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Proposition 5.10. If \( \mathrm{F} \) is a finite dimensional extension field of a finite field \( \mathbf{K} \), then \( \mathbf{F} \) is finite and is Galois over \( \mathrm{K} \) . The Galois group \( {Au}{t}_{\mathrm{K}}\mathrm{F} \) is cyclic. | SKETCH OF PROOF. Let \( {Z}_{p} \) be the prime subfield of \( K \) . Then \( F \) is finite dimensional over \( {Z}_{p} \) (Theorem 1.2), say of dimension \( n \), which implies that \( \left| F\right| = {p}^{n} \) . By the proof of Proposition 5.6 and Exercise \( {3.2F} \) is a splitting field over \( {Z}_{p} \) and ... | No |
Theorem 6.2. Let \( \\mathrm{F} \) be an extension field of \( \\mathrm{K} \) . Then \( \\mathrm{u}\\varepsilon \\mathrm{F} \) is both separable and purely inseparable over \( \\mathbf{K} \) if and only if \( \\mathbf{u}\\varepsilon \\mathbf{K} \) . | PROOF. The element \( {u\\varepsilon F} \) is separable and purely inseparable over \( K \) if and only if its irreducible polynomial is of the form \( {\\left( x - u\\right) }^{m} \) and has \( m \) distinct roots in some splitting field. Clearly this occurs only when \( m = 1 \) so that \( x - u \\in K\\left\\lbrack ... | Yes |
Lemma 6.3. Let \( \mathrm{F} \) be an extension field of \( \mathrm{K} \) with char \( \mathrm{K} = \mathrm{p} \neq 0 \) . If \( \mathrm{u} \in \mathrm{F} \) is algebraic over \( \mathrm{K} \), then \( {\mathrm{u}}^{\mathrm{{pn}}} \) is separable over \( \mathrm{K} \) for some \( \mathrm{n} \geq 0 \) . | SKETCH OF PROOF. Use induction on the degree of \( u \) over \( K \) . If \( \deg u = 1 \) or \( u \) is separable, the lemma is true. If \( f \) is the irreducible polynomial of a nonseparable \( u \) of degree greater than one, then \( {f}^{\prime } = 0 \) (Theorem III.6.10), whence \( f \) is a polynomial in \( {x}^... | No |
Theorem 6.4. If \( \mathrm{F} \) is an algebraic extension field of a field \( \mathrm{K} \) of characteristic \( \mathrm{p} \neq 0 \) , then the following statements are equivalent:\n\n(i) \( \mathrm{F} \) is purely inseparable over \( \mathrm{K} \) ;\n\n(ii) the irreducible polynomial of any \( \mathrm{u}\varepsilon ... | SKETCH OF PROOF OF 6.4. (i) \( \Rightarrow \) (ii) Let \( {\left( x - u\right) }^{m} \) be the irreducible polynomial of \( u \in F \) and let \( m = n{p}^{r} \) with \( \left( {n, p}\right) = 1 \) . Then \( {\left( x - u\right) }^{m} = {\left( x - u\right) }^{{p}^{r}n} \) \( = {\left( {x}^{{p}^{r}} - {u}^{{p}^{r}}\rig... | No |
Corollary 6.5. If \( \mathrm{F} \) is a finite dimensional purely inseparable extension field of \( \mathrm{K} \) and char \( \mathrm{K} = \mathrm{p} \neq 0 \), then \( \left\lbrack {\mathrm{F} : \mathrm{K}}\right\rbrack = {\mathrm{p}}^{\mathrm{n}} \) for some \( \mathrm{n} \geq 0 \) . | PROOF. By Theorem \( {1.11F} = K\left( {{u}_{1},\ldots ,{u}_{m}}\right) \) . By hypothesis each \( {u}_{i} \) is purely inseparable over \( K \) and hence over \( K\left( {{u}_{1},\ldots ,{u}_{i - 1}}\right) \) as well (Exercise 2). Theorems 1.6 and 6.4 (ii) imply that every step in the tower \( K \subset K\left( {u}_{... | No |
Lemma 6.6 If \( \mathrm{F} \) is an extension field of \( \mathrm{K},\mathrm{X} \) is a subset of \( \mathrm{F} \) such that \( \mathrm{F} = \mathrm{K}\left( \mathrm{X}\right) \) , and every element of \( \mathrm{X} \) is separable over \( \mathrm{K} \), then \( \mathrm{F} \) is a separable extension of \( \mathrm{K} \... | PROOF. If \( v \in F \), then there exist \( {u}_{1},\ldots ,{u}_{n} \in X \) such that \( v \in K\left( {{u}_{1},\ldots ,{u}_{n}}\right) \) by Theorem 1.3. Let \( {f}_{i}{\varepsilon K}\left\lbrack x\right\rbrack \) be the irreducible separable polynomial of \( {u}_{i} \) and \( E \) a splitting field of \( \left\{ {{... | Yes |
Theorem 6.7. Let \( \mathrm{F} \) be an algebraic extension field of \( \mathrm{K},\mathrm{S} \) the set of all elements of \( \mathrm{F} \) which are separable over \( \mathbf{K} \), and \( \mathbf{P} \) the set of all elements of \( \mathbf{F} \) which are purely inseparable over \( \mathrm{K} \) .\n\n(i) \( \mathrm{... | SKETCH OF PROOF OF 6.7. (i) If \( u, v \in S \) and \( v \neq 0 \), then \( K\left( {u, v}\right) \) is separable over \( K \) by Lemma 6.6, which implies that \( u - v, u{v}^{-1}{\varepsilon S} \) . Therefore, \( S \) is a subfield. Lemma 6.3 and Theorem 6.4 imply (ii). (iii) is a routine exercise using Exercise III.1... | No |
Corollary 6.8. If \( \mathrm{F} \) is a separable extension field of \( \mathrm{E} \) and \( \mathrm{E} \) is a separable extension field of \( \mathrm{K} \), then \( \mathrm{F} \) is separable over \( \mathrm{K} \) . | PROOF. If \( S \) is as in Theorem 6.7, then \( E \subset S \) and \( F \) is purely inseparable over \( S \) . But \( F \) is separable over \( E \) and hence over \( S \) (Exercise 3.12). Therefore, \( F = S \) by Theorem 6.2. | No |
Corollary 6.9. Let \( \mathrm{F} \) be an algebraic extension field of \( \mathrm{K} \), with char \( \mathrm{K} = \mathrm{p} \neq 0 \) . If \( \mathrm{F} \) is separable over \( \mathrm{K} \), then \( \mathrm{F} = {\mathrm{{KF}}}^{\mathrm{{pn}}} \) for each \( \mathrm{n} \geq 1 \) . If \( \left\lbrack {\mathrm{F} : \m... | SKETCH OF PROOF. Let \( S \) be as in Theorem 6.7. If \( \left\lbrack {F : K}\right\rbrack \) is finite, then \( F = K\left( {{u}_{1},\ldots ,{u}_{m}}\right) = S\left( {{u}_{1},\ldots ,{u}_{m}}\right) \) by Theorem 1.11. Since each \( {u}_{i} \) is purely inseparable over \( S \) (Theorem 6.7), there is an \( n \geq 1 ... | Yes |
Lemma 6.11. Let \( \mathrm{F} \) be an extension field of \( \mathrm{E},\mathrm{E} \) an extension field of \( \mathrm{K} \) and \( \mathrm{N} \) a normal extension field of \( \mathbf{K} \) containing \( \mathbf{F} \) . If \( \mathbf{r} \) is the cardinal number of distinct \( \mathbf{E} \) -mono-morphisms \( \mathrm{... | PROOF. For convenience we assume that \( r, t \) are finite. The same proof will work in the general case with only slight modifications of notation. Let \( {\tau }_{1},\ldots ,{\tau }_{r} \) be all the distinct \( E \) -monomorphisms \( F \rightarrow N \) and \( {\sigma }_{1},\ldots ,{\sigma }_{t} \) all the distinct ... | Yes |
Proposition 6.12. Let \( \mathbf{F} \) be a finite dimensional extension field of \( \mathbf{K} \) and \( \mathbf{N} \) a normal extersion field of \( \mathrm{K} \) containing \( \mathrm{F} \) . The number of distinct \( \mathrm{K} \) -monomorphisms \( \mathrm{F} \rightarrow \mathrm{N} \) is precisely \( {\left\lbrack ... | SKETCH OF PROOF. Let \( S \) be the maximal subfield of \( F \) separable over \( K \) (Theorem 6.7(i)). Every \( K \) -monomorphism \( S \rightarrow N \) extends to a \( K \) -automorphism of \( N \) (Theorems 3.8 and 3.14 and Exercise 3.2) and hence (by restriction) to a \( K \) -monomorphism \( F \rightarrow N \) . ... | No |
Corollary 6.13. If \( \mathrm{F} \) is an extension field of \( \mathrm{E} \) and \( \mathrm{E} \) is an extension field of \( \mathrm{K} \), then\n\n\[{\left\lbrack \mathrm{F} : \mathrm{E}\right\rbrack }_{\mathrm{s}}{\left\lbrack \mathrm{E} : \mathrm{K}\right\rbrack }_{\mathrm{s}} = {\left\lbrack \mathrm{F} : \mathrm{... | PROOF. Exercise; use Lemma 6.11 and Proposition 6.12. | No |
