^{\prime }, A}\right) \) . Since \( m \) is a monomorphism, \( u{i}^{\prime } = ... | Yes |
Proposition 4.34. Let \( \mathcal{C} \) be an additive category. If an element \( u \) of \( \operatorname{Hom}\left( {A, B}\right) \) has a kernel \( \left( {K, i}\right) \), then \( i \) is a monomorphism. Dually if \( u \) has a cokernel \( \left( {C, p}\right) \), then \( p \) is an epimorphism. | Proof. Suppose that \( u \) has a kernel \( \left( {K, i}\right) \). For any object \( {K}^{\prime } \), the zero morphism \( {i}^{\prime } = 0 \) of \( \operatorname{Hom}\left( {{K}^{\prime }, A}\right) \) has the property that \( u{i}^{\prime } = 0 \). The uniqueness property of the kernel says that the \( \varphi \)... | Yes |
Proposition 4.35. Let \( \mathcal{C} \) be an additive category, and let \( u \) be in \( \operatorname{Hom}\left( {A, B}\right) \) . If \( u \) has a kernel and \( \ker u \) has a cokernel, then \( \operatorname{coker}\left( {\ker u}\right) \) is a kernel of \( u \) . Briefly\n\n\[ \ker \left( {\operatorname{coker}\le... | Proof. Let \( \left( {K, i}\right) \) be a kernel of \( u \), and let \( \left( {C, p}\right) \) be a cokernel of \( i \) . We are to show that \( i \) is a kernel of \( p \) . For the existence step, suppose that \( {i}^{\prime } \) in \( \operatorname{Hom}\left( {{K}^{\prime }, A}\right) \) has \( p{i}^{\prime } = 0 ... | Yes |
Proposition 4.36. In any abelian category, every morphism that is both a monomorphism and an epimorphism is an isomorphism. | Proof. If \( f \in \operatorname{Hom}\left( {K, A}\right) \) is a monomorphism, then \( f = \ker g \) for some \( g \) in some \( \operatorname{Hom}\left( {A, B}\right) \) by (v). This fact implies that \( {gf} = g \circ \left( {\ker g}\right) = 0 \) . If \( f \) is also an epimorphism, then the equality \( {gf} = 0 \)... | Yes |
Lemma 4.37. In an abelian category \( \mathcal{C} \), every monomorphism is the kernel of its cokernel, and every epimorphism is the cokernel of its kernel. | Proof. If \( m \) is a monomorphism, then (v) says that \( m = \ker u \) for some \( u \) . Substituting into the first conclusion of Proposition 4.35, we obtain \( \ker \left( {\operatorname{coker}m}\right) = \) \( m \) . If \( e \) is an epimorphism, then (v) says that \( e = \operatorname{coker}u \) for some \( u \)... | No |
Proposition 4.39. In an abelian category \( \mathcal{C} \), let \( X, Y, Z \) be objects, and let \( f \in \operatorname{Hom}\left( {X, Z}\right) \) and \( g \in \operatorname{Hom}\left( {Y, Z}\right) \) be morphisms. Let \( X \oplus Y \) be the direct sum, let \( {p}_{X} \) and \( {p}_{Y} \) be the projections on the ... | Proof. From \( {hm} = h\ker h = 0 \), we obtain \( 0 = f{p}_{X}m - g{p}_{Y}m = f\widetilde{g} - g\widetilde{f} \) , and thus \( f\widetilde{g} = g\widetilde{f} \) . Now suppose that \( {W}^{\prime },{\widetilde{f}}^{\prime } \), and \( {\widetilde{g}}^{\prime } \) are given with \( f{\widetilde{g}}^{\prime } = \) \( g{... | Yes |
Proposition 4.41. Let \( \mathcal{C} \) be an abelian category, let \( X \) be an object in \( \mathcal{C} \) , and define \( x \equiv y \) for two morphisms \( x \) and \( y \) with codomain \( X \) if there exist epimorphisms \( u \) and \( v \) with \( {xu} = {yv} \) . Then the relation \( \equiv \) on the morphisms... | Proof. Assuming that \( x \equiv y \) and \( y \equiv z \), write \( {xu} = {yv} \) and \( {yr} = {zs} \) for epimorphisms \( u, v, r, s \) . Since \( v \) and \( r \) have the same codomain, namely \( \operatorname{domain}\left( y\right) \), the pullback \( \left( {\widetilde{v},\widetilde{r}}\right) \) of \( \left( {... | Yes |
Theorem 4.42. The members of an abelian category satisfy the following properties:\n\n(a) a morphism \( f \in \operatorname{Hom}\left( {X, Y}\right) \) is a monomorphism if and only if every \( x{ \in }_{m}X \) with \( {fx} \equiv 0 \) has \( x \equiv 0 \) ,\n\n(b) a morphism \( f \in \operatorname{Hom}\left( {X, Y}\ri... | Proof. For (a) and (b), if \( f \) is a monomorphism and \( {fx} \equiv f{x}^{\prime } \), then \( {fxu} = \) \( f{x}^{\prime }v \) for suitable epimorphisms \( u \) and \( v \), and cancellation yields \( {xu} = {x}^{\prime }v \) and hence \( x \equiv {x}^{\prime } \) . Conversely suppose \( {fx} \equiv 0 \) only for ... | Yes |
Proposition 5.1. If \( \Gamma \) is a basis of \( \mathbb{K} \) over \( \mathbb{Q} \) whose members all lie in \( R \) , then \( {\left| R/\mathbb{Z}\left( \Gamma \right) \right| }^{2} = D\left( \Gamma \right) /{D}_{\mathbb{K}} \) . In particular, \( \Gamma \) is a \( \mathbb{Z} \) basis of \( R \) if and only if \( D\... | Proof. Let \( \Delta \) and \( \Omega \) be two bases of \( \mathbb{K} \) over \( \mathbb{Q} \) whose members all lie in \( R \) , and suppose that \( \mathbb{Z}\left( \Delta \right) \subseteq \mathbb{Z}\left( \Omega \right) \) . Then the above discussion shows that\n\n\[ \left| {D\left( \Delta \right) }\right| = \left... | No |
Proposition 5.2. Let \( \Gamma = \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) be an ordered \( \mathbb{Q} \) basis of \( \mathbb{K} \) lying in \( R \) . If the \( \mathbb{Z} \) span \( \mathbb{Z}\left( \Gamma \right) \) of \( \Gamma \) is a proper subgroup of \( R \), then there exists a prime number \( p \) such that \... | Proof. Let \( \mathbb{Z}\left( \Gamma \right) \) be a proper subgroup of \( R \), and put \( m = \left| {R/\mathbb{Z}\left( \Gamma \right) }\right| \) . Choose a \( \mathbb{Z} \) basis \( \left( {{w}_{1},\ldots ,{w}_{n}}\right) \) of \( R \), and write \( {v}_{i} = \mathop{\sum }\limits_{{j = 1}}^{n}{c}_{ij}{w}_{j} \) ... | Yes |
