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Lemma 8.57. Let \( R \) be a Dedekind domain, and let \( I \) be a nonzero ideal of \( R \) . Then there exists a finite product \( {P}_{1}\cdots {P}_{k} \) of nonzero prime ideals, possibly empty and not necessarily having distinct factors, such that \( {P}_{1}\cdots {P}_{k} \subseteq I \) . | Proof. We argue by contradiction. Among all nonzero ideals for which there is no such finite product, choose one, say \( J \), that is maximal under inclusion. This choice is possible since \( R \) is Noetherian. The ideal \( J \) cannot be prime since otherwise \( J \subseteq J \) would be the containment asserted by ... | Yes |
Lemma 8.58. Let \( R \) be a Dedekind domain, regard \( R \) as embedded in its field of fractions \( F \), let \( P \) be a nonzero prime ideal in \( R \), and define\n\n\[ \n{P}^{-1} = \{ x \in F \mid {xP} \subseteq R\} .\n\]\n\nThen the set \( P{P}^{-1} \) of sums of products equals \( R \) . | Proof. By definition of \( {P}^{-1}, P \subseteq P{P}^{-1} \subseteq R \) . Since \( P \) is an ideal and \( P{P}^{-1} \) is closed under addition and negatives, \( P{P}^{-1} \) is an ideal. Property (iii) of Dedekind domains shows that \( P \) is a maximal ideal in \( R \), and therefore \( P{P}^{-1} = P \) or \( P{P}... | Yes |
Corollary 8.59. Let \( R \) be a Dedekind domain, and let \( P \) be a nonzero prime ideal in \( R \). Then there exists an element \( \pi \) in \( P \) such that \( \pi \) is not in \( {P}^{2} \), and any such element has the property that \( {\pi }^{k} \) is not in \( {P}^{k + 1} \) for any \( k \geq 1 \). | Proof. Proposition 8.52 shows that \( {P}^{2} \) is a proper subset of \( P \), and therefore we can find an element \( \pi \) in \( P \) that is not in \( {P}^{2} \). Since the principal ideal \( \left( \pi \right) \) has \( \left( \pi \right) \subseteq P \) and \( \left( \pi \right) \subsetneqq {P}^{2} \), the factor... | Yes |
Corollary 8.60. Let \( R \) be a Dedekind domain, and let \( P \) be a nonzero prime ideal in \( R \). For any integer \( e \geq 1 \), the natural action of \( R \) on powers of \( P \) makes \( {P}^{e - 1}/{P}^{e} \) into a vector space over the field \( R/P \), and this vector space is 1-dimensional. | Proof. Since \( R\left( {P}^{e - 1}\right) \subseteq {P}^{e - 1} \) and \( P\left( {P}^{e - 1}\right) \subseteq {P}^{e} \), we obtain\n\n\[ \left( {R/P}\right) \left( {{P}^{e - 1}/{P}^{e}}\right) \subseteq {P}^{e - 1}/{P}^{e}. \]\n\nThus \( {P}^{e - 1}/{P}^{e} \) is a unital \( R/P \) module, i.e., a vector space over ... | Yes |
Lemma 8.61. If \( P \) and \( Q \) are distinct maximal ideals in an integral domain \( R \) and if \( k \) and \( l \) are positive integers, then \( {P}^{k} + {Q}^{l} = R \) . | Proof. We know that \( {P}^{k} + {Q}^{l} \) is an ideal. Arguing by contradiction, assume that it is proper. Then we can find a maximal ideal \( M \) with \( M \supseteq {P}^{k} + {Q}^{l} \) . This \( M \) satisfies \( M \supseteq {P}^{k} \) and \( M \supseteq {Q}^{l} \) . By Lemma 8.56, \( M \supseteq P \) and \( M \s... | Yes |
Corollary 8.62. If \( R \) is a Dedekind domain with only finitely many prime ideals, then \( R \) is a principal ideal domain. | Proof. Let \( {P}_{1},\ldots ,{P}_{n} \) be the distinct nonzero prime ideals. Theorem 8.55 shows that any nonzero ideal \( I \) in \( R \) factors uniquely as \( I = {P}_{1}^{{k}_{1}}\cdots {P}_{n}^{{k}_{n}} \) with each \( {k}_{j} \geq 0 \) . For \( 1 \leq i \leq n \), Corollary 8.59 produces \( {\pi }_{i} \) in \( {... | Yes |
Corollary 8.63. If \( R \) is a Dedekind domain and if \( I = \mathop{\prod }\limits_{{j = 1}}^{n}{P}_{j}^{{k}_{j}} \) is the unique factorization of a nonzero proper ideal \( I \) as the product of positive powers of distinct prime ideals \( {P}_{j} \), then the map \( r \mapsto \mathop{\prod }\limits_{{j = 1}}^{n}{P}... | Proof. Lemma 8.61 shows that \( {P}_{i}^{{k}_{i}} + {P}_{j}^{{k}_{j}} = R \) if \( i \neq j \) . Then the result follows immediately from the Chinese Remainder Theorem (Theorem 8.27). | No |
Proposition 9.1 If \( x \in \mathbb{K} \) is algebraic over \( \mathbb{k} \), then the minimal polynomial of \( x \) over \( \mathbb{k} \) is prime as a polynomial in \( \mathbb{K}\left\lbrack X\right\rbrack \) . | Proof. Suppose that \( F\left( X\right) \) factors nontrivially as \( F\left( X\right) = G\left( X\right) H\left( X\right) \) . Since \( F\left( x\right) = 0 \), either \( G\left( x\right) = 0 \) or \( H\left( x\right) = 0 \), and then we have a contradiction to the fact that \( F \) has minimal degree among all polyno... | Yes |