Corollary 6.14. Let \( {f\varepsilon K}\left\lbrack x\right\rbrack \) be an irreducible monic polynomial over a field \( K, F \) a splitting field of \( \mathbf{f} \) over \( \mathbf{K} \) and \( {\mathrm{u}}_{1} \) a root of \( \mathbf{f} \) in \( \mathbf{F} \) . Then\n\n(i) every root of \( \mathrm{f} \) has multipli... | SKETCH OF PROOF. Assume char \( K = p \neq 0 \) since the case char \( K = 0 \) is trivial. (i) For any \( i > 1 \) there is a \( K \) -isomorphism \( \sigma : K\left( {u}_{1}\right) \cong K\left( {u}_{\mathrm{i}}\right) \) with \( \sigma \left( {u}_{1}\right) = {u}_{\mathrm{i}} \) that extends to a \( K \) -isomorphis... | Yes |
Proposition 6.15. (Primitive Element Theorem) Let \( \mathrm{F} \) be a finite dimensional extension field of \( \mathbf{K} \) .\n\n(i) If \( \mathrm{F} \) is separable over \( \mathrm{K} \), then \( \mathrm{F} \) is a simple extension of \( \mathrm{K} \).\n\n(ii) (Artin) More generally, \( \mathrm{F} \) is a simple ex... | SKETCH OF PROOF OF 6.15. The first paragraph of the proof of Lemma 3.17, which is valid even if the field \( K \) is finite, shows that a separable extension has only finitely many intermediate fields. Thus it suffices to prove (ii). Since (ii) clearly holds if \( K \) is finite (Corollary 5.8), we assume that \( K \) ... | Yes |
Theorem 7.2. If \( \mathrm{F} \) is a finite dimensional Galois extension field of \( \mathrm{K} \) and\n\n\[ \n{Au}{t}_{\mathrm{K}}\mathrm{F} = \\left\\{ {{\\sigma }_{1},\\ldots ,{\\sigma }_{\\mathrm{n}}}\\right\\}\n\]\n\nthen for any \( \\mathrm{u}\\varepsilon \\mathrm{F} \) ,\n\n\[ \n{\\mathrm{N}}_{\\mathrm{K}}{}^{\... | PROOF. Let \( \\bar{K} \) be an algebraic closure of \( K \) which contains \( F \) . Since \( F \) is normal over \( K \) (Corollary 3.15), the \( K \) -monomorphisms \( F \\rightarrow \\bar{K} \) are precisely the elements of \( {\\operatorname{Aut}}_{K}F \) by Theorem 3.14. Since \( F \) is also separable over \( K ... | Yes |
Theorem 7.3. Let \( \mathrm{F} \) be a finite dimensional extension field of \( \mathbf{K} \) . Then for all \( \mathbf{u},\mathbf{v}\varepsilon \mathbf{F} \) :\n\n(i) \( {\mathrm{N}}_{\mathrm{K}}{}^{\mathrm{F}}\left( \mathrm{u}\right) {\mathrm{N}}_{\mathrm{K}}{}^{\mathrm{F}}\left( \mathrm{v}\right) = {\mathrm{N}}_{\ma... | SKETCH OF PROOF. (i) and (ii) follow directly from Definition 7.1 and the facts that \( r = {\left\lbrack F : K\right\rbrack }_{s} \) and \( {\left\lbrack F : K\right\rbrack }_{s}{\left\lbrack F : K\right\rbrack }_{i} = \left\lbrack {F : K}\right\rbrack \) . | No |
Lemma 7.5. If \( \mathrm{S} \) is a set of distinct automorphisms of a field \( \mathrm{F} \), then \( \mathrm{S} \) is linearly independent. | PROOF. If \( S \) is not linearly independent then there exist nonzero \( {a}_{i}{\varepsilon F} \) and distinct \( {\sigma }_{i}{\varepsilon S} \) such that\n\n\[ {a}_{1}{\sigma }_{1}\left( u\right) + {a}_{2}{\sigma }_{2}\left( u\right) + \cdots + {a}_{n}{\sigma }_{n}\left( u\right) = 0\text{ for all }{u\varepsilon F}... | Yes |
Proposition 7.7. Let \( \mathrm{F} \) be a cyclic extension field of \( \mathrm{K} \) of degree \( \mathrm{n} \) and suppose \( \mathrm{n} = {\mathrm{{mp}}}^{\mathrm{t}} \) where \( 0 \neq \mathrm{p} = \) char \( \mathrm{K} \) and \( \left( {\mathrm{m},\mathrm{p}}\right) = 1 \) . Then there is a chain of intermediate f... | SKETCH OF PROOF. By hypothesis \( F \) is Galois over \( K \) and \( {\operatorname{Aut}}_{K}F \) is cyclic (abelian) so that every subgroup is normal. Recall that every subgroup and quotient group of a cyclic group is cyclic (Theorem I.3.5). Consequently, the Fundamental Theorem 2.5(ii) implies that for any intermedia... | No |
Corollary 7.9. If \( \mathrm{K} \) is a field of characteristic \( \mathrm{p} \neq 0 \) and \( {\mathrm{x}}^{\mathrm{p}} - \mathrm{x} - \mathrm{a}\varepsilon \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \), then \( {\mathrm{x}}^{\mathrm{p}} - \mathrm{x} - \mathrm{a} \) is either irreducible or splits in \( \mathrm{K}... | PROOF. We use the notation of Proposition 7.8. In view of the last paragraph of that proof it suffices to prove that if \( {\operatorname{Aut}}_{K}F \cong \operatorname{Im}\theta = {Z}_{p} \), then \( {x}^{p} - x - a \) is irreducible. If \( u \) and \( v = u + i\left( {{i\varepsilon }{Z}_{p} \subset K}\right) \) are r... | Yes |
Lemma 1.10. Let \( \mathrm{n} \) be a positive integer and \( \mathrm{K} \) a field which contains a primitive \( \mathrm{n} \) th root of unity \( \zeta \) . (i) If \( \mathrm{d} \mid \mathrm{n} \), then \( {\zeta }^{\mathrm{n}/\mathrm{d}} = \eta \) is a primitive \( \mathrm{d} \) th root of unity in \( \mathrm{K} \) ... | PROOF. (i) \( \zeta \) generates a multiplicative cyclic group of order \( n \) by definition. If \( d \mid n \), then \( \eta = {\zeta }^{n/d} \) has order \( d \) by Theorem I.3.4, whence \( \eta \) is a primitive \( d \) th root of unity. (ii) If \( u \) is a root of \( {x}^{d} - a \), then so is \( {\eta }^{i}u \) ... | Yes |
Theorem 7.11. Let \( \mathrm{n} \) be a positive integer and \( \mathrm{K} \) a field which contains a primitive nth root of unity \( \zeta \) . Then the following conditions on an extension field \( \mathbf{F} \) of \( \mathbf{K} \) are equivalent.\n\n(i) \( \mathrm{F} \) is cyclic of degree \( \mathrm{d} \), where \(... | PROOF. (ii) \( \Rightarrow \) (i) Lemma 7.10 shows that \( F = K\left( u\right) \) and \( F \) is Galois over \( K \) for any root \( u \) of \( {x}^{n} - a \) . If \( \sigma \in {\operatorname{Aut}}_{K}F = {\operatorname{Aut}}_{K}K\left( u\right) \), then \( \sigma \) is completely determined by \( \sigma \left( u\rig... | Yes |
Theorem 8.1. Let \( \mathrm{n} \) be a positive integer, \( K \) a field such that char \( \mathrm{K} \) does not divide \( \mathrm{n} \) and \( \mathrm{F} \) a cyclotomic extension of \( \mathrm{K} \) of order \( \mathrm{n} \) . (i) \( \mathrm{F} = \mathrm{K}\left( \zeta \right) \), where \( {\zeta \varepsilon }\mathr... | SKETCH OF PROOF OF 8.1. (i) The remarks preceding Lemma 7.10 show that \( F \) contains a primitive \( n \) th root of unity \( \zeta \) . By definition \( {1}_{K},\zeta ,\ldots ,{\zeta }^{n - 1} \in K\left( \zeta \right) \) are the \( n \) distinct roots of \( {x}^{n} - {1}_{K} \), whence \( F = K\left( \zeta \right) ... | Yes |
Proposition 8.2. Let \( \mathrm{n} \) be a positive integer, \( \mathrm{K} \) a field such that char \( \mathrm{K} \) does not divide \( \mathrm{n} \) and \( {\mathrm{g}}_{\mathrm{n}}\left( \mathrm{x}\right) \) the \( \mathrm{n} \) th cyclotomic polynomial over \( \mathrm{K} \) . (i) \( {\mathrm{x}}^{\mathrm{n}} - {1}_... | PROOF. (i) Let \( F \) be a cyclotomic extension of \( K \) of order \( n \) and \( \zeta \in F \) a primitive \( n \) th root of unity. Lemma 7.10 (applied to \( F \) ) shows that the cyclic group \( G = \langle \zeta \rangle \) of all \( n \) th roots of unity contains all \( d \) th roots of unity for every divisor ... | Yes |
Lemma 9.3. If \( \mathrm{F} \) is a radical extension field of \( \mathrm{K} \) and \( \mathrm{N} \) is a normal closure of \( \mathrm{F} \) over \( \mathrm{K} \) (Theorem 3.16), then \( \mathrm{N} \) is a radical extension of \( \mathrm{K} \) . | SKETCH OF PROOF. The proof consists of combining two facts. (i) If \( F \) is any finite dimensional extension of \( K \) (not necessarily radical) and \( N \) is the normal closure of \( F \) over \( K \), then \( N \) is the composite field \( {E}_{1}{E}_{2}\cdots {E}_{r} \), where each \( {E}_{i} \) is a subfield of... | No |
