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Corollary 44. The primary ideals belonging to the isolated primes in a minimal primary decomposition of \( I \) are uniquely defined by \( I \) . | Proof: Let \( P \) be a minimal element in the set \( \left\{ {{P}_{1},\ldots ,{P}_{m}}\right\} \) of primes belonging to \( I \), and take \( D = R - P \) in Proposition 43. Then \( D \cap {P}_{i} = \varnothing \) only for \( P = {P}_{i} \), so the contraction of the localization of \( I \) at \( D \) is precisely the... | Yes |
Proposition 45. Let \( R \) be a commutative ring with 1. Then the following are equivalent:\n\n(1) \( R \) is a local ring with unique maximal ideal \( M \)\n\n(2) if \( M \) is the set of elements of \( R \) that are not units, then \( M \) is an ideal\n\n(3) there is a maximal ideal \( M \) of \( R \) such that ever... | Proof: If \( a \in R \) then the ideal \( \left( a\right) \) is either \( R \), in which case \( a \) is a unit, or is a proper ideal, in which case \( \left( a\right) \) is contained in a maximal ideal (Proposition 11 of Section 7.4). It follows that if \( R \) is a local ring and \( M \) is its unique maximal ideal t... | Yes |
Proposition 46. For any commutative ring \( R \) with 1, let \( {R}_{P} \) be the localization of \( R \) at the prime ideal \( P \) and let \( {}^{e}P \) be the extension of \( P \) to \( {R}_{P} \) . (1) The ring \( {R}_{P} \) is a local ring with unique maximal ideal \( {}^{e}P \) . The contraction of \( {}^{e}P \) ... | Proof: If \( {P}^{\prime } \) is a prime ideal of \( R \), then \( {P}^{\prime } \cap \left( {R - P}\right) = \varnothing \) if and only if \( {P}^{\prime } \subseteq P \) , so (3) is immediate from (3) in Proposition 38, and (4) follows. Since \( {}^{e}P \neq {R}_{P} \) by (2) of Proposition 38, it follows from (3) th... | Yes |
Proposition 47. Let \( M \) be an \( R \) -module. Then the following are equivalent:\n\n(1) \( M = 0 \) ,\n\n(2) \( {M}_{P} = 0 \) for all prime ideals \( P \) of \( R \), and\n\n(3) \( {M}_{\mathrm{m}} = 0 \) for all maximal ideals \( \mathrm{m} \) of \( R \) . | Proof: The implications (1) implies (2) implies (3) are obvious, so it remains to prove that (3) implies (1). Suppose \( m \) is a nonzero element in \( M \), and consider the annihilator \( I \) of \( m \) in \( R \), i.e., the ideal of elements \( r \in R \) with \( {rm} = 0 \) . Since \( m \) is nonzero \( I \) is a... | Yes |
Proposition 48. Let \( R \) be an integral domain. Then \( R \) is the intersection of the localizations of \( R : R = { \cap }_{P}{R}_{P} \) . In fact, \( R = { \cap }_{\mathfrak{m}}{R}_{\mathfrak{m}} \) is the intersection of the localizations of \( R \) at the maximal ideals \( \mathfrak{m} \) of \( R \) . | Proof: As mentioned, \( R \subseteq { \cap }_{\mathfrak{m}}{R}_{\mathfrak{m}} \) . Suppose now that \( a \) is an element of the fraction field \( F \) of \( R \) that is contained in \( {R}_{\mathfrak{m}} \) for every maximal ideal \( \mathfrak{m} \) of \( R \), and consider\n\n\[ \n{I}_{a} = \{ d \in R \mid {da} \in ... | Yes |
Proposition 49. Let \( R \) be an integral domain. Then the following are equivalent:\n\n(1) \( R \) is normal, i.e., \( R \) is integrally closed (in its field of fractions)\n\n(2) \( {R}_{P} \) is normal for all prime ideals \( P \) of \( R \)\n\n(3) \( {R}_{\mathrm{m}} \) is normal for all maximal ideals \( \mathrm{... | Proof: Let \( F \) be the field of fractions of \( R \), so all of the various localizations of \( R \) may be considered as subrings of \( F \) .\n\nAssume first that \( R \) is integrally closed and suppose \( y \in F \) is integral over \( {R}_{P} \) . Then \( y \) is a root of a monic polynomial of degree \( n \) w... | Yes |
Corollary 50. Let \( R \) be a subring of the commutative ring \( S \) with \( 1 \in R \), and assume that \( S \) is integral over \( R \) . If \( P \) is a prime ideal in \( R \), then there is a prime ideal \( Q \) of \( S \) with \( P = Q \cap R \) . | Proof: Let \( D = R - P \) so that \( D \) is a multiplicatively closed subset of both \( R \) and \( S \) . Then the following diagram commutes:\n\n\n\nwhere the vertical maps are inclusions. It is easy to see that ... | Yes |
Proposition 51. If \( V \) is an affine variety over an algebraically closed field \( k \) then the rational functions on \( V \) that are regular at all points of \( V \) are precisely the polynomial functions \( k\left\lbrack V\right\rbrack \) . | Proof: This follows from Proposition 48, which shows that the intersection (in \( k\left( V\right) \) ) of all of the localizations of \( k\left\lbrack V\right\rbrack \) at the maximal ideals of \( k\left\lbrack V\right\rbrack \) is precisely \( k\left\lbrack V\right\rbrack \) . | Yes |
Proposition 53. Let \( R \) be a commutative ring with 1 . The maps \( \mathcal{Z} \) and \( \mathcal{I} \) between \( R \) and Spec \( R \) defined above satisfy\n\n(1) for any ideal \( I \) of \( R,\mathcal{Z}\left( I\right) = \mathcal{Z}\left( {\operatorname{rad}\left( I\right) }\right) = \mathcal{Z}\left( {\mathcal... | Proof: If \( P \) is a prime ideal containing the ideal \( I \) then \( P \) contains rad \( I \) (Exercise 8, Section 2), which implies \( \mathcal{Z}\left( I\right) = \mathcal{Z}\left( {\operatorname{rad}\left( I\right) }\right) \) . Since \( \operatorname{rad}I \) is the intersection of all the prime ideals containi... | No |
Proposition 56. Let \( f \in R \) and let \( {X}_{f} \) be the corresponding principal open set in \( X = \operatorname{Spec}R \) . Then\n\n(1) \( {X}_{f} = X \) if and only if \( f \) is a unit, and \( {X}_{f} = \varnothing \) if and only if \( f \) is nilpotent,\n\n(2) \( {X}_{f} \cap {X}_{g} = {X}_{fg} \),\n\n(3) \(... | Proof: Parts (1), (2) and (7) are left as easy exercises. For (3), observe that, by definition, \( {X}_{{g}_{1}} \cup \cdots \cup {X}_{{g}_{n}} \) consists of the primes \( P \) not containing at least one of \( {g}_{1},\ldots ,{g}_{n} \) . Hence \( {X}_{{g}_{1}} \cup \cdots \cup {X}_{{g}_{n}} \) is the complement of t... | No |
Proposition 58. Let \( X = \operatorname{Spec}R \) and let \( \mathcal{O} = {\mathcal{O}}_{X} \) be its structure sheaf. The stalk of \( \mathcal{O} \) at the point \( P \in X \) is isomorphic to the localization \( {R}_{P} \) of \( R \) at \( P : {\mathcal{O}}_{P} \cong {R}_{P} \) . In particular, the stalk \( {\mathc... | Proof: If \( \left( {s, U}\right) \) represents an element in the stalk \( {\mathcal{O}}_{P} \), then \( s\left( P\right) \) is an element of the localization \( {R}_{P} \) . By the definition of the direct limit, this element does not depend on the choice of representative \( \left( {s, U}\right) \), and so gives a we... | Yes |