Proposition 5.3. If \( I \) is a nonzero ideal in \( R \), then\n\n(a) \( I \) contains a positive \( k \) in \( \mathbb{Z} \) and\n\n(b) \( I \) additively is of the form \( I = \mathbb{Z}\left( \Gamma \right) \) for some \( \mathbb{Q} \) basis \( \Gamma \) of \( \mathbb{K} \) whose members lie in \( R \) . | Proof. Let \( r \) be a nonzero member of \( I \), and let \( P\left( X\right) \) be the field polynomial of \( r \) . Then \( P\left( X\right) \) is of the form \( P\left( X\right) = {X}^{n} + {a}_{n - 1}{X}^{n - 1} + \cdots + {a}_{1}X + {\left( -1\right) }^{n}{N}_{\mathbb{K}/\mathbb{Q}}\left( r\right) \) , has intege... | Yes |
Theorem 5.6 (Kummer’s criterion). Let \( \mathbb{K} \) be a number field, and let \( R \) be its ring of algebraic integers. Suppose that \( F\left( X\right) \) is a monic irreducible polynomial in \( \mathbb{Z}\left\lbrack X\right\rbrack \), that \( \xi \) is a root of \( F\left( X\right) \) in \( \mathbb{C} \), and t... | \[ \bar{F}\left( X\right) = {\bar{F}}_{1}{\left( X\right) }^{{e}_{1}}\cdots {\bar{F}}_{g}{\left( X\right) }^{{e}_{g}} \] be the unique factorization of \( \bar{F}\left( X\right) \) in \( {\mathbb{F}}_{p}\left\lbrack X\right\rbrack \) into a product of powers of distinct irreducible monic polynomials, and let \( {f}_{i}... | Yes |
Lemma 5.7. Suppose that \( A \) is an additive subgroup of finite index \( J \) in \( R \) and that \( m \geq 1 \) is an integer relatively prime to \( J \) . Then for each \( r \in R \), there exists \( a \in A \) with \( r - a \) in \( {mR} \) . | Proof. Let \( \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) be a \( \mathbb{Z} \) basis of \( R \), and let \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a \( \mathbb{Z} \) basis of \( A \) . We can write \( {v}_{j} = \mathop{\sum }\limits_{{i = 1}}^{n}{c}_{ij}{u}_{i} \) for an integer matrix \( \left\lbrack {c}_{i... | Yes |
Proposition 5.8. For a cube-free integer \( m > 1 \), let \( \mathbb{K} = \mathbb{Q}\left( \sqrt[3]{m}\right) \), and let \( R \) be the ring of algebraic integers in \( \mathbb{K} \) . Write \( m = a{b}^{2} \) for positive square-free integers \( a \) and \( b \) with \( \operatorname{GCD}\left( {a, b}\right) = 1 \), ... | Proof. Let \( \omega = {e}^{{2\pi i}/3} \) . The conjugates of \( {\theta }_{1} \) can be taken to be \( {\sigma }_{1}\left( {\theta }_{1}\right) = {\theta }_{1} \) , \( {\sigma }_{2}\left( {\theta }_{1}\right) = \omega {\theta }_{1} \), and \( {\sigma }_{3}\left( {\theta }_{1}\right) = {\omega }^{2}{\theta }_{1} \) . ... | Yes |
Proposition 5.10. Let \( \mathbb{K} = \mathbb{Q}\left( \sqrt[3]{m}\right) \) be a pure cubic extension of Type I, and let \( R \) be its ring of algebraic integers. If \( p \) is a prime number, then the ideal \( \left( p\right) R \) of \( R \) splits into prime ideals as follows:\n\n(a) \( \left( p\right) R = {P}_{1}{... | Proof. The prime divisors of \( {D}_{\mathbb{K}} \) are 3 and the prime divisors of \( a \) and \( b \) . For all other primes Theorem 5.6 shows that all ramification indices are 1. Let \( p \) be a prime of the form \( {6k} \pm 1 \) not dividing \( {D}_{\mathbb{K}} \) . The multiplicative group \( {\mathbb{F}}_{p}^{ \... | Yes |
Proposition 5.12. The subgroup of \( {R}^{ \times } \) of elements of finite order consists of all \( {l}^{\text{th }} \) roots of unity in \( \mathbb{C} \), where \( l \) is an integer depending on \( \mathbb{K} \) that is bounded when the degree \( n = \left\lbrack {\mathbb{K} : \mathbb{Q}}\right\rbrack \) is bounded... | Proof. We are to bound the integers \( k \) for which primitive \( {k}^{\text{th }} \) roots of unity occur in \( \mathbb{K} \) . Let \( k \) have prime decomposition \( k = {p}_{1}^{{m}_{1}}\cdots {p}_{r}^{{m}_{r}} \) . From Section IX. 9 of Basic Algebra, we know that the cyclotomic polynomial \( {\Phi }_{k}\left( X\... | Yes |
Theorem 5.13 (Dirichlet Unit Theorem). Let \( \mathbb{K} \) be a number field of degree \( n \) with \( {r}_{1} + {r}_{2} \) absolute values, and let \( R \) be the ring of algebraic integers in \( \mathbb{K} \) . The kernel of the restriction to \( {R}^{ \times } \) of the function Log is the finite subgroup of roots ... | The proof of Theorem 5.13 will occupy the remainder of this section. We begin by clarifying in Lemma 5.14 the relationship between discrete subgroups and lattices in Euclidean space and by proving in Proposition 5.15 a weak version of Theorem 5.13 that addresses everything except the existence questions. | No |
Proposition 5.15 (weak form of Dirichlet Unit Theorem). The kernel of the restriction to \( {R}^{ \times } \) of Log is the finite subgroup of roots of unity in \( {\mathbb{K}}^{ \times } \), and the image of this restriction of Log is a discrete additive subgroup in the vector subspace of elements \( \left( {{x}_{1},\... | Proof. For \( \alpha \) in \( {R}^{ \times } \), we calculate that\n\n\[ \n\log \parallel \alpha {\parallel }_{1} + \cdots + \log \parallel \alpha {\parallel }_{{r}_{1}} + 2\log \parallel \alpha {\parallel }_{{r}_{1} + 1} + \cdots + 2\log \parallel \alpha {\parallel }_{{r}_{1} + {r}_{2}}\n\]\n\n\[ \n= \log \left( {\lef... | Yes |
Theorem 5.16 (Minkowski Lattice-Point Theorem). \( {}^{15} \) Let \( L \) be a lattice in \( {\mathbb{R}}^{m} \), and let \( {V}_{0} \) be the volume of a fundamental parallelotope. If \( E \) is any compact convex set in \( {\mathbb{R}}^{m} \) containing 0, closed under negatives, and having volume \( \left( E\right) ... | Proof. Without loss of generality, \( L \) is the standard lattice of points with all coordinates in \( \mathbb{Z} \), and \( {V}_{0} \) is 1 . Fix an arbitrarily small positive constant \( \epsilon \), and first assume that the given set \( E \) has volume \( \left( E\right) \geq {\left( 2 + \epsilon \right) }^{m}{V}_... | Yes |