Theorem 9.4. If \( \mathrm{F} \) is a radical extension field of \( \mathrm{K} \) and \( \mathrm{E} \) is an intermediate field, then \( {Au}{t}_{\mathrm{K}}\mathrm{E} \) is a solvable group. | PROOF. If \( {K}_{0} \) is the fixed field of \( E \) relative to the group \( {\operatorname{Aut}}_{K}E \), then \( E \) is Galois over \( {K}_{0},{\operatorname{Aut}}_{{K}_{0}}E = {\operatorname{Aut}}_{K}E \) and \( F \) is a radical extension of \( {K}_{0} \) (Exercise 1). Thus we may assume to begin with that \( E ... | No |
Corollary 9.5. Let \( \mathrm{K} \) be a field and \( \mathrm{f} \varepsilon \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) . If the equation \( \mathrm{f}\left( \mathrm{x}\right) = 0 \) is solvable by radicals, then the Galois group of \( \mathrm{f} \) is a solvable group. | PROOF. Immediate from Theorem 9.4 and Definition 9.2. | No |
Let \( \mathrm{K} \) be a field and \( \mathrm{f}\varepsilon \mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) a polynomial of degree \( \mathrm{n} > 0 \), where char \( \mathrm{K} \) does not divide \( \mathrm{n} \) ! (which is always true when char \( \mathrm{K} = 0 \) ). Then the equation \( \mathrm{f}\left( \mathrm... | SKETCH OF PROOF. ( \( \Leftarrow \) ) Let \( E \) be a splitting field of \( f \) over \( K \) . In view of Proposition 9.6 we need only show that \( E \) is Galois over \( K \) and char \( K \nmid \left\lbrack {E : K}\right\rbrack \) . Since char \( K \nmid n \) ! the irreducible factors of \( f \) are separable by Th... | No |
Theorem 1.2. Let \( \mathrm{F} \) be an extension field of \( \mathrm{K} \) and \( \left\{ {{\mathrm{s}}_{1},\ldots ,{\mathrm{s}}_{\mathrm{n}}}\right\} \) a subset of \( \mathrm{F} \) which is algebraically independent over \( \mathrm{K} \) . Then there is \( \mathrm{K} \) -isomorphism \( \mathrm{K}\left( {{\mathrm{s}}... | SKETCH OF PROOF. The assignment \( {x}_{i} \mapsto {s}_{i} \) defines a \( K \) -epimorphism of rings \( \theta : K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow K\left\lbrack {{s}_{1},\ldots ,{s}_{n}}\right\rbrack \) by Theorems III.5.5 and V.1.3. The algebraic independence of \( \left\{ {{s}_{1},\ldo... | Yes |
For \( \mathrm{i} = 1,2 \) let \( {\mathrm{F}}_{\mathrm{i}} \) be an extension field of \( {\mathrm{K}}_{\mathrm{i}} \) and \( {\mathrm{S}}_{\mathrm{i}} \subset {\mathrm{F}}_{\mathrm{i}} \) with \( {\mathrm{S}}_{\mathrm{i}} \) algebraically independent over \( {\mathrm{K}}_{\mathrm{i}} \) . If \( \varphi : {\mathrm{S}}... | SKETCH OF PROOF OF 1.3. For each \( n \geq {1\sigma } \) induces a monomorphism of rings \( {K}_{1}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow {K}_{2}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) (also denoted \( \sigma \) ; see p. 235). Every element of\n\n\( {K}_{1}\left( {S}_{1}\right) \... | Yes |
Theorem 1.5. Let \( \mathrm{F} \) be an extension field of \( \mathbf{K},\mathrm{S} \) a subset of \( \mathrm{F} \) algebraically independent over \( \mathrm{K} \), and \( \mathrm{u}\varepsilon \mathrm{F} - \mathrm{K}\left( \mathrm{S}\right) \) . Then \( \mathrm{S} \cup \{ \mathrm{u}\} \) is algebraically independent o... | PROOF. ( \( \Leftarrow \) ) If there exist distinct \( {s}_{1},\ldots ,{s}_{n - 1}{\varepsilon S} \) and an \( {f\varepsilon K}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) such that \( f\left( {{s}_{1},\ldots ,{s}_{n - 1}, u}\right) = 0 \), then \( u \) is a root of \( f\left( {{s}_{1},\ldots ,{s}_{n - 1},{x}... | Yes |
Corollary 1.6. Let \( \mathrm{F} \) be an extension field of \( \mathrm{K} \) and \( \mathrm{S} \) a subset of \( \mathrm{F} \) that is algebraically independent over \( \mathbf{K} \). Then \( \mathbf{S} \) is a transcendence base of \( \mathbf{F} \) over \( \mathbf{K} \) if and only if \( \mathbf{F} \) is algebraic ov... | PROOF. Exercise. | No |
Corollary 1.7. If \( \mathrm{F} \) is an extension field of \( \mathrm{K} \) and \( \mathrm{F} \) is algebraic over \( \mathrm{K}\left( \mathrm{X}\right) \) for some subset \( \mathrm{X} \) of \( \mathrm{F} \) (in particular, if \( \mathrm{F} = \mathrm{K}\left( \mathrm{X}\right) \) ), then \( \mathrm{X} \) contains a t... | PROOF. Let \( S \) be a maximal algebraically independent subset of \( X \) ( \( S \) exists by a routine Zorn’s Lemma argument). Then every \( {u\varepsilon X} - S \) is algebraic over \( K\left( S\right) \) by Theorem 1.5, whence \( K\left( X\right) \) is algebraic over \( K\left( S\right) \) by Theorem V.1.12. Conse... | Yes |
Theorem 1.8. Let \( \mathrm{F} \) be an extension field of \( \mathrm{K} \). If \( \mathrm{S} \) is a finite transcendence base of \( \mathrm{F} \) over \( \mathrm{K} \), then every transcendence base of \( \mathrm{F} \) over \( \mathrm{K} \) has the same number of elements as \( \mathrm{S} \). | SKETCH OF PROOF. Let \( S = \left\{ {{s}_{1},\ldots ,{s}_{n}}\right\} \) and let \( T \) be any transcendence base. We claim that some \( {t}_{1}{\varepsilon T} \) is transcendental over \( K\left( {{s}_{2},\ldots ,{s}_{n}}\right) \). Otherwise every element of \( T \) is algebraic over \( K\left( {{s}_{2},\ldots ,{s}_... | Yes |
Theorem 1.11. If \( \mathrm{F} \) is an extension field of \( \mathrm{E} \) and \( \mathrm{E} \) an extension field of \( \mathrm{K} \), then\n\n\[ \operatorname{tr}.d.\mathrm{\;F}/\mathrm{K} = \left( {\operatorname{tr}.d.\mathrm{\;F}/\mathrm{E}}\right) + \left( {\operatorname{tr}.d.\mathrm{E}/\mathrm{K}}\right) . \] | PROOF. Let \( S \) be a transcendence base of \( E \) over \( K \) and \( T \) a transcendence base of \( F \) over \( E \) . Since \( S \subset E, S \) is algebraically dependent over \( E \), whence \( S \cap T = \varnothing \) . It suffices to show that \( S \cup T \) is a transcendence base of \( F \) over \( K \),... | Yes |
Theorem 1.12. Let \( {\mathrm{F}}_{1} \) [resp. \( {\mathrm{F}}_{2} \) ] be an algebraically closed field extension of a field \( {\mathrm{K}}_{1} \) [resp. \( {\mathrm{K}}_{2} \) ]. If tr.d. \( {\mathrm{F}}_{1}/{\mathrm{K}}_{1} = \) tr.d. \( {\mathrm{F}}_{2}/{\mathrm{K}}_{2} \), then every isomorphism of fields \( {\m... | PROOF. Let \( {S}_{i} \) be a transcendence base of \( {F}_{i} \) over \( {K}_{i} \) . Since \( \left| {S}_{1}\right| = \left| {S}_{2}\right| \) , \( \sigma : {K}_{1} \cong {K}_{2} \) extends to an isomorphism \( \bar{\sigma } : {K}_{1}\left( {S}_{1}\right) \cong {K}_{2}\left( {S}_{2}\right) \) by Corollary 1.3. \( {F}... | Yes |
Theorem 2.2. Let \( \mathrm{C} \) be an algebraically closed field with subfields \( \mathrm{K},\mathrm{E},\mathrm{F} \) such that \( \mathrm{K} \subset \mathrm{E} \cap \mathrm{F} \) . Then \( \mathrm{E} \) and \( \mathrm{F} \) are linearly disjoint over \( \mathrm{K} \) if and only if \( \mathrm{F} \) and \( \mathrm{E... | PROOF. It suffices to assume \( E \) and \( F \) linearly disjoint and show that \( F \) and \( E \) are linearly disjoint. Suppose \( X \subset F \) is linearly independent over \( K \), but not over \( E \) so that \( {r}_{1}{u}_{1} + \cdots + {r}_{n}{u}_{n} = 0 \) for some \( {u}_{i}{\varepsilon X} \) and \( {r}_{i}... | No |
Lemma 2.3. Let \( \mathrm{C} \) be an algebraically closed field with subfields \( \mathrm{K},\mathrm{E},\mathrm{F} \) such that \( \mathrm{K} \subset \mathrm{E} \cap F \) . Let \( \mathrm{R} \) be a subring of \( \mathrm{E} \) such that \( \mathrm{K}\left( \mathrm{R}\right) = E \) and \( \mathrm{K} \subset \mathrm{R} ... | PROOF OF 2.3. (i) \( \Rightarrow \) (ii) and (i) \( \Rightarrow \) (iii) are trivial. (ii) \( \Rightarrow \) (i) Let \( X = \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) be a finite subset of \( E \) which is linearly independent over \( K \) . We must show that \( X \) is linearly independent over \( F \) . Since \( {u... | Yes |