Proposition 1. Let \( \mathcal{J} \) be the Jacobson radical of the commutative ring \( R \). (1) If \( I \) is a proper ideal of \( R \), then so is \( \left( {I,\mathcal{J}}\right) \), the ideal generated by \( I \) and \( \mathcal{J} \). (2) The Jacobson radical contains the nilradical of \( R \) : rad \( 0 \subsete... | Proof: If \( I \) is a proper ideal in \( R \), then \( I \subseteq M \) for some maximal ideal \( M \). Since \( \mathcal{J} \subseteq M \), also \( \left( {I,\mathcal{J}}\right) \subseteq M \), which proves (1). Part (2) follows from the definitions of the two radicals and Proposition 12 in Section 15.2 since maximal... | Yes |
Theorem 3. Let \( R \) be an Artinian ring.\n\n(1) There are only finitely many maximal ideals in \( R \) . | Proof: To prove (1), let \( \mathcal{S} \) be the set of all ideals of \( R \) that are the intersection of a finite number of maximal ideals. By Proposition 2, \( \mathcal{S} \) has a minimal element, say \( {M}_{1} \cap {M}_{2} \cap \cdots \cap {M}_{n} \) . Then for any maximal ideal \( M \) we have\n\n\[ M \cap {M}_... | No |
Corollary 4. The ring \( R \) is Artinian if and only if \( R \) is Noetherian and has Krull dimension 0. | Proof: The forward implication was proved in Theorem 3. Suppose now that \( R \) is Noetherian and that \( R \) has Krull dimension 0, i.e., that prime ideals of \( R \) are maximal. Since \( R \) is Noetherian, by Corollary 22(3) in Section 15.2, the ideal \( \left( 0\right) = {P}_{1}\cdots {P}_{n} \) is the product o... | Yes |
Proposition 5. Suppose \( R \) is a Discrete Valuation Ring with respect to the valuation \( v \) , and let \( t \) be any element of \( R \) with \( v\left( t\right) = 1 \) . Then\n\n(1) A nonzero element \( u \in R \) is a unit if and only if \( v\left( u\right) = 0 \) . | \( \textit{Proof: If }u \) is a unit, then \( {uv} = 1 \) for some \( v \in \mathbb{R} \) and then \( v\left( u\right) \mathbf{ + }v\left( v\right) = v\left( {uv}\right) = 1 \) with \( v\left( u\right) \geq 0 \) and \( v\left( v\right) \geq 0 \) shows that \( v\left( u\right) = 0 \) . Conversely, if \( u \) is nonzero ... | Yes |
Corollary 6. Let \( R \) be a Discrete Valuation Ring.\n\n(1) The ring \( R \) is an integrally closed local ring with unique maximal ideal given by the elements with strictly positive valuation: \( M = \{ r \in R \mid v\left( r\right) > 0\} \) . Every nonzero ideal in \( R \) is of the form \( {M}^{n} \) for some inte... | Proof: Any U.F.D. is integrally closed in its fraction field (Example 3 in Section 15.3), so \( R \) is integrally closed. The remainder of the statements follow immediately from the description of the ideals of \( R \) in Proposition 5. | No |
Corollary 8. If \( R \) is any Noetherian, integrally closed, integral domain and \( P \) is a minimal nonzero prime ideal of \( R \), then the localization \( {R}_{P} \) of \( R \) at \( P \) is a Discrete Valuation Ring. | Proof: By results in Section 15.4, the localization \( {R}_{P} \) is a Noetherian (Proposition 38(4)), integrally closed (Proposition 49), integral domain (Proposition 46(2)), that is a local ring with unique nonzero prime ideal (Proposition 46(4)), so \( {R}_{P} \) satisfies (5) in the theorem. | Yes |
Proposition 9. Let \( R \) be an integral domain and let \( A \) be a fractional ideal of \( R \). (1) If \( A \) is a nonzero principal fractional ideal then \( A \) is invertible. | Proof: If \( A = {xR} \) is a nonzero principal fractional ideal, then taking \( B = {x}^{-1}R \) shows that \( A \) is invertible, proving (1). | Yes |
Proposition 11. Suppose the integral domain \( R \) is a local ring that is not a field. Then \( R \) is a Discrete Valuation Ring if and only if every nonzero fractional ideal of \( R \) is invertible. | Proof: If \( R \) is a D.V.R. with uniformizing parameter \( t \), then by Proposition 5 every nonzero ideal of \( R \) is of the form \( \left( {t}^{n}\right) \) for some \( n \geq 0 \) and every element \( d \) in \( R \) can be written in the form \( u{t}^{m} \) for some unit \( u \in R \) and some \( m \geq 0 \) . ... | Yes |
Proposition 12. Let \( v \) be a point on the irreducible affine curve \( C \) over \( k \) . Then \( C \) is nonsingular at \( v \) if and only if the local ring \( {\mathcal{O}}_{v, C} \) is a Discrete Valuation Ring. | Proof: Suppose first that \( v \) is nonsingular. Then \( {\dim }_{k}\left( {{\mathfrak{m}}_{v, C}/{\mathfrak{m}}_{v, C}^{2}}\right) = 1 \), and since \( {\mathcal{O}}_{v, C} \) is Noetherian, it follows from Exercise 12 in Section 1 that \( {\mathfrak{m}}_{v, C} \) is principal. Hence \( {\mathcal{O}}_{v, C} \) is a D... | Yes |
Corollary 13. An irreducible affine curve \( C \) over an algebraically closed field \( k \) is smooth if and only if its coordinate ring \( k\left\lbrack C\right\rbrack \) is integrally closed. | Proof: The curve \( C \) is smooth if and only if every localization \( {\mathcal{O}}_{v, C} \) is a D.V.R. Since \( k\left\lbrack C\right\rbrack \) has Krull dimension 1 (Exercise 11 in Section 1), the same is true for each \( {\mathcal{O}}_{v, C} \) . It then follows by Theorem 7(5) that every localization \( {\mathc... | No |
(1) Every Principal Ideal Domain is a Dedekind Domain. | Proof: A P.I.D. is clearly Noetherian, is integrally closed since it is a U.F.D. (Example 3, Section 15.3), and nonzero prime ideals are maximal (Proposition 7 in Section 8.2), which proves (1). | Yes |
Corollary 16. If \( {\mathcal{O}}_{K} \) is the ring of integers in an algebraic number field \( K \) then every nonzero ideal \( I \) in \( {\mathcal{O}}_{K} \) can be written uniquely as the product of powers of distinct prime ideals: | \[ I = {P}_{1}^{{e}_{1}}{P}_{2}^{{e}_{2}}\cdots {P}_{n}^{{e}_{n}} \] where \( {P}_{1},\ldots ,{P}_{n} \) are distinct prime ideals and \( {e}_{i} \geq 1 \) for \( i = 1,\ldots, n \) . | Yes |
Proposition 18. (Chinese Remainder Theorem) Suppose \( R \) is a Dedekind Domain, \( {P}_{1},{P}_{2},\ldots ,{P}_{n} \) are distinct prime ideals in \( R \) and \( {a}_{i} \geq 0 \) are integers, \( i = 1,\ldots, n \) . Then\n\n\[ R/{P}_{1}^{{a}_{1}}\cdots {P}_{n}^{{a}_{n}} \cong R/{P}_{1}^{{a}_{1}} \times R/{P}_{2}^{{... | Proof: This is immediate from Theorem 17 in Section 7.6 since the previous proposition shows that the \( {P}_{i}^{{a}_{i}} \) are pairwise comaximal ideals. | Yes |
Corollary 19. Suppose \( I \) is an ideal in the Dedekind Domain \( R \). Then (1) there is an ideal \( J \) of \( R \) relatively prime to \( I \) such that the product \( {IJ} = \left( a\right) \) is a principal ideal, (2) if \( I \) is nonzero then every ideal in the quotient \( R/I \) is principal; equivalently, if... | Proof: Suppose \( I = {P}_{1}^{{e}_{1}}\cdots {P}_{n}^{{e}_{n}} \) is the prime ideal factorization of \( I \) in \( R \). For each \( i = 1,\ldots, n \), let \( {r}_{i} \) be an element of \( {P}_{i}^{{e}_{i}} - {P}_{i}^{{e}_{i} + 1} \). By the proposition, there is an element \( a \in R \) with \( a \equiv {r}_{i}{\;... | Yes |