Proposition 5.18. Multiplication of nonzero ideals in \( R \) descends to a well-defined multiplication of equivalence classes of ideals, and the resulting multiplication makes the set of equivalence classes into an abelian group. The identity element of this group is the class of principal ideals. | Proof. If \( I \) is a nonzero ideal, let \( \left\lbrack I\right\rbrack \) denote its equivalence class, and define \( \left\lbrack I\right\rbrack \left\lbrack J\right\rbrack = \left\lbrack {IJ}\right\rbrack \) . Suppose that \( \left( r\right) I = \left( s\right) {I}^{\prime } \) exhibits an equivalence. Then the equ... | Yes |
For a particular number field \( \mathbb{K} \), if there exists a real constant \( C \) with the property that each nonzero ideal \( J \) of \( R \) contains an element \( s \neq 0 \) with\n\n\[ \left| {{N}_{\mathbb{K}/\mathbb{Q}}\left( s\right) }\right| \leq {CN}\left( J\right) \]\n\nthen each equivalence class of ide... | Proof. Let a nonzero ideal \( I \) in \( R \) be given. By the remarks with Proposition 5.18, choose a nonzero element \( r \) in \( R \) and an ideal \( J \) such that \( r{I}^{-1} = J \) . Multiplication by \( I \) and use of the remarks shows that \( \left( r\right) = {JI} \) . By hypothesis for the lemma, choose a ... | Yes |
Theorem 5.21 (Minkowski). For any number field \( \mathbb{K} \) of degree \( n \), each nonzero ideal \( J \) of \( R \) contains an element \( s \neq 0 \) with\n\n\[ \left| {{N}_{\mathbb{K}/\mathbb{Q}}\left( s\right) }\right| \leq {\left( \frac{4}{\pi }\right) }^{{r}_{2}}\frac{n!}{{n}^{n}}{\left| {D}_{\mathbb{K}}\righ... | We shall prove Theorem 5.21 shortly by applying Minkowski's Lattice-Point Theorem to the lattice \( \Phi \left( J\right) \) in \( \Omega = {\mathbb{R}}^{{r}_{1}} \times {\mathbb{C}}^{{r}_{2}} \), where \( \Phi \) is the mapping described after the proof of Theorem 5.16. The particular compact convex set in the applicat... | Yes |
Corollary 5.22 (Minkowski). For any number field \( \mathbb{K} \) of degree \( n \) , \[ {\left| {D}_{\mathbb{K}}\right| }^{1/2} \geq {\left( \frac{\pi }{4}\right) }^{{r}_{2}}\frac{{n}^{n}}{n!}. \] Therefore \( {D}_{\mathbb{K}} > 1 \) if \( n \geq 2 \), and there exists at least one prime number that ramifies in \( \ma... | Proof. Set \( J = R \) in Theorem 5.21, so that \( N\left( J\right) = 1 \) . The nonzero element \( s \) must have \( \left| {{N}_{\mathbb{K}/\mathbb{Q}}\left( s\right) }\right| \geq 1 \) . The theorem says that \( {\left( 4/\pi \right) }^{{r}_{2}}\left( {n!/{n}^{n}}\right) {\left| {D}_{\mathbb{K}}\right| }^{1/2} \geq ... | Yes |
Lemma 5.23. In \( {\mathbb{R}}^{m} \), let \( S \) be the standard simplex with all \( {x}_{i} \geq 0 \) and with \( \mathop{\sum }\limits_{{i = 1}}^{m}{x}_{i} \leq 1 \) . If \( {a}_{1},\ldots ,{a}_{m} \) are positive real numbers, then\n\n\[ \n{\int }_{S}{x}_{1}^{{a}_{1} - 1}{x}_{2}^{{a}_{2} - 1}\cdots {x}_{m}^{{a}_{m... | SKETCH OF PROOF. Let \( I \) be the unit cube, given by \( 0 \leq {u}_{i} \leq 1 \) for \( 1 \leq i \leq m \) . We make the change of variables \( x = \varphi \left( u\right) \) that carries the points \( u \) of the cube \( I \) one-one onto the points \( x \) of the simplex \( S \) and that is given by\n\n\[ \n{x}_{1... | Yes |
Proposition 6.1. With the \( p \) -adic absolute value imposed on \( \mathbb{Q} \), let \( \mathcal{R} \) be the subring of \( \mathop{\prod }\limits_{{j = 1}}^{\infty }\mathbb{Q} \) consisting of all Cauchy sequences, and let \( \mathcal{I} \) be the ideal in \( \mathcal{R} \) consisting of all sequences convergent to... | Proof. First let us prove that \( \mathcal{I} \) is a maximal ideal. Arguing by contradiction, let \( \left\{ {q}_{n}\right\} \) be a Cauchy sequence that is not in \( \mathcal{I} \), i.e., is not convergent to 0 . Then there exists an \( {\epsilon }_{0} > 0 \) such that \( {\left| {q}_{n}\right| }_{p} \geq {\epsilon }... | Yes |
Proposition 6.2. Let \( v \) be a discrete valuation of a field \( F \), let \( {R}_{v} \) be the valuation ring, and let \( {P}_{v} \) be the valuation ideal. Then\n\n(a) \( {R}_{v} \) is a principal ideal domain,\n\n(b) there exists an element \( \pi \) in \( {P}_{v} \) with \( v\left( \pi \right) = 1 \), and any suc... | Proof. The ideal \( {P}_{v} \) contains an element \( \pi \) with \( v\left( \pi \right) = 1 \) because \( v\left( \cdot \right) \) is assumed to be onto \( \mathbb{Z} \cup \{ + \infty \} \) . Suppose that \( x \) is a nonzero member of \( {P}_{v} \) and that \( v\left( x\right) = n > 0 \) . Then \( v\left( {{\pi }^{-n... | Yes |
Lemma 6.3. Let \( R \) be a Dedekind domain regarded as a subring of its field of fractions \( F \), let \( P \) be a nonzero prime ideal in \( R \), and let \( {v}_{P} \) be the valuation of \( F \) defined by \( P \). Then any element \( x \) of \( F \) with \( {v}_{P}\left( x\right) = 0 \) is of the form \( x = a{b}... | Proof. If \( x \) is an element of \( F \) with \( {v}_{P}\left( x\right) = 0 \), write \( x = {a}^{\prime }{b}^{\prime - 1} \) with \( {a}^{\prime } \in R \) and \( {b}^{\prime } \in R \). Then \( {v}_{P}\left( {a}^{\prime }\right) = {v}_{P}\left( {b}^{\prime }\right) = n \) for some integer \( n \geq 0 \). Since \( {... | Yes |
Proposition 6.4. Let \( R \) be a Dedekind domain regarded as a subring of its field of fractions \( F \), let \( P \) be a nonzero prime ideal in \( R \), and let \( {v}_{P}\left( \cdot \right) \) be the corresponding valuation of \( F \). If \( S \) denotes the multiplicative system in \( R \) consisting of the compl... | Proof. The set \( S \) consists exactly of the members \( x \) of \( R \) with \( {v}_{P}\left( x\right) \leq 0 \). Since \( {v}_{P} \) is nonnegative on \( R \), these are the members \( x \) of \( R \) with \( {v}_{P}\left( x\right) = 0 \). Thus each \( x \) in \( {S}^{-1}R \) has \( {v}_{P}\left( x\right) \geq 0 \),... | Yes |
