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Lemma 82.6. Let \( A \) be a commutative \( R \) -algebra and \( u \) a unit in \( A \) . Then \( {u}^{-1} \) is integral over \( R \) if and only if \( {u}^{-1} \) lies in \( R\left\lbrack u\right\rbrack \) . | Proof. The element \( {u}^{-1} \) is integral over \( R \) if and only if there exists an equation\n\n\[ \n{u}^{-n} + {r}_{1}{u}^{-n + 1} + \cdots + {r}_{n} = 0\text{in}A\text{for some}{r}_{1},\ldots ,{r}_{n} \in R \n\]\n\nif and only if\n\n\[ \nu\left( {{r}_{1} + {r}_{2}u + \cdots + {r}_{n}{u}^{n - 1}}\right) = - 1\te... | Yes |
Proposition 82.7. Let \( A \subset B \) be domains with \( B/A \) integral. Then \( A \) is a field if and only if \( B \) is a field. | Proof. ( \( \Leftarrow \) ): Let \( x \in A \) be nonzero. Then \( {x}^{-1} \) lies in the field \( B \) and is integral over \( A \) if and only if \( {x}^{-1} \in A\left\lbrack x\right\rbrack = A \) by the lemma.\n\n\( \left( \Rightarrow \right) \) : Let \( y \in B \) be nonzero. Then there exists \( {a}_{1},\ldots ,... | Yes |
Corollary 82.8. Let \( \varphi : A \rightarrow B \) be an integral ring homomorphism, \( \mathfrak{P} \) a prime ideal in \( B \). Then \( \mathfrak{P} \) is a maximal ideal in \( B \) if and only if \( {}^{a}\varphi \left( \mathfrak{P}\right) \) is a maximal ideal in \( A \). In particular, the restriction of \( {}^{a... | Proof. let \( \mathfrak{P} \) be a prime ideal in \( B \), then \( \varphi \) induces a monomorphism \( \bar{\varphi } : A/{\varphi }^{-1}\left( \mathfrak{P}\right) \rightarrow B/\mathfrak{P} \) which is integral by Remark 82.3(3). | No |
Lemma 82.10. (Nakayama’s Lemma) Let \( \mathfrak{A} \) be an ideal in \( \operatorname{rad}\left( R\right) \) and \( M \) a finitely generated \( R \) -module. If \( M = \mathfrak{A}M \), then \( M = 0 \) . | Proof. Suppose that \( M \neq 0 \) and \( M = \mathop{\sum }\limits_{{i = 1}}^{n}R{m}_{i} \) with \( n \) minimal. Then there exist \( {a}_{1},\ldots ,{a}_{n} \) in \( \mathfrak{A} \) satisfying \( {m}_{1} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{m}_{i} \), as \( M = \) \( \mathfrak{A}M \) . Thus \( \left( {1 - {a}... | Yes |
Corollary 82.11. Let \( \mathfrak{A} \subset \operatorname{rad}\left( R\right) \) be an ideal in \( R, M \) a finitely generated \( R \) -module, and \( N \subset M \) a submodule. If \( M = N + \mathfrak{A}M \), then \( M = N \) . | Proof. We have \( M/N \) is a finitely generated \( R \) -module satisfying \( M/N = \mathfrak{A}\left( {M/N}\right) \), so \( M/N = 0 \) by Nakayama’s Lemma. Hence \( M = \) \( N \) . | Yes |
Theorem 82.14. (Cohen-Seidenberg Theorem)\n\nLet \( B/A \) be integral. Then:\n\n(1) (Incomparability) If \( {\mathfrak{P}}_{1} \subset {\mathfrak{P}}_{2} \) are prime ideals in \( B \), and satisfy \( {\mathfrak{P}}_{1} \cap A = {\mathfrak{P}}_{2} \cap A \), then \( {\mathfrak{P}}_{1} = {\mathfrak{P}}_{2} \) . | Proof. (1): Let \( \mathfrak{p} = {\mathfrak{P}}_{1} \cap A = {\mathfrak{P}}_{2} \cap A \) in \( \operatorname{Spec}\left( A\right) \) and \( S = A \smallsetminus \mathfrak{p} \) . Then \( S \cap {\mathfrak{P}}_{i} = \varnothing \) for \( i = 1,2 \), so \( {S}^{-1}B/{S}^{-1}A \) is integral, \( \left( {{S}^{-1}A,{S}^{-... | Yes |
Corollary 82.15. Let \( B/A \) be an integral extension with \( B \) a domain and \( \mathfrak{B} < B \) a nonzero ideal. Then \( \mathfrak{B} \cap A < A \) is a nonzero ideal. | Proof. Let \( \mathfrak{P} \in V\left( \mathfrak{B}\right) \) . Certainly, \( \mathfrak{P} \cap A < A \) as \( 1 \notin \mathfrak{P} \) . Since \( A \) and \( B \) are domains, the zero ideal in \( B \) lies over the zero ideal in \( A \) . By Incomparability, \( \mathfrak{P} \) does not lie over (0). Suppose that \( \... | Yes |
Lemma 82.16. (Prime Avoidance Lemma) Let \( {\mathfrak{A}}_{1},\ldots ,{\mathfrak{A}}_{n} \) be ideals in \( R \), at least \( n - 2 \) of which are prime. Let \( S \subset R \) be a subrng (it does not have to have a 1) contained in \( {\mathfrak{A}}_{1} \cup \cdots \cup {\mathfrak{A}}_{n} \) . Then there exists a \( ... | Proof. The case \( n = 1 \) is trivial, so assume that \( n > 1 \) and assume that the result is false. By induction, \( S ⊄ {\mathfrak{A}}_{1} \cup \cdots \widehat{{\mathfrak{A}}_{i}} \cup \cdots \cup {\mathfrak{A}}_{n} \) for each \( i \) where \( \frown \) means omit. So for each \( i = 1,\ldots, n \), there exists ... | Yes |
Theorem 82.17. Let \( A \) be a normal domain (i.e., an integrally closed domain) with quotient field \( F \) and \( L/F \) a normal extension of fields. Let \( {\mathfrak{P}}_{1} \) and \( {\mathfrak{P}}_{2} \) be prime ideals in \( {A}_{L} \) lying over \( \mathfrak{p} \) in \( \operatorname{Spec}\left( A\right) \), ... | Proof. If \( b \in {A}_{L} \) and \( \sigma \in G\left( {L/F}\right) \), then \( \sigma \left( b\right) \in L \) is integral over \( \sigma \left( A\right) = A \), i.e., \( \sigma \left( b\right) \in {\left( \sigma \left( A\right) \right) }_{L} = {A}_{L} \), so \( \sigma \left( {A}_{L}\right) = {A}_{L} \) . As \( {\sig... | No |
Corollary 82.18. Let \( A \) be a normal domain with quotient field \( F \) and \( K/F \) a finite field extension. Then \( {}^{a}i : \operatorname{Spec}\left( {A}_{K}\right) \rightarrow \operatorname{Spec}\left( A\right) \) has finite fibers, where \( i \) is the inclusion map of \( A \) in \( {A}_{K} \), i.e., \( {\l... | Proof. Let \( L/F \) be a normal closure of \( K/F \) . Then \( L/F \) is a finite normal extension, hence \( G\left( {L/F}\right) \) is finite. Let \( \mathfrak{p} \) be a prime ideal in \( A \) . By Lying Over, there exists a prime ideal \( \mathfrak{P} \) in \( {A}_{L} \) lying over \( \mathfrak{p} \) . We have \( {... | Yes |
Corollary 82.20. Let \( A \) be a semi-local normal domain with quotient field \( F \) and \( K/F \) a finite field extension. Then \( {A}_{K} \) is semi-local. | Proof. As \( {A}_{K}/A \) is integral, \( {\left. {}^{a}i\right| }_{\operatorname{Max}{\left( A\right) }_{K}} : \operatorname{Max}\left( {A}_{K}\right) \rightarrow \operatorname{Max}\left( A\right) \) , where \( i \) is the inclusion of \( A \) in \( {A}_{K} \) . As \( {}^{a}i \) has finite fibers, the result follows. | No |
Proposition 82.21. Let \( A \) be a normal domain with quotient field \( F \) and \( K/F \) a finite field extension. Suppose that \( \mathfrak{P} \), a prime ideal in \( {A}_{K} \) , lies over the prime ideal \( \mathfrak{p} \) in \( A \) . Let \( \bar{F} \) denote the quotient field of \( A/\mathfrak{p} \) . Then for... | Proof. Let \( - : A\left\lbrack t\right\rbrack \rightarrow \left( {A/\mathfrak{p}}\right) \left\lbrack t\right\rbrack \subset \bar{F} \) be the canonical epimorphism and \( x \in {A}_{K} \) an element satisfying \( \alpha = x + \mathfrak{P} \) in \( {A}_{K}/\mathfrak{P} \) . Then the minimal polynomial \( {m}_{F}\left(... | Yes |
Corollary 82.22. Let \( A \) be a normal domain with quotient field \( F \) and \( K/F \) a finite field extension. Suppose that there exists an element \( x \) in \( {A}_{K} \) satisfying \( {A}_{K} = A\left\lbrack x\right\rbrack \) . Let \( \mathfrak{P} \) be a prime ideal in \( {A}_{K} \) lying over the prime ideal ... | Proof. \( \bar{K} = {qf}\left( {A/\mathfrak{p}}\right) \left\lbrack {x + \mathfrak{P}}\right\rbrack \) | No |