Corollary 20. If \( R \) is a Dedekind Domain then \( R \) is a P.I.D. (i.e., \( R \) has class number 1) if and only if \( R \) is a U.F.D. | Proof: Every P.I.D. is a U.F.D., so suppose that \( R \) is a U.F.D. and let \( P \) be any prime ideal in \( R \) . Then \( P = {Ra} + {Rb} \) for some \( a \neq 0 \) and \( b \) in \( R \) by Corollary 19. We have \( \left( {a}^{\prime }\right) \subseteq P \) for one of the irreducible factors \( {a}^{\prime } \) of ... | Yes |
Proposition 21. Let \( R \) be a Dedekind Domain with fraction field \( K \). (1) Suppose \( I \) and \( J \) are two fractional ideals of \( R \). Then \( I \cong J \) as \( R \)-modules if and only if \( I \) and \( J \) differ by a nonzero principal ideal: \( I = \left( a\right) J \) for some \( 0 \neq a \in K \). | Proof: Multiplication by \( 0 \neq a \in K \) gives an \( R \)-module isomorphism from \( J \) to \( \left( a\right) J \), so if \( I = \left( a\right) J \) we have \( I \cong J \) as \( R \)-modules. For the converse, observe that we may assume \( J \neq 0 \) and then \( I \cong J \) implies \( R \cong {J}^{-1}I \). B... | Yes |
Corollary 23. A finitely generated module over a Dedekind Domain is projective if and only if it is torsion free. | Proof: We showed that a finitely generated torsion free \( R \) -module is projective in the proof of Theorem 22, so by the decomposition of \( M \) in Theorem 22, \( M \) is projective if and only if \( \operatorname{Tor}\left( M\right) \) is projective (cf. Exercise 3 in Section 10.5). To complete the proof it suffic... | No |
Proposition 1. A homomorphism \( \alpha : \mathcal{A} \rightarrow \mathcal{B} \) of cochain complexes induces group homomorphisms from \( {H}^{n}\left( \mathcal{A}\right) \) to \( {H}^{n}\left( \mathcal{B}\right) \) for \( n \geq 0 \) on their respective cohomology groups. | Proof: It is an easy exercise to show that the commutativity of (4) implies that the images and kernels at each stage of the maps in the first row are mapped to the corresponding images and kernels for the maps in the second row, thus giving a well defined map on the respective quotient (cohomology) groups. | No |
Theorem 2. (The Long Exact Sequence in Cohomology) Let \( 0 \rightarrow \mathcal{A}\overset{\alpha }{ \rightarrow }\mathcal{B}\overset{\beta }{ \rightarrow }\mathcal{C} \rightarrow 0 \) be a short exact sequence of cochain complexes. Then there is a long exact sequence of cohomology groups:\n\n\[ 0 \rightarrow {H}^{0}\... | Proof: The details of this proof are somewhat lengthy. For each \( n \) the verification that the sequence \( {H}^{n}\left( \mathcal{A}\right) \rightarrow {H}^{n}\left( \mathcal{B}\right) \rightarrow {H}^{n}\left( \mathcal{C}\right) \) is exact is a straightforward check of the definition of exactness of each map, simi... | No |
Proposition 3. For any \( R \) -module \( A \) we have \( {\operatorname{Ext}}_{R}^{0}\left( {A, D}\right) \cong {\operatorname{Hom}}_{R}\left( {A, D}\right) \) . | Proof: Since the sequence \( {P}_{1}\overset{{d}_{1}}{ \rightarrow }{P}_{0}\overset{\epsilon }{ \rightarrow }A \rightarrow 0 \) is exact, it follows that the corresponding sequence \( 0 \rightarrow {\operatorname{Hom}}_{R}\left( {A, D}\right) \overset{\epsilon }{ \rightarrow }{\operatorname{Hom}}_{R}\left( {{P}_{0}, D}... | Yes |
Proposition 4. Let \( f : A \rightarrow {A}^{\prime } \) be any homomorphism of \( R \) -modules and take projective resolutions of \( A \) and \( {A}^{\prime } \), respectively. Then for each \( n \geq 0 \) there is a lift \( {f}_{n} \) of \( f \) such that the following diagram commutes:\n\n![890496f0-315c-47f0-b859-... | Proof: Given the two rows and map \( f \) in (8), then since \( {P}_{0} \) is projective we may lift the map \( {f\epsilon } : {P}_{0} \rightarrow {A}^{\prime } \) to a map \( {f}_{0} : {P}_{0} \rightarrow {P}_{0}^{\prime } \) in such a way that \( {\epsilon }^{\prime }{f}_{0} = {f\epsilon } \) (Proposition 30(2) in Se... | Yes |
Proposition 5. Let \( f : A \rightarrow {A}^{\prime } \) be a homomorphism of \( R \) -modules and take projective resolutions of \( A \) and \( {A}^{\prime } \) as in Proposition 4. Then for every \( n \) there is an induced group homomorphism \( {\varphi }_{n} : {\operatorname{Ext}}_{R}^{n}\left( {{A}^{\prime }, D}\r... | Proof: The existence of the map on the cohomology groups \( {\operatorname{Ext}}_{R}^{n} \) follows from Proposition 1 applied to the homomorphism of cochain complexes (9). The more difficult part is showing these maps do not depend on the choice of lifts \( {f}_{n} \) in Proposition 4. This is easily seen to be equiva... | No |
Theorem 6. The groups \( {\operatorname{Ext}}_{R}^{n}\left( {A, D}\right) \) depend only on \( A \) and \( D \), i.e., they are independent of the choice of projective resolution of \( A \) . | Proof: In the notation of Proposition 4 let \( {A}^{\prime } = A \), let \( f : A \rightarrow {A}^{\prime } \) be the identity map and let the two rows of (8) be two projective resolutions of \( A \) . For any choice of lifts of the identity map, the resulting homomorphisms on cohomology groups \( {\varphi }_{n} : {\op... | Yes |
Proposition 7. (Simultaneous Resolution) Let \( 0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \) be a short exact sequence of \( R \) -modules, let \( L = A \) have a projective resolution as in (6) above, and let \( N \) have a similar projective resolution where the projective modules are denoted by \( {\... | Proof: The left and right nonzero columns of (11) are exact by hypothesis. The modules in the middle column are projective (cf. Exercise 3, Section 10.5) and the row maps are the obvious ones to make each row a split exact sequence. It remains then to define the vertical maps in the middle column in such a way as to ma... | No |
Theorem 8. Let \( 0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \) be a short exact sequence of \( R \) -modules. Then there is a long exact sequence of abelian groups\n\n\[ 0 \rightarrow {\operatorname{Hom}}_{R}\left( {N, D}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {M, D}\right) \rightarrow {\oper... | Proof: Take a simultaneous projective resolution of the short exact sequence as in Proposition 7 and take homomorphisms into \( D \) . To obtain the cohomology groups \( {\operatorname{Ext}}_{R}^{n} \) from the resulting diagram, as noted in the discussion preceding Proposition 3 we replace the lowest nonzero row in th... | Yes |