Corollary 6.6. Let \( R \) be a Dedekind domain regarded as a subring of its field of fractions \( F \) . If \( x \) is a member of \( \mathbb{F} \) such that \( v\left( x\right) \geq 0 \) for every discrete valuation \( v \) of \( F \) satisfying \( R \subseteq {R}_{v} \), then \( x \) lies in \( R \) . | Proof. We may assume that \( x \neq 0 \) . Write \( x = a{b}^{-1} \) with \( a \) and \( b \) in \( R \) . Theorem 6.5 shows that the valuations in question are the ones determined by the nonzero prime ideals of \( R \) . If the principal ideals \( \left( a\right) \) and \( \left( b\right) \) factor as \( \left( a\righ... | Yes |
Proposition 6.7. Let \( R \) be a Dedekind domain, let \( F \) be its field of fractions, let \( K \) be a finite algebraic extension of \( F \), and let \( T \) be the integral closure of \( R \) in \( K \) . If a discrete valuation \( v \) of \( K \) is \( \geq 0 \) on \( R \), then it is \( \geq 0 \) on \( T \) . | Proof. If \( x \neq 0 \) is in \( T \), then the minimal polynomial of \( x \) over \( R \) is a monic polynomial in \( T\left\lbrack X\right\rbrack \), and thus there exist an integer \( n \) and coefficients \( {a}_{n - 1},\ldots ,{a}_{0} \) in \( R \) such that\n\n\[ \n{x}^{n} = {a}_{n - 1}{x}^{n - 1} + \cdots + {a}... | Yes |
The only discrete valuations of the field \( \mathbb{Q} \) of rationals are the ones leading to the \( p \) -adic absolute value for each prime number \( p \) . If \( K \) is a number field and \( T \) is its the ring of algebraic integers, then the only discrete valuations of \( K \) are the valuations \( {v}_{P} \) c... | Proof. If \( v \) is an arbitrary discrete valuation of \( \mathbb{Q} \), then property (iv) of discrete valuations shows that \( v\left( {-1}\right) = v\left( 1\right) = 0 \), and property (ii) allows us to conclude that \( v \) is nonnegative on all of \( \mathbb{Z} \) . Thus \( \mathbb{Z} \) is contained in the valu... | Yes |
Let \( \mathbb{k} \) be any field, and let \( F = \mathbb{k}\left( X\right) \) be the field of rational expressions in one indeterminate over \( \mathbb{k} \). Regard \( F \) as the field of fractions of the principal ideal domain \( \mathbb{k}\left\lbrack X\right\rbrack \). Then the only discrete valuations of \( F \)... | Proof. Let \( v \) be a discrete valuation of \( F \) that is 0 on \( {\mathbb{k}}^{ \times } \). First suppose that \( v\left( X\right) \geq 0 \). Being 0 on the coefficients, \( v \) is nonnegative on all polynomials. Thus \( \mathbb{k}\left\lbrack X\right\rbrack \) is contained in the valuation ring of \( v \), and ... | Yes |
Corollary 6.10. Let \( \mathbb{k} \) be a field, let \( F = \mathbb{k}\left( X\right) \) be the field of rational expressions in one indeterminate over \( \mathbb{k} \), let \( K \) be a finite algebraic extension of \( \mathbb{k}\left\lbrack X\right\rbrack \), let \( T \) be the integral closure of \( \mathbb{k}\left\... | Proof. The argument is similar to the one for Corollary 6.8, except that we have to take into account what Proposition 6.9 says when \( v\left( X\right) < 0 \) . The conclusion is that either \( v \) is \( \geq 0 \) on \( \mathbb{k}\left\lbrack X\right\rbrack \), and then Proposition 6.7 and Theorem 6.5 show that \( v ... | Yes |
Proposition 6.11. Let \( R \) be a Dedekind domain regarded as a subring of its field of fractions \( F \), suppose that \( \left| \cdot \right| \) is an absolute value on \( F \) defined by means of a discrete valuation \( v \), and suppose that the subset \( {R}_{v} \) of \( F \) for which \( \left| x\right| \leq 1 \... | Proof. In terms of \( v \), the set \( {R}_{v} \) is the valuation ring, and the set \( {P}_{v} \) is the valuation ideal. The hypothesis \( R \subseteq {R}_{v} \) is the hypothesis of Theorem 6.5. Part (a) of that theorem shows that \( P = R \cap {P}_{v} \) is a prime ideal in \( R \) . Conclusions (a) and (b) here fo... | Yes |
Proposition 6.12. Two nontrivial absolute values on a field \( F \) are equivalent if and only if\n\n\[ \left\{ {x \in F\left| {\;{\left| x\right| }_{1} > 1}\right. }\right\} \subseteq \left\{ {x \in F\left| {\;{\left| x\right| }_{2} > 1}\right. }\right\} ,\]\n\nif and only if they induce the same topology on \( F \) . | Proof. If the two absolute values are equivalent, then it is immediate from the definition of equivalent that equality holds in the stated inclusion. Conversely\n\n\( {}^{6} \) In many books an equivalence class of absolute values on a field is called a \ | No |
Proposition 6.13. If \( \\left| \\cdot \\right| \) is an absolute value on the field \( F \), then the topology on \( F \) induced by the associated metric makes \( F \) into a topological field. | Proof. To see that addition, subtraction, and multiplication are continuous on \( F \), let \( \\left\\{ {x}_{n}\\right\\} \) and \( \\left\\{ {y}_{n}\\right\\} \) be convergent sequences in \( F \) with respective limits \( x \) and \( y \) . Use of the triangle inequality on \( F \) gives\n\n\[ \n\\left| {\\left( {{x... | Yes |
Proposition 6.14. If \( \left| \cdot \right| \) is an absolute value on the field \( F \) for which there is some \( c \) with \( \left| n\right| \leq c \) for all integers \( n \in \mathbb{Z} \), i.e., for all additive multiples of 1, then \( \left| \cdot \right| \) is nonarchimedean. In particular, \( \left| \cdot \r... | Proof. If \( x \) and \( y \) are in \( F \) and if \( n \) is any positive integer, then the Binomial Theorem gives \( {\left( x + y\right) }^{n} = \mathop{\sum }\limits_{{j = 0}}^{n}\left( \begin{array}{l} n \\ j \end{array}\right) {x}^{n - j}{y}^{j} \) . Therefore\n\n\[ \n{\left| x + y\right| }^{n} = \mathop{\sum }\... | Yes |
