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Theorem 4.9. Every homomorphism of a ring \( R \) into an algebraically closed field \( L \) can be extended to every integral extension \( E \) of \( R \) . | Proof. If \( R \) is a field, then \( E \) is a field, by 3.3, and \( E \) is an algebraic extension of \( R \) ; we saw that 4.9 holds in that case.\n\nNow, let \( R \) be local and let \( \varphi : R \rightarrow L \) be a homomorphism whose kernel is the maximal ideal \( \mathfrak{m} \) of \( R \) . Then \( \varphi \... | Yes |
Proposition 5.1. Let \( R \) be a domain and let \( Q \) be its quotient field. Every finitely generated submodule of \( Q \) is a fractional ideal of \( R \) . If \( R \) is Noetherian, then every fractional ideal of \( R \) is finitely generated as a submodule. | Proof. If \( n > 0 \) and \( {q}_{1} = {a}_{1}/{c}_{1},\ldots ,{q}_{n} = {a}_{n}/{c}_{n} \in Q \), then\n\n\[ R{q}_{1} + \cdots + R{q}_{n} = R{b}_{1}/c + \cdots + R{b}_{n}/c = \left( {R{b}_{1} + \cdots + R{b}_{n}}\right) /c, \]\n\nwhere \( c = {c}_{1}\cdots {c}_{n} \) ; hence \( R{q}_{1} + \cdots + R{q}_{n} \) is a fra... | Yes |
Proposition 5.3. (1) Every invertible fractional ideal is finitely generated. | Proof. (1). If \( \mathfrak{{AB}} = R \), then \( 1 = {a}_{1}{b}_{1} + \cdots + {a}_{n}{b}_{n} \) for some \( {a}_{1},\ldots ,{a}_{n} \in \mathfrak{A} \) and \( {b}_{1},\ldots ,{b}_{n} \in \mathfrak{B} \) . Then \( R{a}_{1} + \cdots + R{a}_{n} \subseteq \mathfrak{A} \) ; conversely, \( a \in \mathfrak{A} \) implies \( ... | Yes |
Lemma 5.5. If \( \mathfrak{a} = {\mathfrak{p}}_{1}{\mathfrak{p}}_{2}\cdots {\mathfrak{p}}_{r} = {\mathfrak{q}}_{1}{\mathfrak{q}}_{2}\cdots {\mathfrak{q}}_{s} \) is a product of invertible prime ideals \( {\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{r} \) and \( {\mathfrak{q}}_{1},\ldots ,{\mathfrak{q}}_{s} \) of \( R \),... | Proof. By induction on \( r \) . If \( r = 0 \), then \( \mathfrak{a} = R \) and \( s = 0 \) : otherwise, \( \mathfrak{a} \subseteq {\mathfrak{q}}_{1} \subsetneqq R \) . Let \( r > 0 \) . Then \( {\mathfrak{p}}_{r} \), say, is minimal in \( \left\{ {{\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{r}}\right\} \) . Since \( {... | Yes |
Proposition 5.6. Let \( \mathfrak{a} \neq 0 \) be an ideal of a Dedekind domain \( R \) and let \( {\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{n} \) be distinct nonzero prime ideals of \( R \) . There exists a principal ideal \( \mathfrak{b} \) such that \( {e}_{\mathfrak{b}}\left( {\mathfrak{p}}_{i}\right) = {e}_{\math... | Proof. Let \( {\mathfrak{a}}_{i} = {\mathfrak{p}}_{i}^{{e}_{\mathfrak{a}}\left( {\mathfrak{p}}_{i}\right) } \) and let \( {\mathfrak{c}}_{i} = {\mathfrak{p}}_{i}^{{e}_{\mathfrak{a}}\left( {\mathfrak{p}}_{i}\right) + 1} \) . Then \( \mathfrak{a} \subseteq {\mathfrak{a}}_{i} \) and \( \mathfrak{a} \nsubseteq {\mathfrak{c... | Yes |
Proposition 5.8 (Chinese Remainder Theorem). Let \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{n} \) be ideals of a commutative ring \( R \) such that \( {\mathfrak{a}}_{i} + {\mathfrak{a}}_{j} = R \) whenever \( i \neq j \) . For every \( {x}_{1},\ldots ,{x}_{n} \in R \) there exists \( x \in R \) such that \( x + {\m... | Proof. Let \( {\mathfrak{b}}_{j} = \mathop{\prod }\limits_{{i \neq j}}{\mathfrak{a}}_{i} \) . If \( {\mathfrak{a}}_{j} + {\mathfrak{b}}_{j} \neq R \), then \( {\mathfrak{a}}_{j} + {\mathfrak{b}}_{j} \) is contained in a maximal ideal \( \mathfrak{m} \) of \( R,{\mathfrak{a}}_{k} \subseteq \mathfrak{m} \) for some \( k ... | Yes |
Theorem 6.2. For a domain \( R \) the following conditions are equivalent:\n\n(1) \( R \) is a Dedekind domain;\n\n(2) \( R \) is Noetherian and integrally closed, and every nonzero prime ideal of \( R \) is maximal;\n\n(3) \( R \) is Noetherian and \( {R}_{\mathfrak{p}} \) is a PID for every prime ideal \( \mathfrak{p... | Proof. (1) implies (2). Let \( R \) be Dedekind. If \( x \in Q\left( R\right) \) is integral over \( R \) , then \( R\left\lbrack x\right\rbrack \subseteq Q\left( R\right) \) is a finitely generated \( R \) -module, \( R\left\lbrack x\right\rbrack \) is a fractional ideal by 5.1, \( R\left\lbrack x\right\rbrack \) is i... | Yes |
Proposition 6.3. Let \( R \) be an integrally closed domain and let \( E \) be a finite separable field extension of its quotient field \( Q \) . The integral closure of \( R \) in \( E \) is contained in a finitely generated submodule of \( E \) . | Proof. By the primitive element theorem, \( E = Q\left( \alpha \right) \) for some \( \alpha \in E \) . We may assume that \( \alpha \) is integral over \( R \) : by \( {2.6},{r\alpha } \) is integral over \( R \) for some \( 0 \neq r \in R \), and \( E = Q\left( {r\alpha }\right) \) . Since \( E \) is separable over \... | Yes |
Theorem 6.5. Let \( R \) be a Dedekind domain with quotient field \( Q \) . The integral closure of \( R \) in any finite field extension of \( Q \) is a Dedekind domain. | Proof. By 3.10,6.3, the integral closure \( \bar{R} \) of \( R \) in a finite extension of \( Q \) is integrally closed and is contained in a finitely generated \( R \) -module \( M \) . Hence \( \bar{R} \) is Noetherian: its ideals are submodules of \( M \) and satisfy the ascending chain condition, by 6.4. If \( \mat... | Yes |
Corollary 6.6. In every finite field extension of \( \\mathbb{Q} \), the algebraic integers constitute a Dedekind domain. | This follows from Theorem 6.5, since \( \\mathbb{Z} \) is a Dedekind domain. Thus, the ideals of any ring of algebraic integers (over \( \\mathbb{Z} \) ) can be factored uniquely into products of positive powers of prime ideals, even though the algebraic integers themselves may lack a similar property. | Yes |
Proposition 7.1. Let \( R \) be a PID and let \( J \) be the ring of algebraic integers of a finite field extension \( E \) of \( Q = Q\left( R\right) \) . There exists a basis of \( E \) over \( Q \) that also generates \( J \) as an \( R \) -module. Hence \( J \cong {R}^{n} \) (as an \( R \) -module), where \( n = \l... | Proof. By 6.3, \( J \) is contained in a finitely generated submodule \( M \) of \( E \) . Then \( M \) is a torsion free, finitely generated \( R \) -module. In Section VIII. 6 we prove by other methods that \( J \subseteq M \) must have a finite basis \( {\beta }_{1},\ldots ,{\beta }_{n} \), which means that every el... | Yes |
Proposition 7.2. Let \( R \) be an integrally closed domain, let \( J \) be the ring of algebraic integers of a finite Galois extension \( E \) of \( Q\left( R\right) \), and let \( \mathfrak{p} \) be a prime ideal of \( R \) . There are only finitely many prime ideals of \( J \) that lie over \( \mathfrak{p} \), and t... | Proof. Let \( G = \operatorname{Gal}\left( {E : Q\left( R\right) }\right) \) . If \( \alpha \) is integral over \( R \), then \( {\sigma \alpha } \) is integral over \( R \) for every \( \sigma \in G \) ; hence the norm \( \mathrm{N}\left( \alpha \right) = \mathop{\prod }\limits_{{\sigma \in G}}{\sigma \alpha } \) is i... | Yes |