If \( x \in \mathbb{K} \) is algebraic over \( \mathbb{k} \), then the field \( \mathbb{k}\left( x\right) \) coincides with the ring \( \mathbb{k}\left\lbrack x\right\rbrack \) . Moreover, if the minimal polynomial of \( x \) over \( \mathbb{k} \) has degree \( n \) , then each element of \( \mathbb{k}\left( x\right) \... | Proof. Since the substitution ring homomorphism \( {\varphi }_{x} \) carries \( \mathbb{k}\left\lbrack X\right\rbrack \) onto \( \mathbb{k}\left\lbrack x\right\rbrack \) , we have an isomorphism of rings \( \mathbb{k}\left\lbrack x\right\rbrack \cong \mathbb{k}\left\lbrack X\right\rbrack /\ker {\varphi }_{x} = \mathbb{... | Yes |
Corollary 9.3. If \( x \in \mathbb{K} \) is algebraic over \( \mathbb{k} \), then the field \( \mathbb{k}\left( x\right) \), regarded as a vector space over \( \mathbb{k} \), is of dimension \( n \), where \( n \) is the degree of the minimal polynomial of \( x \) over \( \mathbb{k} \). The elements \( 1, x,{x}^{2},\ld... | Proof. This is just a restatement of the second conclusion of Theorem 9.2. \( ▱ \) | Yes |
Proposition 9.4. If the vector-space dimension of \( \mathbb{K} \) over \( \mathbb{k} \) is some finite \( n \) , then \( \mathbb{K} \) is an algebraic extension of \( \mathbb{k} \), and each element \( x \) of \( \mathbb{K} \) has some nonzero polynomial \( F\left( X\right) \) in \( \mathbb{k}\left\lbrack X\right\rbra... | Proof. This is immediate since the elements \( 1, x,{x}^{2},\ldots ,{x}^{n} \) of \( \mathbb{K} \) have to be linearly dependent over \( \mathbb{k} \) . | Yes |
Corollary 9.5. If \( x \) is in \( \mathbb{K} \), then \( x \) is algebraic over \( \mathbb{k} \) if and only if \( \mathbb{k}\left( x\right) \) is a finite algebraic extension of \( \mathbb{k} \) . In this case the minimal polynomial of \( x \) over \( \mathbb{k} \) has degree \( \left\lbrack {\mathbb{k}\left( x\right... | Proof. If \( x \) is algebraic over \( \mathbb{k} \), then \( \left\lbrack {\mathbb{k}\left( x\right) : \mathbb{k}}\right\rbrack \) is finite and is the degree of the minimal polynomial of \( x \) over \( \mathbb{k} \), by Corollary 9.3. Proposition 9.4 shows in this case that \( \mathbb{k}\left( x\right) \) is a finit... | Yes |
Theorem 9.6. Let \( \mathbb{k},\mathbb{K} \), and \( \mathbb{L} \) be fields with \( \mathbb{k} \subseteq \mathbb{K} \subseteq \mathbb{L} \), and suppose that \( \left\lbrack {\mathbb{K} : \mathbb{k}}\right\rbrack = n \) and \( \left\lbrack {\mathbb{L} : \mathbb{K}}\right\rbrack = m \), finite or infinite. Let \( \left... | Proof of spanning. If \( \xi \) is in \( \mathbb{L} \), write \( \xi = \mathop{\sum }\limits_{j}{a}_{j}{\xi }_{j} \) with each \( {a}_{j} \) in \( \mathbb{K} \) and with only finitely many \( {a}_{j} \) ’s not 0 . Then expand each \( {a}_{j} \) in terms of the \( {\omega }_{i} \) ’s, and substitute.\n\nProof of LINEAR ... | Yes |
Corollary 9.7. If \( \mathbb{k},\mathbb{K} \), and \( \mathbb{L} \) are fields with \( \mathbb{k} \subseteq \mathbb{K} \subseteq \mathbb{L} \), then\n\n\( \left\lbrack {\mathbb{L} : \mathbb{k}}\right\rbrack = \left\lbrack {\mathbb{L} : \mathbb{K}}\right\rbrack \left\lbrack {\mathbb{K} : \mathbb{k}}\right\rbrack \) | Proof. This is immediate by counting basis elements in Theorem 9.6. | No |
Theorem 9.8. If \( \mathbb{K}/\mathbb{k} \) is a field extension and if \( {x}_{1},\ldots ,{x}_{n} \) are members of \( \mathbb{K} \) that are algebraic over \( \mathbb{k} \), then \( \mathbb{k}\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) is a finite algebraic extension of \( \mathbb{k} \) . | Proof. Since \( {x}_{i} \) is algebraic over \( \mathbb{k} \), it is algebraic over \( \mathbb{k}\left( {{x}_{1},\ldots ,{x}_{i - 1}}\right) \) . Hence \( \left\lbrack {\mathbb{k}\left( {{x}_{1},\ldots ,{x}_{i}}\right) : \mathbb{k}\left( {{x}_{1},\ldots ,{x}_{i - 1}}\right) }\right\rbrack \) is finite. Applying Corolla... | Yes |
Corollary 9.9 If \( \mathbb{K}/\mathbb{k} \) is a field extension, then the elements of \( \mathbb{K} \) that are algebraic over \( \mathbb{k} \) form a field. | Proof. If \( x \) and \( y \) in \( \mathbb{K} \) are algebraic over \( \mathbb{k} \), then \( \mathbb{k}\left( {x, y}\right) \) is a finite algebraic extension of \( \mathbb{k} \), according to Theorem 9.8. This extension contains \( x \pm y \) and \( {xy} \) , and it contains \( {x}^{-1} \) if \( x \neq 0 \) . The co... | Yes |
Theorem 9.10 (existence theorem for simple algebraic extensions). If \( F\left( X\right) \) is a monic prime polynomial in \( \mathbb{k}\left\lbrack X\right\rbrack \), then there exists a simple algebraic extension \( \mathbb{K} = \mathbb{k}\left( x\right) \) of \( \mathbb{k} \) such that \( x \) is a root of \( F\left... | Proof. Define \( \mathbb{K} = \mathbb{k}\left\lbrack X\right\rbrack /\left( {F\left( X\right) }\right) \) as a ring. Since \( F\left( X\right) \) is prime, \( \left( {F\left( X\right) }\right) \) is a nonzero prime ideal, hence maximal. Therefore \( \mathbb{K} \) is a field, an extension field of \( \mathbb{k} \) . Def... | Yes |