Corollary 6.16. If \( \left| \cdot \right| \) is a nontrivial absolute value on a number field \( F \), then the restriction of \( \left| \cdot \right| \) to \( \mathbb{Q} \) is nontrivial. | Proof. Since \( \left| \cdot \right| \) is nontrivial, there exists \( x \) with \( \left| x\right| > 1 \) . Raising \( x \) to a power if necessary, we may assume that \( \left| x\right| \geq 2 \) . Arguing by contradiction, suppose that \( \left| q\right| = 1 \) for all nonzero \( q \) in \( \mathbb{Q} \) . Since \( ... | Yes |
Corollary 6.17. If \( \left| \cdot \right| \) is a nontrivial discrete absolute value on the field \( F \) , then \( \left| \cdot \right| \) is nonarchimedean, and \( \left| x\right| = {r}^{-v\left( x\right) } \) for some discrete valuation of \( F \) . | Proof. First we show that \( \left| \cdot \right| \) is nonarchimedean. Proposition 6.14 immediately handles the case that \( F \) has nonzero characteristic, and we may therefore take the characteristic to be 0 . Let \( D \) be the discrete image subgroup of \( {F}^{ \times } \) . This \( D \) in particular must conta... | Yes |
Corollary 6.18. If \( \left| \cdot \right| \) is a nontrivial discrete absolute value on a field \( F \), then the corresponding valuation ring \( R = \{ x \in F\left| \right| x \mid \leq 1\} \) and the valuation ideal \( P = \{ x \in F \mid \left| x\right| < 1\} \) are open and closed in \( F \) . | Proof. The definitions of \( R \) and \( P \) in the statement show that \( R \) is closed and \( P \) is open. Let \( D \) be the image of \( {F}^{ \times } \) under \( \left| \cdot \right| \) . A discrete subgroup of positive reals has to be equal \( {}^{7} \) to \( \{ 1\} \) or to the subgroup \( {r}^{\mathbb{Z}} \)... | Yes |
Lemma 6.19. If \( R \) is a Dedekind domain regarded as a subring of its field of fractions \( F \), and if \( \left| \cdot \right| \) is a nonarchimedean absolute value on \( F \) that is \( \leq 1 \) on \( R \), then \( \left| \cdot \right| \) is discrete. Hence either \( \left| \cdot \right| \) is trivial or else it... | Proof. The subset of \( x \in R \) for which \( \left| x\right| < 1 \) is a proper ideal \( I \) in \( R \), and we let \( P \) be a prime ideal containing \( I \) . Since \( R \) is a Dedekind domain, \( P \) defines a corresponding discrete valuation \( {v}_{P} \) . Let \( {\left| x\right| }_{P} = {2}^{-{v}_{P}\left(... | Yes |
Proposition 6.20. Let \( R \) be a Dedekind domain regarded as a subring of its field of fractions \( F \), let \( K \) be a finite algebraic extension of \( F \), and let \( T \) be the integral closure of \( R \) in \( K \) . If \( \left| \cdot \right| \) is a nonarchimedean absolute value on \( K \) that is \( \leq ... | Proof. As with Proposition 6.7, \( T \) is a Dedekind domain. If \( x \neq 0 \) is in \( T \) , then the minimal polynomial of \( x \) over \( R \) is a monic polynomial in \( R\left\lbrack X\right\rbrack \), and thus there exist an integer \( n \) and coefficients \( {a}_{n - 1},\ldots ,{a}_{0} \) in \( R \) such that... | Yes |
Corollary 6.21. If \( K \) is a number field, then every nontrivial nonarchimedean absolute value \( \left| \cdot \right| \) on \( K \) comes from the valuation \( {v}_{P} \) relative to some nonzero prime ideal \( P \) in the ring of algebraic integers in \( K \) . | Proof. Since \( \left| \cdot \right| \) is nonarchimedean, its restriction to \( \mathbb{Q} \) is nonarchimedean. By Ostrowski’s Theorem (or by inspection), it is \( \leq 1 \) on \( \mathbb{Z} \) . The result now follows from Proposition 6.20 if we take \( R \) to be \( \mathbb{Z} \) and \( F \) to be \( \mathbb{Q} \) ... | Yes |
Corollary 6.22. Let \( \mathbb{k} \) be a field, let \( F = \mathbb{k}\left( X\right) \) be the field of rational expressions in one indeterminate over \( \mathbb{k} \), let \( K \) be a finite algebraic extension of \( \mathbb{k}\left\lbrack X\right\rbrack \), let \( T \) be the integral closure of \( \mathbb{k}\left\... | Proof. The argument is similar to the one for Corollary 6.21, except that we have to take into account what happens when \( \left| X\right| > 1 \) . We apply Proposition 6.20 either with \( R = \mathbb{k}\left\lbrack X\right\rbrack \) or with \( R = \mathbb{k}\left\lbrack {X}^{-1}\right\rbrack \) .\n\nSince \( \left| \... | Yes |
Theorem 6.23 (Weak Approximation Theorem). Let \( {\left. \left| \cdot \right| \right| }_{1},\ldots ,{\left| \cdot \right| }_{n} \) be inequivalent nontrivial absolute values on a field \( F \). If \( \epsilon > 0 \) is a real number and \( {x}_{1},\ldots ,{x}_{n} \) are elements of \( F \), then there exists \( y \) i... | Proof. First let us prove that we can find an element \( z \) in \( F \) with\n\n\[{\left| z\right| }_{1} > 1\;\text{ and }\;{\left| z\right| }_{j} < 1\text{ for }2 \leq j \leq n.\]\n\n\( \left( *\right) \)\n\nWe do so by induction on \( n \), the case \( n = 2 \) being Proposition 6.12. Assuming the result for \( n - ... | Yes |
Theorem 6.24. Let \( F \) be a field with a nontrivial absolute value \( {\left| \cdot \right| }_{F} \), let \( d \) be the associated metric on \( F \), let \( \mathcal{R} \) be the subring of \( \mathop{\prod }\limits_{{j = 1}}^{\infty }F \) consisting of all Cauchy sequences relative to \( d \), and let \( \mathcal{... | Proof. The proof of this theorem is almost the same as the first part of the proof of Proposition 6.1, apart from notational changes. The differences occur in spots where the ultrametric inequality was invoked in the proof of Proposition 6.1 and only the triangle inequality is available here. The main such difference i... | Yes |