Proposition 7.4. Let \( J \) be the ring of algebraic integers of a finite Galois extension \( E \) of \( \mathbb{Q} \) and let \( {\mathfrak{P}}_{1},\ldots ,{\mathfrak{P}}_{r} \) be the prime ideals of \( J \) that lie over \( p\mathbb{Z} \), where \( p \) is prime. All \( J/{\mathfrak{P}}_{i} \) are isomorphic; \( \b... | Proof. If \( \sigma \in \operatorname{Gal}\left( {E : \mathbb{Q}}\right) \), then \( {\sigma J} = J \) ; hence all \( J/{\mathfrak{P}}_{i} \) are isomorphic, by 7.2. Moreover, \( {\mathfrak{P}}_{i} \) is maximal by 3.7; hence \( \bar{E} \cong J/{\mathfrak{P}}_{i} \) is a field.\n\nThe projections \( J \rightarrow J/{\m... | Yes |
Proposition 8.1. If \( N \) is a submodule of \( M \), then \( M \) is Artinian if and only if \( N \) and \( M/N \) are Artinian. | Proof. Assume that \( N \) and \( M/N \) are Artinian and let \( {S}_{1} \supseteq \cdots {S}_{n} \supseteq {S}_{n + 1} \supseteq \cdots \) be an infinite descending sequence of submodules of \( M \) . Then \( {S}_{1} \cap N \supseteq \) \( \cdots {S}_{n} \cap N \supseteq {S}_{n + 1} \cap N \supseteq \cdots \) is an in... | No |
Lemma 8.2. If \( \mathfrak{m} \) is a maximal ideal of a Noetherian ring \( R \), then \( R/{\mathfrak{m}}^{n} \) is an Artinian R-module, for every \( n > 0 \) . | Proof. Let \( M = {\mathfrak{m}}^{n - 1}/{\mathfrak{m}}^{n}\left( {M = R/\mathfrak{m}\text{, if}n = 1}\right) \) . We show that \( M \) is an Artinian \( R \) -module; since \( \left( {R/{\mathfrak{m}}^{n}}\right) /\left( {{\mathfrak{m}}^{n - 1}/{\mathfrak{m}}^{n}}\right) \cong R/{\mathfrak{m}}^{n - 1} \), it then foll... | Yes |
Theorem 8.3 (Krull Intersection Theorem [1928]). Let \( \mathfrak{a} \neq R \) be an ideal of a Noetherian ring \( R \) and let \( \mathbf{i} = \mathop{\bigcap }\limits_{{n > 0}}{\mathfrak{a}}^{n} \) . Then \( \mathfrak{a}\mathbf{i} = \mathbf{i} \) and \( \left( {1 - a}\right) \mathbf{i} = 0 \) for some \( a \in \mathf... | Proof. Let \( \mathfrak{q} \) be a primary ideal that contains \( \mathfrak{a}i \) and let \( \mathfrak{p} \) be its radical. Then \( {\mathfrak{p}}^{n} \subseteq \mathfrak{q} \) for some \( n \), by 1.6, and \( \mathfrak{i} \subseteq \mathfrak{q} \) : otherwise, \( \mathfrak{a} \subseteq \mathfrak{p} \), since \( \mat... | Yes |
Lemma 8.4. Let \( \mathfrak{a} \) be an ideal of a commutative ring \( R \) and let \( M \) be a finitely generated \( R \) -module. If \( \mathfrak{a}M = M \), then \( \left( {1 - a}\right) M = 0 \) for some \( a \in \mathfrak{a} \) . | Proof. \( \mathfrak{a}M \) is the set of all sums \( {a}_{1}{x}_{1} + \cdots + {a}_{n}{x}_{n} \) in which \( {a}_{1},\ldots ,{a}_{n} \in \mathfrak{a} \) and \( {x}_{1},\ldots ,{x}_{n} \in M \) . If \( M \) is generated by \( {e}_{1},\ldots ,{e}_{m} \), then every element \( x \) of \( \mathfrak{a}M \) is a sum \( {a}_{... | Yes |
Corollary 8.5 (Nakayama’s Lemma). Let \( \mathfrak{a} \) be an ideal of a commutative ring \( R \) and let \( M \) be a finitely generated \( R \) -module. If \( \mathfrak{a} \) is contained in every maximal ideal of \( R \) and \( \mathfrak{a}M = M \), then \( M = 0 \) . | This makes a fine exercise. A more general version is proved in Section IX.5. | No |
Proposition 8.6. Let \( \mathfrak{a} \neq R \) be an ideal of a Noetherian ring \( R \). There exists a prime ideal of \( R \) that is minimal over \( \mathfrak{a} \); in fact, every prime ideal that contains \( \mathfrak{a} \) contains a prime ideal that is minimal over \( \mathfrak{a} \). Moreover, there are only fin... | Proof. By 1.9, \( \mathfrak{a} \) is the intersection \( \mathfrak{a} = {\mathfrak{q}}_{1} \cap \cdots \cap {\mathfrak{q}}_{r} \) of finitely many primary ideals with radicals \( {\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{r} \). A prime ideal that contains \( \mathfrak{a} \) contains \( {\mathfrak{q}}_{1}\cdots {\mathf... | Yes |
Theorem 8.8 (Krull's Hauptidealsatz [1928]). In a Noetherian ring, a prime ideal that is minimal over a principal ideal has height at most 1. | Proof. First let \( \mathfrak{p} = \mathfrak{m} \) be the maximal ideal of a Noetherian local ring \( R \) . Assume that \( \mathfrak{m} \) is minimal over a principal ideal \( {Ra} \) of \( R \) . Then \( \operatorname{Rad}{Ra} \subseteq \mathfrak{m} \) , \( \mathfrak{m} = \operatorname{Rad}{Ra} \), and \( {\mathfrak{... | Yes |
Lemma 9.1. Let \( {\mathfrak{p}}_{0},{\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{m},{\mathfrak{q}}_{1},\ldots ,{\mathfrak{q}}_{n} \) be prime ideals of a Noetherian ring \( R \) . If \( {\mathfrak{p}}_{0} \supsetneqq {\mathfrak{p}}_{1} \supsetneqq \cdots \supsetneqq {\mathfrak{p}}_{m} \) and \( {\mathfrak{p}}_{0} \nsubs... | Proof. By induction on \( m \) . There is nothing to prove if \( m \leqq 1 \) . If \( m \geqq 2 \), then the induction hypothesis yields \( {\mathfrak{p}}_{0} \supsetneqq {\mathfrak{p}}_{1}^{\prime } \supsetneqq \cdots \supsetneqq {\mathfrak{p}}_{m - 2}^{\prime } \supsetneqq {\mathfrak{p}}_{m - 1} \) for some prime ide... | Yes |
Lemma 9.3. Let \( R \) be a domain and let \( \mathfrak{P} \) be a prime ideal of \( R\left\lbrack X\right\rbrack \) . If \( \mathfrak{P} \cap R = 0 \), then \( \mathfrak{P} \) has height at most 1. | Proof. Let \( Q \) be the quotient field of \( R \) and let \( S = R \smallsetminus \{ 0\} \), so that \( {S}^{-1}R = Q \) . Since every \( r \in R \smallsetminus 0 \) is a unit in \( Q \) and in \( Q\left\lbrack X\right\rbrack \) ,4.2 yields an injective homomorphism \( \theta : {S}^{-1}\left( {R\left\lbrack X\right\r... | Yes |
Theorem 9.4. If \( R \) is a Noetherian domain, then \( \dim R\left\lbrack X\right\rbrack = 1 + \dim R \) . | Proof. First, \( \left( X\right) \) is a prime ideal of \( R\left\lbrack X\right\rbrack \) and \( R\left\lbrack X\right\rbrack /\left( X\right) \cong R \) ; hence \( \dim R\left\lbrack X\right\rbrack \geqq 1 + \dim R\left\lbrack X\right\rbrack /\left( X\right) = \dim R + 1 \) . In particular, \( \dim R\left\lbrack X\ri... | Yes |
Proposition 10.1. Every algebraic set is the zero set of an ideal. | Proof. Let \( S \subseteq K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) and let \( \mathfrak{a} \) be the ideal generated by \( S \) . If \( f\left( x\right) = \) 0 for all \( f \in S \), then \( \left( {{u}_{1}{f}_{1} + \cdots + {u}_{m}{f}_{m}}\right) \left( x\right) = 0 \) for all \( {u}_{1}{f}_{1} + \cdots... | Yes |
Proposition 10.2. Every intersection of algebraic sets is an algebraic set. The union of finitely many algebraic sets is an algebraic set. | Proof. First, \( \mathop{\bigcap }\limits_{{i \in I}}Z\left( {S}_{i}\right) = Z\left( {\mathop{\bigcup }\limits_{{i \in I}}{S}_{i}}\right) \) . Hence \( \mathop{\bigcap }\limits_{{i \in I}}Z\left( {\mathfrak{a}}_{i}\right) = Z\left( {\mathop{\bigcup }\limits_{{i \in I}}{\mathfrak{a}}_{i}}\right) \) \( = Z\left( {\matho... | Yes |