Theorem 2.4. Let \( \mathrm{C} \) be an algebraically closed field with subfields \( \mathrm{K},\mathrm{E},\mathrm{L},\mathrm{F} \) such that \( \mathrm{K} \subset \mathrm{E} \) and \( \mathrm{K} \subset \mathrm{L} \subset \mathrm{F} \) . Then \( \mathrm{E} \) and \( \mathrm{F} \) are linearly disjoint over \( \mathrm{... | \n\n\n\( \left( \Leftarrow \right) \) If a subset \( X \) of \( E \) is linearly independent over \( K \), then \( X \) is linearly independent over \( L \) by (i). Therefore (since \( X \subset E \subset {EL} \) ), ... | Yes |
Lemma 2.6. If \( \mathrm{F} \) is an extension field of \( \mathrm{K} \) of characteristic \( \mathrm{p} \neq 0 \) and \( \mathrm{C} \) is an algebraically closed field containing \( \mathrm{F} \), then for any \( \mathrm{n} \geq 0 \) a subset \( \mathrm{X} \) of \( \mathrm{F} \) is linearly independent over \( {\mathr... | SKETCH OF PROOF. Every \( a \in K \) is of the form \( a = {v}^{{p}^{n}} \) for some \( v \in {K}^{1/{p}^{n}} \) (Exercise 5). For the first statement note that \( \mathop{\sum }\limits_{i}{a}_{i}{u}_{i}{}^{{p}^{n}} = 0\left( {{a}_{i} \in K;{u}_{i} \in X}\right) \Leftrightarrow \) \( \mathop{\sum }\limits_{i}{v}_{i}{}^... | No |
Theorem 2.7. Let \( \mathrm{F} \) be a field contained in an algebraically closed field \( \mathrm{C} \) . If \( \mathrm{F} \) is a purely transcendental extension of a field \( \mathrm{K} \) of characteristic \( \mathrm{p} \neq 0 \), then \( \mathrm{F} \) and \( {\mathbf{K}}^{1/\mathbf{{pn}}} \) are linearly disjoint ... | PROOF. Let \( F = K\left( S\right) \) with \( S \) a transcendence base of \( F \) over \( K \) . If \( S = \varnothing \), then \( F = K \) and every linearly independent subset of \( F \) over \( K \) consists of exactly one nonzero element of \( K \) . Such a nonzero singleton is clearly linearly independent over an... | Yes |
Theorem 2.8. Let \( \mathrm{F} \) be an algebraic extension field of a field \( \mathrm{K} \) of characteristic \( \mathrm{p} \neq 0 \) and \( \mathrm{C} \) an algebraically closed field containing \( \mathrm{F} \) . Then \( \mathrm{F} \) is separable over \( \mathrm{K} \) if and only if \( \mathrm{F} \) and \( {\mathr... | PROOF. We shall prove here only that separability implies that \( F \) and \( {K}^{1/p} \) are linearly disjoint. The other half of the proof will be an easy consequence of a result below (see the Remarks after Theorem 2.10). Let \( X = \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) be a finite subset of \( F \) which is... | No |
Corollary 2.12. (Mac Lane’s Criterion) If \( \mathrm{F} \) is an extension field of a field \( \mathrm{K} \) and \( \mathrm{F} \) is separably generated over \( \mathrm{K} \), then \( \mathrm{F} \) is separable over \( \mathrm{K} \) . Conversely, if \( \mathrm{F} \) is separable and finitely generated over \( \mathrm{K... | SKETCH OF PROOF. The proof of (iv) \( \Rightarrow \) (iii) \( \Rightarrow \) (i) in Theorem 2.10 is valid here with \( F = E \) since it uses only the fact that \( E \) is separably generated. The last two statements are consequences of the proof of (i) \( \Rightarrow \) (iv) in Theorem 2.10. | No |
Corollary 2.13. Let \( \mathrm{F} \) be an extension field of \( \mathrm{K} \) and \( \mathrm{E} \) an intermediate field.\n\n(i) If \( \mathrm{F} \) is separable over \( \mathrm{K} \), then \( \mathrm{E} \) is separable over \( \mathrm{K} \) ;\n\n(ii) if \( \mathrm{F} \) is separable over \( \mathrm{E} \) and \( \math... | SKETCH OF PROOF OF 2.13. (ii) Use Theorems 2.4 and 2.10. (iii) If char \( K \) \( = p \neq 0 \), let \( X \) be a subset of \( F \) which is linearly independent over \( E \) . Extend \( X \) to a basis \( U \) of \( F \) over \( E \) and let \( V \) be a basis of \( E \) over \( K \) . The proof of Theorem IV.2.16 sho... | No |
Theorem 1.1. If \( \mathrm{R} \) is a ring, then the set of all \( \mathrm{n} \times \mathrm{m} \) matrices over \( \mathrm{R} \) forms an \( \mathrm{R} - \mathrm{R} \) bimodule under addition, with the \( \mathrm{n} \times \mathrm{m} \) zero matrix as the additive identity. Multiplication of matrices, when defined, is... | PROOF. Exercise. | No |
Theorem 1.2. Let \( \mathrm{R} \) be a ring with identity. Let \( \mathrm{E} \) be a free left \( \mathrm{R} \) -module with a finite basis of \( \mathrm{n} \) elements and \( \mathrm{F} \) a free left \( \mathrm{R} \) -module with a finite basis of \( \mathrm{m} \) elements. Let \( \mathrm{M} \) be the left \( \mathrm... | PROOF. Let \( \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) be a basis of \( E,\left\{ {{v}_{1},\ldots ,{v}_{m}}\right\} \) a basis of \( F \) and \( {f\varepsilon }{\operatorname{Hom}}_{R}\left( {E, F}\right) \) . There are elements \( {r}_{ij} \) of \( R \) such that\n\n\[ f\left( {u}_{1}\right) = {r}_{11}{v}_{1} + {r... | Yes |
Theorem 1.3. Let \( \mathrm{R} \) be a ring with identity and let \( \mathrm{E},\mathrm{F},\mathrm{G} \), be free left \( \mathrm{R} \) -modules with finite ordered bases \( \mathrm{U} = \left\{ {{\mathrm{u}}_{1},\ldots ,{\mathrm{u}}_{\mathrm{n}}}\right\} ,\mathrm{V} = \left\{ {{\mathrm{v}}_{1},\ldots ,{\mathrm{v}}_{\m... | PROOF. If \( A = \left( {r}_{ij}\right) \) and \( B = \left( {s}_{kj}\right) \), then for each \( i = 1,2,\ldots, n \)\n\n\[ \n{gf}\left( {u}_{i}\right) = g\left( {\mathop{\sum }\limits_{{k = 1}}^{m}{r}_{ik}{v}_{k}}\right) = \mathop{\sum }\limits_{{k = 1}}^{m}{r}_{ik}g\left( {v}_{k}\right) = \mathop{\sum }\limits_{{k =... | Yes |
Theorem 1.4. Let \( \mathrm{R} \) be a ring with identity and \( \mathrm{E} \) a free left \( \mathrm{R} \) -module with a finite basis of \( \mathrm{n} \) elements. Then there is an isomorphism of rings:\n\n\[ \n{\operatorname{Hom}}_{\mathrm{R}}\left( {\mathrm{E},\mathrm{E}}\right) \cong {\operatorname{Mat}}_{\mathrm{... | SKETCH OF PROOF OF 1.4. Let \( \phi : {\operatorname{Hom}}_{R}\left( {E, E}\right) \rightarrow {\operatorname{Mat}}_{n}R \) be the anti-isomorphism that assigns to each map \( f \) its matrix relative to the given basis. Verify that the map \( \psi : {\operatorname{Mat}}_{n}R \rightarrow {\operatorname{Mat}}_{n}{R}^{op... | No |
Lemma 1.5. Let \( \mathrm{R} \) be a ring with identity and \( \mathrm{E},\mathrm{F} \) free left \( \mathrm{R} \) -modules with ordered bases \( \mathrm{U},\mathrm{V} \) respectively such that \( \left| \mathrm{U}\right| = \mathrm{n} = \left| \mathrm{V}\right| \) . Let \( \mathrm{A}\varepsilon {\operatorname{Mat}}_{\m... | SKETCH OF PROOF. An \( R \) -module homomorphism \( f : E \rightarrow F \) is an isomorphism if and only if there exists an \( R \) -module homomorphism \( {f}^{-1} : F \rightarrow E \) such that \( {f}^{-1}f = {1}_{E} \) and \( f{f}^{-1} = {1}_{F} \) (see Theorem I.2.3). Suppose \( f \) is an isomorphism with matrix \... | No |
Theorem 1.6. Let \( \mathrm{R} \) be a ring with identity. Let \( \mathrm{E} \) and \( \mathrm{F} \) be free left \( \mathrm{R} \) -modules with finite ordered bases \( \mathrm{U} \) and \( \mathrm{V} \) respectively such that \( \left| \mathrm{U}\right| = \mathrm{n},\left| \mathrm{V}\right| = \mathrm{m} \) . Let \( \m... | PROOF. ( \( \Rightarrow \) ) If \( B \) is the \( n \times m \) matrix of \( f \) relative to the bases \( {U}^{\prime } \) of \( E \) and \( {V}^{\prime } \) of \( F \), then \( \left| {U}^{\prime }\right| = n \) and \( \left| {V}^{\prime }\right| = m \) . Let \( P \) be the \( n \times n \) matrix of the identity map... | Yes |