Theorem 6.25. If \( \iota : \left( {F,{\left| \cdot \right| }_{F}}\right) \rightarrow \left( {K,{\left| \cdot \right| }_{K}}\right) \) is a completion of the valued field \( \left( {F,{\left| \cdot \right| }_{F}}\right) \) and if \( \varphi : \left( {F,{\left| \cdot \right| }_{F}}\right) \rightarrow \left( {L,{\left| \... | Proof. The theory of completion of a metric space produces a unique continuous function \( \Phi : K \rightarrow L \) such that \( \varphi = \Phi \circ \iota \), and this continuous function respects the metrics. It is necessary to check only that \( \Phi \) respects addition and multiplication.\n\nThe argument is the s... | Yes |
Theorem 6.26. Let \( \iota : \left( {F,{\left| \cdot \right| }_{F}}\right) \rightarrow \left( {\bar{F},{\left| \cdot \right| }_{\bar{F}}}\right) \) be a completion of a valued field, and suppose that \( {\left. \mid \cdot \right| }_{F} \) is nontrivial and discrete. Let \( v\left( \cdot \right) \) be the discrete valua... | Proof. Write \( {\left| {F}^{ \times }\right| }_{F} \) in the form \( {r}^{\mathbb{Z}} \) for a unique real number \( r > 1 \) . For (a), since \( {\left| \iota \left( x\right) \right| }_{\bar{F}} = {\left| x\right| }_{F} \) and since \( \iota \left( F\right) \) is dense in \( \bar{F} \), the continuity of the absolute... | Yes |
Proposition 6.27. Let \( F \) be a number field with \( \left\lbrack {F : \mathbb{Q}}\right\rbrack = n \), and let there be \( {r}_{1} \) distinct field maps of \( F \) into \( \mathbb{R} \) and \( {r}_{2} \) complex conjugate pairs of distinct field maps of \( F \) into \( \mathbb{C} \), with \( {r}_{1} + 2{r}_{2} = n... | Proof. The remarks above show everything except that these \( {r}_{1} + {r}_{2} \) absolute values are mutually inequivalent. To prove this fact, suppose that \( \sigma \) and \( {\sigma }^{\prime } \) are two field maps of \( F \) into the same field, \( \mathbb{R} \) or \( \mathbb{C} \), such that \( x \mapsto \left|... | Yes |
Theorem 6.28 (Hensel’s Lemma). Let \( F \) be a field with a nontrivial discrete absolute value \( \left| \cdot \right| \) , necessarily nonarchimedean, and assume that \( F \) is complete. Let \( R \) be the valuation ring, and let \( f\left( X\right) \) be a polynomial in \( R\left\lbrack X\right\rbrack \) . Suppose ... | Proof. Put \( c = \left| {f\left( {a}_{0}\right) }\right| /{\left| {f}^{\prime }\left( {a}_{0}\right) \right| }^{2} < 1 \) . We prove the following three statements together by induction on \( n \) :\n\n(i) \( {a}_{n} \) is well defined and is in \( R \) ,\n\n(ii) \( \left| {{f}^{\prime }\left( {a}_{n}\right) }\right| ... | Yes |
Corollary 6.29 (Hensel’s Lemma). Let \( F \) be a field with a nontrivial discrete absolute value, necessarily nonarchimedean, and assume that \( F \) is complete. Let \( R \) be the valuation ring, let \( \mathfrak{p} \) be the unique maximal ideal, and let \( f\left( X\right) \) be a polynomial in \( R\left\lbrack X\... | Proof. Let \( {a}_{0} \) be any member of \( R \) whose image in \( R/\mathfrak{p} \) is \( \bar{a} \) . The assumptions imply that \( f\left( {a}_{0}\right) \) is in \( \mathfrak{p} \) and that \( {f}^{\prime }\left( {a}_{0}\right) \) is in \( R \) but not \( \mathfrak{p} \) . Thus the hypotheses of Theorem 6.28 are s... | Yes |
Lemma 6.35. In the above notation the two determinations of \( {\left| \cdot \right| }_{{P}_{i}} \) on \( {K}_{{P}_{j}} \) coincide-one by using Theorem 6.33 to insist that \( {\varphi }_{j}{\psi }_{0} \) in Figure 6.2 be the composition of homomorphisms of valued fields, and the other by using Proposition 6.34 to insi... | Proof. Let us give two different names to the two possible absolute values on \( {K}_{{P}_{j}} \), writing \( {\left| \cdot \right| }^{\prime } \) for the one that makes \( {\left| {\psi }_{j}\left( k\right) \right| }^{\prime } = {\left| k\right| }_{{P}_{j}} \) for \( k \in K \) and writing \( {\left| \cdot \right| }^{... | Yes |
Lemma 6.36. In the above notation and that of Theorem 6.31, the ramification index \( {e}_{j}^{ * } \) corresponding to \( {K}_{{P}_{j}}/{F}_{\mathfrak{p}} \) for the closure of the ideal \( {\psi }_{j}\left( {P}_{j}\right) \) coincides with the ramification index \( {e}_{j} \) corresponding to \( K/F \) for the ideal ... | Proof. Let \( {v}_{\mathfrak{p}, F},{v}_{{P}_{j}, K},{v}_{\mathfrak{p},{F}_{\mathfrak{p}}} \), and \( {v}_{{P}_{j},{K}_{{P}_{i}}} \) be the valuations corresponding to the absolute values on \( F, K,{F}_{\mathfrak{p}} \), and \( {K}_{{P}_{j}} \), respectively. The last of these is well defined by Lemma 6.35. Propositio... | Yes |
Corollary 6.37. Let \( R \) be a Dedekind domain regarded as a subring of its field of fractions \( F \), let \( K \) be a finite separable extension of \( F \) with \( \left\lbrack {K : F}\right\rbrack = n \), and let \( T \) be the integral closure of \( R \) in \( K \) . Let \( \mathfrak{p} \) be a nonzero prime ide... | Proof. Conclusion (a) was proved in the paragraph before the statement of the corollary, and (b) and (c) follow immediately from (a).\n\nUnder the assumption that \( K = F\left( \xi \right) \), the minimal polynomial \( m\left( X\right) \) of \( \xi \) and the characteristic polynomial \( \det \left( {{X1} - l\left( \x... | Yes |
Proposition 6.38. Let \( F \) be a complete valued field under a nonarchimedean discrete valuation \( v \), let \( R \) and \( \mathfrak{p} \) be the valuation ring and valuation ideal for \( v \), let \( K \) be a finite separable extension of \( F \) of degree \( n \), let \( T \) be the integral closure of \( R \) i... | The proof is carried out in Problems 15-16 at the end of the chapter. | No |
