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Proposition 1. If \( I \) is an ideal of the Noetherian ring \( R \), then the quotient \( R/I \) is a Noetherian ring. Any homomorphic image of a Noetherian ring is Noetherian. | Proof: If \( R \) is a ring and \( I \) is an ideal in \( R \), then any infinite ascending chain of ideals in the quotient \( R/I \) would correspond by the Lattice Isomorphism Theorem to an infinite ascending chain of ideals in \( R \) . This gives the first statement, and the second follows by the first Isomorphism ... | Yes |
Theorem 2. The following are equivalent:\n\n(1) \( R \) is a Noetherian ring.\n\n(2) Every nonempty set of ideals of \( R \) contains a maximal element under inclusion.\n\n(3) Every ideal of \( R \) is finitely generated. | Proof: The proof is identical to that of Theorem 1 in Section 12.1 in the special case where the \( R \) -module \( M \) is \( R \) itself (and submodules are ideals). | No |
Corollary 5. The ring \( R \) is a finitely generated \( k \) -algebra if and only if there is some surjective \( k \) -algebra homomorphism\n\n\[ \varphi : k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \rightarrow R \]\n\nfrom the polynomial ring in a finite number of variables onto \( R \) that is the ... | Proof: If \( R \) is generated as a \( k \) -algebra by \( {r}_{1},\ldots ,{r}_{n} \), then we may define the map \( \varphi : k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow R \) by \( \varphi \left( {x}_{i}\right) = {r}_{i} \) for all \( i \) and \( \varphi \left( a\right) = a \) for all \( a \in k \... | Yes |
Theorem 6. Let \( V \subseteq {\mathbb{A}}^{n} \) and \( W \subseteq {\mathbb{A}}^{m} \) be affine algebraic sets. Then there is a bijective correspondence\n\n\[ \left\{ \begin{matrix} \text{ morphisms from }V\text{ to }W \\ \text{ as algebraic sets } \end{matrix}\right\} \leftrightarrow \left\{ \begin{matrix} k\text{-... | Proof: The proof of (3) is left as an exercise and (4) is then immediate. | No |
Proposition 8. With notation as above, let \( R = k\left\lbrack {{y}_{1},\ldots ,{y}_{m},{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and let \( \mathcal{A} \) be the ideal generated by \( {y}_{1} - {\varphi }_{1},\ldots ,{y}_{m} - {\varphi }_{m} \) together with generators for \( I \) . Let \( G \) be the reduced Gröbner ... | Proof: If we show \( \ker \Phi = \mathcal{A} \cap k\left\lbrack {{y}_{1},\ldots ,{y}_{m}}\right\rbrack \) modulo \( J \) then (a) follows by Proposition 30 in Section 9.6. Suppose first that \( f \in \mathcal{A} \cap k\left\lbrack {{y}_{1},\ldots ,{y}_{m}}\right\rbrack \) . If \( {f}_{1},\ldots ,{f}_{s} \) are generato... | Yes |
Corollary 9. The map \( \Phi \) is surjective if and only if for each \( i,1 \leq i \leq n \), the reduced Gröbner basis \( G \) contains a polynomial \( {x}_{i} - {h}_{i} \) where \( {h}_{i} \in k\left\lbrack {{y}_{1},\ldots ,{y}_{m}}\right\rbrack \) . | (1) Let \( \Phi : \mathbb{Q}\left\lbrack {u, v}\right\rbrack \rightarrow \mathbb{Q}\left\lbrack x\right\rbrack \) be defined by \( \Phi \left( u\right) = {x}^{2} + x \) and \( \Phi \left( v\right) = {x}^{3} \) . The reduced Gröbner basis \( G \) for the ideal \( \mathcal{A} = \left( {u - {x}^{2} - x, v - {x}^{3}}\right... | Yes |
Proposition 10. Suppose \( \alpha \) is a root of the irreducible polynomial \( p\left( x\right) \in k\left\lbrack x\right\rbrack \) and \( \beta \in k\left( \alpha \right) \), say \( \beta = f\left( \alpha \right) \) for the polynomial \( f \in k\left\lbrack x\right\rbrack \) . Let \( G \) be the reduced Gröbner basis... | Proof: The kernel of the \( k \) -algebra homomorphism \( k\left\lbrack y\right\rbrack \rightarrow k\left\lbrack x\right\rbrack /\left( p\right) \cong k\left( \alpha \right) \) defined by mapping \( y \) first to \( f \) and then to \( \beta \) is the principal ideal generated by the minimal polynomial of \( \beta \) i... | Yes |