Corollary 1.7 Let \( \mathrm{R} \) be a ring with identity and \( \mathrm{E} \) a free left \( \mathrm{R} \) -module with an ordered basis \( \mathrm{U} \) of finite cardinality \( \mathrm{n} \) . Let \( \mathrm{A} \) be the \( \mathrm{n} \times \mathrm{n} \) matrix of \( \mathrm{f}\varepsilon {\operatorname{Hom}}_{\ma... | SKETCH OF PROOF. If \( E = F, U = V \), and \( {U}^{\prime } = {V}^{\prime } \) in the proof of Theorem 1.6, then \( Q = {P}^{-1} \) by Lemma 1.5. | No |
Theorem 1.9. Let \( \mathrm{R} \) be a ring with identity and \( \mathrm{E},\mathrm{F} \) free right \( \mathrm{R} \) -modules with finite bases \( \mathrm{U} \) and \( \mathrm{V} \) of cardinality \( \mathrm{n} \) and \( \mathrm{m} \) respectively. Let \( \mathrm{N} \) be the right \( \mathrm{R} \) -module of all \( \... | PROOF. Exercise; see Theorems 1.2-1.4. Note that for right modules (iii) is actually an isomorphism rather than an anti-isomorphism. | No |
Proposition 1.10. Let \( \mathrm{R} \) be a ring with identity and \( \mathrm{f} : \mathrm{E} \rightarrow \mathrm{F} \) a homomorphism of finitely generated free left \( \mathbf{R} \) -modules. If \( \mathbf{A} \) is the matrix of \( \mathbf{f} \) relative to (ordered) bases \( \mathrm{U} \) and \( \mathrm{V} \), then ... | PROOF OF 1.10. Recall that the dual basis \( {V}^{ * } = \left\{ {{v}_{1}*,\ldots ,{v}_{m} * }\right\} \) of \( {F}^{ * } = {\operatorname{Hom}}_{R}\left( {F, R}\right) \) is determined by:\n\n\[ \n{v}_{i} * \left( {v}_{j}\right) = {\delta }_{ij}\;\text{ (Kronecker delta; }1 \leq i, j \leq m\text{ ),} \n\]\n\nand simil... | Yes |
Theorem 2.3. Let \( \mathrm{f} : \mathrm{E} \rightarrow \mathrm{F} \) be a linear transformation of finite dimensional left [resp. right] vector spaces over a division ring D. If \( \mathrm{A} \) is the matrix of \( \mathrm{f} \) relative to some pair of ordered bases, then the rank of \( \mathrm{f} \) is equal to the ... | PROOF OF 2.3. Let \( A \) be the \( n \times m \) [resp. \( m \times n \) ] matrix of \( f \) relative to ordered bases \( U = \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) of \( E \) and \( V = \left\{ {{v}_{1},\ldots ,{v}_{m}}\right\} \) of \( F \) . Then under the usual isomorphism \( F \cong {D}^{m} \) given by \( \... | Yes |
Proposition 2.4. Any linear transformation \( \mathrm{f} : \mathrm{E} \rightarrow \mathrm{F} \) of finite dimensional left vector spaces over a division ring \( \mathrm{D} \) has the same rank as its dual map \( \bar{\mathrm{f}} : {\mathrm{F}}^{ * } \rightarrow {\mathrm{E}}^{ * } \) . | PROOF OF 2.4. Let rank \( f = r \) . By Corollary IV.2.14 there is a basis \( X = \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) such that \( \left\{ {{u}_{r + 1},\ldots ,{u}_{n}}\right\} \) is a basis of Ker \( f \) and \( {Y}_{1} = \) \( \left\{ {f\left( {u}_{1}\right) ,\ldots, f\left( {u}_{r}\right) }\right\} \) is a ... | Yes |
Corollary 2.5. If \( \\mathrm{A} \) is an \( \\mathrm{n} \\times \\mathrm{m} \) matrix over a division ring \( \\mathrm{D} \), then row rank \( \\mathrm{A} = \) column rank \( \\mathrm{A} \) . | PROOF. Let \( f : {D}^{n} \\rightarrow {D}^{m} \) be a linear transformation of left vector spaces with matrix \( A \) relative to the standard bases. Then the dual map \( \\bar{f} \) of right vector spaces also has matrix \( A \) (Proposition 1.10). By Theorem 2.3 and Proposition 2.4 row rank \( A = \\operatorname{ran... | Yes |
Theorem 2.6. Let \( \mathrm{M} \) be the set of all \( \mathrm{n} \times \mathrm{m} \) matrices over a division ring \( \mathrm{D} \) and let A, B ε M.\n\n(i) \( \mathrm{A} \) is equivalent to \( {\mathrm{E}}_{\mathrm{r}}^{\mathrm{n},\mathrm{m}} \) if and only if rank \( \mathrm{A} = \mathrm{r} \) . | SKETCH OF PROOF. (i) \( A \) is the matrix of some linear transformation \( f : {D}^{n} \rightarrow {D}^{m} \) relative to some pair of bases by Theorem 1.2. If rank \( A = r \), then Corollary IV.2.14 implies that there exist bases \( U = \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) of \( {D}^{n} \) and \( V = \left\{... | No |
Corollary 2.9. Every \( \mathrm{n} \times \mathrm{n} \) elementary matrix \( \mathrm{E} \) over a ring \( \mathrm{R} \) with identity is invertible and its inverse is an elementary matrix. | SKETCH OF PROOF. Verify that \( {I}_{n} \) may be obtained from \( E \) by performing a single elementary row operation \( T \) . If \( F \) is the elementary matrix obtained by performing \( T \) on \( {I}_{n} \), then \( {FE} = {I}_{n} \) by Theorem 2.8. Verify directly that \( {EF} = {I}_{n} \) . | No |
Corollary 2.10. If \( \mathrm{B} \) is the matrix obtained from an \( \mathrm{n} \times \mathrm{m} \) matrix \( \mathrm{A} \) over a ring \( \mathrm{R} \) with identity by performing a finite sequence of elementary row and column operations, then \( \mathrm{B} \) is equivalent to \( \mathrm{A} \) . | PROOF. Since each row [column] operation used to obtain \( B \) from \( A \) is given by left [right] multiplication by an appropriate elementary matrix (Theorem 2.8), we have \( B = \left( {{E}_{p}\cdots {E}_{1}}\right) A\left( {{F}_{1}\cdots {F}_{q}}\right) = {PAQ} \) with each \( {E}_{i}{F}_{j} \) an elementary matr... | Yes |
Proposition 2.11. If \( \mathrm{A} \) is an \( \mathrm{n} \times \mathrm{m} \) matrix of rank \( \mathrm{r} > 0 \) over a principal ideal domain \( \mathrm{R} \), then \( \mathrm{A} \) is equivalent to a matrix of the form \( \left( \begin{array}{ll} {\mathrm{L}}_{\mathrm{r}} & 0 \\ 0 & 0 \end{array}\right) \), where \... | SKETCH OF PROOF OF 2.11. (i) Recall that \( a, b \in R \) are associates if \( a \mid b \) and \( b \mid a \) . By Theorem III.3.2 \( a \) and \( b \) are associates if and only if \( a = {bu} \) with \( u \) a unit. We say that \( c \in R \) is a proper divisor of \( a \in R \) if \( c \mid a \) and \( c \) is not an ... | No |
Proposition 2.12. The following conditions on an \( \mathrm{n} \times \mathrm{n} \) matrix \( \mathrm{A} \) over a division ring \( \mathrm{D} \) are equivalent:\n\n(i) \( \operatorname{rank}\mathrm{A} = \mathrm{n} \) ;\n\n(ii) \( \mathrm{A} \) is equivalent to the identity matrix \( {\mathrm{I}}_{\mathrm{n}} \) ;\n\n(... | SKETCH OF PROOF. (i) \( \Leftrightarrow \) (ii) by Theorem 2.6 since \( {E}_{n}^{n, n} = {I}_{n} \) . (i) \( \Rightarrow \) (iii) The rows of any matrix of rank \( n \) are necessarily linearly independent (see Theorem IV.2.5 and Definition 2.2.) Consequently, the first row of \( A = \left( {a}_{ij}\right) \) is not th... | No |
Theorem 3.2. If \( \mathrm{B} \) and \( \mathrm{C} \) are modules over a commutative ring \( \mathrm{R} \) with identity, then every alternating \( \mathrm{R} \) -multilinear function \( \mathrm{f} : {\mathrm{B}}^{\mathrm{n}} \rightarrow \mathrm{C} \) is skew-symmetric. | SKETCH OF PROOF. In the special case when \( n = 2 \) and \( \sigma = \left( {12}\right) \), we have:\n\n\[ 0 = f\left( {{b}_{1} + {b}_{2},{b}_{1} + {b}_{2}}\right) = f\left( {{b}_{1},{b}_{1}}\right) + f\left( {{b}_{1},{b}_{2}}\right) + f\left( {{b}_{2},{b}_{1}}\right) + f\left( {{b}_{2},{b}_{2}}\right) \]\n\n\[ = 0 + ... | No |
Theorem 3.3. If \( \mathrm{R} \) is a commutative ring with identity and \( \mathrm{r} \in \mathrm{R} \), then there exists a unique alternating \( \mathrm{R} \) -multilinear form \( \mathrm{f} : {\left( {\mathrm{R}}^{\mathrm{n}}\right) }^{\mathrm{n}} \rightarrow \mathrm{R} \) such that \( \mathrm{f}\left( {{\varepsilo... | PROOF OF 3.3. (Uniqueness) If such an alternating \( n \) -linear form \( f \) exists and if \( \left( {{X}_{1},\ldots ,{X}_{n}}\right) \in {\left( {R}^{n}\right) }^{n} \), then for each \( i \) there exist \( {a}_{ij} \in R \) such that \( {X}_{i} = \left( {{a}_{i1},{a}_{i2},\ldots ,{a}_{in}}\right) \) \( = \mathop{\s... | Yes |