Lemma 6.39. In the above setting with \( K/F \) Galois, let \( {P}_{i} \) be one of the ideals \( {P}_{1},\ldots ,{P}_{g} \) . Then a member \( \sigma \) of the Galois group \( G = \operatorname{Gal}\left( {K/F}\right) \) extends to a field automorphism of \( {K}_{{P}_{i}} \) fixing \( {F}_{\mathfrak{p}} \) if and only... | Proof. If \( \sigma \) is an isometry from \( K \) into itself in the metric determined by \( {\left| \cdot \right| }_{{P}_{i}} \) , then \( \sigma \) is uniformly continuous as a function from \( K \) into the complete space \( {\dot{K}}_{{P}_{i}} \) and therefore extends to a continuous function from the completion \... | Yes |
Proposition 6.40. In the above setting with \( K/F \) Galois, let \( P \) be one of the ideals \( {P}_{1},\ldots ,{P}_{g} \), let \( G = \operatorname{Gal}\left( {K/F}\right) \) be the Galois group, and let \( {G}_{P} \) be the decomposition group at \( P \) . Then \( {K}_{P} \) is a Galois extension of \( {F}_{\mathfr... | Proof. Since \( {K}_{P} \) is generated by \( {F}_{\mathfrak{p}} \) and \( K \), it is obtained by adjoining to \( {F}_{\mathfrak{p}} \) the same roots of the same polynomials over \( F \) that are used to generate \( K \) . Therefore \( {K}_{P}/{F}_{\mathfrak{p}} \) is a Galois extension.\n\nLemma 6.39 gives us the ma... | Yes |
Lemma 6.41. In the above setting, if \( M \) is a nonzero fractional ideal of \( K \) , then so is its dual \( \widehat{M} \) . | Proof. Since \( T \) has \( K \) as its field of fractions, there exists an \( F \) vector space basis \( \left\{ {{t}_{1},\ldots ,{t}_{n}}\right\} \) of \( K \) consisting of members of \( T \) . If \( {m}_{0} \) is a nonzero member of \( M \) and \( {m}_{j} = {t}_{j}{m}_{0} \), then \( \left\{ {{m}_{1},\ldots ,{m}_{n... | Yes |
Proposition 6.42. In the above setting, the dual \( \widehat{T} \) of \( T \) is of the form \( \widehat{T} = \) \( \mathcal{D}{\left( K/F\right) }^{-1} \) for an ideal \( \mathcal{D}\left( {K/F}\right) \) of \( T \) . This ideal \( \mathcal{D}\left( {K/F}\right) \) has the property that\n\n\[ \widehat{M} = {M}^{-1}\ma... | Proof. From the definition, \( \widehat{T} \) consists of all \( x \) in \( K \) for which \( {\operatorname{Tr}}_{K/F}\left( {xt}\right) \) is in \( R \) ; any member \( x \) of \( T \) has this property, and thus \( T \subseteq \widehat{T} \) . Lemma 6.41 shows that \( \widehat{T} \) is a fractional ideal of \( K \) ... | Yes |
Proposition 6.43. In the above setting, if \( L \) is a field with \( F \subseteq L \subseteq K \), then\n\n\[ \mathcal{D}\left( {K/F}\right) = \mathcal{D}\left( {K/L}\right) \mathcal{D}\left( {L/F}\right) \]\n\nas an equality of fractional ideals in \( K \) . | Proof. We use the fact that traces can be computed in stages. An element \( x \) of \( K \) is in \( \mathcal{D}{\left( K/F\right) }^{-1} \) if and only if \( {\operatorname{Tr}}_{K/F}\left( {xT}\right) \subseteq R \), if and only if \( {\operatorname{Tr}}_{L/F}\left( {{\operatorname{Tr}}_{K/L}\left( {xT}\right) }\righ... | Yes |
Theorem 6.44 (Strong Approximation Theorem). Let \( F \) be a number field, let \( R \) be its ring of algebraic integers, let \( {P}_{1},\ldots ,{P}_{r} \) be distinct nonzero prime ideals in \( R \), and let \( {v}_{{P}_{i}} \) for each \( j \) be the valuation of \( F \) and of its completion that corresponds to \( ... | Proof. \( {}^{23} \) We may assume that each \( {l}_{j} \) satisfies \( {l}_{j} \geq 0 \) . Recall that for each prime number \( p \), there are only finitely many prime ideals \( P \) in \( R \) with \( P \cap \mathbb{Z} = \) \( p\mathbb{Z} \) . Possibly by moving some of the conditions \( {v}_{Q}\left( y\right) \geq ... | Yes |
Theorem 6.45 (Dedekind’s Theorem on Differents). Let \( R \) be a Dedekind domain regarded as a subring of its field of fractions \( F \), let \( K \) be a finite separable extension of \( F \) with \( \left\lbrack {K : F}\right\rbrack = n \), and let \( T \) be the integral closure of \( R \) in \( K \). Suppose that ... | \[ {e}_{j}^{\prime } = \left\{ \begin{array}{ll} {e}_{j} - 1 & \text{ if }p\text{ does not divide }{e}_{j}, \\ {\bar{e}}_{j} & \text{ with }{\bar{e}}_{j} \geq {e}_{j}\text{ if }p\text{ divides }{e}_{j}. \end{array}\right. \] | Yes |
Proposition 6.46. Let \( R \) be a Dedekind domain regarded as a subring of its field of fractions \( F \), let \( K \) be a finite separable extension of \( F \) with \( \left\lbrack {K : F}\right\rbrack = n \) , and let \( T \) be the integral closure of \( R \) in \( K \) . Suppose that \( T \) has the strong approx... | Proof. We actually will show equality of the inverses of the two sides of the displayed formula. By the first conclusion of Proposition 6.42, we are to show that a member \( x \) of \( K \) has\n\n\[ {\operatorname{Tr}}_{K/F}\left( {xT}\right) \subseteq R\;\text{if and only if}\;{\operatorname{Tr}}_{{K}_{P}/{F}_{\mathf... | Yes |
Lemma 6.47. Let \( F \) be a complete valued field with respect to a discrete nonarchimedean valuation, let \( R \) be its valuation ring, let \( \mathfrak{p} \) be its valuation ideal, let \( K \) be a finite separable extension of \( F \) with \( \left\lbrack {K : F}\right\rbrack = n \), let \( T \) be the integral c... | \[ {e}^{\prime } = \left\{ \begin{array}{ll} e - 1 & \text{ if }p\text{ does not divide }e, \\ \bar{e} & \text{ with }\bar{e} \geq e\text{ if }p\text{ divides }e. \end{array}\right. \] | Yes |