Theorem 82.23. (Going Down Theorem) Let \( B/A \) be integral with \( A \) a normal domain and \( B \) a domain. Let \( {\mathfrak{p}}_{1} \subset {\mathfrak{p}}_{2} \) be prime ideals in \( A \) and \( {\mathfrak{P}}_{2} \) a prime ideal in \( B \) lying over \( {\mathfrak{p}}_{2} \). Then there exists a prime ideal \... | Proof. Let \( F \) be the quotient field of \( A \) and \( K \) the quotient field of \( B \). Since \( B/A \) is integral, \( K/F \) is algebraic. Let \( L/K \) be a normal closure of \( K/F \). As \( {A}_{L}/A \) is integral and \( B \subset {A}_{L} \), we have \( {A}_{L}/B \) is also integral. Let \( {\mathfrak{Q}}_... | Yes |
Lemma 83.2. Let \( \mathfrak{Q} \) be a primary ideal in \( R \), then \( \sqrt{\mathfrak{Q}} \) is a prime ideal in \( R \) and is the smallest prime ideal in \( R \) containing \( \mathfrak{Q} \) . | Proof. Suppose that \( {xy} \) lies in \( \mathfrak{Q} \) . Then there exists a positive integer \( n \) such that \( {x}^{n}{y}^{n} \) lies in \( \mathfrak{Q} \) . Hence either \( {x}^{n} \in \mathfrak{Q} \) or \( {y}^{nm} \in \mathfrak{Q} \) i.e., either \( x \in \sqrt{\mathfrak{Q}} \) or \( y \in \sqrt{\mathfrak{Q}}... | Yes |
Lemma 83.4. Let \( \\mathfrak{A} < R \) be an ideal satisfying \( \\sqrt{\\mathfrak{A}} \) is a maximal ideal in \( R \) . Then \( \\mathfrak{A} \) is a \( \\sqrt{\\mathfrak{A}} \) -primary ideal. In particular, if \( \\mathfrak{m} \) is a maximal ideal in \( R \), then \( {\\mathfrak{m}}^{n} \) is an \( \\mathfrak{m} ... | Proof. Let \( \\mathfrak{m} = \\sqrt{\\mathfrak{A}} \) . Then \( \\mathfrak{m} = \\mathop{\\bigcap }\\limits_{{V\\left( \\mathfrak{A}\\right) }}\\mathfrak{p} \), hence \( \\mathfrak{m} \\subset \\mathfrak{P} \) for every prime ideal in \( V\\left( \\mathfrak{A}\\right) \) . Since \( \\mathfrak{m} \) is maximal, \( V\\l... | Yes |
Theorem 83.11. Suppose that \( \mathfrak{A} < R \) is an ideal that has an irredundant primary decomposition\n\n\[ \mathfrak{A} = {\mathfrak{Q}}_{1} \cap \cdots \cap {\mathfrak{Q}}_{n}\text{with}{\mathfrak{p}}_{i} = \sqrt{{\mathfrak{Q}}_{i}}\text{for}i = 1,\ldots, n\text{.} \]\n\nThen\n\n\[ \left\{ {{\mathfrak{p}}_{1},... | Proof. Let \( x \in R \) . We have \( \left( {\mathfrak{A} : x}\right) = \left( {\bigcap {\mathfrak{Q}}_{i} : x}\right) = \bigcap \left( {{\mathfrak{Q}}_{i} : x}\right) \), so by the Computation 83.8,\n\n\[ \sqrt{\left( \mathfrak{A} : x\right) } = \bigcap \sqrt{\left( {\mathfrak{Q}}_{i} : x\right) } = \mathop{\bigcap }... | No |
Proposition 83.14. Suppose that the ideal \( \mathfrak{A} \) has an irredundant primary decomposition. Then a prime ideal \( \mathfrak{p} \) in \( R \) is an isolated prime of \( \mathfrak{A} \) if and only if \( \mathfrak{p} \in V\left( \mathfrak{A}\right) \) is minimal if and only if \( \mathfrak{p}/\mathfrak{A} \in ... | Proof. Let \( \mathfrak{A} = {\mathfrak{Q}}_{1} \cap \cdots \cap {\mathfrak{Q}}_{n} \) be an irredundant primary decomposition of \( \mathfrak{A} \) in \( R \) with \( {\mathfrak{p}}_{i} = \sqrt{{\mathfrak{Q}}_{i}} \) for \( i = 1,\ldots, n \) . Suppose that \( \mathfrak{p} \) lies in \( V\left( \mathfrak{A}\right) \),... | No |
Proposition 83.15. Suppose that the ideal \( \mathfrak{A} \) in \( R \) has an irredundant primary decomposition \( \mathfrak{A} = {\mathfrak{Q}}_{1} \cap \cdots \cap {\mathfrak{Q}}_{n} \) with \( {\mathfrak{p}}_{i} = \sqrt{{\mathfrak{Q}}_{i}} \) for \( i = 1,\ldots, n \). Then\n\n\[ \bigcup {\mathfrak{p}}_{i} = \{ x \... | Proof. Let \( - : R \rightarrow R/\mathfrak{A} \) be the canonical epimorphism, then Check. \( \left( \overline{0}\right) = \overline{{\mathfrak{Q}}_{1}} \cap \cdots \cap \overline{{\mathfrak{Q}}_{n}} \) is an irredundant primary decomposition in \( \bar{R} \). So it suffices to prove \( \operatorname{zd}\left( R\right... | No |
Proposition 83.18. Suppose that the ideal \( \mathfrak{A} \) in \( R \) has an irredundant primary decomposition \( \mathfrak{A} = {\mathfrak{Q}}_{1} \cap \cdots \cap {\mathfrak{Q}}_{n} \) with \( {\mathfrak{p}}_{i} = \sqrt{{\mathfrak{Q}}_{i}} \) for \( i = 1,\ldots, n \) and \( S \) is a multiplicative set in \( R \) ... | Proof. We have\n\n\[ \n{S}^{-1}\mathfrak{A} = {S}^{-1}\left( {\mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{Q}}_{i}}\right) = \mathop{\bigcap }\limits_{{i = 1}}^{n}{S}^{-1}{\mathfrak{Q}}_{i} = \mathop{\bigcap }\limits_{{i = 1}}^{m}{S}^{-1}{\mathfrak{Q}}_{i} \n\]\n\nwith \( {S}^{-1}{\mathfrak{Q}}_{i} \) being \( {S}^... | Yes |
Theorem 83.22. Suppose that the ideal \( \mathfrak{A} \) in \( R \) has an irredundant primary decomposition \( \mathfrak{A} = {\mathfrak{Q}}_{1} \cap \cdots \cap {\mathfrak{Q}}_{n} \) with \( {\mathfrak{p}}_{i} = \sqrt{{\mathfrak{Q}}_{i}} \) for \( i = 1,\ldots, n \) and \( I = \left\{ {{\mathfrak{p}}_{{i}_{1}},\ldots... | Proof. Let \( S = R \smallsetminus \mathop{\bigcup }\limits_{{j = 1}}^{m}{\mathfrak{p}}_{{i}_{j}} \) . Then by the above remark,\n\n\[{\mathfrak{Q}}_{{i}_{1}} \cap \cdots \cap {\mathfrak{Q}}_{{i}_{m}} = {\varphi }_{R}^{-1}\left( {{S}^{-1}\mathfrak{A}}\right)\]\n\nwhere \( {\varphi }_{R} : R \rightarrow {S}^{-1}R \) is ... | Yes |
Let \( \mathfrak{A} < R \) be a radical ideal, i.e., \( \mathfrak{A} = \sqrt{\mathfrak{A}} \). Suppose that \( \left| {\operatorname{Min}\left( {R/\mathfrak{A}}\right) }\right| \) is finite. Then \( \mathfrak{A} \) has an irredundant primary decomposition and \( {\operatorname{Ass}}_{R}V\left( \mathfrak{A}\right) = \{ ... | We have\n\n\[\n\mathfrak{A} = \sqrt{\mathfrak{A}} = \mathop{\bigcap }\limits_{{V\left( \mathfrak{A}\right) }}\mathfrak{p} = \mathop{\bigcap }\limits_{{\mathfrak{p} \in V\left( \mathfrak{A}\right) }}\mathfrak{p}\n\]\n\nwith the right hand side an irredundant primary decomposition of \( \mathfrak{A} \) . If \( R \) is No... | Yes |
Lemma 83.25. Let \( R \) be Noetherian ring and \( \mathfrak{A} < R \) an irreducible ideal. Then \( \mathfrak{A} \) is a primary ideal. | Proof. Since \( R/\mathfrak{A} \) is also Noetherian, by the Correspondence Theorem, it suffices to show that if (0) is irreducible, then it is primary. So suppose that (0) is irreducible and \( {xy} = 0,\;x, y \in R \), with \( y \neq 0 \) . By the ascending chain condition,\n\n\[ \left( {0 : x}\right) \subset \left( ... | No |
Lemma 83.28. Let \( R \) be a Noetherian ring and \( \mathfrak{A} < R \) an ideal. Then there exists a positive integer \( n \) satisfying \( {\left( \sqrt{\mathfrak{A}}\right) }^{n} \subset \mathfrak{A} \) . | Proof. Since ideals in a Noetherian ring are finitely generated, \( \mathfrak{A} = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) for some \( {a}_{i} \in \sqrt{\mathfrak{A}} \) . It follows that there exists a positive integer \( N \) such that \( {a}_{i}^{N} \in \mathfrak{A} \) for all \( i \), so \( {\left( \sqrt{\mathfr... | Yes |