Proposition 11. Let \( I \) be an ideal in the commutative ring \( R \) . Then rad \( I \) is an ideal containing \( I \), and \( \left( {\operatorname{rad}I}\right) /I \) is the nilradical of \( R/I \) . In particular, \( R/I \) has no nilpotent elements if and only if \( I = \operatorname{rad}I \) is a radical ideal. | Proof: It is clear that \( I \subseteq \operatorname{rad}I \) . By definition, the nilradical of \( R/I \) consists of the elements in the quotient some power of which is 0 . Under the Lattice Isomorphism Theorem for rings this collection of elements corresponds to the elements of \( R \) some power of which lie in \( ... | Yes |
The radical of a proper ideal \( I \) is the intersection of all prime ideals containing \( I \) . In particular, the nilradical is the intersection of all the prime ideals in \( R \) . | Proof: Passing to \( R/I \), Proposition 11 shows that it suffices to prove this result for \( I = 0 \), and in this case the statement is that the nilradical \( N \) of \( R \) is the intersection of all the prime ideals in \( R \) . Let \( {N}^{\prime } \) denote the intersection of all the prime ideals in \( R \) .\... | Yes |
Corollary 13. Prime (and hence also maximal) ideals are radical. | Proof: If \( P \) is a prime ideal, then \( P \) is clearly the intersection of all the prime ideals containing \( P \), so \( P = \operatorname{rad}P \) by the proposition. | Yes |
Proposition 14. If \( R \) is a Noetherian ring then for any ideal \( I \) some positive power of \( \operatorname{rad}I \) is contained in \( I \) . In particular, the nilradical, \( N \), of a Noetherian ring is a nilpotent ideal: \( {N}^{k} = 0 \) for some \( k \geq 1 \) . | Proof: For any ideal \( I \), the ideal rad \( I \) is finitely generated since \( R \) is Noetherian. If \( {a}_{1},\ldots ,{a}_{m} \) are generators of rad \( I \), then by definition of the radical, for each \( i \) we have \( {a}_{i}^{{k}_{i}} \in I \) for some positive integer \( {k}_{i} \) . Let \( k \) be the ma... | Yes |
Proposition 16. Suppose \( \varphi : V \rightarrow W \) is a morphism of algebraic sets and \( \widetilde{\varphi } : k\left\lbrack W\right\rbrack \rightarrow \)\n\n\( k\left\lbrack V\right\rbrack \) is the associated \( k \) -algebra homomorphism of coordinate rings. Then\n\n(1) The kernel of \( \widetilde{\varphi } \... | Proof: Since \( \widetilde{\varphi } = f \circ \varphi \), we have \( \widetilde{\varphi }\left( f\right) = 0 \) if and only if \( \left( {f \circ \varphi }\right) \left( P\right) = 0 \) for all \( P \in V \), i.e., \( f\left( Q\right) = 0 \) for all \( Q = \varphi \left( P\right) \in \varphi \left( V\right) \), which ... | Yes |
(1) The affine algebraic set \( V \) is irreducible if and only if \( \mathcal{I}\left( V\right) \) is a prime ideal. | Proof: Let \( I = \mathcal{I}\left( V\right) \) and suppose first that \( V = {V}_{1} \cup {V}_{2} \) is reducible, where \( {V}_{1} \) and \( {V}_{2} \) are proper closed subsets. Since \( {V}_{1} \neq V \), there is some function \( {f}_{1} \) that vanishes on \( {V}_{1} \) but not on \( V \), i.e., \( {f}_{1} \in \m... | Yes |
Corollary 18. An affine algebraic set \( V \) is a variety if and only if its coordinate ring \( k\left\lbrack V\right\rbrack \) is an integral domain. | Proof: This follows immediately since \( \mathcal{I}\left( V\right) \) is a prime ideal if and only if the quotient \( k\left\lbrack V\right\rbrack = k\left\lbrack {\mathbb{A}}^{n}\right\rbrack /\mathcal{I}\left( V\right) \) is an integral domain (Proposition 13 of Chapter 7). | Yes |