Theorem 10.3 (Hilbert’s Nullstellensatz). Let \( K \) be a field, let \( \mathfrak{a} \) be an ideal of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \), and let \( f \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . If every zero of \( f \) in \( {\bar{K}}^{n} \) is a zero of \( \mathfrak{a} \), ... | Proof. Assume that \( \mathfrak{a} \) contains no power of \( f \) . By 1.4 there is a prime ideal \( \mathfrak{p} \) of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) that contains \( \mathfrak{a} \) but not \( f \) . Then \( R = K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /\mathfrak{p} \) is a do... | Yes |
The mappings \( \mathfrak{I} \) and \( z \) induce an order reversing one-to-one correspondence between algebraic sets in \( {\bar{K}}^{n} \) and semiprime ideals of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . | For every \( A \subseteq {\bar{K}}^{n},\mathfrak{I}\left( A\right) \) is a semiprime ideal, since \( {f}^{n}\left( x\right) = 0 \) implies \( f\left( x\right) = 0 \) . By definition, \( A \subseteq \mathfrak{Z}\left( {\mathfrak{I}\left( A\right) }\right) \) and \( \mathfrak{a} \subseteq \mathfrak{I}\left( {\mathfrak{Z}... | Yes |
Proposition 11.1. If \( K \) is a field and \( A \subseteq {\bar{K}}^{n} \) is an affine algebraic set, then \( C\left( A\right) \cong K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /\mathfrak{I}\left( A\right) \) . Hence \( C\left( A\right) \) is a commutative, finitely generated ring extension of \( K \) ; its ... | Proof. Two polynomials \( f, g \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) induce the same polynomial function on \( A \) if and only if \( f\left( x\right) - g\left( x\right) = 0 \) for all \( x \in A \), if and only if \( f - g \in \mathfrak{I}\left( A\right) \) . Therefore \( C\left( A\right) \cong K... | Yes |
Proposition 11.2. Let \( A \) be an algebraic set. For every \( x \in A \) let \( {\mathfrak{m}}_{x} = \{ f \in \) \( C\left( A\right) \mid f\left( x\right) = 0\} \) . The mapping \( x \mapsto {\mathfrak{m}}_{x} \) is a bijection of \( A \) onto the set of all maximal ideals of \( C\left( A\right) \) . | Proof. By 10.5 the mapping \( x \mapsto \Im \left( {\{ x\} }\right) = \left\{ {f \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \mid f\left( x\right) = 0}\right\} \) is a bijection of \( {\bar{K}}^{n} \) onto the set of all maximal ideals of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . By 10.6... | Yes |
Proposition 11.3. For every algebraic variety \( A,\dim A = \dim C\left( A\right) \) . | Proof. First, \( C\left( A\right) \) is Noetherian, by III.11.4. By 10.7, \( \dim A \) is the length of the longest strictly descending sequence \( {\mathfrak{P}}_{0} \supsetneqq {\mathfrak{P}}_{1} \supsetneqq \cdots \supsetneqq {\mathfrak{P}}_{m} = \mathfrak{I}\left( A\right) \) of prime ideals of \( K\left\lbrack {{X... | Yes |
Proposition 1.1. Let \( A \) be an abelian group and let \( R \) be a ring. There is a one-to-one correspondence between left \( R \) -module structures \( R \times A \rightarrow A \) on \( A \) and ring homomorphisms \( R \rightarrow {\operatorname{End}}_{\mathbb{Z}}\left( A\right) \) ; and unital left \( R \) -module... | Proof. Let \( A \) is an \( R \) -module. The action \( {\alpha }_{r} \) of \( r \in R \) on \( A,{\alpha }_{r}\left( x\right) = {rx} \) , is an endomorphism of \( A \), since \( r\left( {x + y}\right) = {rx} + {ry} \) for all \( x, y \in A \) . Then \( r \mapsto {\alpha }_{r} \) is a ring homomorphism of \( R \) into ... | Yes |
Let \( A \) be an abelian group and let \( R \) be a ring. There is a one-to-one correspondence between left \( R \) -module structures on \( A \) and unital left \( {R}^{1} \) -module structures on \( A \). | Thus every left \( R \) -module can be made uniquely into a unital left \( {R}^{1} \) -module. Consequently, the general study of modules can be limited to unital modules (over rings with identity). We do so in later sections. | No |
Proposition 1.3. Every right \( R \) -module is a left \( {R}^{\mathrm{{op}}} \) -module, and conversely. Every unital right \( R \) -module is a unital left \( {R}^{\mathrm{{op}}} \) -module, and conversely. | Proof. Let \( M \) be a right \( R \) -module. Define a left action of \( {R}^{\text{op }} \) on \( M \) by \( {rx} = {xr} \), for all \( r \in R \) and \( x \in M \) . Then \( r\left( {sx}\right) = \left( {xs}\right) r = x\left( {sr}\right) = \left( {sr}\right) x = \) \( \left( {r * s}\right) x \), and (1) follows fro... | Yes |
Proposition 1.6. Let \( M \) be a unital left \( R \) -module. The submodule of \( M \) generated by a subset \( S \) of \( M \) is the set of all linear combinations of elements of \( S \) with coefficients in \( R \) . | In particular, when \( M \) is a unital left \( R \) -module, the submodule of \( M \) generated by a finite subset \( \left\{ {{a}_{1},\ldots ,{a}_{n}}\right\} \) of elements of \( M \) is the set of all linear combinations \( {r}_{1}{a}_{1} + \cdots + {r}_{n}{a}_{n} \) with \( {r}_{1},\ldots ,{r}_{n} \in R \) ; such ... | No |
Let \( M \) be a left \( R \) -module and let \( A \) be a submodule of \( M \) . The quotient group \( M/A \) is a left \( R \) -module, in which \( r\left( {x + A}\right) = {rx} + A \) for all \( r \in R \) and \( x \in M \) . If \( M \) is unital, then \( M/A \) is unital. The projection \( x \mapsto x + A \) is a h... | Proof. Since every subgroup of the abelian group \( M \) is normal, there is a quotient group \( M/A \), in which cosets of \( A \) are added as subsets, in particular \( \left( {x + A}\right) + \left( {y + A}\right) = \left( {x + y}\right) + A \) for all \( x, y \in M \) . Since \( A \) is a submodule, the action of \... | Yes |
Proposition 2.3. If \( A \) is a submodule of a module \( M \), then \( C \mapsto C/A \) is an inclusion preserving, one-to-one correspondence between submodules of \( M \) that contain \( A \) and submodules of \( M/A \) . | Proof. We saw (Proposition I.4.10) that direct and inverse image under the projection induce a one-to-one correspondence, which preserves inclusion, between subgroups of \( M \) that contain \( A \), and subgroups of \( M/A \) . Direct and inverse image preserve submodules, by 2.1. \( ▱ \) | Yes |
Theorem 2.4 (Factorization Theorem). Let \( A \) be a left \( R \) -module and let \( B \) be a submodule of \( A \) . Every homomorphism of left \( R \) -modules \( \varphi : A \rightarrow C \) whose kernel contains \( B \) factors uniquely through the canonical projection \( \pi : A \rightarrow \) \( A/B\left( {\varp... | Proof. By the corresponding property of abelian groups (Theorem I.5.1), \( \varphi = \psi \circ \pi \) for some unique homomorphism \( \psi : A/B \rightarrow C \) of abelian groups. Then \( \psi \left( {x + B}\right) = \varphi \left( x\right) \) for all \( x \in A \) . Hence \( \psi \) is a module homomorphism. | Yes |
Theorem 2.7 (First Isomorphism Theorem). If \( A \) is a left \( R \) -module and \( B \supseteq C \) are submodules of \( A \), then\n\n\[ A/B \cong \left( {A/C}\right) /\left( {B/C}\right) ; \] | in fact, there is a unique isomorphism \( \theta : A/B \rightarrow \left( {A/C}\right) /\left( {B/C}\right) \) such that \( \theta \circ \rho = \tau \circ \pi \), where \( \pi : A \rightarrow A/C,\rho : A \rightarrow A/B \), and \( \tau : A/C \rightarrow \) \( \left( {A/C}\right) /\left( {B/C}\right) \) are the canonic... | Yes |