Proposition 9. For an \( R \) -module \( Q \) the following are equivalent:\n\n(1) \( Q \) is injective,\n\n(2) \( {\operatorname{Ext}}_{R}^{1}\left( {A, Q}\right) = 0 \) for all \( R \) -modules \( A \), and\n\n(3) \( {\operatorname{Ext}}_{R}^{n}\left( {A, Q}\right) = 0 \) for all \( R \) -modules \( A \) and all \( n... | Proof: We showed (2) implies (1) above, and (3) implies (2) is trivial, so it remains to show that if \( Q \) is injective then \( {\operatorname{Ext}}_{R}^{n}\left( {A, Q}\right) = 0 \) for all \( R \) -modules \( A \) and all \( n \geq 1 \) . Take a projective resolution\n\n\[ \n\cdots \rightarrow {P}_{n} \rightarrow... | Yes |
Theorem 10. Let \( 0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \) be a short exact sequence of \( R \) -modules. Then there is a long exact sequence of abelian groups\n\n\[ 0 \rightarrow {\mathrm{{Hom}}}_{R}\left( {D, L}\right) \rightarrow {\mathrm{{Hom}}}_{R}\left( {D, M}\right) \rightarrow {\mathrm{{Hom... | Proof: Let \( 0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \) be a short exact sequence of \( R \) -modules. By taking a projective resolution of \( D \) and then applying \( {\operatorname{Hom}}_{R}\left( {\_, L}\right) ,{\operatorname{Hom}}_{R}\left( {\_, M}\right) \) and \( {\operatorname{Hom}}_{R}\left... | Yes |
Corollary 18. If \( A \) is an abelian group then \( A \) is torsion free if and only if \( {\mathrm{{Tor}}}_{1}\left( {A, B}\right) = 0 \) for every abelian group \( B \) (in which case \( A \) is flat as a \( \mathbb{Z} \) -module). | Proof: By the proposition, if \( A \) has no elements of finite order then we have \( {\operatorname{Tor}}_{1}\left( {A, B}\right) = {\operatorname{Tor}}_{1}\left( {t\left( A\right), B}\right) = {\operatorname{Tor}}_{1}\left( {0, B}\right) = 0 \) for every abelian group \( B \) . Conversely, if \( {\operatorname{Tor}}_... | Yes |
Proposition 20. Suppose \( {mA} = 0 \) for some integer \( m \geq 1 \) (i.e., the \( G \) -module \( A \) has exponent dividing \( m \) as an abelian group). Then\n\n\[ m{Z}^{n}\left( {G, A}\right) = m{B}^{n}\left( {G, A}\right) = m{H}^{n}\left( {G, A}\right) = 0\;\text{ for all }n \geq 0. \] | Proof: If \( f \in {C}^{n}\left( {G, A}\right) \) is an \( n \) -cochain then \( f \in A \) (if \( n = 0 \) ), in which case \( {mf} = 0 \), or \( f \) is a function from \( {G}^{n} \) to \( A \) (if \( n \geq 1 \) ), in which case \( {mf} \) is a function from \( {G}^{n} \) to \( {mA} = 0 \), so again \( {mf} = 0 \) .... | Yes |
Corollary 22. (Dimension Shifting) Suppose \( 0 \rightarrow A \rightarrow M \rightarrow C \rightarrow 0 \) is a short exact sequence of \( G \) -modules and that \( M \) is cohomologically trivial for \( G \) . Then there is an exact sequence\n\n\[ 0 \rightarrow {A}^{G} \rightarrow {M}^{G} \rightarrow {C}^{G} \rightarr... | Proof: Since \( M \) is cohomologically trivial for \( G \), the portion\n\n\[ {H}^{n}\left( {G, M}\right) \rightarrow {H}^{n}\left( {G, C}\right) \rightarrow {H}^{n + 1}\left( {G, A}\right) \rightarrow {H}^{n + 1}\left( {G, M}\right) \]\n\nof the long exact sequence in Theorem 21 reduces to\n\n\[ 0 \rightarrow {H}^{n}... | Yes |
Proposition 23. (Shapiro’s Lemma) For any subgroup \( H \) of \( G \) and any \( H \) -module \( A \) we have \( {H}^{n}\left( {G,{M}_{H}^{G}\left( A\right) }\right) \cong {H}^{n}\left( {H, A}\right) \) for \( n \geq 0 \) . | Proof: Let \( \cdots \rightarrow {P}_{n} \rightarrow \cdots \rightarrow {P}_{0} \rightarrow \mathbb{Z} \rightarrow 0 \) be a resolution of \( \mathbb{Z} \) by projective \( G \) -modules (for example, the standard resolution). The cohomology groups \( {H}^{n}\left( {G,{M}_{H}^{G}\left( A\right) }\right) \) are computed... | Yes |
For any \( G \) -module \( A \) the module \( {M}_{1}^{G}\left( A\right) \) is cohomologically trivial for \( G \), i.e., \( {H}^{n}\left( {G,{M}_{1}^{G}\left( A\right) }\right) = 0 \) for all \( n \geq 1 \) . | This follows immediately from the proposition applied with \( H = 1 \) together with the computation of the cohomology of the trivial group in Example 2 preceding Proposition 20. | No |
Proposition 26. Suppose \( H \) is a subgroup of \( G \) of index \( m \) . Then Cor \( \circ \) Res \( = m \), i.e., if \( c \) is a cohomology class in \( {H}^{n}\left( {G, A}\right) \) for some \( G \) -module \( A \), then\n\n\[ \operatorname{Cor}\left( {\operatorname{Res}\left( c\right) }\right) = {mc} \in {H}^{n}... | Proof: This follows from the explicit formula for corestriction in Example 4 above, as follows. If \( f \in {\operatorname{Hom}}_{\mathbb{Z}H}\left( {{P}_{n}, A}\right) \) were in \( {\operatorname{Hom}}_{\mathbb{Z}G}\left( {{P}_{n}, A}\right) \), i.e., if \( f \) were also a \( G \) - module homomorphism, then \( {g}_... | Yes |
Corollary 27. Suppose the finite group \( G \) has order \( m \) . Then \( m{H}^{n}\left( {G, A}\right) = 0 \) for all \( n \geq 1 \) and any \( G \) -module \( A \) . | Proof: Let \( H = 1 \), so that \( \left\lbrack {G : H}\right\rbrack = m \), in Proposition 26. Then for any class \( c \in {H}^{n}\left( {G, A}\right) \) we have \( {mc} = \operatorname{Cor}\left( {\operatorname{Res}\left( c\right) }\right) \) . Since \( \operatorname{Res}\left( c\right) \in {H}^{n}\left( {H, A}\right... | Yes |
Corollary 28. If \( G \) is a finite group then \( {H}^{n}\left( {G, A}\right) \) is a torsion abelian group for all \( n \geq 1 \) and all \( G \) -modules \( A \) . | Proof: This is immediate from the previous corollary. | No |
Corollary 29. Suppose \( G \) is a finite group whose order is relatively prime to the exponent of the \( G \) -module \( A \) . Then \( {H}^{n}\left( {G, A}\right) = 0 \) for all \( n \geq 1 \) . In particular, if \( A \) is a finite abelian group with \( \left( {\left| G\right| ,\left| A\right| }\right) = 1 \) then \... | Proof: This follows since the abelian group \( {H}^{n}\left( {G, A}\right) \) is annihilated by \( \left| G\right| \) by the previous corollary and is annihilated by the exponent of \( A \) by Proposition 20 . | No |
Proposition 31. Let \( A \) be a \( G \) -module and let \( E \) be the semidirect product \( A \rtimes G \) . For each cocycle \( f \in {Z}^{1}\left( {G, A}\right) \) define \( {\sigma }_{f} : E \rightarrow E \) by\n\n\[ \n{\sigma }_{f}\left( \left( {a, g}\right) \right) = \left( {a + f\left( g\right), g}\right) .\n\]... | Proof: It is an exercise to see that the cocycle condition implies \( {\sigma }_{f} \) is an automorphism of \( E \) that stabilizes the chain \( 1 \trianglelefteq A \trianglelefteq E \) . Likewise one checks directly that \( {\sigma }_{{f}_{1} + {f}_{2}} = {\sigma }_{{f}_{1}} \circ {\sigma }_{{f}_{2}} \), so the map \... | No |
Corollary 35. If \( A \) is a finite abelian group whose order is relatively prime to \( \left| G\right| \) then all complements to \( A \) in any semidirect product \( E = A \rtimes G \) are conjugate in \( E \) . | ## Examples\n\n(1) Let \( A = \langle a\rangle \) and \( G = \langle g\rangle \) both be cyclic of order 2 . The group \( G \) must act trivially on \( A \), hence \( A \rtimes G = A \times G \) is a Klein 4-group. Here \( A \rtimes G \) is abelian, so every subgroup is conjugate only to itself, and since \( {H}^{1}\le... | No |