Prime ideals are primary. | The first two statements are immediate from the definition of a primary ideal. | No |
Proposition 20. Let \( R \) be a Noetherian ring. Then\n\n(1) every irreducible ideal is primary, and\n\n(2) every proper ideal in \( R \) is a finite intersection of irreducible ideals. | Proof: To prove (1) let \( Q \) be an irreducible ideal and suppose that \( {ab} \in Q \) and \( b \notin Q \) . It is easy to check that for any fixed \( n \) the set of elements \( x \in R \) with \( {a}^{n}x \in Q \) is an ideal, \( {A}_{n} \), in \( R \) . Clearly \( {A}_{1} \subseteq {A}_{2} \subseteq \ldots \) an... | Yes |
Theorem 21. (Primary Decomposition Theorem) Let \( R \) be a Noetherian ring. Then every proper ideal \( I \) in \( R \) has a minimal primary decomposition. If\n\n\[ I = \mathop{\bigcap }\limits_{{i = 1}}^{m}{Q}_{i} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{Q}_{i}^{\prime } \]\n\nare two minimal primary decompositions ... | Proof: The proof of the uniqueness of the set of associated primes is outlined in the exercises, and the proof of the uniqueness of the primary components associated to the minimal primes will be given in Section 4. | No |
Corollary 22. Let \( I \) be a proper ideal in the Noetherian ring \( R \). (1) A prime ideal \( P \) contains the ideal \( I \) if and only if \( P \) contains one of the associated primes of \( I \), hence if and only if \( P \) contains one of the isolated primes of \( I \), i.e., the isolated primes of \( I \) are ... | Proof: The first statement in (1) is an exercise (cf. Exercise 37), and the remainder of (1) follows. | No |
Proposition 23. Let \( R \) be a subring of the commutative ring \( S \) with \( 1 \in R \) and let \( s \in S \) . Then the following are equivalent: (1) \( s \) is integral over \( R \) , (2) \( R\left\lbrack s\right\rbrack \) is a finitely generated \( R \) -module (where \( R\left\lbrack s\right\rbrack \) is the ri... | Proof: Suppose first that (1) holds and let \( s \) be a root of the monic polynomial \( {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} \in R\left\lbrack x\right\rbrack \) . Then \[ {s}^{n} = - \left( {{a}_{n - 1}{s}^{n - 1} + {a}_{n - 2}{s}^{n - 2} + \cdots + {a}_{0}}\right) \] and so \( {s}^{n} \), and then all ... | Yes |
Let \( R \subseteq S \) be as in Proposition 23 and let \( s, t \in S \). (1) If \( s \) and \( t \) are integral over \( R \) then so are \( s \pm t \) and \( {st} \). | Proof: Let \( s \) and \( t \) be integral over \( R \). By Proposition 23 both \( R\left\lbrack s\right\rbrack \) and \( R\left\lbrack t\right\rbrack \) are finitely generated \( R \) -modules, say \[ R\left\lbrack s\right\rbrack = R{s}_{1} + R{s}_{2} + \cdots + R{s}_{n} \] \[ R\left\lbrack t\right\rbrack = R{t}_{1} +... | No |
Theorem 26. Let \( R \) be a subring of the commutative ring \( S \) with \( 1 \in R \) and suppose that \( S \) is integral over \( R \). (1) Assume that \( S \) is an integral domain. Then \( R \) is a field if and only if \( S \) is a field. | Proof: To prove (1) assume first that \( R \) is a field and let \( s \) be a nonzero element of \( S \). Then \( s \) is integral over \( R \), so\n\n\[ \n{s}^{n} + {a}_{n - 1}{s}^{n - 1} + \cdots + {a}_{1}s + {a}_{0} = 0 \n\]\n\nfor some \( {a}_{0},{a}_{1},\ldots ,{a}_{n - 1} \) in \( R \). Since \( S \) is an integr... | Yes |