Corollary 83.30. Let \( R \) be a Noetherian ring, \( \mathfrak{m} \in \operatorname{Max}\left( R\right) \), and \( \mathfrak{Q} < \) \( R \) an ideal. Then the following are equivalent:\n\n(1) \( \mathfrak{Q} \) is \( \mathfrak{m} \) -primary.\n\n(2) \( \sqrt{\Omega } = \mathfrak{m} \) .\n\n(3) There exists a positive... | Proof. We have shown all but the implication \( \left( 3\right) \Rightarrow \left( 2\right) \) which follows from \( \mathfrak{m} = \sqrt{{\mathfrak{m}}^{n}} = \sqrt{\mathfrak{Q}} = \sqrt{\mathfrak{m}} = \mathfrak{m} \) . | No |
Proposition 83.31. Let \( R \) be a Noetherian ring and \( \mathfrak{A} < R \) an ideal. Then\n\n\[{\operatorname{Ass}}_{R}V\left( \mathfrak{A}\right) = \{ \left( {\mathfrak{A} : x}\right) \mid x \in R\} \cap \operatorname{Spec}\left( R\right)\] | Proof. Replacing \( R \) with \( R/\mathfrak{A} \), we may assume that \( \mathfrak{A} = 0 \) . Let \( \mathfrak{A} = {\mathfrak{Q}}_{1} \cap \cdots \cap {\mathfrak{Q}}_{n} \) be an irredundant primary decomposition with \( {\mathfrak{p}}_{i} = \sqrt{{\mathfrak{Q}}_{i}} \) for \( i = 1,\ldots, n \) . As the decompositi... | Yes |
Lemma 84.5. Let \( M \) be an \( R \) -module. Then \( M \) has finite length if and only if it is both Noetherian and Artinian. | Proof. If \( M \) has finite length, all proper chains of submodules of \( M \) have length bounded by the length of \( M \) by the Jordan-Hölder Theorem. Conversely, suppose that \( M \) is both Noetherian and Artinian. Since it is Noetherian, there exists a maximal proper submodule \( {M}_{1} < M \), i.e., \( M/{M}_{... | Yes |
Corollary 84.7. Suppose \( {\mathfrak{m}}_{1},\ldots ,{\mathfrak{m}}_{n} \) are maximal ideals in \( R \) (not necessarily distinct) and further that \( {\mathfrak{m}}_{1}\cdots {\mathfrak{m}}_{n} = 0 \) . Then \( R \) is Noetherian if and only if \( R \) is Artinian. | Proof. Each \( {\mathfrak{m}}_{1}\cdots {\mathfrak{m}}_{i}/{\mathfrak{m}}_{1}\cdots {\mathfrak{m}}_{i + 1} \) is an \( \left( {R/{\mathfrak{m}}_{i + 1}}\right) \) -vector space so is Artinian if and only if it is Noetherian. We know that \( {\mathfrak{m}}_{1}\cdots {\mathfrak{m}}_{i} \) is Noetherian (resp., Artinian) ... | No |
Proposition 84.8. Let \( R \) be a nonzero commutative Artinian ring. Then \( R \) is semi-local of dimension zero. In particular, \( \operatorname{rad}\left( R\right) = \operatorname{nil}\left( R\right) \) . | Proof. Let \( \mathfrak{p} \) be a prime ideal of \( R \) . Then \( R/\mathfrak{p} \) is an Artinian domain. Thus to show that \( \dim R \) is zero, i.e., \( \operatorname{Spec}\left( R\right) = \operatorname{Max}\left( R\right) \) , we need only show that any Artinian domain is a field. But if \( R \) is an Artinian d... | Yes |
Theorem 84.9. (Akizuki) Let \( R \) be a commutative ring. Then the following are equivalent.\n\n(1) \( R \) is Artinian.\n\n(2) \( R \) is Noetherian of dimension zero.\n\n(3) Every finitely generated \( R \) -module has finite length. | Proof. Every module over an Artinian (resp., Noetherian) ring \( R \) is Artinian (resp., Noetherian), since it is a quotient of \( {R}^{n} \) for some \( n \) . Thus by the above, we know that \( R \) satisfies both (1) and (2) if and only if \( R \) satisfies (3). So we need only show that \( R \) satisfies (1) if an... | Yes |
Corollary 84.11. Let \( R \) be a domain. Then the following are equivalent:\n\n(1) \( R \) is Noetherian of dimension at most one.\n\n(2) If \( 0 < \mathfrak{A} < R \) is a ideal then \( R/\mathfrak{A} \) has finite length.\n\n(3) If \( 0 < \mathfrak{A} < R \) is a ideal then \( R/\mathfrak{A} \) is Artinian. | Proof. If (1) holds and \( 0 < \mathfrak{A} < R \) is a ideal then \( R/\mathfrak{A} \) is Noetherian of dimension zero so (2) holds. Clearly, (2) implies (3), so we need only show that (3) implies (1).\n\nIf (3) holds then \( R/\mathfrak{A} \) is a Noetherian ring of dimension zero for any ideal \( 0 < \mathfrak{A} < ... | Yes |
Lemma 84.12. Let \( M \) be a non-trivial \( R \) -module and \( \mathfrak{p} \) a prime ideal in \( R \) containing \( {\operatorname{ann}}_{R}\left( M\right) \) and a minimal such prime ideal, i.e., no prime ideal containing \( {\operatorname{ann}}_{R}\left( M\right) \) properly lies in \( \mathfrak{p} \) . Then \( \... | Proof. Let \( S \) be the multiplicative set in \( R \) defined by\n\n\[ S \mathrel{\text{:=}} \{ {ab} \mid a \in R \smallsetminus \mathfrak{p}\text{ and }b \in R \smallsetminus \operatorname{zd}\left( M\right) \} .\n\nClaim 84.13. \( S \cap {\operatorname{ann}}_{R}\left( M\right) = \varnothing \).\n\nSuppose not. Then... | Yes |
Corollary 84.14. If \( \dim R = 0 \) then \( {R}^{ \times } = R \smallsetminus \operatorname{zd}\left( R\right) \) . | Proof. We have\n\n(*)\n\n\[ \operatorname{zd}\left( R\right) \supset \mathop{\bigcup }\limits_{{\operatorname{Min}\left( R\right) }}\mathfrak{p} = \mathop{\bigcup }\limits_{{\operatorname{Spec}\left( R\right) }}\mathfrak{p} = \mathop{\bigcup }\limits_{{\operatorname{Max}\left( R\right) }}\mathfrak{m}. \]\n\nSince \( {R... | No |
Lemma 84.15. Let \( R \) be a Noetherian domain of dimension one. Let a and \( c \) be non-zero elements of \( R \). Let\n\n\[ \mathfrak{A} = \mathop{\bigcup }\limits_{{n = 0}}^{\infty }\left( {{Rc} : R{a}^{n}}\right) \mathrel{\text{:=}} \left\{ {x \in R \mid x{a}^{n} \in {Rc}\text{ for some integer }n}\right\} .\n\]\n... | Proof. Let \( {\mathfrak{A}}_{k} = \left( {{Rc} : R{a}^{k}}\right) \mathrel{\text{:=}} \left\{ {x \in R \mid x{a}^{k} \in {Rc}}\right\} \). Since \( {\mathfrak{A}}_{k} \subset \) \( {\mathfrak{A}}_{k + 1} \), for all \( k \), we know that \( \mathfrak{A} \) is an ideal. Since \( R \) is Noetherian, there exists an inte... | Yes |
Theorem 85.1. (Noether Normalization Theorem) Let \( F \) be a field and \( A = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) an affine \( F \) -algebra that is also a domain. Let \( r = \operatorname{tr}{\deg }_{F}{qfA} \) . Then there exist \( {y}_{1},\ldots ,{y}_{r} \) in \( A \) algebraically independent ... | Proof. Let \( A = F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) . If \( {qf}\left( A\right) /F \) is a finite field extension, then it is integral. Consequently, we may assume that \( {qf}\left( A\right) /F \) is not algebraic. Relabeling, we may also assume that \( {x}_{1} \) is transcendental over \( F \) .... | Yes |
Corollary 85.2. Let \( F \) be a field, then\n\n\[ \dim F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack = n = \operatorname{tr}{\deg }_{F}F\left( {{t}_{1},\ldots ,{t}_{n}}\right) . \] | Proof. Since \( \left( 0\right) < \left( {t}_{1}\right) < \cdots < \left( {{t}_{1},\ldots ,{t}_{n}}\right) \) is a chain of prime ideals in \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \), we have \( \dim F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \geq n \) . Suppose that \( \left( 0\right) < {\mat... | Yes |
Corollary 85.4. Let \( A \) be an affine \( F \) -algebra that is also a domain. Then \( \dim A = \operatorname{tr}{\deg }_{F}A \) . | Proof. Let \( {r}^{\prime } = \operatorname{tr}{\deg }_{F}A \) . By Noether Normalization there exist \( {y}_{1},\ldots ,{y}_{r} \) in \( A \) algebraically independent over \( F \) with \( A/F\left\lbrack {{y}_{1},\ldots ,{y}_{r}}\right\rbrack \) integral. Therefore, \( {qf}\left( A\right) /F\left( {{y}_{1},\ldots ,{y... | Yes |