Corollary 38. If \( A \) is a finite abelian group and \( \left( {\left| A\right| ,\left| G\right| }\right) = 1 \) then every extension of \( G \) by \( A \) splits. | Proof: This follows immediately from Corollary 29 in Section 2. | No |
Theorem 39. (Schur’s Theorem) If \( E \) is any finite group containing a normal subgroup \( N \) whose order and index are relatively prime, then \( N \) has a complement in \( E \) . | Proof: We use induction on the order of \( E \) . Since we may assume \( N \neq 1 \), let \( p \) be a prime dividing \( \left| N\right| \) and let \( P \) be a Sylow \( p \) -subgroup of \( N \) . Let \( {E}_{0} \) be the normalizer in \( E \) of \( P \) and let \( {N}_{0} = N \cap {E}_{0} \) . By Frattini’s Argument ... | Yes |
Proposition 40. The \( F \) -algebra \( {B}_{f} \) with \( K \) -vector space basis \( {u}_{\sigma } \) in (39) and multiplication defined by (40) is a central simple \( F \) -algebra. | Proof: It remains to show that the center of \( {B}_{f} \) is \( F \) and that \( {B}_{f} \) contains no nonzero proper ideals. Suppose \( x = \mathop{\sum }\limits_{{\sigma \in G}}{\alpha }_{\sigma }{u}_{\sigma } \) is an element in the center of \( {B}_{f} \) . Then \( {x\beta } = {\beta x} \) for \( \beta \in K \) s... | Yes |
Proposition 41. The crossed product algebra for the trivial cohomology class in \( {H}^{2}\left( {G,{K}^{ \times }}\right) \) is isomorphic to the matrix algebra \( {M}_{n}\left( F\right) \) where \( n = \left\lbrack {K : F}\right\rbrack \) . | Proof: If \( \alpha \in K \) then multiplication by \( \alpha \) defines a linear transformation \( {T}_{\alpha } \) of \( K \) viewed as an \( n \) -dimensional vector space over \( F \) . Similarly, every automorphism \( \sigma \in G \) defines an \( F \) -linear transformation \( {T}_{\sigma } \) of \( K \), and we ... | Yes |
Theorem 1. (Maschke’s Theorem) Let \( G \) be a finite group and let \( F \) be a field whose characteristic does not divide \( \left| G\right| \) . If \( V \) is any \( {FG} \) -module and \( U \) is any submodule of \( V \), then \( V \) has a submodule \( W \) such that \( V = U \oplus W \) (i.e., every submodule is... | Proof: The idea of the proof of Maschke’s Theorem is to produce an \( {FG} \) -module homomorphism\n\n\[ \pi : V \rightarrow U \]\n\nwhich is a projection onto \( U \), i.e., which satisfies the following two properties:\n\n(i) \( \pi \left( u\right) = u\; \) for all \( u \in U \)\n\n(ii) \( \pi \left( {\pi \left( v\ri... | No |
Corollary 2. If \( G \) is a finite group and \( F \) is a field whose characteristic does not divide \( \left| G\right| \), then every finitely generated \( {FG} \) -module is completely reducible (equivalently, every \( F \) -representation of \( G \) of finite degree is completely reducible). | Proof: Let \( V \) be a finitely generated \( {FG} \) -module. As noted above, \( V \) is finite dimensional over \( F \), so we may proceed by induction on its dimension. If \( V \) is irreducible, it is completely reducible and the result holds. Suppose therefore that \( V \) has a proper, nonzero \( {FG} \) -submodu... | Yes |
Corollary 3. Let \( G \) be a finite group, let \( F \) be a field whose characteristic does not divide \( \left| G\right| \) and let \( \varphi : G \rightarrow {GL}\left( V\right) \) be a representation of \( G \) of finite degree. Then there is a basis of \( V \) such that for each \( g \in G \) the matrix of \( \var... | Proof: By Corollary 2 we may write \( V = {U}_{1} \oplus {U}_{2} \oplus \cdots \oplus {U}_{m} \), where \( {U}_{i} \) is an irreducible \( {FG} \) -submodule of \( V \) . Let \( {\mathcal{B}}_{i} \) be a basis of \( {U}_{i} \) and let \( \mathcal{B} \) be the union of the \( {\mathcal{B}}_{i} \) ’s. For each \( g \in G... | Yes |
Theorem 4. (Wedderburn’s Theorem) Let \( R \) be a nonzero ring with 1 (not necessarily commutative). Then the following are equivalent:\n\n(1) every \( R \) -module is projective\n\n(2) every \( R \) -module is injective\n\n(3) every \( R \) -module is completely reducible\n\n(4) the ring \( R \) considered as a left ... | Proof: A proof of Wedderburn's Theorem is outlined in Exercises 1 to 10 | No |
Lemma 7. Let \( R \) be an arbitrary nonzero ring.\n\n(1) If \( M \) and \( N \) are simple \( R \) -modules and \( \varphi : M \rightarrow N \) is a nonzero \( R \) -module homomorphism, then \( \varphi \) is an isomorphism.\n\n(2) (Schur’s Lemma) If \( M \) is a simple \( R \) -module, then \( {\operatorname{Hom}}_{R... | Proof of Lemma 7: To prove (1) note that since \( \varphi \) is nonzero, \( \ker \varphi \) is a proper submodule of \( M \) . By simplicity of \( M \) we have \( \ker \varphi = 0 \) . Similarly, the image of \( \varphi \) is a nonzero submodule of the simple module \( N \), hence \( \varphi \left( M\right) = N \) . Th... | Yes |
Proposition 8. Let \( R = {R}_{1} \times {R}_{2} \times \cdots \times {R}_{r} \), where \( {R}_{i} \) is the ring of \( {n}_{i} \times {n}_{i} \) matrices over the division ring \( {\Delta }_{i} \), for \( i = 1,2,\ldots, r \) . (1) Identify \( {R}_{i} \) with the \( {i}^{\text{th }} \) component of the direct product.... | Proof: In part (1) since multiplication in the direct product of rings is componentwise it is clear that \( {z}_{i} \) times the element \( \left( {{a}_{1},\ldots ,{a}_{r}}\right) \) of \( R \) is the \( r \) -tuple with \( {a}_{i} \) in position \( i \) and zeros elsewhere. Thus \( {R}_{i} = {z}_{i}R,{z}_{i} \) is the... | Yes |
Proposition 9. If \( \Delta \) is a division ring that is a finite dimensional vector space over an algebraically closed field \( F \) and \( F \subseteq Z\left( \Delta \right) \), then \( \Delta = F \) . | Proof: Since \( F \subseteq Z\left( \Delta \right) \), for each \( \alpha \in \Delta \) the division ring generated by \( \alpha \) and \( F \) is a field. Also, since \( \Delta \) is finite dimensional over \( F \) the field \( F\left( \alpha \right) \) is a finite extension of \( F \) . Because \( F \) is algebraical... | Yes |
(1) Let \( A \) be a finite abelian group. Every irreducible complex representation of \( A \) is 1-dimensional (i.e., is a homomorphism from \( A \) into \( {\mathbb{C}}^{ \times } \) ) and \( A \) has \( \left| A\right| \) inequivalent irreducible complex representations. Furthermore, every finite dimensional complex... | Proof: If \( A \) is abelian, \( \mathbb{C}A \) is a commutative ring. Since a \( k \times k \) matrix ring is not commutative whenever \( k > 1 \) we must have each \( {n}_{i} = 1 \) . Thus \( r = \left| A\right| \) (= the number of conjugacy classes of \( A \) ). Since every \( \mathbb{C}A \) -module is a direct sum ... | Yes |