Corollary 27. Suppose \( R \) is a subring of the ring \( S \) with \( 1 \in R \) and assume \( S \) is integral and finitely generated (as a ring) over \( R \) . If \( P \) is a maximal ideal in \( R \) then there is a nonzero and finite number of maximal ideals \( Q \) of \( S \) with \( Q \cap R = P \) . | Proof: There exists at least one maximal ideal \( Q \) lying over \( P \) by (2) of the theorem, so we must see why there are only finitely many such maximal ideals in \( S \) . If \( Q \) is a maximal ideal of \( S \) with \( Q \cap R = P \) then \( S/Q \) is a field containing the field \( R/P \) . To prove that ther... | Yes |
Proposition 28. An element \( \alpha \) in some field extension of \( \mathbb{Q} \) is an algebraic integer if and only if \( \alpha \) is algebraic over \( \mathbb{Q} \) and its minimal polynomial \( {m}_{\alpha ,\mathbb{Q}}\left( x\right) \) has integer coefficients. In particular, the algebraic integers in \( \mathb... | Proof: If \( \alpha \) is algebraic over \( \mathbb{Q} \) with \( {m}_{\alpha .\mathbb{Q}}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \), then by definition \( \alpha \) is integral over \( \mathbb{Z} \) . Conversely, assume \( \alpha \) is integral over \( \mathbb{Z} \), and let \( f\left( x\right) \) be... | Yes |
Theorem 29. Let \( K \) be a number field of degree \( n \) over \( \mathbb{Q} \) . (1) The ring \( {\mathcal{O}}_{K} \) of integers in \( K \) is a Noetherian ring and is a free \( \mathbb{Z} \) -module of rank \( n \) . (2) For every \( \beta \in K \) there is some nonzero \( d \in \mathbb{Z} \) such that \( {d\beta ... | Proof: Note first that any \( \mathbb{Z} \) -linear dependence relation among elements in \( {\mathcal{O}}_{K} \) is a \( \mathbb{Q} \) -linear dependence relation in \( K \), and multiplying a \( \mathbb{Q} \) -linear dependence relation of elements of \( {\mathcal{O}}_{K} \) in \( K \) by a common denominator for the... | Yes |
Theorem 30. (Noether’s Normalization Lemma) Let \( k \) be a field and suppose that \( A = k\left\lbrack {{r}_{1},{r}_{2},\ldots ,{r}_{m}}\right\rbrack \) is a finitely generated \( k \) -algebra. Then for some \( q,0 \leq q \leq m \) , there are algebraically independent elements \( {y}_{1},{y}_{2},\ldots ,{y}_{q} \in... | Proof: Proceed by induction on \( m \) . If \( {r}_{1},\ldots ,{r}_{m} \) are algebraically independent over \( k \) then take \( {y}_{i} = {r}_{i}, i = 1,\ldots, m \) . Otherwise, there exists \( f\left( {{x}_{1},\ldots ,{x}_{m}}\right) \in \) \( k\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack \) such that \( f\l... | No |
Theorem 31. (Hilbert’s Nullstellensatz — Weak Form) Let \( k \) be an algebraically closed field. Then \( M \) is a maximal ideal in the polynomial ring \( k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) if and only if \( M = \left( {{x}_{1} - {a}_{1},\ldots ,{x}_{n} - {a}_{n}}\right) \) for some \( {a}... | Proof: Certainly \( \left( {{x}_{1} - {a}_{1},\ldots ,{x}_{n} - {a}_{n}}\right) \) is a maximal ideal in \( k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) . Conversely, for any maximal ideal \( M \) in \( k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) , let \( E = k\left\lbrack {{x}_{1... | Yes |
Corollary 33. (Variant of Hilbert’s Nullstellensatz) If \( k \) is any field with algebraic closure \( \bar{k} \) and \( I \) is an ideal in \( k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \), then \( {\mathcal{I}}_{k}\left( {{\mathcal{Z}}_{\bar{k}}\left( I\right) }\right) = \operatorname{rad}I \), wher... | Proof: Since \( \bar{k}\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) is an integral extension of \( k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) (generated by the integral elements \( \bar{k} \) ), the corollary follows immediately from Theorem 32 and the remarks on radicals above. | Yes |