Theorem 2.8 (Second Isomorphism Theorem). If \( A \) and \( B \) are submodules of a left \( R \) -module, then\n\n\[ \left( {A + B}\right) /B \cong A/\left( {A \cap B}\right) \]\n\nin fact, there is an isomorphism \( \theta : A/\left( {A \cap B}\right) \rightarrow \left( {A + B}\right) /B \) unique such that \( \theta... | Proofs. In Theorem 2.6, there is by Theorem I.5.2 a unique isomorphism \( \theta \) of abelian groups such that \( \varphi = \iota \circ \theta \circ \pi \) . Then \( \theta \left( {a + \operatorname{Ker}\varphi }\right) = \varphi \left( a\right) \) for all \( a \in A \) . Therefore \( \theta \) is a module homomorphis... | No |
Proposition 2.9. A unital left \( R \) -module is cyclic if and only if it is isomorphic to \( R/L\left( { = {}_{R}R/L}\right) \) for some left ideal \( L \) of \( R \) . If \( M = {Rm} \) is cyclic, then \( M \cong R/\operatorname{Ann}\left( m\right) \), where\n\n\[\n\operatorname{Ann}\left( m\right) = \{ r \in R \mid... | Proof. Let \( M = {Rm} \) be cyclic. Then \( \varphi : r \mapsto {rm} \) is a module homomorphism of \( {}_{R}R \) onto \( M \) . By 2.6, \( M \cong R/\operatorname{Ker}\varphi \), and we see that \( \operatorname{Ker}\varphi = \operatorname{Ann}\left( m\right) \) . In particular, \( \operatorname{Ann}\left( m\right) \... | Yes |
Proposition 3.2. Let \( M \) and \( {\left( {A}_{i}\right) }_{i \in I} \) be left \( R \) -modules. For every family \( {\left( {\varphi }_{i}\right) }_{i \in I} \) of module homomorphisms \( {\varphi }_{i} : {A}_{i} \rightarrow M \) there exists a unique module homomorphism \( \varphi : {\bigoplus }_{i \in I}{A}_{i} \... | Proof. First we prove a quick lemma. | No |
Lemma 3.3. If \( x = {\left( {x}_{i}\right) }_{i \in I} \in {\bigoplus }_{i \in I}{A}_{i} \), then \( x = \mathop{\sum }\limits_{{i \in I}}{\iota }_{i}\left( {x}_{i}\right) \) ; moreover, \( x \) can be written uniquely in the form \( x = \mathop{\sum }\limits_{{i \in I}}{\iota }_{i}\left( {y}_{i}\right) \), where \( {... | Proof. Let \( x = {\left( {x}_{i}\right) }_{i \in I} \in {\bigoplus }_{i \in I}{A}_{i} \), so that \( {x}_{i} \in {A}_{i} \) for all \( i \in I \) and \( J = \{ i \in \) \( \left. {I \mid {x}_{i} \neq 0}\right\} \) is finite. Then \( y = \mathop{\sum }\limits_{{i \in I}}{\iota }_{i}\left( {x}_{i}\right) \) is defined a... | Yes |
Proposition 3.5. Let \( {\left( {M}_{i}\right) }_{i \in I} \) be left \( R \) -modules. For a left \( R \) -module \( M \) the following conditions are equivalent:\n\n(1) \( M \cong {\bigoplus }_{i \in I}{M}_{i} \) ;\n\n(2) \( M \) contains submodules \( {\left( {A}_{i}\right) }_{i \in I} \) such that \( {A}_{i} \cong ... | Proof. (1) implies (2). By 3.3, \( {\bigoplus }_{i \in I}{M}_{i} \) contains submodules \( {M}_{i}^{\prime } = \) \( {\iota }_{i}\left( {M}_{i}\right) \cong {M}_{i} \) such that every element of \( {\bigoplus }_{i \in I}{M}_{i} \) can be written uniquely as a sum \( \mathop{\sum }\limits_{{i \in I}}{a}_{i} \), where \(... | Yes |
Proposition 4.1. A family \( {\left( {e}_{i}\right) }_{i \in I} \) of elements of a [unital] left \( R \) -module \( M \) is a basis of \( M \) if and only if every element of \( M \) can be written uniquely as a linear combination \( x = \mathop{\sum }\limits_{{i \in I}}{x}_{i}{e}_{i} \) (with \( {x}_{i} \in R \) for ... | Proof. By 1.6, \( {\left( {e}_{i}\right) }_{i \in I} \) generates \( M \) if and only if every element of \( M \) is a linear combination \( x = \mathop{\sum }\limits_{{i \in I}}{x}_{i}{e}_{i} \) . If this expression is unique for all \( x \), then \( \mathop{\sum }\limits_{{i \in I}}{r}_{i}{e}_{i} = 0 \) implies \( {r... | Yes |
Proposition 4.2. Let \( M \) be a [unital] left \( R \) -module.\n\n(1) If \( {\left( {e}_{i}\right) }_{i \in I} \) is a basis of \( M \), then there is an isomorphism \( M \cong {\bigoplus }_{i \in I}{}_{R}R \) that assigns to every element of \( M \) its coordinates in the basis \( {\left( {e}_{i}\right) }_{i \in I} ... | Proof. (1). If \( x = \mathop{\sum }\limits_{{i \in I}}{x}_{i}{e}_{i} \) in \( M \), then \( {\left( {x}_{i}\right) }_{i \in I} \in {\bigoplus }_{i \in I}{}_{R}R \) by 4.1. The mapping \( {\left( {x}_{i}\right) }_{i \in I} \mapsto x \) is bijective, by 4.1; the inverse bijection \( x \mapsto {\left( {x}_{i}\right) }_{i... | Yes |
Every free left \( R \) -module \( M \) has a right \( R \) -module structure, which depends on the choice of a basis \( {\left( {e}_{i}\right) }_{i \in I} \) of \( M \), in which \( \left( {\mathop{\sum }\limits_{{i \in I}}{x}_{i}{e}_{i}}\right) r \) \( = \mathop{\sum }\limits_{{i \in I}}{x}_{i}\;r\;{e}_{i}\;. | Proof. First, \( {\bigoplus }_{i \in I}{}_{R}R \) is a right \( R \) -module, on which \( R \) acts componentwise, \( \left( {\left( {x}_{i}\right) }_{i \in I}\right) r = {\left( {x}_{i}r\right) }_{i \in I} \) . If \( {\left( {e}_{i}\right) }_{i \in I} \) is a basis of \( M \), then the isomorphism \( {\bigoplus }_{i \... | Yes |
Proposition 4.5. Let \( X = {\left( {e}_{i}\right) }_{i \in I} \) be a basis of a left \( R \) -module \( M \) . Every mapping \( f \) of \( X \) into a left \( R \) -module \( N \) extends uniquely to a module homomorphism \( \varphi \) of \( M \) into \( N \), namely \( \varphi \left( {\mathop{\sum }\limits_{{i \in I... | Proof. If \( \varphi \) extends \( f \) to \( M \), then \( \varphi \left( {\mathop{\sum }\limits_{{i \in I}}{x}_{i}{e}_{i}}\right) = \mathop{\sum }\limits_{{i \in I}}{x}_{i}\varphi \left( {e}_{i}\right) = \) \( \mathop{\sum }\limits_{{i \in I}}{x}_{i}f\left( {e}_{i}\right) \) for all \( x = \mathop{\sum }\limits_{{i \... | Yes |
Proposition 4.7. Let \( A \) and \( B \) be free right \( R \) -modules with bases \( {e}_{1},\ldots ,{e}_{n} \) and \( {f}_{1},\ldots ,{f}_{m} \) respectively. There is a one-to-one correspondence between module homomorphisms of \( A \) into \( B \) and \( m \times n \) matrices with entries in \( R \) . | Proof. Let \( \varphi : A \rightarrow B \) be a module homomorphism. Then \( \varphi \left( {e}_{j}\right) = \mathop{\sum }\limits_{i}{f}_{i}{r}_{ij} \) for some unique \( {r}_{ij} \in R, i = 1,\ldots, m, j = 1,\ldots, n \) . This defines an \( m \times n \) matrix \( M\left( \varphi \right) = \left( {r}_{ij}\right) \)... | Yes |
Proposition 4.8. If \( \varphi ,\psi : A \rightarrow B \) and \( \chi : B \rightarrow C \) are homomorphisms of free right \( R \) -modules with finite bases, then, in any given bases of \( A, B \), and \( C \) , \( M\left( {\varphi + \psi }\right) = M\left( \varphi \right) + M\left( \psi \right) \) and \( M\left( {\ch... | Proof. We prove the last equality. Let \( {e}_{1},\ldots ,{e}_{n},{f}_{1},\ldots ,{f}_{m},{g}_{1},\ldots ,{g}_{\ell } \) be bases of \( A, B \), and \( C \), respectively. Let \( \varphi \left( {e}_{j}\right) = \mathop{\sum }\limits_{i}{f}_{i}{r}_{ij} \) and \( \chi \left( {f}_{i}\right) = \) \( \mathop{\sum }\limits_{... | Yes |