Corollary 85.5. Let \( A \) be an affine \( F \) -algebra and \( \mathfrak{p} \) a prime ideal in A. Then \( \dim V\left( \mathfrak{p}\right) = \operatorname{tr}{\deg }_{F}A/\mathfrak{p} \) . | Proof. \( \dim V\left( \mathfrak{p}\right) = \dim A/\mathfrak{p} \) . | No |
Corollary 85.6. (Zariski's Lemma) Let \( A \) be an affine \( F \) -algebra. If \( A \) is a field, then \( A/F \) is a finite field extension. | Proof. We have \( 0 = \dim A = \operatorname{tr}{\deg }_{F}A \), so the finitely generated field extension \( A/F \) is algebraic, hence finite. | Yes |
Corollary 85.7. Let \( \varphi : A \rightarrow B \) be an \( F \) -algebra homomorphism of affine \( F \) -algebras. Then the restriction of \( {}^{a}\varphi \) to \( \operatorname{Max}\left( B\right) \) satisfies | Proof. Let \( \mathfrak{m} \) be a maximal ideal in \( B \), then \( \mathfrak{p} = {}^{a}\varphi \left( \mathfrak{m}\right) \) is a prime ideal in \( A \) and \( \varphi \) induces a ring monomorphism \( A/\mathfrak{p} \rightarrow B/\mathfrak{m} \) . As the quotient of a affine \( F \) -algebra is an affine \( F \) -a... | Yes |
Theorem 85.10. (Hilbert Nullstellensatz) (Algebraic Form) Let \( A \) be an affine \( F \) -algebra and \( \mathfrak{A} < A \) an ideal. Then\n\n\[ \sqrt{\mathfrak{A}} = \mathop{\bigcap }\limits_{{V\left( \mathfrak{A}\right) \cap \operatorname{Max}\left( A\right) }}\mathfrak{m} \]\n\ni.e., the closed points in \( \oper... | Proof. Let \( \mathfrak{B} = \mathop{\bigcap }\limits_{{V\left( \mathfrak{A}\right) \cap \operatorname{Max}\left( A\right) }}\mathfrak{m} \) . Therefore, \( \sqrt{\mathfrak{A}} \subset \mathfrak{B} \) . Suppose that \( \sqrt{\mathfrak{A}} < \mathfrak{B} \) . Let \( b \in \mathfrak{B} \smallsetminus \sqrt{\mathfrak{A}} ... | Yes |
Corollary 85.19. Let \( R \) be a Noetherian domain. Then \( R \) is a UFD if and only if every height one prime in \( R \) is principal if and only if the set of height one primes is equal to the set of nonzero principal prime ideals. | Proof. We know by Kaplansky’s Theorem 28.1 that \( R \) is a UFD if and only if every nonzero prime ideal in \( R \) contains a prime element. As every nonzero prime ideal in \( R \) contains a prime ideal of height one by Corollary 85.17, the result follows. | No |
Corollary 85.20. Let \( R \) be a Noetherian ring, \( \mathfrak{A} < R \) an ideal, and \( \mathfrak{p} < R \) a prime ideal. Then\n\n(1) If \( \operatorname{ht}\mathfrak{A} = n \geq 1 \), then there exist \( {a}_{1},\ldots ,{a}_{n} \in \mathfrak{A} \) satisfying \( \operatorname{ht}\left( {{a}_{1},\ldots ,{a}_{i}}\rig... | Proof. (1): As ht \( \mathfrak{A} \geq 1 \) and \( \operatorname{Min}\left( R\right) \) is finite, there exists an element \( {a}_{1} \in \mathfrak{A} \smallsetminus \mathop{\bigcup }\limits_{{\operatorname{Min}\left( R\right) }}\mathfrak{p} \) by the Prime Avoidance Lemma. Therefore, \( \mathfrak{p} \notin V\left( {a}... | Yes |
Let \( F \) be a field and \( R = F\left\lbrack \left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \right\rbrack \), the formal power series in \( {t}_{1},\ldots ,{t}_{n} \) . Then \( R \) is a local ring with maximal ideal \( \mathfrak{m} = \left( {{t}_{1},\ldots ,{t}_{n}}\right) \) by Exercise \( {23.21}\left( 3\rig... | By Theorem 34.9, the regular local ring \( R = F\left\lbrack \left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \right\rbrack \) with \( F \) a field is a UFD. More generally, a deeper result says any regular local ring is a domain and a UFD. The proof requires homological algebra and was one of that subjects first de... | No |
Lemma 86.1. Let \( A \) be a Noetherian domain with quotient field \( F \) . Suppose that for all finite, purely inseparable field extensions \( E \) of \( F \) that \( {A}_{E} \) is a finitely generated \( A \) -module. Then \( A \) is Japanese. | Proof. Let \( K/F \) be a finite field extension and \( L/K \) a normal closure of \( K/F \) . As \( {A}_{K} \subset {A}_{L} \) and \( A \) is a Noetherian ring, it suffices to show that \( {A}_{L} \) is a finitely generated \( A \) -module by Theorem 37.6. So we may assume that \( K/F \) is a normal extension. Let \( ... | Yes |
Theorem 86.2. Every field is universally Japanese. | Proof. Let \( A \) be an affine \( F \) -algebra that is also a domain with quotient field \( K \) and \( L/K \) a finite field extension. We must show that \( {A}_{L} \) is a finitely generated \( A \) -module. By the Noether Normalization Theorem 85.1, there exist \( {x}_{1},\ldots ,{x}_{n} \) in \( A \) algebraicall... | Yes |
Lemma 87.3. Let \( \widetilde{M} \) be a nonzero \( R \) -module and \( {M}_{i} \) irreducible submodules of \( \widetilde{M} \) for \( i \) in \( I \) (not necessarily non-isomorphic). Let \( M \) be the R-module \( \mathop{\sum }\limits_{I}{M}_{i} \) . Then there exists a subset \( J \) of \( I \) satisfying \( M \) ... | Proof. Using Zorn’s Lemma, we see that there exists a subset \( J \) of \( I \) maximal such that \( N = \mathop{\sum }\limits_{J}{M}_{j} = {\bigoplus }_{J}{M}_{j} \) . Let \( {i}_{o} \) be an element in \( I \smallsetminus J \) . Then the \( R \) -module \( {M}_{{i}_{o}} \) is irreducible and contains the submodule \(... | Yes |
Proposition 87.4. Let \( M \) be a nonzero \( R \) -module. Then the following are equivalent:\n\n(1) \( M \) is a sum of irreducible \( R \) -submodules.\n\n(2) \( M \) is a direct sum of irreducible \( R \) -modules.\n\n(3) \( M \) is a completely reducible \( R \) -module. | Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) follows from the lemma.\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) : Let \( N \) be a submodule of \( M \) and \( M = {\bigoplus }_{I}{M}_{i} \) with every \( {M}_{i}, i \in I \), irreducible. Using Zorn’s Lemma, there exists a maximal subset \( J \) of... | Yes |
Corollary 87.9. Let \( M \) be a completely reducible \( R \) -module, and \( N \) a submodule of \( M \) . Then \( N \) and \( M/N \) are completely reducible. | Proof. We may assume that \( N \neq 0 \) . By Claim 87.5, there exists a submodule \( {N}_{0} \) of \( N \) that is the sum of all the irreducible submodules of \( N \) . As \( M \) is completely reducible, \( M = {N}_{0} \oplus {N}_{1} \) for some submodule \( {N}_{1} \) . As in the proof of the proposition, we see th... | Yes |
Proposition 87.12. Let \( R \) be a nonzero ring. Then the following are equivalent:\n\n(1) Every R-module is completely reducible.\n\n(2) Every short exact sequence in \( R \) -modules splits.\n\n(3) The ring \( R \) is semisimple.\n\n(4) \( R = {\bigoplus }_{i = 1}^{n}{\mathfrak{A}}_{i} \) for some \( n \) and some l... | Proof. The equivalence of (1) and (2) follows from the corollaries above and \( \left( 4\right) \Rightarrow \left( 3\right) \) from the proposition. So we need only show \( \left( 3\right) \Rightarrow \left( 4\right) \) .\n\nAs \( R \) is completely reducible, \( R = {\bigoplus }_{I}{\mathfrak{A}}_{i} \) for some minim... | Yes |