Proposition 34. Suppose \( k \) is any field. If \( I = \left( {{f}_{1},\ldots ,{f}_{s}}\right) \) is a proper ideal in \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), then \( f \in \operatorname{rad}I \) if and only if \( \left( {{f}_{1},\ldots ,{f}_{s},1 - {yf}}\right) = k\left\lbrack {{x}_{1},\ldots ,{x}... | Proof: By Corollary 33, \( \left( {{f}_{1},\ldots ,{f}_{s},1 - {yf}}\right) = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}, y}\right\rbrack \) if and only if the equations\n\n\[ 1 - {yf}\left( {{x}_{1},\ldots ,{x}_{n}}\right) = 0,\;{f}_{1}\left( {{x}_{1},\ldots ,{x}_{n}}\right) = 0,\;\ldots ,\;{f}_{s}\left( {{x}_{1},\ldots ,... | Yes |
Corollary 35. Suppose \( I = \left( {{f}_{1},\ldots ,{f}_{s}}\right) \) in \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then \( f \in \operatorname{rad}I \) if and only if \( \left\{ 1\right\} \) is the reduced Gröbner basis for the ideal \( \left( {{f}_{1},\ldots ,{f}_{s},1 - {yf}}\right) \) in \( k\le... | ## Example\n\nConsider \( I = \left( {{x}^{2} - {y}^{2},{xy}}\right) \) in \( k\left\lbrack {x, y}\right\rbrack \) . The reduced Gröbner basis for \( \left( {{x}^{2} - {y}^{2},{xy},1 - {tx}}\right) \) in \( k\left\lbrack {x, y, t}\right\rbrack \) with respect to the order \( x > y > t \) is \( \{ 1\} \), showing \( x \... | Yes |
Corollary 37. In the notation of Theorem 36,\n\n(1) \( \ker \pi = \{ r \in R \mid {xr} = 0 \) for some \( x \in D\} \) ; in particular, \( \pi : R \rightarrow {D}^{-1}R \) is an injection if and only if \( D \) contains no zero divisors of \( R \), and\n\n(2) \( {D}^{-1}R = 0 \) if and only if \( 0 \in D \), hence if a... | Proof: By definition, we have \( \pi \left( r\right) = 0 \) if and only if \( \left( {r,1}\right) \sim \left( {0,1}\right) \), i.e., if and only if \( {xr} = 0 \) for some \( x \in D \), which is (1). For (2), note that \( {D}^{-1}R = 0 \) if and only\n\nif the 1 of this ring is zero, i.e., \( \left( {1,1}\right) \sim ... | Yes |
Proposition 38. In the preceding notation we have\n\n(1) For any ideal \( J \) of \( {D}^{-1}R \) we have \( J = {}^{e}\left( {{}^{c}J}\right) \) . In particular, every ideal of \( {D}^{-1}R \) is the extension of some ideal of \( R \), and distinct ideals of \( {D}^{-1}R \) have distinct contractions in \( R \) . | Proof: We always have \( {}^{e}\left( {{}^{c}J}\right) \subseteq J \) . For the reverse inclusion let \( a/d \in J \) . Then \( a/1 = d\left( {a/d}\right) \in J \), and so \( a \in {\pi }^{-1}\left( J\right) = {}^{c}J \) . Thus \( a/1 \in {}^{e}\left( {{}^{c}J}\right) \), so we also have \( \left( {a/1}\right) \left( {... | Yes |
Proposition 39. Suppose \( R \) is a commutative ring with 1 and \( I \) is an ideal in \( R\\left\\lbrack x\\right\\rbrack \) . Then \( I \) is a prime ideal in \( R\\left\\lbrack x\\right\\rbrack \) if and only if\ni. \( J = I \\cap R \) is a prime ideal in \( R \), i.e., \( S = R/J \) is an integral domain, and\nii.... | Proof: Suppose \( I \) is a prime ideal in \( R\\left\\lbrack x\\right\\rbrack \), so that \( J = I \\cap R \) is a prime ideal in \( R \) and \( S = R/J \) is an integral domain. By Proposition 2 in Chapter 9, the kernel of the reduction homomorphism \( R\\left\\lbrack x\\right\\rbrack \\mapsto S\\left\\lbrack x\\righ... | Yes |
Proposition 40. Let \( S \) be an integral domain with fraction field \( F \) and let \( A \) be a nonzero ideal in \( S\left\lbrack x\right\rbrack \) . Suppose \( {AF}\left\lbrack x\right\rbrack = \left( {h\left( x\right) }\right) \) where \( h\left( x\right) \) is a polynomial in \( S\left\lbrack x\right\rbrack \) wi... | Proof: We first show \( {AF}\left\lbrack x\right\rbrack \cap {S}_{a}\left\lbrack x\right\rbrack = A{S}_{a}\left\lbrack x\right\rbrack \) . Since \( {S}_{a} \subseteq F \), the containment \( A{S}_{a}\left\lbrack x\right\rbrack \subseteq {AF}\left\lbrack x\right\rbrack \cap {S}_{a}\left\lbrack x\right\rbrack \) is immed... | Yes |