Proposition 4.11. Let \( M \) be a free left \( R \) -module with an infinite basis. All bases of \( M \) have the same number of elements. | Proof. The proof is a cardinality argument. Let \( X \) and \( Y \) be bases of \( M \) . Every \( x \in X \) is a linear combination of elements of a finite subset \( {Y}_{x} \) of \( Y \) . Hence the submodule of \( M \) generated by \( \mathop{\bigcup }\limits_{{x \in X}}{Y}_{x} \) contains \( X \) and is all of \( ... | Yes |
Proposition 4.12. Let \( M \) be a free left \( R \) -module. If \( R \) is commutative, then all bases of \( M \) have the same number of elements. | Proof. The proof uses quotient rings of \( R \) . Let \( {\left( {e}_{i}\right) }_{i \in I} \) be a basis of \( M \) and let \( \mathfrak{a} \) be an ideal of \( R \) . Then \( \mathfrak{a}M \) is generated by products \( {rx} \) with \( r \in \mathfrak{a} \) , \( x \in M \), whose coordinates are all in \( \mathfrak{a... | Yes |
Lemma 5.1. Let \( X \) be a linearly independent subset of a vector space \( V \). If \( y \in V \smallsetminus X \), then \( X \cup \{ y\} \) is linearly independent if and only if \( y \) is not a linear combination of elements of \( X \). | Proof. If \( X \cup \{ y\} \) is not linearly independent, then \( {r}_{y}y + \mathop{\sum }\limits_{{x \in X}}{r}_{x}x = 0 \) , where \( {r}_{x},{r}_{y} \) are not all zero. Then \( {r}_{y} \neq 0 \) : otherwise, \( X \) is not linearly independent. Therefore \( y \) is a linear combination of elements of \( X \). \( ... | No |
Theorem 5.2. Every vector space has a basis. | Proof. Let \( V \) be a left \( D \) -module, where \( D \) is a division ring. The union \( X = \mathop{\bigcup }\limits_{{i \in I}}{X}_{i} \) of a nonempty chain \( {\left( {X}_{i}\right) }_{i \in I} \) of linearly independent subsets of \( V \) is linearly independent: if \( \mathop{\sum }\limits_{{x \in X}}{r}_{x}x... | Yes |
Lemma 5.4. Let \( V \) be a vector space and let \( X, Y \) be bases of \( V \) . For every \( x \in X \) there exists \( y \in Y \) such that \( \left( {X\smallsetminus \{ x\} }\right) \cup \{ y\} \) is a basis of \( V \) . | Proof. If \( x \in Y \), then \( y = x \) serves. Assume that \( x \notin Y \), and let \( S \) be the subspace (submodule) of \( V \) generated by \( X \smallsetminus \{ x\} \) . If \( Y \subseteq S \), then \( S = V, x \in S \) , and \( x \) is a linear combination of elements of \( X \smallsetminus \{ x\} \), causin... | Yes |
Theorem 5.5. All bases of a vector space have the same number of elements. | Proof. Let \( X \) and \( Y \) be bases of a vector space \( V \) . If \( X \) is infinite, or if \( Y \) is infinite, then \( \left| X\right| = \left| Y\right| \), by 4.11. Now, assume that \( X \) and \( Y \) are finite. Repeated applications of Lemma 5.4 construct a basis of \( V \) in which every element of \( X \)... | Yes |
Proposition 6.4. If \( R \) is a domain, then the torsion elements of an \( R \) -module \( M \) constitute a submodule \( T\left( M\right) \) of \( M \), and \( M/T\left( M\right) \) is torsion-free. | The proof is an exercise. | No |
Proposition 6.5. Let \( R \) be a principal ideal domain and let \( P \) be a set of representative prime elements of \( R \) . Every torsion \( R \) -module \( M \) is a direct sum \( M = {\bigoplus }_{p \in P}M\left( p\right) \), where \[ M\left( p\right) = \left\{ {x \in M \mid {p}^{k}x = 0\text{ for some }k > 0}\ri... | Proof. Let \( x \in M \) . If \( {ax} = 0 \) and \( a = {bc} \neq 0 \), where \( b, c \in R \) are relatively prime, then \( 1 = {ub} + {vc} \) for some \( u, v \in R \) and \( x = {ubx} + {vcx} \), where \( c\left( {ubx}\right) = 0 \) and \( b\left( {vcx}\right) = 0 \) . If now \( a = u{p}_{1}^{{k}_{1}}\cdots {p}_{r}^... | Yes |
Proposition 6.6. Let \( R \) be a principal ideal domain and let \( M \cong R/{Ra} \) be a cyclic \( R \) -module, where \( a = u{p}_{1}^{{k}_{1}}\cdots {p}_{r}^{{k}_{r}} \in R \) is the product of a unit and positive powers of distinct representative prime elements of \( R \) . Then \( M\left( {p}_{i}\right) \cong R/R... | Proof. We have \( {ax} = 0 \) in \( M = R/{Ra} \), for all \( x \in M \) . Let \( x \in M\left( p\right) \), with \( {p}^{k}x = 0 \) . If \( p \neq {p}_{1},\ldots ,{p}_{r} \), then \( a \) and \( {p}^{k} \) are relatively prime, \( 1 = {ua} + v{p}^{k} \) for some \( u, v \in R \), and \( x = {uax} + v{p}^{k}x = 0 \) . ... | Yes |
Lemma 6.7. Let \( R \) be a PID and let \( p \in R \) be a prime element. If \( M \cong R/R{p}^{{k}_{1}} \oplus \cdots \oplus R/R{p}^{{k}_{t}} \) and \( 0 < {k}_{1} \leqq \cdots \leqq {k}_{t} \), then the numbers \( t \) and \( {k}_{1},\ldots ,{k}_{t} \) are uniquely determined by \( M \) . | Proof. If \( A \) is an \( R \) -module, then \( A/{pA} \) is an \( R/{Rp} \) -module, since \( \left( {Rp}\right) \left( {A/{pA}}\right) = 0 \), and \( A/{pA} \) is a vector space over the field \( R/{Rp} \) . Moreover, \( p\left( {A \oplus B}\right) = {pA} \oplus {pB} \), so that \( \left( {A \oplus B}\right) /p\left... | Yes |
Proposition 7.1. Let \( V \) be a vector space over a field \( K \) . Every linear transformation \( T : V \rightarrow V \) induces a \( K\left\lbrack X\right\rbrack \) -module structure on \( V \), in which \( \left( {{a}_{0} + {a}_{1}X + \cdots + {a}_{n}{X}^{n}}\right) v = {a}_{0}v + {a}_{1}{Tv} + \cdots + {a}_{n}{T}... | Proof. The ring \( {\operatorname{End}}_{K}\left( V\right) \) of all linear transformations of \( V \) comes with a homomorphism \( \sigma : K \rightarrow {\operatorname{End}}_{K}\left( V\right) \) that assigns to \( a \in K \) the scalar transformation \( \sigma \left( a\right) : v \mapsto {av} \) . Now, \( T \) commu... | Yes |
Proposition 7.2. In the \( K\left\lbrack X\right\rbrack \) -module structure on \( V \) induced by \( T \), a submodule of \( V \) is a subspace \( S \) of \( V \) such that \( {TS} \subseteq S \), and then the \( K\left\lbrack X\right\rbrack \) -module structure on \( S \) induced by \( V \) coincides with the \( K\le... | Proof. Either way, \( {Xs} = {Ts} = {T}_{\mid S}s \) for all \( s \in S \) . (Then \( f{\left( T\right) }_{\mid S} = f\left( {T}_{\mid S}\right) \) for all \( f \in K\left\lbrack X\right\rbrack \) .) \( ▱ \) | Yes |
Proposition 7.3. If \( V \) is finite-dimensional, then the ideal \( \operatorname{Ann}\left( V\right) \) of \( K\left\lbrack X\right\rbrack \) is generated by a unique monic polynomial \( {m}_{T} \in K\left\lbrack X\right\rbrack \) ; then \( {m}_{T}\left( T\right) = 0 \), and \( f\left( T\right) = 0 \) if and only if ... | Proof. \( \operatorname{Ann}\left( V\right) = \{ f \in K\left\lbrack X\right\rbrack \mid f\left( T\right) = 0\} \) is an ideal of \( K\left\lbrack X\right\rbrack \) . Moreover, \( \operatorname{Ann}\left( V\right) \neq 0 \) : \( \operatorname{Ann}\left( V\right) \) is the kernel of the homomorphism \( \varphi : f \maps... | Yes |