Lemma 87.13. Let \( M \) be an \( R \) -module, then the rings \( {\operatorname{End}}_{R}\left( {\mathop{\coprod }\limits_{{i = 1}}^{n}M}\right) \) and \( {\mathbb{M}}_{n}\left( {{\operatorname{End}}_{R}\left( M\right) }\right) \) are isomorphic. | Proof. Let \( N = \mathop{\coprod }\limits_{{i = 1}}^{n}{M}_{i} \) with \( M = {M}_{i} \) for all \( i \) . Then we have the usual maps\n\n\[{\iota }_{i} : {M}_{i} \rightarrow N\\text{ given by }m \mapsto \\left( {0,\\ldots ,\\underset{i}{\\underbrace{m}},\\ldots ,0}\\right)\]\n\nand\n\n\[{\pi }_{j} : N \rightarrow {M}... | Yes |
Lemma 87.14. Let \( e \) be an idempotent in the ring \( R \) and define\n\n\[ \n\\rho : {eRe} \\rightarrow {\\operatorname{End}}_{R}\\left( {Re}\\right) \\text{by eae} \\mapsto {\\rho }_{\\text{eae }} : {xe} \\mapsto {xe} \\cdot \\text{eae.}\n\]\n\nView Re as a right \( \\left( {{\\operatorname{End}}_{R}\\left( \\oper... | Proof. Check that \( {\\rho }_{\\text{eae }} \) is an \( R \) -homomorphism of \( {Re} \) when we write endomorphisms on the right, i.e., \( \\left( {re}\\right) {\\rho }_{\\text{eae }} = r\\left( {e{\\rho }_{\\text{eae }}}\\right) \) . [Notice when we write it in this way it looks like the associative law.] It is also... | Yes |
Theorem 87.16. (Wedderburn's Theorem) Let \( R \) be a nonzero ring. Then the following are equivalent:\n\n(1) \( R \) is simple and left Artinian.\n\n(2) \( R \) is simple and semisimple.\n\n(3) There exists a division ring \( D \) and a positive integer \( n \) such that \( R \cong {\mathbb{M}}_{n}\left( D\right) \) ... | Proof. We have already shown that \( \left( 3\right) \Rightarrow \left( 1\right) \) and \( \left( 3\right) \Rightarrow \left( 2\right) \) .\n\n\( \left( 1\right) \Rightarrow \left( 2\right) \) : As \( R \) is left Artinian, there exists minimal left ideal \( \mathfrak{A} \) of \( R \) by the Minimal Principle. Let\n\n\... | Yes |
Corollary 87.17. Let \( R \) be a simple ring. then \( R \) is left Artinian if and only if \( R \) is right Artinian if and only if \( R \) is left semisimple if and only if \( R \) is right semisimple. | Proof. This is true for \( {\mathbb{M}}_{n}\left( D\right) \) with \( D \) a division ring. | No |
Corollary 87.19. Let \( F \) be a field and \( A \) a finite dimensional \( F \) - algebra. Suppose that \( A \) is also a simple ring. Then \( A \cong {\mathbb{M}}_{n}\left( D\right) \) for some division ring \( D \) containing \( F \) in its center. The division ring \( D \) is a finite dimensional \( F \) -algebra, ... | Proof. The ring \( A \) is left Artinian since a finite dimensional vector space over \( F \) . | No |
Lemma 88.1. Let \( {M}_{1},\ldots ,{M}_{r} \) and \( {N}_{1},\ldots ,{N}_{s} \) be two finite collections of non-isomorphic irreducible R-modules. If\n\n\[ {M}_{1}^{{m}_{1}} \coprod \cdots \coprod {M}_{r}^{{m}_{r}} \cong {N}_{1}^{{n}_{1}} \coprod \cdots \coprod {N}_{s}^{{n}_{s}} \]\n\nfor some positive integers \( {m}_... | Proof. Let \( M = {M}_{1}^{{m}_{1}} \coprod \cdots \coprod {M}_{r}^{{m}_{r}}, N = {N}_{1}^{{n}_{1}} \coprod \cdots \coprod {N}_{s}^{{n}_{s}} \), and \( \varphi : M \rightarrow N \) an \( R \) -isomorphism. Clearly, \( \varphi \) and \( {\varphi }^{-1} \) sets up a bijection\n\n\[ \left\{ {{M}_{1},\ldots ,{M}_{r}}\right... | Yes |
Lemma 88.2. Let \( R \) be a semisimple ring. Then \( R \) is left Artinian. More precisely, if \( \mathfrak{B} \) is a non-trivial left ideal of \( R \), then there exist unique minimal left ideals \( {\mathfrak{A}}_{1},\ldots ,{\mathfrak{A}}_{m} \) in \( R \) satisfying \( \mathfrak{B} = {\mathfrak{A}}_{1} \oplus \cd... | Proof. We have shown that \( R = {\mathfrak{A}}_{1} \oplus \cdots \oplus {\mathfrak{A}}_{n} \) for minimal left ideals \( {\mathfrak{A}}_{i} \) and if \( \mathfrak{A} \) is a minimal left ideal, then \( \mathfrak{A} = {\mathfrak{A}}_{i} \) for some \( i \) . It follows, as before, that \[ \mathfrak{B} = {\bigoplus }_{i... | Yes |
Proposition 90.5. Let \( R = \left( {K/F,\sigma, a}\right) \) be a cyclic algebra over \( F \) of degree \( n \) . Then \( R \) is a central simple algebra over \( F \) . Moreover, the centralizer \( {Z}_{R}\left( K\right) \mathrel{\text{:=}} \{ x \in R \mid {xb} = {bx} \) for all \( b \in K\} \) is \( K \) and the onl... | Proof. We first show that \( R \) is simple. Write\n\n\[ R = K{1}_{R} \oplus \cdots \oplus K{x}^{n - 1} \]\n\n\[ {x}^{n} = a\text{with}a \in F\text{and}{x\alpha } = \sigma \left( \alpha \right) x\text{for all}\alpha \in K\text{.} \]\n\nLet \( 0 < \mathfrak{A} \subset R \) be a 2-sided ideal. Choose a nonzero element \(... | Yes |
Corollary 90.7. Let \( R = \left( {K/F,\sigma, a}\right) \) be a cyclic algebra over \( F \) of prime degree \( p \) . Then \( R \) is either a division ring or \( R \cong {\mathbb{M}}_{p}\left( F\right) \) . | Proof. By Wedderburn’s Theorem 87.16, \( R \cong {\mathbb{M}}_{r}\left( D\right) \) for some division ring \( D \), so \( {p}^{2} = {r}^{2}{\dim }_{F}\left( D\right) \) . As \( p \) is a prime number, \( p = r \) and \( {\dim }_{F}\left( D\right) = 1 \) splits,(i.e., \( R \) splits) or \( R \) is a division algebra (i.... | Yes |
Theorem 90.8. Let \( R = \left( {K/F,\sigma, a}\right) \) be a cyclic algebra of degree \( n \) over \( F \) . Then \( R \cong {\mathbb{M}}_{n}\left( F\right) \) if and only if \( a \in {\mathrm{N}}_{K/F}\left( {K}^{ \times }\right) \) . | Proof. ( \( \Leftarrow \) ): Suppose that \( a \in {\mathrm{N}}_{K/F}\left( {K}^{ \times }\right) \), then there exists an \( \alpha \in {K}^{ \times } \) such that \( a{N}_{K/F}\left( \alpha \right) = 1 \) . Let \( y = {\alpha x} \) where \( x \) is the image of \( t \) in \( K\left\lbrack {t,\sigma }\right\rbrack \) ... | Yes |
Proposition 91.1. Let \( D \) be a division ring that satisfies \( {x}^{n} = x \) for each \( x \) in \( D \) and some integer \( n > 1 \) depending on \( x \) . Then \( D \) is a field. | Proof. We know that the center of \( D \) contains a prime field, so \( n{1}_{D} \) lies in the center for all integers \( n \) . Let \( x \) be a nonzero element of \( D \) . Then there exist positive integers \( n \) and \( {n}^{\prime } \) such that \( {x}^{n} = x \) and \( {\left( 2x\right) }^{{n}^{\prime }} = {2x}... | Yes |
Corollary 91.2. Let \( D \) be a division algebra that is algebraic over a finite field. Then \( D \) is commutative. In particular, \( D/F \) is an algebraic extension of fields. | Proof. If \( F \) is the finite field in \( D \) and of characteristic \( p \), then \( F\left( x\right) \) is a finite field for each \( x \in D \), hence \( {x}^{{p}^{n}} = x \) for some \( n \) . It follows that \( D \) is commutative by the proposition. | Yes |
Lemma 92.4. Let \( R \) be a commutative ring and \( G \) a finite group. Suppose that \( M,{M}^{\prime }, N,{N}^{\prime } \) are \( R\left\lbrack G\right\rbrack \) -modules and \( f : M \rightarrow N \) an \( R \) -homomorphism. If \( \varphi : {M}^{\prime } \rightarrow \dot{M} \) and \( \psi : N \rightarrow {N}^{\pri... | Proof. By definition, \( \left( {\sigma f}\right) \left( m\right) = \sigma \left( {f\left( {{\sigma }^{-1}m}\right) }\right) \), so\n\n\[ \n{\operatorname{Tr}}_{G}\left( {\psi f\varphi }\right) = \mathop{\sum }\limits_{G}\sigma \circ \left( {\psi f\varphi }\right) = \mathop{\sum }\limits_{G}\left( {\sigma \circ \psi }\... | Yes |