Theorem 3.5. Let \( \mathrm{R} \) be a commutative ring with identity and \( \mathrm{A},\mathrm{B}\varepsilon {\operatorname{Mat}}_{\mathrm{n}}\mathrm{R} \) . (i) Every alternating \( \mathrm{R} \) -multilinear form \( \mathrm{f} \) on \( {\mathrm{{Mat}}}_{\mathrm{n}}\mathrm{R} \) is a unique scalar multiple of the det... | SKETCH OF PROOF. (i) Let \( f\left( {I}_{n}\right) = {r\varepsilon R} \) . Let \( d \) be the determinant function. Verify that the function \( {rd} : {\operatorname{Mat}}_{n}R \rightarrow R \) given by \( A \mapsto r\left| A\right| = {rd}\left( A\right) \) is also an alternating \( R \) -multilinear form on \( {\opera... | No |
If \( \mathrm{A} \) is an \( \mathrm{n} \times \mathrm{n} \) matrix over a commutative ring \( \mathrm{R} \) with identity, then for each \( \mathrm{i} = 1,2,\ldots ,\mathrm{n} \) , \[ \left| \mathrm{A}\right| = \mathop{\sum }\limits_{{j = 1}}^{n}{\left( -1\right) }^{i + j}{a}_{ij}\left| {A}_{ij}\right| \] and for each... | PROOF OF 3.6. We let \( j \) be fixed and prove the second statement. By Theorem 3.3 and Definition 3.4 it suffices to show that the map \( \phi : {\operatorname{Mat}}_{n}R \rightarrow R \) given by \( A = \left( {a}_{ij}\right) \left| { \rightarrow \mathop{\sum }\limits_{{i = 1}}^{n}{\left( -1\right) }^{i + j}{a}_{ij}... | Yes |
Proposition 3.7. If \( \mathrm{A} = \left( {\mathrm{a}}_{\mathrm{i}}\right) \) is an \( \mathrm{n} \times \mathrm{n} \) matrix over a commutative ring \( \mathrm{R} \) with identity and \( {\mathrm{A}}^{\mathrm{a}} = \left( {\mathrm{b}}_{\mathrm{{ij}}}\right) \) is the \( \mathrm{n} \times \mathrm{n} \) matrix with \( ... | PROOF OF 3.7. The \( \left( {i, j}\right) \) entry of \( A{A}^{a} \) is \( {c}_{ij} = \mathop{\sum }\limits_{{k = 1}}^{n}{\left( -1\right) }^{j + k}{a}_{ik}\left| {A}_{jk}\right| \) . If \( i = j \) , then \( {c}_{ii} = \left| A\right| \) by Proposition 3.6. If \( i \neq j \) (say \( i < j \) ) and \( A \) has rows \( ... | Yes |
Corollary 3.8. (Cramer’s Rule) Let \( \mathrm{A} = \left( {\mathrm{a}}_{\mathrm{{ij}}}\right) \) be the matrix of coefficients of the system of \( \mathrm{n} \) linear equations in \( \mathrm{n} \) unknowns\n\n\[ \n{\mathrm{a}}_{11}{\mathrm{x}}_{1} + {\mathrm{a}}_{12}{\mathrm{x}}_{2} + \cdots + {\mathrm{a}}_{1\mathrm{n... | PROOF. Clearly the given system has a solution if and only if the matrix equation \( {AX} = B \) has a solution, where \( X \) and \( B \) are the column vectors \( X = {\left( {x}_{1}\cdots {x}_{n}\right) }^{t} \) , \( B = {\left( {b}_{1}\cdots {b}_{n}\right) }^{t} \) . Since \( \left| A\right| \neq 0, A \) is inverti... | Yes |
Theorem 4.1. Let \( \mathrm{E} \) be an \( \mathrm{n} \) -dimensional vector space over a field \( \mathrm{K},\phi : \mathrm{E} \rightarrow \mathrm{E} \) a linear transformation and \( \mathrm{A} \) an \( \mathrm{n} \times \mathrm{n} \) matrix over \( \mathrm{K} \) . (i) There exists a unique monic polynomial of positi... | PROOF. (i) By Theorem III.5.5 there is a unique (nonzero) ring homomorphism \( \zeta = {\zeta }_{\phi } : K\left\lbrack x\right\rbrack \rightarrow {\operatorname{Hom}}_{K}\left( {E, E}\right) \) such that \( x \mapsto \phi \) and \( k \mapsto k{1}_{E} \) for all \( k \in K \) . Consequently, if \( {f\varepsilon K}\left... | Yes |
Theorem 4.2. Let \( \phi : \mathrm{E} \rightarrow \mathrm{E} \) be a linear transformation of an \( \mathrm{n} \) -dimensional vector space \( \mathrm{E} \) over a field \( \mathrm{K} \) .\n\n(i) There exist monic polynomials of positive degree \( {\mathrm{q}}_{1},{\mathrm{q}}_{2},\ldots ,{\mathrm{q}}_{\mathrm{t}} \in ... | SKETCH OF PROOF OF 4.2. (i) As indicated above \( E \) is a left module over the principal ideal domain \( K\left\lbrack x\right\rbrack \) with \( {fu} = f\left( \phi \right) \left( u\right) \left( {f \in K\left\lbrack x\right\rbrack, u \in E}\right) \) . Since \( E \) is finite dimensional over \( K \) and \( K \subse... | Yes |
Theorem 4.3. Let \( \phi : \mathrm{E} \rightarrow \mathrm{E} \) be a linear transformation of a finite dimensional vector space \( \mathrm{E} \) over a field \( \mathrm{K} \) . Then \( \mathrm{E} \) is a \( \phi \) -cyclic space and \( \phi \) has minimal polynomial \( \mathrm{q} = {\mathrm{x}}^{\mathrm{r}} + {\mathrm{... | PROOF OF 4.3. ( \( \Rightarrow \) ) If \( E \) is \( \phi \) -cyclic, then the remarks preceding Theorem 4.2 show that for some \( v \in E, E \) is the cyclic \( K\left\lbrack x\right\rbrack \) -module \( K\left\lbrack x\right\rbrack v \), with the \( K\left\lbrack x\right\rbrack \) -module structure induced by \( \phi... | Yes |
Corollary 4.4. Let \( \psi : \mathrm{E} \rightarrow \mathrm{E} \) be a linear transformation of a finite dimensional vector space \( \mathrm{E} \) over a field \( \mathrm{K} \) . Then \( \mathrm{E} \) is a \( \psi \) -cyclic space and \( \psi \) has minimal polynomial \( \mathrm{q} = {\left( \mathrm{x} - \mathrm{b}\rig... | SKETCH OF PROOF OF 4.4. Let \( \phi = \psi - b{1}_{E}\varepsilon {\operatorname{Hom}}_{K}\left( {E, E}\right) \) . Then \( q = {\left( x - b\right) }^{r} \) is the minimal polynomial of \( \psi \) if and only if \( {x}^{r} \) is the minimal polynomial of \( \phi \) (for example, \( {\phi }^{r} = {\left( \psi - b{1}_{E}... | Yes |
Lemma 4.5. Let \( \phi : \mathrm{E} \rightarrow \mathrm{E} \) be a linear transformation of an \( \mathrm{n} \) -dimensional vector space \( \mathrm{E} \) over a field \( \mathrm{K} \) . For each \( \mathrm{i} = 1,\ldots \), t let \( {\mathrm{M}}_{\mathrm{i}} \) be an \( {\mathrm{n}}_{\mathrm{i}} \times {\mathrm{n}}_{\... | SKETCH OF PROOF OF 4.5. ( \( \Rightarrow \) ) For each \( i \) let \( {V}_{i} \) be an ordered basis of \( {E}_{i} \) such that the matrix of \( \phi \mid {E}_{i} \) relative to \( {V}_{i} \) is \( {M}_{i} \) . Since \( E = {E}_{1} \oplus \cdots \oplus {E}_{t} \), it follows easily that \( V = \mathop{\bigcup }\limits_... | Yes |
Corollary 4.8. Let \( \phi : \mathrm{E} \rightarrow \mathrm{E} \) be a linear transformation of an \( \mathrm{n} \) -dimensional vector space \( \mathrm{E} \) over a field \( \mathrm{K} \) .\n\n(i) If \( \phi \) has matrix \( \mathrm{A}\varepsilon {\operatorname{Mat}}_{\mathrm{n}}\mathrm{K} \) relative to some basis, t... | PROOF. Exercise. | No |
Proposition 4.9. Let \( \mathrm{A} \) be an \( \mathrm{n} \times \mathrm{n} \) matrix over a field \( \mathrm{K} \) . Then the matrix of polynomials \( {\mathrm{{xI}}}_{\mathrm{n}} - \mathrm{A}\varepsilon {\operatorname{Mat}}_{\mathrm{n}}\mathrm{K}\left\lbrack \mathrm{x}\right\rbrack \) is equivalent (over \( \mathrm{K... | SKETCH OF PROOF OF 4.9. Let \( \phi : {K}^{n} \rightarrow {K}^{n} \) be the \( K \) -linear transformation with matrix \( A = \left( {a}_{ij}\right) \) relative to the standard basis \( \left\{ {\varepsilon }_{i}\right\} \) of \( {K}^{n} \) . As usual \( {K}^{n} \) is a \( K\left\lbrack x\right\rbrack \) -module with s... | No |
Lemma 5.1. (i) If \( {A}_{1},{A}_{2},\ldots ,{A}_{r} \) are square matrices (of various sizes) over a commutative ring \( \mathbf{K} \) with identity and \( {\mathbf{p}}_{\mathrm{i}} \in \mathbf{K}\left\lbrack \mathbf{x}\right\rbrack \) is the characteristic polynomial of \( {\mathbf{A}}_{\mathrm{i}} \), then \( {\math... | SKETCH OF PROOF. (i) If \( {A\varepsilon }{\operatorname{Mat}}_{n}K \) and \( {B\varepsilon }{\operatorname{Mat}}_{m}K \), then\n\n\[ \left( \begin{array}{ll} A & 0 \\ 0 & B \end{array}\right) = \left( \begin{array}{ll} A & 0 \\ 0 & {I}_{m} \end{array}\right) \left( \begin{array}{ll} {I}_{n} & 0 \\ 0 & B \end{array}\ri... | No |