Lemma 7.5. If \( {\dim }_{K}V = n \) is finite and \( V = K\left\lbrack X\right\rbrack \) e is a cyclic \( K\left\lbrack X\right\rbrack \) -module, then \( \deg {m}_{T} = n \) and \( e,{Te},\ldots ,{T}^{n - 1}e \) is a basis of \( V \) over \( K \) . | Proof. Let \( m = \deg {m}_{T} \) . Every element of \( V \) has the form \( f\left( X\right) e \) for some \( f \in K\left\lbrack X\right\rbrack \) . Now, \( f = {m}_{T}q + r \), where \( \deg r < m \), and \( f\left( X\right) e = r\left( X\right) e \) . Hence every element of \( V \) is a linear combination of \( e,{... | Yes |
Lemma 7.6. If \( {\dim }_{K}V = n \) is finite, \( V \) is a cyclic \( K\left\lbrack X\right\rbrack \) -module, and \( {m}_{T} = \) \( {\left( X - \lambda \right) }^{m} \) for some \( \lambda \in K \), then \( m = n \) and \( V \) has a basis \( {e}_{1},\ldots ,{e}_{n} \) over \( K \) such that \( T{e}_{1} = \lambda {e... | Proof. First, \( m = n \), by 7.5. Let \( V = K\left\lbrack X\right\rbrack e \), where \( e \in V \) . Let \( {e}_{i} = \) \( {\left( T - \lambda \right) }^{n - i}e \) for all \( i = 1,2,\ldots, n \) . By the binomial theorem, \( {e}_{n - i} = {\left( T - \lambda \right) }^{i}e \) is a linear combination of \( {T}^{i}e... | Yes |
Theorem 7.10 (Cayley-Hamilton). If \( V \) is a finite-dimensional vector space over a field \( K \), then \( {c}_{T}\left( T\right) = 0 \) for every linear transformation \( T \) of \( V \) . | Proof. If \( K \) is algebraically closed, then \( V \) has a basis in which the matrix of \( T \) is in Jordan form. Then \( {c}_{T}\left( X\right) = {\left( X - {\lambda }_{1}\right) }^{{n}_{1}}\cdots {\left( X - {\lambda }_{r}\right) }^{{n}_{r}} \), where \( {\lambda }_{1},\ldots ,{\lambda }_{r} \) are the distinct ... | Yes |
Proposition 8.2. If \( N \) is a submodule of a module \( M \), then \( M \) is Noetherian if and only if \( N \) and \( M/N \) are Noetherian. | Proof. Let \( N \) and \( M/N \) be Noetherian and let \( {A}_{1} \subseteq {A}_{2} \subseteq \cdots \subseteq {A}_{n} \subseteq \) \( {A}_{n + 1} \subseteq \cdots \) be an infinite ascending sequence of submodules of \( M \) . Then\n\n\[ \n{A}_{1} \cap N \subseteq {A}_{2} \cap N \subseteq \cdots \subseteq {A}_{n} \cap... | No |
Proposition 8.3. If \( R \) is left Noetherian, then every finitely generated left \( R \) -module is Noetherian. | Proof. A finitely generated free left \( R \) -module \( F \) is the direct sum \( F = {R}^{\left( n\right) } \) of \( n \) copies of \( {}_{R}R \) and is Noetherian, by induction on \( n : {R}^{\left( 1\right) } = {}_{R}R \) is Noetherian, and if \( {R}^{\left( n\right) } \) is Noetherian, then so is \( {R}^{\left( n ... | Yes |
Proposition 8.8. The following conditions on a module \( M \) are equivalent:\n\n(1) \( M \) is both Noetherian and Artinian;\n\n(2) every chain of submodules of \( M \) is finite;\n\n(3) \( M \) is of finite length ( \( M \) has a composition series).\n\nThen (4) all chains of submodules of \( M \) have length at most... | Proof. (1) implies (2). Suppose that \( M \) is Artinian and contains an infinite chain \( C \) of submodules. Since \( M \) is Artinian, \( C \) has a minimal element \( {A}_{1} \) , which is in fact the least element of \( C \) since \( C \) is a chain. Then \( C \smallsetminus \left\{ {A}_{1}\right\} \) is an infini... | Yes |
Proposition 9.1. Let \( < \) be a monomial order on \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . A monomial order on \( F \) is defined by \( {X}^{a}{\varepsilon }_{i} < {X}^{b}{\varepsilon }_{j} \) if and only if either \( {X}^{a} < {X}^{b} \) in \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack ... | The proof is an exercise. Note that, in any monomial order, \( \alpha \geqq \beta \) whenever \( \alpha = {X}^{m}\beta \) is a multiple of \( \beta \) . | No |
Proposition 9.2. Every monomial order on \( F \) satisfies the descending chain condition. | This is proved like Proposition III.12.2, using the following lemma. | No |
Lemma 9.3. An submodule of \( F \) that is generated by a set \( S \) of monomials is generated by a finite subset of \( S \) . | Proof. First, \( F \) is a Noetherian module, by 8.3. Hence the submodule \( M \) generated by \( S \) is generated by finitely many \( {f}_{1},\ldots ,{f}_{t} \in F \) . Every nonzero term of \( {f}_{i} \) is a multiple of some \( \sigma \in S \) . Let \( T \) be the set of all \( \sigma \in S \) that divide a nonzero... | Yes |
Proposition 9.4. Let \( f,{g}_{1},\ldots ,{g}_{k} \in F,{g}_{1},\ldots ,{g}_{k} \neq 0 \) . There exist \( {q}_{1},\ldots ,{q}_{k} \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) and \( r \in F \) such that \[ f = {q}_{1}{g}_{1} + \cdots + {q}_{k}{g}_{k} + r \] \( \operatorname{ldm}\left( {{q}_{i}{g}_{i}}\r... | This is proved like Proposition III.12.4. Division in \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is a particular case. | No |
Proposition 9.5. Let \( M \) be a submodule of \( F \) . If \( {g}_{1},\ldots ,{g}_{k} \in M \) and \( \operatorname{ldm}{g}_{1},\ldots ,\operatorname{ldm}{g}_{k} \) generate \( \operatorname{ldm}M \), then \( {g}_{1},\ldots ,{g}_{k} \) is a Gröbner basis of \( M \) . | Proof. We need only show that \( {g}_{1},\ldots ,{g}_{k} \) generate \( M \) . Let \( f \in M \) . By 9.4, \( f = \mathop{\sum }\limits_{i}{q}_{i}{g}_{i} + r \), where none of \( \operatorname{ldm}{g}_{1},\ldots ,\operatorname{ldm}{g}_{k} \) divides a nonzero term of \( r \) . Then \( r = 0 \) : otherwise, \( r \in M,\... | Yes |
Proposition 9.6. Let \( {g}_{1},\ldots ,{g}_{k} \) be a Gröbner basis of a submodule \( M \) of \( F \) . All divisions of \( f \in F \) by \( {g}_{1},\ldots ,{g}_{k} \) yield the same remainder \( r \), and \( f \in M \) if and only if \( r = 0 \) . | Proof. This is proved like Proposition III.12.5. Let \( r \) be the remainder in a division of \( f \) by \( {g}_{1},\ldots ,{g}_{k} \) . If \( r = 0 \), then \( f \in M \) . Conversely, if \( f \in M \) , then \( r = 0 \) : otherwise, \( r \in M,\operatorname{ldm}r \in \operatorname{ldm}M \) is a linear combination of... | Yes |
Proposition 9.8 (Buchberger’s Algorithm). Let \( {g}_{1},\ldots ,{g}_{k} \neq 0 \) generate a submodule \( M \) of \( F \) . Compute a sequence \( B \) of elements of \( F \) as follows. Start with \( B = {g}_{1},\ldots ,{g}_{k} \) . Compute all polynomials \( {r}_{i, j} \) with \( i < j \) of \( B \) as in 9.7 and add... | Proof. Let \( L \) be the submodule generated by \( \operatorname{ldm}{g}_{1},\ldots ,\operatorname{ldm}{g}_{k} \) . Since \( {r}_{i, j} \) is the remainder of some \( {d}_{i, j} \in M \) in a division by \( {g}_{1},\ldots ,{g}_{k} \) we have \( {r}_{i, j} \in M \), but, if \( {r}_{i, j} \neq 0 \), then no \( \operator... | Yes |
Lemma 9.9. The syzygy submodule \( S \) of monomials \( {\alpha }_{1},\ldots ,{\alpha }_{k} \in F \) is generated by all \( {s}_{i, j} = \left( {{\lambda }_{ij}/{\alpha }_{i}}\right) {\zeta }_{i} - \left( {{\lambda }_{ij}/{\alpha }_{j}}\right) {\zeta }_{j} \), where \( {\lambda }_{ij} = \operatorname{lcm}\left( {{\alph... | Proof. First, \( {s}_{i, j} \in S \), since \( \left( {{\lambda }_{ij}/{\alpha }_{i}}\right) {\alpha }_{i} - \left( {{\lambda }_{ij}/{\alpha }_{j}}\right) {\alpha }_{j} = 0 \) . For every monomial \( \gamma \) of \( F \) let \[ {S}_{\gamma } = \left\{ {\mathop{\sum }\limits_{h}{a}_{h}{X}^{{m}_{h}}{\zeta }_{h} \in S \mi... | Yes |