Theorem 9.11 (uniqueness theorem for simple algebraic extensions). If \( F\left( X\right) \) is a monic prime polynomial in \( \mathbb{k}\left\lbrack X\right\rbrack \) and if \( \mathbb{K} = \mathbb{k}\left( x\right) \) and \( {\mathbb{K}}^{\prime } = \mathbb{k}\left( y\right) \) are two simple algebraic extensions suc... | Proof. In view of the proof of Theorem 9.10, there is no loss of generality in assuming that \( \mathbb{K} = \mathbb{k}\left\lbrack X\right\rbrack /\left( {F\left( X\right) }\right) \) . Since \( y \) is algebraic over \( \mathbb{k} \), we can form the substitution homomorphism \( {\varphi }_{y} : \mathbb{k}\left\lbrac... | Yes |
Theorem 9.12 (existence of splitting field). If \( F\\left( X\\right) \) is a nonconstant polynomial in \( \\mathbb{k}\\left\\lbrack X\\right\\rbrack \), then there exists a splitting field of \( F\\left( X\\right) \) over \( \\mathbb{k} \) . | Proof. We begin by constructing a certain extension field \( \\mathbb{K} \) of \( \\mathbb{k} \) in which \( F\\left( X\\right) \) factors completely into degree-one factors in \( \\mathbb{K}\\left\\lbrack X\\right\\rbrack \) . We do so by induction on \( n = \\deg F\\left( X\\right) \) . For \( n = 1 \), there is noth... | Yes |
Theorem 9.13 (uniqueness of splitting field). If \( F\\left( X\\right) \) is a nonconstant polynomial in \( \\mathbb{k}\\left\\lbrack X\\right\\rbrack \), then any two splitting fields of \( F\\left( X\\right) \) over \( \\mathbb{k} \) are \( \\mathbb{k} \) isomorphic. | The idea of the proof is simple enough, but carrying out the idea runs into a technical complication. The idea is to proceed by induction, using the uniqueness result for simple algebraic extensions (Theorem 9.11) repeatedly until all the roots have been addressed. The difficulty is that after one step the coefficients... | No |
Proposition 9.15. Differentiation on \( \mathbb{k}\left\lbrack X\right\rbrack \) satisfies the product rule: \( F = {GH} \) implies \( {F}^{\prime } = {G}^{\prime }H + G{H}^{\prime } \) . | Proof. Because of the \( \mathbb{k} \) linearity, it is enough to prove the result for monomials. Thus let \( G\left( X\right) = {X}^{m} \) and \( H\left( X\right) = {X}^{n} \), so that \( F\left( X\right) = {X}^{m + n} \) . Then \( {F}^{\prime }\left( X\right) = \left( {m + n}\right) {X}^{m + n - 1},{G}^{\prime }\left... | Yes |
Corollary 9.16. If \( n \) is a positive integer, if \( r \) is in \( \mathbb{k} \), and if \( F\left( X\right) = {\left( X - r\right) }^{n} \) in \( \mathbb{k}\left\lbrack X\right\rbrack \), then \( {F}^{\prime }\left( X\right) = n{\left( X - r\right) }^{n - 1} \) . | Proof. This is immediate by induction from Proposition 9.15 since the derivative of \( X - r \) is 1 . | No |
Corollary 9.17. Let \( r \) be in \( \mathbb{k} \), and let \( F\left( X\right) \) be in \( \mathbb{k}\left\lbrack X\right\rbrack \) . If \( {\left( X - r\right) }^{2} \) divides \( F\left( X\right) \), then \( F\left( r\right) = {F}^{\prime }\left( r\right) = 0 \) . | Proof. Write \( F\left( X\right) = {\left( X - r\right) }^{2}G\left( X\right) \) . If we substitute \( r \) for \( X \), we see that \( F\left( r\right) = 0 \) . If instead we differentiate, using Proposition 9.15 and Corollary 9.16, then we obtain \( {F}^{\prime }\left( X\right) = 2\left( {X - r}\right) G\left( X\righ... | Yes |
Lemma 9.18. If \( \mathbb{k} \) is a field of characteristic \( p \neq 0 \), then the map \( \varphi : \mathbb{k} \rightarrow \mathbb{k} \) given by \( \varphi \left( x\right) = {x}^{p} \) is a field mapping. | Proof. The computation \( \varphi \left( {uv}\right) = {\left( uv\right) }^{p} = {u}^{p}{v}^{p} = \varphi \left( u\right) \varphi \left( v\right) \) shows that \( \varphi \) respects products. If \( u \) and \( v \) are in \( \mathbb{k} \), then\n\n\[ \varphi \left( {u + v}\right) = {\left( u + v\right) }^{p} = \varphi... | Yes |
If \( q \) and \( r \) are integers with \( 2 \leq q \leq r \), then the finite field \( {\mathbb{F}}_{q} \) is isomorphic to a subfield of the finite field \( {\mathbb{F}}_{r} \) if and only if \( r = {q}^{n} \) for some integer \( n \geq 1 \) . | Proof. If \( {\mathbb{F}}_{q} \) is isomorphic to a subfield of \( {\mathbb{F}}_{r} \), then we may consider \( {\mathbb{F}}_{r} \) as a vector space over \( {\mathbb{F}}_{q} \), say of dimension \( n \) . In this case, \( {\mathbb{F}}_{r} \) has \( {q}^{n} \) elements.\n\nConversely let \( r = {q}^{n} \), and regard \... | Yes |
Proposition 9.20. The following conditions on the field \( \mathbb{k} \) are equivalent:\n\n(a) \( \mathbb{k} \) has no nontrivial algebraic extensions,\n\n(b) every irreducible polynomial in \( \mathbb{k}\left\lbrack X\right\rbrack \) has degree 1,\n\n(c) every polynomial in \( \mathbb{k}\left\lbrack X\right\rbrack \)... | Proof. If (a) holds, then (b) holds since any irreducible polynomial of degree greater than 1 would give a nontrivial simple algebraic extension (Theorem 9.10). If (b) holds and a polynomial of positive degree is given, apply (b) to an irreducible factor to see that the given polynomial has a root; thus (c) holds. Cond... | Yes |