Theorem 92.5. (Maschke’s Theorem) Let \( F \) be a field and \( G \) a finite group. Suppose that either \( \operatorname{char}F = 0 \) or \( \operatorname{char}F/\left| G\right| \) . Then \( F\left\lbrack G\right\rbrack \) is semisimple. | Proof. Let \( V \) be an \( F\left\lbrack G\right\rbrack \) -module and \( W \) an \( F\left\lbrack G\right\rbrack \) -submodule. We must show that \( W \) is a direct summand of \( V \) as an \( F\left\lbrack G\right\rbrack \) -module. Let \( {i}_{W} : W \rightarrow V \) denote the inclusion map. We must show that thi... | No |
Lemma 92.7. Let \( R \) be a commutative ring and \( G \) a finite group. If \( {C}_{{g}_{1}},\ldots ,{C}_{{g}_{r}} \) are the distinct class sums of \( G \) in \( R\left\lbrack G\right\rbrack \), then the center \( Z\left( {R\left\lbrack G\right\rbrack }\right) \) is a free \( R \) -module on basis \( \mathcal{B} \mat... | Proof. Let \( \sigma \) be an element of \( G \) . Then for all \( g \) in \( G \), we have \( \sigma {C}_{g}{\sigma }^{-1} = {C}_{g} \), so \( {C}_{g} \) lies in \( Z\left( {R\left\lbrack G\right\rbrack }\right) \) . Since \( G = \bigvee \mathcal{C}\left( {g}_{i}\right) \) is an \( R \) -basis for \( R\left\lbrack G\r... | Yes |
Theorem 92.8. Let \( G \) be a finite group and \( F \) an algebraically closed field of characteristic zero or positive characteristic not dividing the order of \( G \) . Let \( {\mathfrak{A}}_{1},\ldots ,{\mathfrak{A}}_{r} \) be a basic set for \( F\left\lbrack G\right\rbrack \) and \( {n}_{i} = {\dim }_{F}{\mathfrak... | Proof. There are no finite dimensional \( F \) -division rings over \( F \) except for \( F \), as any such would contain an element \( x \) not in \( F \), but then \( F\left( x\right) \) would be a commutative division ring distinct from \( F \) . It also follows that if \( {B}_{i} \) is the Wedderburn component corr... | Yes |
Let \( G \) be a finite group of order \( n \) . We give some examples of complex representations, i.e., representations \( \varphi : G \rightarrow \) \( {\mathrm{{GL}}}_{n}\left( \mathbb{C}\right) \) . | 1. We interpret Theorem 92.8 in the case of complex representations. Let \( {\mathfrak{A}}_{1},\ldots ,{\mathfrak{A}}_{r} \) be a basic set for \( \mathbb{C}\left\lbrack G\right\rbrack \) and \( {n}_{i} = {\dim }_{\mathbb{C}}{\mathfrak{A}}_{i}, i = 1,\ldots, r \) . Let \( {\varphi }_{i} \) be the irreducible representa... | Yes |
Theorem 93.4. (Burnside) Let \( F \) be a field, \( A \) an \( F \) -algebra (not necessarily finitely generated), and \( M \) an irreducible \( A \) -module that is also a finite-dimensional vector space over \( F \) (so cyclic as finitely generated).\n\n(1) If \( F \) is algebraically closed, then \( F = {\operatorna... | Proof. \( {\operatorname{End}}_{A}\left( M\right) \) is a finite dimensional vector space over \( F \) as \( M \) is. By Schur’s Lemma 87.15, \( D = {\operatorname{End}}_{A}\left( M\right) \) is a division ring.\n\n(1): Let \( x \in D \) . Then \( F \subset Z\left( D\right) \), so \( F\left( x\right) \subset D \) is a ... | Yes |
Proposition 95.3. Let \( V,{V}^{\prime } \), and \( {V}^{\prime \prime } \) be \( F\left\lbrack G\right\rbrack \) -modules, finite dimensional over \( F \) . Then\n\n(1) \( {\chi }_{V}\left( {{\sigma x}{\sigma }^{-1}}\right) = {\chi }_{V}\left( x\right) \) for all \( \sigma \) in \( G \) .\n\n(2) If\n\n\[ 0 \rightarrow... | Proof. We already showed (1). As for (2), we may assume that \( {V}^{\prime } \subset V \) with \( {\mathcal{B}}^{\prime } \) a basis for \( {V}^{\prime } \) extended to a basis to a basis \( \mathcal{B} \) for \( V \) . If \( {\varphi }_{V} : G \rightarrow {\mathrm{{GL}}}_{n}\left( F\right) \), then with the obvious n... | Yes |
Let \( G \) be a finite group with char \( F \) not dividing \( \left| G\right| \) , so \( F\left\lbrack G\right\rbrack \) is semi-simple by Maschke’s Theorem 92.5. Let\n\n\[ \lambda : G \rightarrow {\operatorname{Aut}}_{F}\left( {F\left\lbrack G\right\rbrack }\right) \text{be given by}x \mapsto {\lambda }_{x} : y \map... | Note: \( {\lambda }_{x} \) with \( x \) not \( {e}_{G} \) in \( G \) permutes \( G \) without fixed points and \( G \) is a basis for \( F\left\lbrack G\right\rbrack \), so\n\n\[ \operatorname{trace}{\varphi }_{\lambda }\left( x\right) = {\chi }_{\lambda }\left( x\right) = 0\text{ for all }x \neq {e}_{G}. \] | Yes |
Proposition 96.2. Let \( G \) be a torsion group, \( V \) an \( n \) -dimension vector space over \( F, V \) an \( F\left\lbrack G\right\rbrack \) -module. Let \( x \) be an element of \( G \) and \( N \) the order of the cyclic subgroup \( \langle x\rangle \) in \( G \) . Then \( {\chi }_{V}\left( x\right) \) is a sum... | Proof. Let \( N = {\dim }_{F}V \) and \( \varphi : G \rightarrow {\mathrm{{GL}}}_{N}\left( F\right) \) afford \( V \) . Set \( \alpha = \varphi \left( x\right) \) and \( n = \left| {\langle x\rangle }\right| \) . Taking a Jordan canonical form of \( \alpha \) in \( {\mathrm{{GL}}}_{N}\left( \widetilde{F}\right) \) we s... | Yes |
Proposition 96.5. Let \( G \) be a finite group, \( F \) a field of characteristic zero, and \( \varphi : G \rightarrow {\mathrm{{GL}}}_{n}\left( F\right) \) a representation. Then \( \ker {\chi }_{\varphi } = \ker \varphi \) . In particular, \( \ker {\chi }_{\varphi } \) is a normal subgroup of \( G \) . | Proof. Certainly, \( \ker \varphi \subset \ker {\chi }_{\varphi } \), so we need only show the reverse inclusion. So suppose that \( {\chi }_{\varphi }\left( x\right) = {\chi }_{\varphi }\left( 1\right) \), then by the proposition, we have\n\n\[ n = {\chi }_{\varphi }\left( 1\right) = {\varepsilon }_{1} + \cdots + {\va... | Yes |
Theorem 96.6. Let \( G \) be a finite group and \( \varphi : G \rightarrow {\mathrm{{GL}}}_{n}\left( F\right) \) and \( {\varphi }^{\prime } : G \rightarrow {\mathrm{{GL}}}_{m}\left( F\right) \) two irreducible representations. Let\n\n\[ \n{a}_{ij},{a}_{kl}^{\prime } : G \rightarrow F\text{ satisfy }\varphi \left( x\ri... | Proof. Let \( V \) with (ordered) basis \( \mathcal{B} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be afforded by \( \varphi \) and \( {V}^{\prime } \) with (ordered) basis \( {\mathcal{B}}^{\prime } = \left\{ {{v}_{1}^{\prime },\ldots ,{v}_{m}^{\prime }}\right\} \) be afforded by \( {\varphi }^{\prime } \) . Check ... | Yes |
Theorem 96.7. (Orthogonal Relations) Let \( G \) be a finite group and \( \chi ,{\chi }^{\prime } \) two inequivalent characters.\n\n(1) If both \( \chi \) and \( {\chi }^{\prime } \) are irreducible, then\n\n\[ \mathop{\sum }\limits_{G}\chi \left( x\right) {\chi }^{\prime }\left( {x}^{-1}\right) = 0 \] | Proof. Let \( \chi = {\chi }_{\varphi } \) and \( {\chi }^{\prime } = {\chi }_{{\varphi }^{\prime }} \) with \( \varphi = \left( {a}_{ij}\right) \) and \( {\varphi }^{\prime } = \left( {a}_{ij}^{\prime }\right) \) the representations giving \( \chi \) and \( {\chi }^{\prime } \) respectively. (1): By the theorem,\n\n\[... | Yes |