Proposition 13. Let \( {z}_{1},\ldots ,{z}_{r} \) be the orthogonal primitive central idempotents in \( \mathbb{C}G \) labelled in such a way that \( {z}_{i} \) acts as the identity on the irreducible \( \mathbb{C}G \) -module \( {M}_{i} \), and let \( {\chi }_{i} \) be the character afforded by \( {M}_{i} \). Then\n\n... | Proof: Let \( z = {z}_{i} \) and write\n\n\[ \nz = \mathop{\sum }\limits_{{g \in G}}{\alpha }_{g}g \n\]\n\nRecall from Example 4 in this section that if \( \rho \) is the regular character of \( G \) then\n\n\[ \n\rho \left( g\right) = \left\{ \begin{array}{ll} 0 & \text{ if }g \neq 1 \\ \left| G\right| & \text{ if }g ... | Yes |
Proposition 14. If \( \psi \) is any character of \( G \) then \( \psi \left( x\right) \) is a sum of roots of 1 in \( \mathbb{C} \) and \( \psi \left( {x}^{-1}\right) = \overline{\psi \left( x\right) } \) for all \( x \in G \) . | Proof: Let \( \varphi \) be a representation whose character is \( \psi \), fix an element \( x \in G \) and let \( \left| x\right| = k \) . Since the minimal polynomial of \( \varphi \left( x\right) \) divides \( {X}^{k} - 1 \) (hence has distinct roots), there is a basis of the underlying vector space such that the m... | Yes |
Theorem 15. (The First Orthogonality Relation for Group Characters) Let \( G \) be a finite group and let \( {\chi }_{1},\ldots ,{\chi }_{r} \) be the irreducible characters of \( G \) over \( \mathbb{C} \) . Then with respect to the inner product \( \left( {,\text{ }}\right) \) above we have\n\n\[ \left( {{\chi }_{i},... | Proof: We have just established that the irreducible characters form an orthonormal basis for the space of class functions. If \( \theta \) is any class function, write \( \theta = \mathop{\sum }\limits_{{i = 1}}^{r}{a}_{i}{\chi }_{i} \) , for some \( {a}_{i} \in \mathbb{C} \) . It follows from linearity of the Hermiti... | Yes |
Proposition 17. If \( {\psi }_{1} \) and \( {\psi }_{2} \) are characters, then so is their product \( {\psi }_{1}{\psi }_{2} \) . | Proof: Let \( {V}_{1} \) and \( {V}_{2} \) be \( \mathbb{C}G \) -modules affording characters \( {\psi }_{1} \) and \( {\psi }_{2} \) and define \( W = {V}_{1}{ \otimes }_{\mathbb{C}}{V}_{2} \) . Since each \( g \in G \) acts as a linear transformation on \( {V}_{1} \) and \( {V}_{2} \), the action of \( g \) on simple... | Yes |
Proposition 2. Let \( \alpha \in \mathbb{C} \) .\n\n(1) The following are equivalent:\n\n(i) \( \alpha \) is an algebraic integer,\n\n(ii) \( \alpha \) is algebraic over \( \overline{\mathbb{Q}} \) and the minimal polynomial of \( \alpha \) over \( \mathbb{Q} \) has integer coefficients, and\n\n(iii) \( \mathbb{Z}\left... | Proof: These are established in Section 15.3. (The portion of Section 15.3 consisting of integral extensions and properties of algebraic integers may be read independently from the rest of Chapter 15.) | No |
Corollary 3. For every character \( \psi \) of the finite group \( G,\psi \left( x\right) \) is an algebraic integer for all \( x \in G \) . | Proof: By Proposition 14 in Section 18.3, \( \psi \left( x\right) \) is a sum of roots of 1 . Each root of 1 is an algebraic integer, so the result follows immediately from Proposition 2(2). | Yes |
Corollary 5. The degree of each complex irreducible representation of a finite group \( G \) divides the order of \( G \), i.e., \( {\chi }_{i}\left( 1\right) \left| \right| G \mid \) for \( i = 1,2,\ldots, r \) . | Proof: Under the notation of Proposition 4 and with \( {g}_{j} \in {\mathcal{K}}_{j} \) we have\n\n\[ \frac{\left| G\right| }{{\chi }_{i}\left( 1\right) } = \frac{\left| G\right| }{{\chi }_{i}\left( 1\right) }\left( {{\chi }_{i},{\chi }_{i}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{j = 1}}^{r}\frac{\left| {\mathcal{K}... | Yes |
Lemma 6. If \( G \) is any group that has a conjugacy class \( \mathcal{K} \) and an irreducible matrix representation \( \varphi \) with character \( \chi \) such that \( \left( {\left| \mathcal{K}\right| ,\chi \left( 1\right) }\right) = 1 \), then for \( g \in \mathcal{K} \) either \( \chi \left( g\right) = 0 \) or \... | Proof: By hypothesis there exist \( s, t \in \mathbb{Z} \) such that \( s\left| \mathcal{K}\right| + {t\chi }\left( 1\right) = 1 \) . Thus\n\n\[ s\left| \mathcal{K}\right| \chi \left( g\right) + {t\chi }\left( 1\right) \chi \left( g\right) = \chi \left( g\right) . \]\n\nDivide both sides of this by \( \chi \left( 1\rig... | Yes |
Lemma 7. If \( \left| \mathcal{K}\right| \) is a power of a prime for some nonidentity conjugacy class \( \mathcal{K} \) of \( G \) , then \( G \) is not a non-abelian simple group. | Proof: Suppose to the contrary that \( G \) is a non-abelian simple group and let \( \left| \mathcal{K}\right| = {p}^{c} \) . Let \( g \in \mathcal{K} \) . If \( c = 0 \) then \( g \in Z\left( G\right) \), contrary to a non-abelian simple group having a trivial center. As above, let \( {\chi }_{1},\ldots ,{\chi }_{r} \... | Yes |
Lemma 9. If \( G \) is solvable of order \( > 1 \), then there exists \( P \trianglelefteq G \) with \( P \) a nontrivial \( p \) -group for some prime \( p \) . | Proof: This is a special case of the exercise on minimal normal subgroups of solvable groups at the end of Section 6.1. One can see this easily by letting \( P \) be a nontrivial Sylow subgroup of the last nontrivial term, \( {G}^{\left( n - 1\right) } \), in the derived series of \( G \) (where \( G \) has solvable le... | No |
Lemma 10. Let \( G \) be a group of order \( {p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}\cdots {p}_{t}^{{\alpha }_{t}} \) where \( {p}_{1},\ldots ,{p}_{t} \) are distinct primes. Suppose there are subgroups \( H \) and \( \bar{K} \) of \( G \) such that for each \( i \in \{ 1,\ldots, t\} \) , either \( {p}_{i}^{{\al... | Proof: Fix some \( i \in \{ 1,\ldots, t\} \) and suppose first that \( {p}_{i}^{{\alpha }_{i}} \) divides the order of \( H \) . Since \( {HK} \) is a disjoint union of right cosets of \( H \) and each of these right cosets has order equal to \( \left| H\right| \), it follows that \( {p}_{i}^{{\bar{\alpha }}_{i}} \) di... | Yes |
Theorem 11. Let \( H \) be a subgroup of the finite group \( G \) and let \( {g}_{1},\ldots ,{g}_{m} \) be representatives for the distinct left cosets of \( H \) in \( G \) . Let \( V \) be an \( {FH} \) -module affording the matrix representation \( \varphi \) of \( H \) of degree \( n \) . The \( {FG} \) -module \( ... | Proof: First note that \( {FG} \) is a free right \( {FH} \) -module:\n\n\[ {FG} = {g}_{1}{FH} \oplus {g}_{2}{FH} \oplus \cdots \oplus {g}_{m}{FH}. \]\n\nSince tensor products commute with direct sums (Theorem 17, Section 10.4), as abelian groups we have\n\n\[ W = {FG}{ \otimes }_{FH}V \cong \left( {{g}_{1} \otimes V}\... | Yes |