Lemma 9.21. If \( \mathbb{K}/\mathbb{k} \) is an algebraic extension of fields and if every nonconstant polynomial in \( \mathbb{k}\left\lbrack X\right\rbrack \) splits into degree-one factors in \( \mathbb{K} \), then \( \mathbb{K} \) is algebraically closed. | Proof. Let \( {\mathbb{K}}^{\prime } \) be an algebraic extension of \( \mathbb{K} \), and let \( x \) be in \( {\mathbb{K}}^{\prime } \) . Let \( G\left( X\right) \) be the minimal polynomial of \( x \) over \( \mathbb{K} \), and write \( G\left( X\right) \) as\n\n\[ G\left( X\right) = {X}^{n} + {c}_{n - 1}{X}^{n - 1}... | Yes |
Theorem 9.22 (Steinitz). Every field \( \mathbb{k} \) has an algebraic closure, and this is unique up to \( \mathbb{k} \) isomorphism. | Proof OF EXISTENCE. With \( \mathbb{k} \) as the given field, let \( S \) be the set of nonconstant polynomials \( s\left( X\right) \) in \( \mathbb{k}\left\lbrack X\right\rbrack \), and introduce a well ordering into \( S \) by means of Zermelo’s Well-Ordering Theorem (Section A5 of the appendix). Let us write \( \pre... | No |
Theorem 9.24. The set \( \mathcal{C} \) of \( x \) coordinates that can be constructed from \( x = 1 \) and \( x = 0 \) by straightedge and compass forms a subfield of \( \mathbb{R} \) such that the square root of any positive element of the field lies in the field. Conversely the members of \( \mathcal{C} \) are those... | Proof of CONVERSE. Suppose we have a subfield \( F = {F}_{n} \) of \( \mathbb{R} \) of the kind described in the statement of the theorem. The possibilities for obtaining a new constructible point from \( F \) by an additional construction arise from three situations: the intersection of two lines, each passing through... | Yes |
Theorem 9.25 (Gauss). \( {}^{3} \) A regular \( n \) -gon is constructible with straightedge and compass if and only if \( n \) is the product of distinct Fermat primes and a power of 2. | For the necessity let \( n \) be prime, and suppose that a regular \( n \) -gon is constructible. Returning from degrees to radians, we observe that each central angle is \( {2\pi }/n \) . Thus the constructibility implies the constructibility of \( \cos {2\pi }/n \), and it follows that \( {e}^{{2\pi i}/n} = \cos {2\p... | Yes |
Lemma 9.26. A polynomial \( F\left( X\right) \) in \( \mathbb{k}\left\lbrack X\right\rbrack \) has no repeated roots in its splitting field \( \mathbb{K} \) if and only if \( \operatorname{GCD}\left( {F,{F}^{\prime }}\right) = 1 \), where \( {F}^{\prime }\left( X\right) \) is the derivative of \( F\left( X\right) \) . | Proof. The polynomial \( F\left( X\right) \) has repeated roots in \( \mathbb{K} \) if and only if \( F\left( X\right) \) is divisible by \( {\left( X - r\right) }^{2} \) for some \( r \in \mathbb{K} \), if and only if some \( r \in \mathbb{K} \) has \( F\left( r\right) = \) \( {F}^{\prime }\left( r\right) = 0 \) (by C... | Yes |
Proposition 9.27. An irreducible polynomial \( F\left( X\right) \) in \( \mathbb{k}\left\lbrack X\right\rbrack \) is separable if and only if \( {F}^{\prime }\left( X\right) \neq 0 \) . In particular, every irreducible (necessarily nonconstant) polynomial is separable if \( \mathbb{k} \) has characteristic 0 . | Proof. Since the polynomial \( F\left( X\right) \) is irreducible and \( \operatorname{GCD}\left( {F,{F}^{\prime }}\right) \) divides \( F\left( X\right) ,\operatorname{GCD}\left( {F,{F}^{\prime }}\right) \) equals 1 or \( F\left( X\right) \) in all cases. If \( {F}^{\prime }\left( X\right) = 0 \), then \( \operatornam... | Yes |
Theorem 9.28. Let \( \mathbb{k} \subseteq \mathbb{L} \subseteq \mathbb{K} \) be an inclusion of fields such that \( \mathbb{K} \) is a simple algebraic extension of \( \mathbb{L} \) of the form \( \mathbb{K} = \mathbb{L}\left( \alpha \right) \), let \( \overline{\mathbb{K}} \) be an algebraic closure of \( \mathbb{K} \... | Proof. Any field mapping \( \varphi : \mathbb{K} \rightarrow \overline{\mathbb{K}} \) is uniquely determined by \( {\left. \varphi \right| }_{\mathbb{L}} \) and \( \varphi \left( \alpha \right) \) . If \( \sigma = {\left. \varphi \right| }_{\mathbb{L}} \), then the equality \( M\left( \alpha \right) = 0 \) implies that... | Yes |
Corollary 9.29. Let \( \mathbb{K} = \mathbb{k}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) be a finite algebraic extension of the field \( \mathbb{k} \), and let \( \overline{\mathbb{K}} \) be an algebraic closure of \( \mathbb{K} \). Then the number of field mappings of \( \mathbb{K} \) into \( \overline{\ma... | Proof. For \( 1 \leq j \leq n \), let \( {M}_{j}\left( X\right) \) be the minimal polynomial of \( {\alpha }_{j} \) over \( \mathbb{k}\left( {{\alpha }_{1},\ldots ,{\alpha }_{j - 1}}\right) \), let \( {d}_{j} \) be the degree of \( {M}_{j}\left( X\right) \), and let \( {s}_{j} \) be the number of distinct roots of \( {... | Yes |