Theorem 6.53. If \( K \) is a global field, then the image of \( {K}^{ \times } \) in the subgroup \( {\mathbb{I}}^{1} \) of the ideles \( \mathbb{I} \) under the diagonal mapping \( \iota : {K}^{ \times } \rightarrow \mathbb{I} \) is discrete, and the quotient group \( {\mathbb{I}}^{1}/\iota \left( {K}^{ \times }\righ... | Proof of discreteness of \( \iota \left( {K}^{ \times }\right) \) in Theorem 6.53. The set \( U \) of all \( {\left( {x}_{v}\right) }_{v} \in \mathbb{I} \) such that \( {\left| {x}_{v} - 1\right| }_{v} < 1 \) for all archimedean places and \( {\left| {x}_{v} - 1\right| }_{v} \leq 1 \) for all nonarchimedean places is a... | Yes |
Corollary 7.2. Let \( K \) be a field, let \( \bar{K} \) be an algebraic closure, let \( n \) be a positive integer, and let \( I \) be a prime ideal in \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . Then \( I \) contains every polynomial in \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) that ... | Proof. If \( f \) is a member of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) that vanishes on the locus of common zeros of \( I \), then (b) in the theorem shows that \( {f}^{k} \) is in \( I \) for some \( k \) . Since \( I \) is prime, one of the factors of \( {f}^{k} = f\cdots f \) lies in \( I \) . | Yes |
Lemma 7.3. If \( K \) is a field and \( L \) is an extension field that is generated as a \( K \) algebra by \( n \) elements \( {x}_{1},\ldots ,{x}_{n} \), i.e., if \( L = K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), then every \( {x}_{j} \) is algebraic over \( K \) . | Proof. We proceed by induction on \( n \) . For \( n = 1 \), if \( L = K\left\lbrack {x}_{1}\right\rbrack \), then we know from the elementary theory of fields that \( {x}_{1} \) is algebraic over \( K \) .\n\nFor the inductive step, suppose that \( L = K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Since \(... | Yes |
Lemma 7.4. Let \( K \) be a field, and let \( L \) be an algebraic extension of \( K \) . If \( I \) is a proper ideal in \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \), then \( {IL}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is a proper ideal in \( L\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\r... | Proof. First let us identify the integral closure of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) in the field \( L\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) as \( L\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . The ring \( L\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is a unique fact... | Yes |
Lemma 7.6. If \( L/K \) is a field extension, then\n\n(a) any transcendence set of \( L \) over \( K \) can be extended to a transcendence basis of \( L \) over \( K \) ,\n\n(b) any subset of \( L \) that generates \( L \) as a field over \( K \) has a subset that is a transcendence basis of \( L \) over \( K \) . | Proof. For (a), order by inclusion upward the transcendence sets containing the given one. To apply Zorn's Lemma, we need only show that the union of a chain of transcendence sets in \( L \) over \( K \) is again a transcendence set. Thus let finitely many elements of the union of the sets in the chain be given. Since ... | Yes |
Theorem 7.7. If \( L/K \) is a field extension, then there exists an intermediate field \( {K}^{\prime } \) such that \( {K}^{\prime }/K \) is purely transcendental and \( L/{K}^{\prime } \) is algebraic. | Proof. Lemma 7.6 produces a transcendence basis \( S \) for \( L/K \) . Define \( {K}^{\prime } \) to be the intermediate field \( K\left( S\right) \) generated by \( K \) and \( S \) . Then \( {K}^{\prime } \) is purely transcendental over \( K \) by definition. If \( x \) is a member of \( L \) that is not in \( {K}^... | Yes |
Lemma 7.8 (Exchange Lemma). Let \( L/K \) be a field extension. If \( E \) is any subset of \( L \), let \( K\left( E\right) \) be the subfield of \( L \) generated by \( K \) and \( E \), and let \( \overline{K\left( E\right) } \) be the subfield of all elements in \( L \) that are algebraic over \( K\left( E\right) \... | Proof. The condition that \( x \) lie in \( \overline{K\left( {E\cup \{ y\} }\right) } \) implies that there exist a finite subset \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) of \( E \) and a member \( f \) of \( K\left( {{X}_{1},\ldots ,{X}_{n}, Y}\right) \left\lbrack Z\right\rbrack \) such that\n\n\[ f\left( {{x}... | Yes |
Proposition 7.10. If \( K \) is a field of characteristic \( p \) and if \( \alpha \) is a member of \( K \) such that \( \sqrt[p]{\alpha } \) is not in \( K \), then \( {X}^{{p}^{\mu }} - \alpha \) is irreducible in \( K\left\lbrack X\right\rbrack \) for every \( \mu \geq 0 \) . | Proof. Let \( L \) be a splitting field of \( {X}^{{p}^{\mu }} - \alpha \) over \( K \) . If \( \beta \) is a root of \( {X}^{{p}^{\mu }} - \alpha \) , then \( {\beta }^{{p}^{\mu }} = \alpha \), and hence \( {X}^{{p}^{\mu }} - \alpha = {X}^{{p}^{\mu }} - {\beta }^{{p}^{\mu }} = {\left( X - \beta \right) }^{{p}^{\mu }} ... | Yes |
Corollary 7.11. If \( L/K \) is an algebraic extension in characteristic \( p \), if \( \alpha \) is a purely inseparable element of \( L \) over \( K \), and if \( e \) is the smallest nonnegative integer such that \( {\alpha }^{{p}^{e}} \) lies in \( K \), then the minimal polynomial of \( \alpha \) over \( K \) is \... | Proof. This is immediate from Proposition 7.10. | No |
Corollary 7.12. If \( L/K \) is an algebraic extension in characteristic \( p \) and if \( \alpha \) is an element of \( L \) that is separable and purely inseparable over \( K \), then \( \alpha \) lies in \( K \) . | Proof. Since \( \alpha \) is purely inseparable over \( K \), Corollary 7.11 says that the minimal polynomial of \( \alpha \) over \( K \) is \( {X}^{{p}^{e}} - {\alpha }^{{p}^{e}} \), where \( e \) is the smallest nonnegative integer such that \( {\alpha }^{{p}^{e}} \) lies in \( K \) . The separability of \( \alpha \... | Yes |
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