In the notation of Theorem 11, if \( \psi \) is the character afforded by \( V \) then the induced character is given by \[ {\operatorname{Ind}}_{H}^{G}\left( \psi \right) \left( g\right) = \mathop{\sum }\limits_{{i = 1}}^{m}\psi \left( {{g}_{i}^{-1}g{g}_{i}}\right) \] where \( \psi \left( {{g}_{i}^{-1}g{g}_{i}}\right)... | Proof: From the matrix of \( g \) computed above, the blocks \( \varphi \left( {{g}_{i}^{-1}g{g}_{i}}\right) \) down the diagonal of \( \Phi \left( g\right) \) are zero except when \( {g}_{i}^{-1}g{g}_{i} \in H \). Thus the trace of the block matrix \( \Phi \left( g\right) \) is the sum of the traces of the matrices \(... | Yes |
Proposition 13. Let \( G \) be a Frobenius group of order \( {q}^{a}p \), where \( p \) and \( q \) are distinct primes, such that the Frobenius kernel \( Q \) is an elementary abelian \( q \) -group of order \( {q}^{a} \) and the cyclic group \( G/Q \) acts irreducibly by conjugation on \( Q \) . Then the following ho... | Proof: Note that \( {QP} \) equals \( G \) by order consideration. By definition of a Frobenius group and because \( Q \) is abelian, \( {C}_{G}\left( h\right) = Q \) for every nonidentity element \( h \) of \( Q \) . If \( x \) were an element of order \( {pq} \), then \( {x}^{p} \) would be an element of order \( q \... | Yes |
Proposition 14. Let \( G \) be a group, let \( H \) be a subgroup of \( G \) and let \( \psi \) and \( {\psi }^{\prime } \) be characters of \( H \) .\n\n(1) (Induction of characters is additive) \( {\operatorname{Ind}}_{H}^{G}\left( {\psi + {\psi }^{\prime }}\right) = {\operatorname{Ind}}_{H}^{G}\left( \psi \right) + ... | It follows from part (1) of Proposition 14 that if \( \mathop{\sum }\limits_{{i = 1}}^{s}{n}_{i}{\psi }_{i} \) is any integral linear combination of characters of \( H \) with \( {n}_{i} \geq 0 \) for all \( i \) then\n\n\[ {\operatorname{Ind}}_{H}^{G}\left( {\mathop{\sum }\limits_{{i = 1}}^{s}{n}_{i}{\psi }_{i}}\right... | Yes |
For any \( i \in \{ 1,2,3,4\} \) let \( q = {q}_{i} \), let \( Q = {Q}_{i} \), let \( N = {N}_{i} \) and let \( p = \left| {N : Q}\right| \). Let \( {\psi }_{1},\ldots ,{\psi }_{4} \) be any irreducible characters of \( N \) of degree \( p \) (not necessarily distinct) and let \( \alpha = {\psi }_{1} - {\psi }_{2} \) a... | Proof: By Proposition 13, there are nonprincipal characters \( {\lambda }_{1},\ldots ,{\lambda }_{4} \) of \( Q \) of degree 1 such that \( {\psi }_{j} = {\operatorname{Ind}}_{Q}^{N}\left( {\lambda }_{j}\right) \) for \( j = 1,\ldots ,4 \). By Corollary 12 therefore, each \( {\psi }_{j} \) vanishes on \( N - Q \), henc... | Yes |
For any \( i \in \{ 1,2,3,4\} \) let \( q = {q}_{i} \), let \( Q = {Q}_{i} \), let \( N = {N}_{i} \) and let \( p = \left| {N : Q}\right| \) . Let \( {\psi }_{1},\ldots ,{\psi }_{k} \) be the distinct irreducible characters of \( N \) of degree \( p \) . Then there are distinct irreducible characters \( {\chi }_{1},\ld... | Proof: Let \( {\alpha }_{j} = {\psi }_{1} - {\psi }_{j} \) for \( j = 2,3,\ldots, k \) so \( {\alpha }_{j} \) satisfies the hypothesis of Lemma 15. Since \( {\psi }_{1} \neq {\psi }_{j} \), by Lemma 15\n\n\[ 2 = \left| \right| {\alpha }_{j}{\left| \right| }^{2} = {\left( {\alpha }_{j},{\alpha }_{j}\right) }_{N} = {\lef... | Yes |
Lemma 17. The exceptional characters associated to \( {Q}_{i} \) are all distinct from the exceptional characters associated to \( {Q}_{j} \) for \( i \) and \( j \) distinct elements of \( \{ 1,2,3,4\} \) . | Proof: Let \( \chi \) be an exceptional character associated to \( {Q}_{i} \) and let \( \theta \) be an exceptional character associated to \( {Q}_{j} \) . By construction, there are distinct irreducible characters \( \psi \) and \( {\psi }^{\prime } \) of \( {Q}_{i} \) such that \( {\psi }^{ * } - {\psi }^{\prime * }... | Yes |
Proposition 1. Let \( I \) be a nonempty countable set and for each \( i \in I \) let \( {A}_{i} \) be a set. The cardinality of the Cartesian product is the product of the cardinalities of the sets \( {A}_{i} \), i.e., \[ \left| {\mathop{\prod }\limits_{{i \in I}}{A}_{i}}\right| = \mathop{\prod }\limits_{{i \in I}}\le... | Proof: In order to count the number of choice functions note that each \( i \in I \) may be mapped to any of the \( \left| {A}_{i}\right| \) elements of \( {A}_{i} \) and for \( i \neq j \) the values of choice functions at \( i \) and \( j \) may be chosen completely independently. Thus the number of choice functions ... | Yes |
Theorem 2. Assuming the usual (Zermelo-Fraenkel) axioms of set theory, the following are equivalent: (1) Zorn's Lemma (2) the Axiom of Choice (3) the Well Ordering Principle. | Proof: This follows from elementary set theory. We refer the reader to Real and Abstract Analysis by Hewitt and Stromberg, Springer-Verlag, 1965, Section 3 for these equivalences and some others. | No |
定理1.1 设 \( f\left( x\right) 是定义在 \( \left\lbrack {0,1}\right\rbrack \) 上的连续函数,且满足 \( 0 \leq f\left( x\right) \leq 1 \) ,则必存在 \( {x}_{0} \in \lbrack 0, 1],使 \( f\left( {x}_{0}\right) = {x}_{0} \) . | 证 作函数 \( F\left( x\right) = f\left( x\right) - x \) . 如 \( {F}^{\prime }\left( 0\right) = 0 \) 或 \( F\left( 1\right) = 0 \) ,则 0 或 1 即为定理要求的 \( {x}_{0} \) ,定理已成立 (此时将有 \( f\left( 0\right) = 0 \) 或 \( f\left( 1\right) = 1 \) ). 由条件 \( 0 \leq f\left( x\right) \leq 1 \) 知 \( F\left( 0\right) = f\left( 0\right) \geq 0, F\l... | Yes |
Distance, function to spline space (距离, 函数到样条空间) | Theorem on(定理), 368 | No |
\[ P\left( {A \cup B}\right) = P\left( A\right) + P\left( B\right) - P\left( {AB}\right) . \] | Thus, the probability that the outcome of the experiment is either in \( A \) or in \( B \) equals the probability that it is in \( A \), plus the probability that it is in \( B \), minus the probability that it is in both \( A \) and \( B \) . | Yes |
Proposition 1.3.1 For random variables \( {X}_{1},\ldots ,{X}_{k} \) ,\n\n\[ E\left\lbrack {\mathop{\sum }\limits_{{j = 1}}^{k}{X}_{j}}\right\rbrack = \mathop{\sum }\limits_{{j = 1}}^{k}E\left\lbrack {X}_{j}\right\rbrack \] | Example 1.3d Consider \( n \) independent trials, each of which is a success with probability \( p \) . The random variable \( X \), equal to the total number of successes that occur, is called a binomial random variable with parameters \( n \) and \( p \) . We can determine its expectation by using the representation\... | Yes |
Proposition 5.1.1 (The Law of One Price) Consider two investments, the first of which costs the fixed amount \( {C}_{1} \) and the second the fixed amount \( {C}_{2} \) . If the (present value) payoff from the first investment is always identical to that of the second investment, then either \( {C}_{1} = {C}_{2} \) or ... | The proof of the law of one price is immediate, because if their costs are unequal then an arbitrage is obtained by buying the cheaper investment and selling the more expensive one. | No |