Corollary 9.30. Let \( \mathbb{K} = \mathbb{k}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) be a finite algebraic extension of the field \( \mathbb{k} \) . If each \( {\alpha }_{j} \) for \( 1 \leq j \leq n \) is separable over \( \mathbb{k} \), then \( \mathbb{K}/\mathbb{k} \) is a separable extension. | Proof. Let \( \beta \) be in \( \mathbb{K} \), We apply the equivalence of (a) and (c) in Corollary 9.29 once to the set of generators \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \) and once to the set of generators \( \left\{ {\beta ,{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \), and the result is immed... | No |
Corollary 9.31. If \( \mathbb{K}/\mathbb{k} \) is an algebraic field extension, then the subset \( \mathbb{L} \) of elements of \( \mathbb{K} \) that are separable over \( \mathbb{k} \) is a subfield of \( \mathbb{K} \) . | Proof. If \( \alpha \) and \( \beta \) are given in \( \mathbb{L} \), we apply Corollary 9.30 to the extension \( \mathbb{k}\left( {\alpha ,\beta }\right) \) of \( \mathbb{k} \) to see that \( \mathbb{L} \) contains the subfield generated by \( \mathbb{k} \) and the elements \( \alpha \) and \( \beta \) . | Yes |
Proposition 9.32. If \( \mathbb{K}/\mathbb{k} \) is a separable algebraic extension and if \( \mathbb{L} \) is a field with \( \mathbb{k} \subseteq \mathbb{L} \subseteq \mathbb{K} \), then \( \mathbb{K} \) is separable over \( \mathbb{L} \), and \( \mathbb{L} \) is separable over \( \mathbb{k} \) . | Proof. The separability assertion about \( \mathbb{L}/\mathbb{k} \) says the same thing about elements of \( \mathbb{L} \) that separability of \( \mathbb{K}/\mathbb{k} \) says about those same elements, and it is therefore immediate that \( \mathbb{L}/\mathbb{k} \) is separable.\n\nNext let us consider \( \mathbb{K}/\... | Yes |
Proposition 9.35. Let \( \mathbb{K} \) be a finite separable algebraic extension of a field \( \mathbb{k} \) , so that \( \left| {\operatorname{Gal}\left( {\mathbb{K}/\mathbb{k}}\right) }\right| \leq \left\lbrack {\mathbb{K} : \mathbb{k}}\right\rbrack \) . Then the following are equivalent.\n\n(a) \( \mathbb{K} \) is t... | Proof. By separability and Theorem 9.34 we can write \( \mathbb{K} = \mathbb{k}\left( \gamma \right) \) throughout the proof for some \( \gamma \) in \( \mathbb{K} \) . Let \( M\left( X\right) \) be the minimal polynomial of \( \gamma \) over \( \mathbb{k} \) .\n\nSuppose (a) holds. We prove (c). Write \( \mathbb{K} = ... | Yes |
Corollary 9.36. If \( \mathbb{K} \) is a finite normal separable extension of \( \mathbb{k} \) and if \( \mathbb{L} \) is a field with \( \mathbb{k} \subseteq \mathbb{L} \subseteq \mathbb{K} \), then \( \mathbb{K} \) is a finite normal separable extension of \( \mathbb{L} \), and the subgroup \( H = \operatorname{Gal}\... | Proof. The field \( \mathbb{K} \) is a separable extension of the intermediate field \( \mathbb{L} \) by Proposition 9.32, and it is a normal extension by Proposition 9.35a. Therefore Proposition 9.35c gives \( \left| {\operatorname{Gal}\left( {\mathbb{K}/\mathbb{L}}\right) }\right| = \left\lbrack {\mathbb{K} : \mathbb... | Yes |
Proposition 9.40. Let \( \mathbb{K} \) be a finite extension of the finite field \( {\mathbb{F}}_{q} \), where \( q = {p}^{a} \) and \( p \) is prime, and suppose that \( \left\lbrack {\mathbb{K} : {\mathbb{F}}_{q}}\right\rbrack = n \) . Then \( \mathbb{K} \) is a Galois extension of \( {\mathbb{F}}_{q} \), the Galois ... | Proof. Theorem 9.14 shows that \( \mathbb{K} \) is a splitting field for \( {X}^{{q}^{n}} - X \) over \( {\mathbb{F}}_{p} \) . Hence it is a splitting field for \( {X}^{{q}^{n}} - X \) over \( {\mathbb{F}}_{q} \), and \( \mathbb{K}/{\mathbb{F}}_{q} \) is a normal extension. The polynomial \( {X}^{{q}^{n}} - X \) has no... | Yes |
Each cyclotomic polynomial \( {\Phi }_{n}\left( X\right) \) lies in \( \mathbb{Z}\left\lbrack X\right\rbrack \), and the degree of \( {\Phi }_{n}\left( X\right) \) is \( \varphi \left( n\right) \), where \( \varphi \) is the Euler \( \varphi \) function defined just before Corollary 1.10. | Proof. We know that \( {\Phi }_{n}\left( X\right) \) is in \( \mathbb{C}\left\lbrack X\right\rbrack \), and we begin by showing by induction on \( n \) that \( {\Phi }_{n}\left( X\right) \) is in \( \mathbb{Q}\left\lbrack X\right\rbrack \) . For \( n = 1 \), we have \( {\Phi }_{1}\left\lbrack X\right\rbrack = X - 1 \),... | Yes |