Theorem 5.2. Let \( \phi : \mathrm{E} \rightarrow \mathrm{E} \) be a linear transformation of an \( \mathrm{n} \) -dimensional vector space over a field \( \mathbf{K} \) with characteristic polynomial \( {\mathrm{p}}_{\phi } \in \mathbf{K}\left\lbrack \mathbf{x}\right\rbrack \), minimal polynomial \( {\mathrm{q}}_{\phi... | PROOF. By Theorem \( {4.6\phi } \) has a basis relative to which \( \phi \) has the matrix \( D \) that is the direct sum of the companion matrices of \( {q}_{1},\ldots ,{q}_{t} \) . Therefore, \( {p}_{\phi } = {p}_{D} \) \( = {q}_{1}{q}_{2}\cdots {q}_{t} \) by Lemma 5.1. Furthermore, \( {q}_{\phi } = {q}_{t} \) by The... | Yes |
Theorem 5.4. Let \( \phi : \mathrm{E} \rightarrow \mathrm{E} \) be a linear transformation of a finite dimensional vector space \( \mathrm{E} \) over a field \( \mathrm{K} \). Then the eigenvalues of \( \phi \) are the roots in \( \mathrm{K} \) of the characteristic polynomial \( {\mathrm{p}}_{\phi } \) of \( \phi \). | SKETCH OF PROOF OF 5.4. Let \( A \) be the matrix of \( \phi \) relative to some ordered basis. If \( k \in K \), then \( k{I}_{n} - A \) is the matrix of \( k{1}_{E} - \phi \) relative to the same basis. If \( \phi \left( u\right) = {ku} \) for some nonzero \( {u\varepsilon E} \), then \( \left( {k{1}_{E} - \phi }\rig... | Yes |
Theorem 5.5. Let \( \phi : \mathrm{E} \rightarrow \mathrm{E} \) be a linear transformation of a finite dimensional vector space \( \mathrm{E} \) over a field \( \mathrm{K} \) . Then \( \phi \) has a diagonal matrix \( \mathrm{D} \) relative to some ordered basis of \( \mathrm{E} \) if and only if the eigenvectors of \(... | PROOF. By Theorem IV.2.5 the eigenvectors of \( \phi \) span \( E \) if and only if \( E \) has a basis consisting of eigenvectors. Clearly \( U = \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) is a basis of eigenvectors with corresponding eigenvalue \( {k}_{1},\ldots ,{k}_{n}{\varepsilon K} \) if and only if the matrix ... | Yes |
Proposition 5.6. Let \( \mathrm{K} \) be a commutative ring with identity. Let \( \phi \) be an endomorphism of a free \( \mathrm{K} \) -module of rank \( \mathrm{n} \) and let \( \mathrm{A} = \left( {\mathrm{a}}_{\mathrm{{ij}}}\right) \varepsilon {\operatorname{Mat}}_{\mathrm{n}}\mathrm{K} \) be the matrix of \( \phi ... | PROOF. \( {c}_{0} = {p}_{\phi }\left( 0\right) = \left| {0{I}_{n} - A}\right| = \left| {-A}\right| = {\left( -1\right) }^{n}\left| A\right| \) by Theorem 3.5(viii). Expand \( {p}_{\phi } = \left| {x{I}_{n} - A}\right| \) along the first row. One term of this expansion is \( \left( {x - {a}_{11}}\right) \left( {x - {a}_... | Yes |
Theorem 1.4. A module A satisfies the ascending [resp. descending] chain condition on submodules if and only if \( \mathrm{A} \) satisfies the maximal [resp. minimal] condition on submodules. | PROOF. Suppose \( A \) satisfies the minimal condition on submodules and \( {A}_{1} \supset {A}_{2} \supset \cdots \) is a chain of submodules. Then the set \( \left\{ {{A}_{i} \mid i \geq 1}\right\} \) has a minimal element, say \( {A}_{n} \) . Consequently, for \( i \geq n \) we have \( {A}_{n} \supset {A}_{i} \) by ... | Yes |
Theorem 1.5. Let \( 0 \rightarrow \mathrm{A}\overset{\mathrm{f}}{ \rightarrow }\mathrm{B}\overset{\mathrm{g}}{ \rightarrow }\mathrm{C} \rightarrow 0 \) be a short exact sequence of modules. Then B satisfies the ascending [resp. descending] chain condition on submodules if and only if A and \( \mathrm{C} \) satisfy it. | SKETCH OF PROOF. If \( B \) satisfies the ascending chain condition, then so does its submodule \( f\left( A\right) \) . By exactness \( A \) is isomorphic to \( f\left( A\right) \), whence \( A \) satisfies the ascending chain condition. If \( {C}_{1} \subset {C}_{2} \subset \cdots \) is a chain of submodules of \( C ... | No |
Corollary 1.6. If \( \mathrm{A} \) is a submodule of a module \( \mathrm{B} \) , then \( \mathrm{B} \) satisfies the ascending [resp. descending] chain condition if and only if \( \mathrm{A} \) and \( \mathrm{B}/\mathrm{A} \) satisfy it. | PROOF. Apply Theorem 1.5 to the sequence \( 0 \rightarrow A\overset{ \subset }{ \rightarrow }B \rightarrow B/A \rightarrow 0 \) . | Yes |
Corollary 1.7. If \( {\mathrm{A}}_{1},\ldots ,{\mathrm{A}}_{\mathrm{n}} \) are modules, then the direct sum \( {\mathrm{A}}_{1} \oplus {\mathrm{A}}_{2} \oplus \cdots \oplus {\mathrm{A}}_{\mathrm{n}} \) satisfies the ascending [resp. descending] chain condition on submodules if and only if each \( {\mathrm{A}}_{\mathrm{... | SKETCH OF PROOF. Use induction on \( n \) . If \( n = 2 \), apply Theorem 1.5 to the sequence \( 0 \rightarrow {A}_{1}\overset{{\iota }_{1}}{ \rightarrow }{A}_{1} \oplus {A}_{2}\overset{{\pi }_{2}}{ \rightarrow }{A}_{2} \rightarrow 0 \) . | No |
Theorem 1.8. If \( \mathrm{R} \) is a left Noetherian [resp. Artinian] ring with identity, then every finitely generated unitary left R-module A satisfies the ascending [resp. descending] chain condition on submodules. | PROOF OF 1.8. If \( A \) is finitely generated, then by Corollary IV.2.2 there is a free \( R \) -module \( F \) with a finite basis and an epimorphism \( \pi : F \rightarrow A \) . Since \( F \) is a direct sum of a finite number of copies of \( R \) by Theorem IV.2.1, \( F \) is left Noetherian [resp. Artinian] by Co... | Yes |
Theorem 1.9. A module A satisfies the ascending chain condition on submodules if and only if every submodule of \( \mathrm{A} \) is finitely generated. In particular, a commutative ring \( \mathrm{R} \) is Noetherian if and only if every ideal of \( \mathrm{R} \) is finitely generated. | PROOF. \( \left( \Rightarrow \right) \) If \( B \) is a submodule of \( A \), let \( S \) be the set of all finitely generated submodules of \( B \) . Since \( S \) is nonempty \( \left( {0\varepsilon S}\right), S \) contains a maximal element \( C \) by Theorem 1.4. \( C \) is finitely generated by \( {c}_{1},{c}_{2},... | Yes |
Theorem 1.10. Any two normal series of a module A have refinements that are equivalent. Any two composition series of \( \mathrm{A} \) are equivalent. | PROOF. See the corresponding results for groups (Lemma II.8.9 and Theorems II.8.10 and II.8.11). | No |
Theorem 1.11. A nonzero module \( \mathrm{A} \) has a composition series if and only if \( \mathrm{A} \) satisfies both the ascending and descending chain conditions on submodules. | PROOF. ( \( \Rightarrow \) ) Suppose \( A \) has a composition series \( S \) of length \( n \) . If either chain condition fails to hold, one can find submodules\n\n\[ A = {A}_{0}\underset{ \neq }{\underbrace{ \supset }}{A}_{1}\underset{ \neq }{\underbrace{ \supset }}{A}_{2}\underset{ \neq }{\underbrace{ \supset }}\cd... | Yes |
Corollary 1.12. If \( \mathrm{D} \) is a division ring, then the ring \( {Ma}{t}_{\mathrm{n}}\mathrm{D} \) of all \( \mathrm{n} \times \mathrm{n} \) matrices over \( \mathbf{D} \) is both Artinian and Noetherian. | SKETCH OF PROOF. In view of Definition 1.2 and Theorem 1.11 it suffices to show that \( R = {\operatorname{Mat}}_{n}D \) has a composition series of left \( R \) -modules and a composition of right \( R \) -modules. For each \( i \) let \( {e}_{i}{\varepsilon R} \) be the matrix with \( {1}_{D} \) in position \( \left(... | No |