Proposition 1.1. A left R-module is simple if and only if it is isomorphic to \( {}_{R}R/L \) for some maximal left ideal \( L \) of \( R \) . | Proof. A simple module is necessarily cyclic, generated by any nonzero element. If \( M \) is cyclic, \( M \cong {}_{R}R/L \) for some left ideal \( L \) of \( R \), the submodules of \( M \) correspond to the submodules \( L \subseteq {L}^{\prime } \subseteq R \) of \( {}_{R}R \) ; hence \( M \) is simple if and only ... | Yes |
Proposition 1.2 (Schur’s lemma). If \( S \) and \( T \) are a simple left \( R \) -modules, then every homomorphism of \( S \) into \( T \) is either 0 or an isomorphism. In particular, \( {\operatorname{End}}_{R}\left( S\right) \) is a division ring. | Proof. If \( \varphi : S \rightarrow T \) is not 0, then \( \operatorname{Ker}\varphi \subsetneqq S \) and \( 0 \subsetneqq \operatorname{Im}\varphi \subseteq T \), whence \( \operatorname{Ker}\varphi = 0,\operatorname{Im}\varphi = T \), and \( \varphi \) is an isomorphism. In particular, every nonzero endomorphism of ... | Yes |
Proposition 1.3. If \( S \) is a simple left \( R \) -module, \( L \) is a minimal left ideal of \( R \), and \( {LS} \neq 0 \), then \( S \cong L \) . If \( R \) is a sum of minimal left ideals \( {\left( {L}_{i}\right) }_{i \in I} \), then every simple left \( R \) -module is isomorphic to some \( {L}_{i} \) . | Proof. If \( {LS} \neq 0 \), then \( {Ls} \neq 0 \) for some \( s \in S,\ell \mapsto \ell s \) is a nonzero homomorphism of \( L \) into \( S \), and \( L \cong S \) by 1.2. Now, let \( R = \mathop{\sum }\limits_{{i \in I}}{L}_{i} \) be a sum of minimal left ideals \( {L}_{i} \) . If \( {L}_{i}S = 0 \) for all \( i \),... | Yes |
Proposition 1.4. Every two-sided ideal of \( {M}_{n}\left( R\right) \) has the form \( {M}_{n}\left( I\right) \) for some unique two-sided ideal \( I \) of \( R \) . | Proof. If \( I \) is an ideal of \( R \), then \( {M}_{n}\left( I\right) \), which consists of all matrices with entries in \( I \), is an ideal of \( {M}_{n}\left( R\right) \) . Conversely, let \( J \) be an ideal of \( {M}_{n}\left( R\right) \) . Let \( I \) be the set of all \( \left( {1,1}\right) \) entries of matr... | Yes |
Proposition 1.6. For every ring \( R,{M}_{n}{\left( R\right) }^{\text{op }} \cong {M}_{n}\left( {R}^{\text{op }}\right) \) . | Proof. Matrices \( A, B \) over a field can be transposed, and then \( {\left( AB\right) }^{t} = {B}^{t}{A}^{t} \) . Matrices \( A, B \) over an arbitrary ring \( R \) can also be transposed, and if \( {A}^{t},{B}^{t} \) are regarded as matrices over \( {R}^{\mathrm{{op}}} \), then \( {\left( AB\right) }^{t} = {B}^{t}{... | Yes |
Proposition 1.7. If \( D \) is a division ring, then \( {M}_{n}\left( D\right) \) is a direct sum of \( n \) minimal left ideals; hence \( {M}_{n}\left( D\right) \) is left Noetherian and left Artinian. | Proof. Let \( {L}_{i} \) be the set of all \( n \times n \) matrices \( M \in {M}_{n}\left( D\right) \) whose entries are all 0 outside the \( i \) th column. We see that \( {L}_{i} \) is a left ideal of \( {M}_{n}\left( D\right) \) and that \( {M}_{n}\left( D\right) = {\bigoplus }_{i}{L}_{i} \) . Readers will verify t... | No |
If \( R = {M}_{n}\left( D\right) \), where \( D \) is a division ring, then all simple left \( R \) -modules are isomorphic; every simple left \( R \) -module \( S \) is faithful and has dimension \( n \) over \( D \) ; moreover, \( {\operatorname{End}}_{R}\left( S\right) \cong {D}^{\text{op }} \) . | Identify \( D \) with the subring of \( R = {M}_{n}\left( D\right) \) that consists of all scalar matrices (scalar multiples of the identity matrix). Then every left \( R \) -module becomes a \( D \) -module. In particular, every left ideal of \( R \) is a \( D \) -module, on which \( D \) acts by scalar multiplication... | Yes |
Proposition 1.9. Let \( D \) and \( {D}^{\prime } \) be division rings. If \( {M}_{n}\left( D\right) \cong {M}_{{n}^{\prime }}\left( {D}^{\prime }\right) \), then \( n = {n}^{\prime } \) and \( D \cong {D}^{\prime } \) . | Proof. Let \( R = {M}_{n}\left( D\right) \) and \( {R}^{\prime } = {M}_{{n}^{\prime }}\left( {D}^{\prime }\right) \) . By \( {1.7},{}_{R}R \) is of length \( n \) ; therefore \( {}_{{R}^{\prime }}{R}^{\prime } \) is of length \( n \) and \( n = {n}^{\prime } \) . If \( \theta : {R}^{\prime } \rightarrow R \) is an isom... | Yes |
Proposition 2.3. Let \( {A}_{1},\ldots ,{A}_{m},{B}_{1},\ldots ,{B}_{n},{C}_{1},\ldots ,{C}_{p} \) be left \( R \) -modules. There is a one-to-one correspondence between module homomorphisms \( \varphi : {\bigoplus }_{j}{B}_{j} \rightarrow {\bigoplus }_{i}{A}_{i} \) and \( m \times n \) matrices \( \left( {\varphi }_{i... | Proof. Let \( {\iota }_{j} : {B}_{j} \rightarrow {\bigoplus }_{j}{B}_{j} \) be the \( j \) th injection and let \( {\pi }_{i} : {\bigoplus }_{i}{A}_{i} = \) \( \mathop{\prod }\limits_{i}{A}_{i} \rightarrow {A}_{i} \) be the \( i \) th projection. By the universal properties of \( {\bigoplus }_{j}{B}_{j} \) and \( \math... | Yes |
Proposition 2.5. Let \( M \) be a left \( R \) -module. If \( M \) is a direct sum of finitely many simple submodules, then \( {\operatorname{End}}_{R}\left( M\right) \) is isomorphic to the direct product of finitely many rings of matrices \( {M}_{{n}_{i}}\left( {D}_{i}\right) \) over division rings \( {D}_{i} \) . | Proof. Let \( M = {\bigoplus }_{k}{S}_{k} \) be the direct sum of finitely many simple submodules \( {S}_{k} \) . Grouping together the modules \( {S}_{j} \) that are isomorphic to each other rewrites \( M \) as a direct sum \( M \cong {S}_{1}^{{n}_{1}} \oplus \cdots \oplus {S}_{r}^{{n}_{r}} \), where \( {n}_{i} > 0 \)... | Yes |
Proposition 2.6. Let \( R \) be a ring [with identity]. If \( R \) is isomorphic to a direct product \( {R}_{1} \times \cdots \times {R}_{n} \) of finitely many rings, then \( R \) has two-sided ideals \( {A}_{1},\ldots ,{A}_{n} \neq 0 \) such that \( {A}_{i} \cong {R}_{i} \) (as a ring) for all \( i, R = \mathop{\sum ... | Proof. If \( R = {R}_{1} \times \cdots \times {R}_{n} \), then the sets\n\n\[ \n{A}_{i} = \left\{ {\left( {{x}_{1},\ldots ,{x}_{n}}\right) \in {R}_{1} \times \cdots \times {R}_{n} \mid {x}_{j} = 0\text{ for all }j \neq i}\right\} \n\]\n\nhave all the properties in the statement.\n\nConversely, let \( {\left( {R}_{i}\ri... | Yes |
In a direct product \( R = {R}_{1} \times \cdots \times {R}_{n} \) of rings [with identity], every left (right, two-sided) ideal of \( {R}_{i} \) is a left (right, two-sided) ideal of \( R \) ; every minimal left (right, two-sided) ideal of \( {R}_{i} \) is a minimal left (right, two-sided) ideal of \( R \) ; every min... | Proof. If \( L \) is a left ideal of \( {R}_{j} \), then \( {RL} = \left( {\mathop{\sum }\limits_{i}{R}_{i}}\right) L = \mathop{\sum }\limits_{i}{R}_{i}L = {R}_{j}L \subseteq L \) and \( L \) is a left ideal of \( R \) . Conversely, a left ideal of \( R \) that is contained in \( {R}_{i} \) is a left ideal of \( {R}_{j... | Yes |