There are no finite extensions of \( \mathbb{R} \) of odd degree greater than 1, the only extension of \( \mathbb{R} \) of degree 2 up to \( \mathbb{R} \) isomorphism is \( \mathbb{C} \), and there are no finite extensions of \( \mathbb{C} \) of degree 2. | Proof. If \( \mathbb{K} \) is a finite extension of \( \mathbb{R} \) of odd degree and if \( x \) is in \( \mathbb{K} \), then \( \left\lbrack {\mathbb{R}\left( x\right) : \mathbb{R}}\right\rbrack \) is odd, and consequently the minimal polynomial \( F\left( X\right) \) of \( x \) over \( \mathbb{R} \) is irreducible o... | Yes |
Lemma 9.45. Let \( \mathbb{k} \) be a field of any characteristic, and let \( p \) be a prime number. If \( a \) is a member of \( \mathbb{k} \) such that \( {X}^{p} - a \) has no root in \( \mathbb{k} \), then \( {X}^{p} - a \) is irreducible in \( \mathbb{k} \) . | Proof. First suppose that \( p \) is different from the characteristic. Let \( \mathbb{L} \) be a splitting field for \( {X}^{p} - a \) . The derivative of \( {X}^{p} - a \), evaluated at any root of \( {X}^{p} - a \) in \( \mathbb{L} \), is nonzero, and Corollary 9.17 shows that \( {X}^{p} - a \) splits as the product... | Yes |
Lemma 9.46. The number 3 is a generator of \( {\mathbb{F}}_{n}^{ \times } \) when \( n \) is prime of the form \( {2}^{{2}^{N}} + 1 \) with \( N > 0 \) . | REMARKS. We verified this assertion for \( n = {17} \) in Section 6, and in principle one could verify the lemma in any particular case in the same way. Here is a general argument using the law of quadratic reciprocity, whose full statement and proof will be given in Advanced Algebra. For a prime number \( n \) that is... | No |
Proposition 9.47. Let \( \\mathbb{K} \) be a finite normal extension of a field \( \\mathbb{k} \) of characteristic 0, suppose that \( \\operatorname{Gal}\\left( {\\mathbb{K}/\\mathbb{k}}\\right) \) is cyclic of order \( n \) with \( \\sigma \) as a generator, and suppose that \( {X}^{n} - 1 \) splits in \( \\mathbb{k}... | PROOF. For \( x \) in \( \\mathbb{K} \), we compute\n\n\\[ \n{E}_{r}{E}_{s}x = {n}^{-2}\\mathop{\\sum }\\limits_{{k{\\;\\operatorname{mod}\\;n}}}{\\omega }^{-{kr}}{\\sigma }^{k}\\left( {\\mathop{\\sum }\\limits_{{l{\\;\\operatorname{mod}\\;n}}}{\\omega }^{-{ls}}{\\sigma }^{l}x}\\right)\n\\]\n\n\\[ \n= {n}^{-2}\\mathop{... | Yes |
Corollary 9.48. Let \( \mathbb{K} \) be a finite normal extension of a field \( \mathbb{k} \) of characteristic 0, suppose that \( \operatorname{Gal}\left( {\mathbb{K}/\mathbb{k}}\right) \) is cyclic of prime order \( p \), and suppose that \( {X}^{p} - 1 \) splits in \( \mathbb{k} \) . Then there exist \( a \) in \( \... | Proof. We apply Proposition 9.47 with \( n = p \) . Since \( \left\lbrack {\mathbb{K} : \mathbb{k}}\right\rbrack = p > 1 \), (d) shows that \( {E}_{0} \) is not the identity. By (b), some \( {E}_{r} \) with \( r \neq 0 \) is not the 0 operator. Let \( x \) be a nonzero element in image \( {E}_{r} \) . Since the generat... | Yes |
Lemma 9.49. Let \( \mathbb{k} \) be a field of characteristic 0, let \( n > 0 \) be an integer, and let \( \mathbb{K} \) be a splitting field for \( \mathop{\prod }\limits_{{r = 1}}^{n}\left( {{X}^{r} - 1}\right) \) over \( \mathbb{k} \) . Then \( \mathbb{K}/\mathbb{k} \) is a Galois extension, the Galois group of \( \... | Proof. Being a splitting field in characteristic \( 0,\mathbb{K} \) is a finite Galois extension of \( \mathbb{k} \) . For \( 1 \leq r \leq n \), let \( {\omega }_{r} \) be a primitive \( {r}^{\text{th }} \) root of 1 in \( \mathbb{K} \) . The primitive \( {r}^{\text{th }} \) roots of 1 are parametrized by the group \(... | Yes |
Theorem 9.50 (Cardan’s formula). Let \( \mathbb{k} \) be a field of characteristic 0 containing \( \sqrt{-3} \), and let \( {X}^{3} + {pX} + q \) be an irreducible cubic in \( \mathbb{k}\left\lbrack X\right\rbrack \) . For this polynomial the discriminant \( D \) is given by\n\n\[ D = - 4{p}^{3} - {27}{q}^{2} \]\n\nThe... | Proof. Define \( {\sigma }_{k} = {r}_{1}^{k} + {r}_{2}^{k} + {r}_{3}^{k} \) for \( 1 \leq k \leq 4 \) . By inspection we have\n\n\[ \left( \begin{matrix} 1 & 1 & 1 \\ {r}_{1} & {r}_{2} & {r}_{3} \\ {r}_{1}^{2} & {r}_{2}^{2} & {r}_{3}^{2} \end{matrix}\right) \left( \begin{array}{lll} 1 & {r}_{1} & {r}_{1}^{2} \\ 1 & {r}... | Yes |
Lemma 9.52. Let \( f\left( X\right) \) in \( \mathbb{C}\left\lbrack X\right\rbrack \) be given by \( f\left( X\right) = \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}{X}^{k} \), and define \( F\left( X\right) \) to be the sum of the derivatives of \( f\left( X\right) \) :\n\n\[ F\left( X\right) = \mathop{\sum }\limits_{{l ... | PROOF. We calculate directly that\n\n\[ F\left( z\right) = \mathop{\sum }\limits_{{l = 0}}^{n}\mathop{\sum }\limits_{{k = l}}^{n}\frac{{a}_{k}k!}{\left( {k - l}\right) !}{z}^{k - l} = \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}\mathop{\sum }\limits_{{l = 0}}^{k}\frac{k!}{\left( {k - l}\right) !}{z}^{k - l} = \mathop{\su... | Yes |