Proposition 41. Let \( D \) be a multiplicatively closed subset of \( R \) containing 1 and let \( M \) be an \( R \) -module. Then \( {D}^{-1}M \cong {D}^{-1}R{ \otimes }_{R}M \) as \( {D}^{-1}R \) -modules, i.e., \( {D}^{-1}M \) is the \( {D}^{-1}R \) -module obtained by extension of scalars from the \( R \) -module ... | Proof: The map from \( {D}^{-1}R \times M \) to \( {D}^{-1}M \) defined by mapping \( \left( {r/d, m}\right) \) to \( {rm}/d \) is well defined and \( R \) -balanced, so induces a homomorphism from \( {D}^{-1}R{ \otimes }_{R}M \) to \( {D}^{-1}M \) . The map sending \( m/d \) to \( \left( {1/d}\right) \otimes m \) give... | Yes |
Proposition 42. Let \( R \) be a commutative ring with 1 and let \( {D}^{-1}R \) be its localization with respect to the multiplicatively closed subset \( D \) of \( R \) containing 1 . (1) Localization commutes with finite sums and intersections of ideals: If \( I \) and \( J \) are ideals of \( R \), then \[ {D}^{-1}... | Proof: We first prove (6). Suppose that \( 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0 \) is a short exact sequence of \( R \) -modules. Every element of \( {D}^{-1}N \) is of the form \( n/d \) for some \( n \in N \) and \( d \in D \) . Since \( \varphi \) is surjecti... | No |
Proposition 43. Let \( R \) be a Noetherian ring and let\n\n\[ I = {Q}_{1} \cap \cdots \cap {Q}_{m} \]\n\nbe a minimal primary decomposition of the proper ideal \( I \), where \( {Q}_{i} \) is a \( {P}_{i} \)-primary ideal. Suppose \( D \) is a multiplicatively closed set of \( R \) containing 1 and the primary ideals ... | Proof: By (3) of Proposition \( {42},{D}^{-1}{Q}_{i} = {D}^{-1}R \) for \( t + 1 \leq i \leq m \), and \( {D}^{-1}{Q}_{i} \) is a \( {D}^{-1}{P}_{i} \)-primary ideal with pullback \( {Q}_{i} \) for \( 1 \leq i \leq t \). By (1) of the same proposition, \( {D}^{-1}I = {D}^{-1}{Q}_{1} \cap \cdots \cap {D}^{-1}{Q}_{t} \),... | Yes |
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) \( {M}_{\mathfrak{m}} = 0 \) for all maximal ideals \( \mathfrak{m} \) 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 \( \mathcal{J} \). | 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). | 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 \( R \) .\n\n(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 |
Proposition 14.\n\n(1) Every Principal Ideal Domain is a Dedekind Domain.\n\n(2) The ring of integers in an algebraic number field 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). Let \( {\mathcal{O}}_{K} \) be the ring of integers in the number field \( K \), i.e., the integral closure of \( \mathbb{Z} ... | 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\n\n(1) there is an ideal \( J \) of \( R \) relatively prime to \( I \) such that the product \( {IJ} = \left( a\right) \) is a principal ideal,\n\n(2) if \( I \) is nonzero then every ideal in the quotient \( R/I \) is principal; equivalent... | 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![e63577fb-02f0-43d4-8fbf-... | 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 \( {\text{Ext}}_{R}^{n} \) from the resulting diagram, as noted in the discussion preceding Proposition 3 we replace the lowest nonzero row in the transf... | 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 \[ m{Z}^{n}\left( {G, A}\right) = m{B}^{n}\left( {G, A}\right) = m{H}^{n}\left( {G, A}\right) = 0\;\text{ for all }n \geq 0. \] | 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 \( {n}_{l} = 1 \) . 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 |
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