Theorem 2.1. An ideal \( \mathrm{P}\left( { \neq \mathrm{R}}\right) \) in a commutative ring \( \mathrm{R} \) is prime if and only if \( \mathrm{R} - \mathrm{P} \) is a multiplicative set. | PROOF. This is simply a restatement of Theorem III.2.15; see Definition III.4.1. | No |
Theorem 2.2. If \( \mathrm{S} \) is a multiplicative subset of a ring \( \mathrm{R} \) which is disjoint from an ideal I of \( \mathbf{R} \), then there exists an ideal \( \mathbf{P} \) which is maximal in the set of all ideals of \( \mathrm{R} \) disjoint from \( \mathrm{S} \) and containing \( \mathrm{I} \) . Further... | SKETCH OF PROOF OF 2.2. The set \( \mathcal{S} \) of all ideals of \( R \) that are disjoint from \( S \) and contain \( I \) is nonempty since \( I \in \mathbb{S} \) . Since \( S \neq \varnothing \) (Definition III.4.1) every ideal in \( \mathcal{S} \) is properly contained in \( R \) . \( \mathcal{S} \) is partially ... | Yes |
Theorem 2.3. Let \( \mathrm{K} \) be a subring of a commutative ring \( \mathrm{R} \) . If \( {\mathrm{P}}_{1},\ldots ,{\mathrm{P}}_{\mathrm{n}} \) are prime ideals of \( \mathrm{R} \) such that \( \mathrm{K} \subset {\mathrm{P}}_{1} \cup {\mathrm{P}}_{2} \cup \cdots \cup {\mathrm{P}}_{\mathrm{n}} \), then \( \mathrm{K... | PROOF OF 2.3. Assume \( K ⊄ {P}_{i} \) for every \( i \) . It then suffices to assume that \( n > 1 \) and \( n \) is minimal; that is, for each \( i, K ⊄ \mathop{\bigcup }\limits_{{j \neq i}}{P}_{j} \) . For each \( i \) there exists \( {a}_{i} \in K - \mathop{\bigcup }\limits_{{j \neq i}}{P}_{j} \) . Since \( K \subs... | Yes |
Proposition 2.4. If \( \mathrm{R} \) is a commutative ring with identity and \( \mathrm{P} \) is an ideal which is maximal in the set of all ideals of \( \mathbf{R} \) which are not finitely generated, then \( \mathbf{P} \) is prime. | PROOF. Suppose \( {ab} \in P \) but \( a \notin P \) and \( b \notin P \) . Then \( P + \left( a\right) \) and \( P + \left( b\right) \) are ideals properly containing \( P \) and therefore finitely generated (by maximality). Consequently \( P + \left( a\right) = \left( {{p}_{1} + {r}_{1}a,\ldots ,{p}_{n} + {r}_{n}a}\r... | Yes |
Theorem 2.6. If \( \mathrm{I} \) is an ideal in a commutative ring \( \mathrm{R} \), then \( \operatorname{Rad}\mathrm{I} = \left\{ {\mathrm{r} \in \mathrm{R} \mid {\mathrm{r}}^{\mathrm{n}} \varepsilon \mathrm{I}}\right. \) for some \( \mathrm{n} > 0\} \) . | PROOF. If Rad \( I = R \), then \( \left\{ {r \in R \mid {r}^{n} \in I}\right\} \subset \operatorname{Rad}I \) . Assume Rad \( I \neq R \) . If \( {r}^{n}{\varepsilon I} \) and \( P \) is any prime ideal containing \( I \), then \( {r}^{n}{\varepsilon P} \) whence \( {r\varepsilon P} \) by Theorem III.2.15. Thus \( \le... | Yes |
Theorem 2.7. If \( \mathrm{I},{\mathrm{I}}_{1},{\mathrm{I}}_{2},\ldots ,{\mathrm{I}}_{\mathrm{n}} \) are ideals in a commutative ring \( \mathrm{R} \), then:\n\n(i) \( \operatorname{Rad}\left( {\operatorname{Rad}\mathrm{I}}\right) = \operatorname{Rad}\mathrm{I} \) ;\n\n(ii) \( \operatorname{Rad}\left( {{\mathrm{I}}_{1}... | SKETCH OF PROOF. In each case we prove one of the two required containments. (i) If \( r \in \operatorname{Rad}\left( {\operatorname{Rad}I}\right) \), then \( {r}^{n} \in \operatorname{Rad}I \) and hence \( {r}^{nm} = {\left( {r}^{n}\right) }^{m} \in I \) for some \( n, m > 0 \) . Therefore, \( {r\varepsilon }\operator... | No |
Theorem 2.9. If \( \mathrm{Q} \) is a primary ideal in a commutative ring \( \mathrm{R} \), then \( R \) ad \( \mathrm{Q} \) is a prime ideal. | PROOF. Suppose \( {ab\varepsilon } \) Rad \( Q \) and \( a \notin \operatorname{Rad}Q \) . Then \( {a}^{n}{b}^{n} = {\left( ab\right) }^{n}{\varepsilon Q} \) for some \( n \) . Since \( a \notin \operatorname{Rad}Q,{a}^{n} \notin Q \) . Since \( Q \) is a primary, there is an integer \( m > 0 \) such that \( {\left( {b... | No |
Theorem 2.10. Let \( \mathrm{Q} \) and \( \mathrm{P} \) be ideals in a commutative ring \( \mathrm{R} \) . Then \( \mathrm{Q} \) is primary for \( \mathrm{P} \) if and only if:\n\n(i) \( \mathrm{Q} \subset \mathrm{P} \subset \operatorname{Rad}\mathrm{Q} \) ; and\n\n(ii) if \( \mathrm{{ab}}\varepsilon \mathrm{Q} \) and ... | SKETCH OF PROOF. Suppose (i) and (ii) hold. If \( {ab} \in Q \) with \( a \in Q \), then \( {b\varepsilon P} \subset \operatorname{Rad}Q \), whence \( {b}^{n}{\varepsilon Q} \) for some \( n > 0 \) . Therefore \( Q \) is primary. To show that \( Q \) is primary for \( P \) we need only show \( P = \operatorname{Rad}Q \... | No |
Theorem 2.11. If \( {\mathrm{Q}}_{1},{\mathrm{Q}}_{2},\ldots ,{\mathrm{Q}}_{\mathrm{n}} \) are primary ideals in a commutative ring \( \mathrm{R} \), all of which are primary for the prime ideal \( \mathrm{P} \), then \( \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathrm{Q}}_{\mathrm{i}} \) is also a primary ideal belongin... | PROOF. Let \( Q = \mathop{\bigcap }\limits_{{i = 1}}^{n}{Q}_{i} \) . Then by Theorem 2.7(ii), Rad \( Q = \mathop{\bigcap }\limits_{{i = 1}}^{n}\operatorname{Rad}{Q}_{i} \n\n\( = \mathop{\bigcap }\limits_{{i = 1}}P = P \) ; in particular, \( Q \subset P \subset \operatorname{Rad}Q \) . If \( {ab\varepsilon Q} \) and \( ... | Yes |
Theorem 2.13. Let \( \\mathrm{I} \) be an ideal in a commutative ring \( \\mathrm{R} \) . If \( \\mathrm{I} \) has a primary decomposition, then \( \\mathrm{I} \) has a reduced primary decomposition. | PROOF. If \( I = {Q}_{1} \\cap \\cdots \\cap {Q}_{n}\\left( {Q}_{i}\\right. \) primary \( ) \) and some \( {Q}_{i} \) contains \( {Q}_{1} \\cap \\cdots \\cap \) \( {Q}_{i - 1} \\cap {Q}_{i + 1} \\cap \\cdots \\cap {Q}_{n} \), then \( I = {Q}_{1} \\cap \\cdots \\cap {Q}_{i - 1} \\cap {Q}_{i + 1} \\cap \\cdots \\cap {Q}_... | Yes |
Theorem 3.2. Let \( \mathrm{R} \) be a commutative ring with identity and \( \mathrm{A} \) a primary submodule of an R-module B. Then \( {\mathrm{Q}}_{\mathrm{A}} = \{ \mathrm{r}\varepsilon \mathrm{R} \mid \mathrm{{rB}} \subset \mathrm{A}\} \) is a primary ideal in \( \mathrm{R} \) . | PROOF. Since \( A \neq B,{1}_{R} \notin {Q}_{A} \), whence \( {Q}_{A} \neq R \) . If \( {rs} \in {Q}_{A} \) and \( s \notin {Q}_{A} \), then \( {sB} ⊄ A \) . Consequently, for some \( b \in B,{sb} \in A \) but \( r\left( {sb}\right) \in A \) . Since \( A \) is primary \( {r}^{n}B \subset A \) for some \( n \) ; that is... | Yes |
Theorem 3.4. Let \( \mathrm{R} \) be a commutative ring with identity and \( \mathrm{B} \) an \( \mathrm{R} \) -module. If a submodule \( \mathrm{C} \) of \( \mathrm{B} \) has a primary decomposition, then \( \mathrm{C} \) has a reduced primary decomposition. | SKETCH OF PROOF. The proof is similar to that of Theorem 2.13. Note that if \( {Q}_{A} = \{ r \in R \mid {rB} \subset A\} \), then \( \mathop{\bigcap }\limits_{{i = 1}}^{n}{Q}_{{A}_{i}} = {Q}_{\cap {A}_{i}} \) . Thus if \( {A}_{1},\ldots ,{A}_{r} \) are all \( P \) -primary submodules for the same prime ideal \( P \), ... | No |
Theorem 3.5. Let \( \mathrm{R} \) be a commutative ring with identity and \( \mathrm{B} \) an \( \mathrm{R} \) -module. Let \( \mathrm{C}\left( { \neq \mathrm{B}}\right) \) be a submodule of \( \mathrm{B} \) with two reduced primary decompositions,\n\n\[{\mathrm{A}}_{1} \cap {\mathrm{A}}_{2} \cap \cdots \cap {\mathrm{A... | PROOF. By changing notation if necessary we may assume that \( {P}_{1} \) is maximal in the set \( \left\{ {{P}_{1},\ldots ,{P}_{k},{P}_{1}{}^{\prime },\ldots ,{P}_{s}{}^{\prime }}\right\} \) . We shall first show that \( {P}_{1} = {P}_{j}{}^{\prime } \) for some \( j \) . Suppose, on the contrary, that \( {P}_{1} \neq... | Yes |
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