Proposition 9.53. If \( \mathbb{K}/\mathbb{k} \) is a finite extension of fields with \( n = \left\lbrack {\mathbb{K} : \mathbb{k}}\right\rbrack \), then norms and traces relative to \( \mathbb{K}/\mathbb{k} \) have the following properties:\n\n(a) \( N\left( {ab}\right) = N\left( a\right) N\left( b\right) \),\n\n(b) \... | Proof. Properties (a) and (b) follow from properties of the determinant in combination with the identities \( M\left( {ab}\right) = M\left( a\right) M\left( b\right) \) and \( M\left( {ca}\right) = {cM}\left( a\right) \) . Properties (c) and (d) follow from properties of the trace in combination with the identities \( ... | Yes |
Proposition 9.54. Let \( \mathbb{K}/\mathbb{k} \) and \( \mathbb{L}/\mathbb{K} \) be finite extensions of fields with \( \left\lbrack {\mathbb{K} : \mathbb{k}}\right\rbrack = n \) and \( \left\lbrack {\mathbb{L} : \mathbb{K}}\right\rbrack = m \), and let \( a \) be in \( \mathbb{K} \) . The element \( a \) acts by mult... | Proof. We choose the bases as in Theorem 7.6. Thus let \( \Gamma = \left( {{\omega }_{1},{\omega }_{2},\ldots }\right) \) be an ordered basis of \( \mathbb{K} \) over \( \mathbb{k} \), and let \( \Delta = \left( {{\xi }_{1},{\xi }_{2},\ldots }\right) \) be a basis of \( \mathbb{L} \) over \( \mathbb{K} \) . Theorem 7.6... | Yes |
Corollary 9.55. Let \( \mathbb{K}/\mathbb{k} \) and \( \mathbb{L}/\mathbb{K} \) be finite extensions of fields with \( \left\lbrack {\mathbb{L} : \mathbb{K}}\right\rbrack = m \), and let \( a \) be in \( \mathbb{K} \) . Let \( {M}_{\mathbb{K}/\mathbb{k}}\left( a\right) \) and \( {M}_{\mathbb{L}/\mathbb{k}}\left( a\righ... | Proof. Proposition 9.54 shows that the matrix of \( {XI} - {M}_{\mathbb{L}/\mathbb{k}}\left( a\right) \) may be taken to be block diagonal with each of the \( m \) diagonal blocks equal to the matrix of \( {XI} - {M}_{\mathbb{K}/\mathbb{k}}\left( a\right) \) . The determinant of \( {XI} - {M}_{\mathbb{L}/\mathbb{k}}\le... | Yes |
Let \( \mathbb{K}/\mathbb{k} \) be a finite extension of fields, and let \( a \) be in \( \mathbb{K} \) . Then the field polynomial of \( a \) relative to \( \mathbb{K}/\mathbb{k} \) is a power of the minimal polynomial of \( a \) over \( \mathbb{k} \), the power being \( \left\lbrack {\mathbb{K} : \mathbb{k}\left( a\r... | Proof. If \( F\left( X\right) \) is in \( \mathbb{k}\left\lbrack X\right\rbrack \), then the operation \( M \) of multiplication has\n\n\[ M\left( {F\left( a\right) }\right) b = F\left( a\right) b = F\left( {M\left( a\right) }\right) b\;\text{ for }b \in \mathbb{K}, \]\n\n\( \left( *\right) \)\n\nas we see by first con... | Yes |
Proposition 9.57. Let \( \mathbb{k} \) be a field, let \( \mathbb{k}\left( a\right) \) be an algebraic extension of \( \mathbb{k} \), and suppose that the minimal polynomial \( F\left( X\right) \) of \( a \) over \( \mathbb{k} \) is separable. Let \( \mathbb{K} \) be a splitting field of \( F\left( X\right) \), and fac... | Proof. Corollary 9.56 shows that \( F\left( X\right) \) equals the field polynomial of \( a \) relative to \( \mathbb{k}\left( a\right) /\mathbb{k} \), i.e., is the characteristic polynomial of the multiplication operator \( {M}_{\mathbb{k}\left( a\right) /\mathbb{k}}\left( a\right) \) . Let \( A \) be the matrix of \(... | Yes |
Corollary 9.59. If \( \mathbb{K} \) is a finite separable extension of the field \( \mathbb{k} \), then the trace function \( {\operatorname{Tr}}_{\mathbb{K}/\mathbb{k}} \) is not identically 0 . | Proof. By the Theorem of the Primitive Element (Theorem 9.34), we can write \( \mathbb{K} = \mathbb{k}\left( a\right) \) for some \( a \neq 0 \) . Let \( {\mathbb{K}}^{\prime } \) be a splitting field for the minimal polynomial of \( a \) over \( \mathbb{k} \) . Then \( {\mathbb{K}}^{\prime }/\mathbb{k} \) is a separab... | Yes |
Theorem 9.60. Let \( R \) be a Dedekind domain, let \( F \) be its field of fractions, let \( K \) be a finite separable extension of \( F \) with \( \left\lbrack {K : F}\right\rbrack = n \), and let \( T \) be the integral closure of \( R \) in \( K \). If \( \mathfrak{p} \) is a nonzero prime ideal in \( R \) and \( ... | PROOF. Corollary 8.63 gives a ring isomorphism \[ T/\left( {\mathfrak{p}T}\right) \cong T/{P}_{1}^{{e}_{1}} \times \cdots \times T/{P}_{g}^{{e}_{g}}. \] \( \left( *\right) \) Recall from the definition of residue class degree that we have a field mapping of \( R/\mathfrak{p} \) into each \( T/{P}_{i} \). Since \( \math... | Yes |
Lemma 9.61. Let \( R \) be a Dedekind domain, let \( F \) be its field of fractions, let \( K \) be a finite separable extension of \( F \), and let \( T \) be the integral closure of \( R \) in \( K \) . Suppose that \( K \) is Galois over \( F \) . If \( \mathfrak{p} \) is a nonzero prime ideal in \( R \) and \( \mat... | Proof. Arguing by contradiction, suppose that \( {P}_{j} \) is not of the form \( \sigma \left( {P}_{1}\right) \) for some \( \sigma \) in \( \operatorname{Gal}\left( {K/F}\right) \) . By the Chinese Remainder Theorem we can choose an element \( t \) of \( T \) with \( t \equiv 0{\;\operatorname{mod}\;{P}_{j}} \) and \... | Yes |