Theorem 96.8. Let \( G \) be a finite group. Suppose that \( \operatorname{char}F \) is zero or does not divide the order of \( G \) and \( F\left\lbrack G\right\rbrack \) is \( F \) -split. If \( {\chi }_{1},\ldots ,{\chi }_{r} \) are all the irreducible characters of \( G \) and \( x, y \) lie in \( G \), then\n\n\[ ... | Proof. Let \( {x}_{1},\ldots {x}_{{r}^{\prime }} \) be a system of representatives for the conjugacy classes of \( G \) . As \( F\left\lbrack G\right\rbrack \) is \( F \) -split, \( r = {r}^{\prime } \) . Let \( C = \left( {c}_{ij}\right) \) with \( \left. {{c}_{ij} = {\chi }_{i}\left( {x}_{j}\right) \text{ (independen... | Yes |
Theorem 97.2. Let \( F \) be a field of characteristic zero and \( G \) a finite group such that \( F\left\lbrack G\right\rbrack \) is \( F \) -split. If \( \chi : G \rightarrow F \) is an irreducible character, then \( \chi \left( 1\right) \left| \right| G \mid \) in \( \mathbb{Z} \) . | Proof. Let \( {x}_{1},\ldots ,{x}_{r} \) be a system of representatives for the conjugacy classes of \( G \) and \( \widetilde{\mathbb{Q}} \) an algebraic closure of \( \mathbb{Q} \) . Then \( \frac{\left| \mathcal{C}\left( {x}_{i}\right) \right| }{\chi \left( 1\right) }\chi \left( {x}_{i}\right) \) and \( \chi \left( ... | Yes |
Corollary 97.3. Let \( F \) be a field of characteristic zero and \( G \) a finite group such that \( F\left\lbrack G\right\rbrack \) is \( F \) -split, say \( F\left\lbrack G\right\rbrack \cong \mathop{\sum }\limits_{{i = 1}}^{r}{\mathbb{M}}_{{n}_{i}}\left( F\right) \) . Then\n\n(1) \( r \) is the number of conjugacy ... | Proof. We have previously shown (1) - (3) and (4) follows from the theorem. | No |
Theorem 97.5. (Burnside) Let \( G \) be a finite group and \( p \) a prime. If there exists an element \( x \) in \( G \) satisfying \( \mathcal{C}\left( x\right) = {p}^{e} > 1 \), then \( G \) is not a simple group. | Proof. Let \( {\chi }_{1},\ldots ,{\chi }_{r} : G \rightarrow \mathbb{C} \) be all the distinct irreducible complex characters. One of these must be the trivial character, say it is \( {\chi }_{1} \) . Let \( {n}_{i} = {\chi }_{i}\left( 1\right) \), so \( {n}_{1} = 1 \) (and \( {\chi }_{1}\left( x\right) = 1 \) ). Sinc... | Yes |
Theorem 97.6. (Burnside’s \( {p}^{a}{q}^{b} \) -Theorem) Let \( G \) be a finite group of order \( {p}^{a}{q}^{b} \) with \( p, q \) distinct primes and \( a, b \) non-negative integers. Then \( G \) is a solvable group. | Proof. By our previous work and induction, it suffices to show that \( G \) is not simple if \( a \) and \( b \) are both positive. Let \( Q \) be a Sylow \( q \) -subgroup of \( G \) and \( {e}_{G} \neq x \) an element in \( Z\left( Q\right) \) . Then \( Q \subset {Z}_{G}\left( x\right) \) , hence \( \left| {\mathcal{... | Yes |
Lemma 98.1. (Frobenius-Schur) Let \( F \) be an algebraically closed field and \( G \) a (not necessarily finite) group, \( \varphi : G \rightarrow {\mathrm{{GL}}}_{n}\left( F\right) \) an irreducible representation, and \( {a}_{ij} : G \rightarrow F,1 \leq i, j \leq n \), the coordinate functions of \( \varphi \), i.e... | Proof. (1): Extending \( \varphi \) to the \( F \) -algebra map \( \varphi : F\left\lbrack G\right\rbrack \rightarrow {\mathbb{M}}_{n}\left( F\right) \) yields (1) by Burnside's Theorem 93.4.\n\n(2): If \( \mathcal{B} \) is \( F \) -linearly dependent, then it is \( F \) -linearly dependent as a set of \( F \) -functio... | Yes |
Theorem 98.6. (Burnside) Let \( F \) be an arbitrary field and \( G \) a subgroup of \( {\mathrm{{GL}}}_{n}\left( F\right) \). Then \( G \) is finite if and only if \( G \) has finitely many conjugacy classes. | Proof. Certainly if \( G \) is finite, then it has finitely many conjugacy classes, so we need only show the converse. We use the notation as in the proof of the previous theorem. So \( \iota \) is the inclusion map. If the character \( \chi \) associated to \( \iota \) is irreducible then the proof of Case 1 for Theor... | Yes |
Lemma 98.7. Let \( G \) be a finitely generated torsion group. Suppose that \( G \) contains an abelian subgroup \( H \) of finite index. Then \( G \) is a finite group. | Proof. Let \( G = \mathop{\bigvee }\limits_{{1 = 1}}^{m}{g}_{i}H \) be a coset decomposition. As \( G \) is finitely generated, there exists a finite subset\n\n\[ \mathcal{S} = \left\{ {{g}_{1},\ldots ,{g}_{m},{g}_{m + 1},\ldots ,{g}_{n}}\right\} \subset G \]\n\ngenerating \( G \) and closed under taking inverses. For ... | Yes |
Theorem 98.9. (Schur) Let \( F \) be a field and \( G \) a finitely generated torsion subgroup of \( {\mathrm{{GL}}}_{n}\left( F\right) \) . Then \( G \) is a finite group. | Proof. By Lemma 98.8, \( G \) has bounded period, so by Lemma 98.7, it suffices to show that \( G \) contains an abelian subgroup of finite index. By Remark 98.5, the only part of the proof of Burnside's Theorem 98.4 in the characteristic zero case that must be modified is the case that the inclusion \( \iota : G \righ... | Yes |
Lemma 99.6. An R-module \( A \) is a coproduct of the R-submodules \( {A}_{1},\ldots ,{A}_{n} \) if and only if there exist \( R \) -homomorphisms \( {\iota }_{j} : {A}_{j} \rightarrow A \) and \( {\pi }_{j} : A \rightarrow {A}_{j} \) for \( j, k = 1,\ldots, n \) satisfying \( {\pi }_{k}{\iota }_{j} = {\delta }_{kj}{1}... | Proof. \( \left( \Rightarrow \right) \) : The definition of coproduct gives rise to the \( {\iota }_{j} \) ’s. For a fixed \( k \) and and each \( j \), we have a diagram \n\nBy the universal property of coproduct, t... | Yes |
Lemma 100.2. Let \( R \) be a commutative ring and \( M, N \) two \( R \) -modules. Then a tensor product of \( M \) and \( N \) exists. | Proof. Let \( P \) be the free \( R \) -module on basis \( \mathcal{B} = M \times N \) . Let \( W \) be the submodule of \( P \) generated by the following:\n\n\[ \left( {m + {m}^{\prime }, n}\right) - \left( {m, n}\right) - \left( {{m}^{\prime }, n}\right) \]\n\n\[ \left( {m, n + {n}^{\prime }}\right) - \left( {m, n}\... | Yes |
Proposition 100.7. Let \( V \) and \( W \) be finite dimensional \( F \) -vector spaces, \( {V}^{ * } = {\operatorname{Hom}}_{F}\left( {V, F}\right) \), the dual space of \( V \) (or finitely generated \( R \) -free modules with \( R \) commutative). Then the natural map \[ \varphi : {V}^{ * }{ \otimes }_{F}W \rightarr... | Proof. Although the map in the proposition is independent of any basis, we use bases to prove it. Let \( \mathcal{B} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be an \( F \) -basis for \( V \) and \( {\mathcal{B}}^{ * } = \left\{ {{f}_{1},\ldots ,{f}_{n}}\right\} \) the dual \( F \) -basis for \( {V}^{ * } \), i.e.... | No |
Let \( V \) be an \( n \) -dimensional \( F \) -vector space. Then \( S\left( V\right) \cong F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) (An analogous statement holds if \( V \) is an \( R \) -free module of rank \( n \) with \( R \) commutative): | Let \( \mathcal{B} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be an \( F \) -basis for \( V \) and \( {}^{ - } : {T}^{m}\left( V\right) \rightarrow {S}^{m}\left( V\right) \) the canonical \( R \) -module epimorphism. Then\n\n\[ \left\{ {\overline{{v}_{{i}_{1}} \otimes \cdots \otimes {v}_{{i}_{m}}} \mid 1 \leq {i}_{... | Yes |