Proposition 5.2.1 One should never exercise an American style call option before its expiration time \( t \) . | Proof. Suppose that the present price of the stock is \( S \), that you own an option to buy one share of the stock at a fixed price \( K \), and that the option expires after an additional time \( t \) . If you exercise the option at this moment, you will realize the amount \( S - K \) . However, consider what would t... | Yes |
Proposition 5.2.2 Let \( C \) be the price of a call option that enables its holder to buy one share of a stock at an exercise price \( K \) at time \( t \) ; also, let \( P \) be the price of a European put option that enables its holder to sell one share of the stock for the amount \( K \) at time \( t \) . Let \( S ... | Proof. If\n\n\[ S + P - C < K{e}^{-{rt}} \]\nthen we can effect a sure win by initially buying one share of the stock, buying one put option, and selling one call option. This initial payout of \( S + P - C \) is borrowed from a bank to be repaid at time \( t \) . Let us now consider the value of our holdings at time \... | Yes |
Proposition 5.2.3 (The Generalized Law of One Price) Consider two investments, the first of which costs the fixed amount \( {C}_{1} \) and the second the fixed amount \( {C}_{2} \) . If \( {C}_{1} < {C}_{2} \) and the (present value) payoff from the first investment is always at least as large as that from the second i... | The arbitrage is clearly obtained by simultaneously buying investment 1 and selling investment 2. | No |
Proposition 5.2.4 Let \( C\left( {K, t}\right) \) be the cost of a call option on a specified security that has strike price \( K \) and expiration time \( t \). (a) For fixed expiration time \( t, C\left( {K, t}\right) \) is a convex and nonincreasing function of \( K \). | Proof. If \( S\left( t\right) \) denotes the price of the security at time \( t \), then the payoff at time \( t \) from a \( \left( {K, t}\right) \) call option is \[ \text{ payoff of option } = \left\{ \begin{array}{ll} S\left( t\right) - K & \text{ if }S\left( t\right) \geq K, \\ 0 & \text{ if }S\left( t\right) < K.... | No |
Theorem 6.1.1 (The Arbitrage Theorem) Exactly one of the following is true: Either\n\n(a) there is a probability vector \( \\mathbf{p} = \\left( {{p}_{1},{p}_{2},\\ldots ,{p}_{m}}\\right) \) for which\n\n\[ \n\\mathop{\\sum }\\limits_{{j = 1}}^{m}{p}_{j}{r}_{i}\\left( j\\right) = 0\\;\\text{ for all }i = 1,\\ldots, n, ... | Proof. See Section 6.3. | No |
Lemma 7.5.1 Using the representations (7.3) and (7.4), \[ I = \left\{ \begin{array}{ll} 1 & \text{ if }Z > \sigma \sqrt{t} - \omega , \\ 0 & \text{ otherwise,} \end{array}\right. \] where \[ \omega = \frac{{rt} + {\sigma }^{2}t/2 - \log \left( {K/s}\right) }{\sigma \sqrt{t}}. \] | Proof. \[ S\left( t\right) > K \Leftrightarrow \exp \left\{ {\left( {r - {\sigma }^{2}/2}\right) t + \sigma \sqrt{t}Z}\right\} > K/s \] \[ \Leftrightarrow Z > \frac{\log \left( {K/s}\right) - \left( {r - {\sigma }^{2}/2}\right) t}{\sigma \sqrt{t}} \] \[ \Leftrightarrow Z > \sigma \sqrt{t} - \omega \] | Yes |
Lemma 7.5.2\n\n\[ E\left\lbrack I\right\rbrack = P\{ S\left( t\right) > K\} = \Phi \left( {\omega - \sigma \sqrt{t}}\right) ,\] | Proof. It follows from its definition that\n\n\[ E\left\lbrack I\right\rbrack = P\{ S\left( t\right) > K\} \]\n\n\[ = P\{ Z > \sigma \sqrt{t} - \omega \} \;\text{ (from Lemma 7.5.1) } \]\n\n\[ = P\{ Z < \omega - \sigma \sqrt{t}\} \]\n\n\[ = \Phi \left( {\omega - \sigma \sqrt{t}}\right) \] | Yes |
Lemma 7.5.3\n\n\[ \n{e}^{-{rt}}E\left\lbrack {{IS}\left( t\right) }\right\rbrack = {s\Phi }\left( \omega \right) \n\] | Proof. With \( c = \sigma \sqrt{t} - \omega \), it follows from the representation (7.3) and Lemma 7.5.1 that\n\n\[ \nE\left\lbrack {{IS}\left( t\right) }\right\rbrack = {\int }_{c}^{\infty }s\exp \left\{ {\left( {r - {\sigma }^{2}/2}\right) t + \sigma \sqrt{t}x}\right\} \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2}{dx} \n\]\... | Yes |
Theorem 7.5.1 (The Black-Scholes Pricing Formula)\n\n\[ C\left( {s, t, K,\sigma, r}\right) = {s\Phi }\left( \omega \right) - K{e}^{-{rt}}\Phi \left( {\omega - \sigma \sqrt{t}}\right) . \] | Proof.\n\n\[ C\left( {s, t, K,\sigma, r}\right) = {e}^{-{rt}}E\left\lbrack {\left( S\left( t\right) - K\right) }^{ + }\right\rbrack \]\n\n\[ = {e}^{-{rt}}E\left\lbrack {I\left( {S\left( t\right) - K}\right) }\right\rbrack \]\n\n\[ = {e}^{-{rt}}E\left\lbrack {I(S\left( t\right) \rbrack - K{e}^{-{rt}}E\left\lbrack I\righ... | Yes |
Proposition 7.5.1\n\n\[ \n\frac{\partial C}{\partial K} = - {e}^{-{rt}}\Phi \left( {\omega - \sigma \sqrt{t}}\right) \n\] | Proof. Because \( S\left( t\right) \) does not depend on \( K \) ,\n\n\[ \n\frac{\partial }{\partial K}{e}^{-{rt}}\left( {S\left( t\right) - K}\right) = - {e}^{-{rt}} \n\]\n\nUsing Equation (7.5), this gives\n\n\[ \n\frac{\partial C}{\partial K} = E\left\lbrack {-I{e}^{-{rt}}}\right\rbrack \n\]\n\n\[ \n= - {e}^{-{rt}}E... | Yes |
Proposition 7.5.2\n\n\[ \frac{\partial C}{\partial s} = \Phi \left( \omega \right) \] | Proof. Using the representation of Equation (7.3), we see that\n\n\[\frac{\partial }{\partial s}{e}^{-{rt}}\left( {S\left( t\right) - K}\right) = {e}^{-{rt}}\frac{\partial S\left( t\right) }{\partial s} = \frac{S\left( t\right) }{s}{e}^{-{rt}}.\]\n\nHence, by Equation (7.5),\n\n\[\frac{\partial C}{\partial s} = \frac{{... | Yes |
Proposition 7.5.3\n\n\[ \frac{\partial C}{\partial r} = {Kt}{e}^{-{rt}}\Phi \left( {\omega - \sigma \sqrt{t}}\right) \] | Proof.\n\n\[ \frac{\partial }{\partial r}\left\lbrack {{e}^{-{rt}}\left( {S\left( t\right) - K}\right) }\right\rbrack = - t{e}^{-{rt}}\left( {S\left( t\right) - K}\right) + {e}^{-{rt}}\frac{\partial S\left( t\right) }{\partial r} \]\n\n\[ = - t{e}^{-{rt}}\left( {S\left( t\right) - K}\right) + {e}^{-{rt}}{tS}\left( t\ri... | Yes |
Lemma 7.5.4 With \( S\left( t\right) \) as given by Equation (7.3), \[ {e}^{-{rt}}E\left\lbrack {{IS}\left( t\right) Z}\right\rbrack = s\left( {{\Phi }^{\prime }\left( \omega \right) + \sigma \sqrt{t}\Phi \left( \omega \right) }\right) . \] | Proof. With \( c = \sigma \sqrt{t} - \omega \), it follows from Lemma 7.5.1 that \( E\left\lbrack {{IZS}\left( t\right) }\right\rbrack \) \[ = {\int }_{c}^{\infty }{xs}\exp \left\{ {\left( {r - {\sigma }^{2}/2}\right) t + \sigma \sqrt{t}x}\right\} \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2}{dx} \] \[ = \frac{1}{\sqrt{2\pi }... | Yes |
\[ \frac{\partial C}{\partial \sigma } = s\sqrt{t}{\Phi }^{\prime }\left( \omega \right) \] | Proof. Equation (7.3) yields that \[ \frac{\partial }{\partial \sigma }\left\lbrack {{e}^{-{rt}}\left( {S\left( t\right) - K}\right) }\right\rbrack = {e}^{-{rt}}S\left( t\right) \left( {-{t\sigma } + \sqrt{t}Z}\right) . \] Hence, by Equation (7.5), \[ \frac{\partial C}{\partial \sigma } = E\left\lbrack {{e}^{-{rt}}{IS}... | Yes |
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