If \( R = {R}_{1} \times \cdots \times {R}_{n} \) is a direct product of rings, then the simple left \( R \) -modules are the simple left \( {R}_{i} \) -modules of the rings \( {R}_{i} \) . | Proof. Let \( S \) be a simple \( R \) -module. Then \( \mathop{\sum }\limits_{i}{R}_{i}S = {RS} = S \neq 0,{R}_{i}S \neq 0 \) for some \( i,{R}_{i}S = R{R}_{i}S \) is a submodule of \( S \), and \( {R}_{i}S = S \) . Let \( {e}_{i} \) be the identity element of \( {R}_{i} \) . If \( j \neq i \), then \( {R}_{j}S = {R}_... | Yes |
Proposition 3.1. A ring \( R \) is semisimple if and only if the module \( {}_{R}R \) is semisimple, if and only if \( R \) is a direct sum of minimal left ideals, and then \( R \) is a direct sum of finitely many minimal left ideals. | Proof. If every left \( R \) -module is semisimple, then so is \( {}_{R}R \) . Conversely, if \( {}_{R}R \) is semisimple, then, by 2.2, every free left \( R \) -module \( F \cong {\bigoplus }_{R}R \) is semisimple, and every left \( R \) -module is semisimple, since it is isomorphic to a quotient module of a free modu... | Yes |
Proposition 3.2. A direct product of finitely many semisimple rings is a semisimple ring. | Proof. If \( {R}_{1},\ldots ,{R}_{n} \) are semisimple, then every \( {R}_{i} \) is a sum of minimal left ideals of \( {R}_{i} \), which by 2.7 are minimal left ideals of \( R = {R}_{1} \times \cdots \times {R}_{n} \) ; hence \( R \) is a sum of minimal left ideals. \( ▱ \) | Yes |
Theorem 3.3 (Artin-Wedderburn). A ring \( R \) is semisimple if and only if it is isomorphic to a direct product \( {M}_{{n}_{1}}\left( {D}_{1}\right) \times \cdots \times {M}_{{n}_{s}}\left( {D}_{s}\right) \) of finitely many matrix rings over division rings \( {D}_{1},\ldots ,{D}_{s} \) . | Proof. If \( R \) is semisimple, then \( {R}^{\mathrm{{op}}} \cong {\operatorname{End}}_{R}\left( {{}_{R}R}\right) \cong {M}_{{n}_{1}}\left( {D}_{1}\right) \times \cdots \times \) \( {M}_{{n}_{s}}\left( {D}_{s}\right) \) for some division rings \( {D}_{1},\ldots ,{D}_{s} \), by VIII.4.10 and 2.5. Hence\n\n\[ R \cong {M... | Yes |
Corollary 3.4. A ring \( R \) is semisimple if and only if \( {R}^{\text{op }} \) is semisimple. | Proof. If \( R \cong {M}_{{n}_{1}}\left( {D}_{1}\right) \times \cdots \times {M}_{{n}_{r}}\left( {D}_{s}\right) \) is semisimple, where \( {D}_{1},\ldots ,{D}_{s} \) are division rings, then \( {R}^{\mathrm{{op}}} \cong {M}_{{n}_{1}}\left( {D}_{1}^{\mathrm{{op}}}\right) \times \cdots \times {M}_{{n}_{s}}\left( {D}_{s}^... | Yes |
Corollary 3.5. Every semisimple ring is left Noetherian, left Artinian, right Noetherian, and right Artinian. | Proof. By 3.1, \( {}_{R}R \) is a finite direct sum \( {}_{R}R = {L}_{1} \oplus \cdots \oplus {L}_{n} \) of finitely many simple submodules. Hence \( {}_{R}R \) has a composition series\n\n\[ 0 \subsetneqq {L}_{1} \subsetneqq {L}_{1} \oplus {L}_{2} \subsetneqq \cdots \subsetneqq {L}_{1} \oplus \cdots \oplus {L}_{n} = R... | Yes |
Proposition 3.6. If \( R \cong {M}_{{n}_{1}}\left( {D}_{1}\right) \times \cdots \times {M}_{{n}_{s}}\left( {D}_{s}\right) \) is a semisimple ring, where \( {D}_{1},\ldots ,{D}_{s} \) are division rings, then every simple left \( R \) -module is isomorphic to a minimal left ideal of some \( {M}_{{n}_{i}}\left( {D}_{i}\r... | Proof. By 2.6,3.3, \( R = {R}_{1} \times \cdots \times {R}_{s} \), where \( {R}_{i} \cong {M}_{{n}_{i}}\left( {D}_{i}\right) \) . By 2.7, \( L \) is a minimal left ideal of \( R \) if and only if \( L \) is a minimal left ideal of some \( {R}_{i} \) . By 1.8, all minimal left ideals of \( {R}_{i} \) are isomorphic as \... | Yes |
Corollary 3.7. Let \( R \) be semisimple and let \( {S}_{1},\ldots ,{S}_{s} \) are, up to isomorphism, all the distinct simple left R-modules (so that every simple left R-module is isomorphic to exactly one of \( {S}_{1},\ldots ,{S}_{s} \) ). Every left \( R \) -module is isomorphic to a direct sum \( {S}_{1}^{{m}_{1}}... | Proof. Up to isomorphism, \( R = {R}_{1} \times \cdots \times {R}_{s} \), where \( {R}_{i} \cong {M}_{{n}_{i}}\left( {D}_{i}\right) \) , and \( {S}_{1},\ldots ,{S}_{s} \) can be numbered so that \( {S}_{i} \) is isomorphic to a minimal left ideal of \( {R}_{i} \) . Then \( {R}_{j}{S}_{i} = 0 \) whenever \( i \neq j \) ... | Yes |
Theorem 3.8. For a ring \( R \) the following properties are equivalent:\n\n(1) \( R \) is simple and semisimple;\n\n(2) \( R \) is semisimple and all simple left \( R \) -modules are isomorphic;\n\n(3) \( R \cong {M}_{n}\left( D\right) \) for some \( n > 0 \) and some division ring \( D \cong {\operatorname{End}}_{R}^... | Proof. (1) implies (3), and (2) implies (3). Let \( R \cong {M}_{{n}_{1}}\left( {D}_{1}\right) \times \cdots \times \) \( {M}_{{n}_{s}}\left( {D}_{s}\right) \) be semisimple, where \( {D}_{1},\ldots ,{D}_{s} \) are division rings. If \( R \) is simple,\nthen \( s = 1 \) . If all simple left \( R \) -modules are isomorp... | Yes |
Proposition 3.9. Let \( R \) be semisimple and let \( {S}_{1},\ldots ,{S}_{s} \) be, up to isomorphism, all the distinct simple left \( R \) -modules. Let \( {R}_{i} \) be the sum of all the minimal left ideals \( L \cong {S}_{i} \) of \( R \) . Then \( {R}_{i} \) is a two-sided ideal of \( R,{R}_{i} \) is a simple lef... | Proof. Let \( L \) be a minimal left ideal of \( R \) . If \( a \in R \) and \( {La} \neq 0 \), then the left ideal \( {La} \) is a minimal left ideal: if \( A \subseteq {La} \) is a nonzero left ideal, then so is \( {L}^{\prime } = \{ x \in L \mid {xa} \in A\} \), whence \( {L}^{\prime } = L \) and \( A = {L}^{\prime ... | Yes |
Proposition 4.1. For every left \( R \) -module \( M \) there is a canonical homomorphism \( \Phi : R \rightarrow {\operatorname{End}}_{E}\left( M\right) \), defined by \( \Phi \left( r\right) : x \mapsto {rx} \) . | Proof. Since \( \left( {rx}\right) \eta = r\left( {x\eta }\right) \) for all \( r \in R, x \in M,\eta \in E \), the action \( \Phi \left( r\right) : x \mapsto {rx} \) of \( r \) on \( M \) is an \( E \) -endomorphism. It is immediate that \( \Phi \) is a ring homomorphism. | Yes |
Theorem 4.2 (Jacobson Density Theorem). Let \( M \) be a semisimple left \( R \) -module and let \( E = {\operatorname{End}}_{R}^{\mathrm{{op}}}\left( M\right) \) . For every \( \xi \in {\operatorname{End}}_{E}\left( M\right) \) and \( {x}_{1},\ldots ,{x}_{n} \in M \) there exists \( r \in R \) such that \( \xi {x}_{i}... | Proof. We first prove 4.2 when \( n = 1 \) . Since \( M \) is semisimple we have \( M = {Rx} \oplus N \) for some submodule \( N \) of \( M \) . Then the projection \( \pi : M \rightarrow {Rx} \) may be viewed as an \( R \) -endomorphism of \( M \), such that \( \left( {rx}\right) \pi = {rx} \) for all \( {rx} \in {Rx}... | Yes |
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