Corollary 102.5. The sign map \( \operatorname{sgn} : {S}_{n} \rightarrow \{ \pm 1\} \) given by\n\n\( \operatorname{sgn}\sigma = \begin{cases} 1 & \text{ if }\sigma \text{ is a product of an even number of transpositions } \\ - 1 & \text{ otherwise. } \end{cases} \)\n\nis a well-defined group homomorphism. | Proof. Let \( R \) be a commutative ring and \( M \) a free \( R \) -module on basis \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \), then \( \mathop{\bigwedge }\limits^{n}\left( M\right) \) is \( R \) -free on basis \( \left\{ {{x}_{1} \land \cdots \land {x}_{n}}\right\} \) . Hence if \( \sigma \in {S}_{n} \) . there ... | No |
Proposition 103.3. Let \( V \) be an \( R \) -free module of rank \( n \) . Let \( \mathcal{B} \) and \( {\mathcal{B}}^{\prime } \) be ordered bases for \( V \) . Then \( {\left\lbrack Id\right\rbrack }_{\mathcal{B},{\mathcal{B}}^{\prime }} \) is an invertible matrix and | \[ {\left\lbrack Id\right\rbrack }_{\mathcal{B},{\mathcal{B}}^{\prime }}^{-1} = {\left\lbrack Id\right\rbrack }_{{\mathcal{B}}^{\prime },\mathcal{B}} \] | Yes |
Theorem 103.7. (Change of Basis Theorem.) Let \( V \) and \( W \) be finitely generated \( R \) -free modules. Let \( \mathcal{B} \) and \( {\mathcal{B}}^{\prime } \) be ordered bases for \( V \) and \( \mathcal{C} \) and \( {\mathcal{C}}^{\prime } \) be ordered basis for \( W \) . Let \( T : V \rightarrow W \) be an \... | The Change of Basis Theorem states that the following diagram commutes\n\n\[ \n\begin{matrix} {V}_{\mathcal{B}}\xrightarrow[]{{\left\lbrack T\right\rbrack }_{\mathcal{B},\mathcal{C}}}{W}_{\mathcal{C}} \\ {\left\lbrack Id\right\rbrack }_{\mathcal{B},{\mathcal{B}}^{\prime }}\mathop{\downarrow }\limits^{{{\left\lbrack T\r... | Yes |
Lemma 105.2. Let \( p \) be a (positive) odd prime and \( n \geq 2 \) . Then for every integer \( a \), we have \( {\left( 1 + ap\right) }^{{p}^{n - 2}} \equiv 1 + a{p}^{n - 1}{\;\operatorname{mod}\;{p}^{n}} \) . | Proof. We induct on \( n \) . The case \( n = 2 \) is trivial, so we may assume that \( n > 2 \) . In particular, \( 2\left( {n - 1}\right) > n + 1 \), so \( p\left( {n - 1}\right) > n + 1 \) . By induction, we may assume the result for \( n - 2 \) and show the result for \( n - 1 \) . By the lemma, induction, and the ... | Yes |
Lemma 105.3. Let \( p \) be a (positive) odd prime and a an integer not divisible by \( p \) . Then for every integer \( n \geq 2 \), the congruence class of \( 1 + {ap} \) in \( \mathbb{Z}/{p}^{n}\mathbb{Z} \) has order \( {p}^{n - 1} \) . | Proof. Let \( - : \mathbb{Z} \rightarrow \mathbb{Z}/{p}^{n}\mathbb{Z} \) the canonical epimorphism. By the previous lemma, we know that \( {\left( 1 + ap\right) }^{{p}^{n - 1}} \equiv 1 + a{p}^{n}{\;\operatorname{mod}\;{p}^{n + 1}} \), hence \( {\left( 1 + ap\right) }^{{p}^{n - 1}} \equiv 1{\;\operatorname{mod}\;{p}^{n... | Yes |
Proposition 105.4. Let \( p \) be a (positive) odd prime, then \( {\left( \mathbb{Z}/{p}^{n}\mathbb{Z}\right) }^{ \times } \) is cyclic for all \( n \) . | Proof. Let \( a \) be an integer relatively prime to \( p \) . We know that \( {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{ \times } \) is cyclic of order \( p - 1 \), so generated by the residue class of \( a \) .\n\nClaim 1. We may assume that \( {a}^{p - 1} ≢ 1{\;\operatorname{mod}\;{p}^{2}} \) :\n\nSuppose that \( {a}... | Yes |
Proposition 105.5. Let \( n \geq 3 \) be an integer. Then \( {\left( \mathbb{Z}/{2}^{n}\mathbb{Z}\right) }^{ \times } \cong \) \( \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/{2}^{n - 2}\mathbb{Z} \) | Proof. Let \( n \geq 3 \). We begin with the following:\n\nClaim. \( {5}^{{2}^{n - 3}} \equiv 1 + {2}^{n - 1}{\;\operatorname{mod}\;{2}^{n}} \):\n\nThis is true for \( n = 3 \), so we proceed by induction. As \( n \geq 3 \), we have \( {2n} - 2 \geq n + 1 \). Assuming the result for \( n - 3 \) we show it holds for \( ... | Yes |
Let \( \\alpha \) be an irrational number. Then there exist infinitely many relatively prime integers \( x, y \) satisfying\n\n\[ \n\\left| {\\frac{x}{y} - \\alpha }\\right| < \\frac{1}{{y}^{2}}\n\] | Proof. Let \( n \) be a positive integer. Partition the half-open interval \( \\lbrack 0,1) \) into \( n \) half-open subintervals,\n\n(†)\n\n\[ \n\\left\\lbrack {0,1}\\right\\rbrack = \\mathop{\\bigvee }\\limits_{{j = 0}}^{{n - 1}}\\left\\lbrack {\\frac{j}{n},\\frac{j + 1}{n}}\\right)\n\]\n\nRecall if \( \\beta \) is ... | Yes |
Lemma 106.2. Let \( d \) be a positive square-free integer. Then there exists a constant \( c \) satisfying\n\n\[ \left| {{x}^{2} - d{y}^{2}}\right| < c \]\n\nhas infinitely many solutions in integers \( x, y \) . | Proof. We have \( {x}^{2} - d{y}^{2} = \left( {x - y\sqrt{d}}\right) \left( {x + y\sqrt{d}}\right) \) in \( \mathbb{Z}\left\lbrack d\right\rbrack \) . By the lemma, there exist infinitely many relatively prime integers \( x, y \) with \( y > 0 \) and satisfying \( \left| {x - y\sqrt{d}}\right| < 1/y \) . Therefore,\n\n... | Yes |
The functional\n\n\\[ J\\left\\lbrack y\\right\\rbrack = {\\int }_{-1}^{1}\\cos {ydx} \\]\n\n(3.12)\n\nhas the Euler-Lagrange equation\n\n\\[ - \\sin y = 0\\text{.}\\]\n\n(3.13) | An extremum may occur along one of the horizontal lines\n\n\\[ y = {n\\pi },\\;n = 0, \\pm 1, \\pm 2,\\ldots ,\\]\n\n(3.14)\n\nbut only if the horizontal line satisfies the boundary conditions. | No |
Example 4.4 (Plateau's problem).\n\nConsider the surface area integral\n\n\\[ \nJ\\left\\lbrack u\\right\\rbrack = {\\iint }_{A}\\sqrt{1 + {u}_{x}^{2} + {u}_{y}^{2}}{dxdy}. \n\\]\n\n(4.139)\n\nThis leads to the partial differential equation\n\n\\[ \n\\frac{\\partial }{\\partial x}\\left( \\frac{{u}_{x}}{\\sqrt{1 + {u}_... | In Chapter 1, we also saw that the minimal surface equation can be given a geometric interpretation. At each point \\( P \\) of our surface, choose a vector normal to the surface, cut the surface with normal planes (that contain the normal vector but that differ in orientation), and obtain a series of plane curves. For... | Yes |
Find (see Figure 5.1), among all curves of length \( l \) in the upper half-plane passing through \( \\left( {-a,0}\\right) \) and \( \\left( {a,0}\\right) \), the one that, together with interval \( \\left\\lbrack {-a, a}\\right\\rbrack \), encloses the largest area. | To solve this problem, we must maximize\n\n\[ J\\left\\lbrack y\\right\\rbrack = {\\int }_{-a}^{+a}{ydx} \]\n\n(5.4)\n\nsubject to the boundary conditions\n\n\[ y\\left( {-a}\\right) = 0,\\;y\\left( {+a}\\right) = 0 \]\n\n(5.5)\n\nand the isoperimetric constraint\n\n\[ K\\left\\lbrack y\\right\\rbrack = {\\int }_{-a}^{... | Yes |
Example 6.2 (Geodesics in the plane).\n\nLet\n\n\[ J\left\lbrack y\right\rbrack = {\int }_{a}^{b}\sqrt{1 + {{y}^{\prime }}^{2}}{dx} \] | Since\n\n\[ \frac{{\partial }^{2}f}{\partial {y}^{\prime 2}} = \frac{1}{{\left( 1 + {y}^{\prime 2}\right) }^{3/2}} = \frac{1}{{\left( 1 + {m}^{2}\right) }^{3/2}} \]\n\nfor lines of slope \( m \), it follows that\n\n\[ {\left. \frac{{\partial }^{2}f}{\partial {y}^{\prime 2}}\right| }_{\widehat{y}\left( x\right) } > 0 \]... | No |
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