Theorem 9.62. Let \( R \) be a Dedekind domain, let \( F \) be its field of fractions, let \( K \) be a finite separable extension of \( F \) with \( \left\lbrack {K : F}\right\rbrack = n \), and let \( T \) be the integral closure of \( R \) in \( K \). Suppose that \( K \) is Galois over \( F \). If \( \mathfrak{p} \... | Proof. For \( \sigma \) in \( \operatorname{Gal}\left( {K/F}\right) \), apply \( \sigma \) to the factorization \( \mathfrak{p}T = \mathop{\prod }\limits_{{i = 1}}^{g}{P}_{i}^{{e}_{i}} \), obtaining \[ \mathfrak{p}T = \sigma {\left( {P}_{1}\right) }^{{e}_{1}}\mathop{\prod }\limits_{{i = 2}}^{g}\sigma {\left( {P}_{i}\ri... | Yes |
Proposition 9.63. Let \( \mathbb{K}/\mathbb{k} \) be a finite Galois extension, and suppose that \( \mathbb{K} \) is the splitting field of a separable polynomial \( F\left( X\right) \) in \( \mathbb{k}\left\lbrack X\right\rbrack \) of degree \( d \) . Let \( D \) be the discriminant of \( F\left( X\right) \), and rega... | Proof. Let \( {r}_{1},\ldots ,{r}_{d} \) be the roots of \( F\left( X\right) \), and put \( \Delta = \mathop{\prod }\limits_{{i < j}}\left( {{r}_{j} - {r}_{i}}\right) \) . Under the identification of \( G \) with a subgroup of the permutation group \( {\mathfrak{S}}_{d} \) on \( \{ 1,\ldots, d\} \), each \( \sigma \) i... | Yes |
Theorem 9.64. Let \( R \) be a Dedekind domain, let \( F \) be its field of fractions, let \( K \) be a finite separable extension of \( F \) with \( \left\lbrack {K : F}\right\rbrack = n \), and let \( T \) be the integral closure of \( R \) in \( K \) . Suppose that \( K \) is Galois over \( F \) . Let \( \mathfrak{p... | Proof. Let \( {K}^{d} \) be the fixed field of \( {G}_{P} \) within \( K \) ; Theorem 9.38 shows that \( \operatorname{Gal}\left( {K/{K}^{d}}\right) = {G}_{P} \) . Let \( {T}^{d} \) be the integral closure of \( R \) in \( {K}^{d} \) ; this is a Dedekind domain, and \( T \) is the integral closure of \( {T}^{d} \) in \... | Yes |
Proposition 10.1. If the unital left \( R \) module \( M \) is semisimple, then \( M \) is the direct sum of some family of simple \( R \) submodules. In more detail if \( \left\{ {{M}_{s} \mid s \in S}\right\} \) is a family of simple \( R \) submodules of the unital left \( R \) module \( M \) whose sum is \( M \), t... | Proof. Call a subset \( U \) of \( S \) \ | No |
Proposition 10.2. Let \( M \) be a semisimple left \( R \) module, and suppose that \( M = {\bigoplus }_{s \in S}{M}_{s} \) is the direct sum of simple \( R \) modules \( {M}_{s} \) . Let \( N \) be any \( R \) submodule of \( M \) . Then\n\n(a) the quotient module \( M/N \) is semisimple. In more detail there is a sub... | Proof. Each simple \( R \) submodule \( {M}_{s} \) of \( M \) maps to an \( R \) submodule \( {\bar{M}}_{s} \) of \( M/N \) . This image either is simple (and then is \( R \) isomorphic to \( {M}_{s} \) ) or is zero. We let \( U \) be the subset of \( S \) for which it is simple. Then \( M/N \) is evidently the sum of ... | Yes |
Lemma 10.3. Suppose that \( E \) is a simple left \( R \) module and that \( M = \) \( {\bigoplus }_{a \in A}{M}_{a} \) is a direct-sum decomposition of the unital left \( R \) module \( M \) into arbitrary \( R \) submodules, not necessarily simple. Then\n\n\[ \n{\operatorname{Hom}}_{R}\left( {E, M}\right) \cong {\big... | Proof. Suppose \( \varphi \) is in \( {\operatorname{Hom}}_{R}\left( {E, M}\right) \) . Write \( {\varphi }_{a} \) for the composition of \( \varphi \) with the projection \( M \rightarrow {M}_{a} \) . The map from left to right in the displayed isomorphism is to be \( \varphi \mapsto {\left\{ {\varphi }_{a}\right\} }_... | Yes |
Proposition 10.4 (Schur’s Lemma). Suppose that \( M \) and \( N \) are simple left \( R \) modules.\n\n(a) If \( M \) and \( N \) are not \( R \) isomorphic, then \( {\operatorname{Hom}}_{R}\left( {M, N}\right) = 0 \) .\n\n(b) \( {\operatorname{End}}_{R}\left( M\right) \) is a division ring.\n\n(c) (Dixmier) If \( R \)... | Proof. Suppose that \( \varphi \) is nonzero in \( {\operatorname{Hom}}_{R}\left( {M, N}\right) \) . Then \( \ker \varphi \) is a proper \( R \) submodule of \( M \), and we must have \( \ker \varphi = 0 \) since \( M \) is simple. Similarly image \( \varphi \) is a nonzero \( R \) submodule of \( N \), and we must hav... | Yes |
Lemma 10.5 (Zassenhaus). Let \( {M}_{1},{M}_{2},{M}_{1}^{\prime } \), and \( {M}_{2}^{\prime } \) be \( R \) submodules of a unital left \( R \) module \( M \) with \( {M}_{1}^{\prime } \subseteq {M}_{1} \) and \( {M}_{2}^{\prime } \subseteq {M}_{2} \) . Then\n\n\( \left( {\left( {{M}_{1} \cap {M}_{2}}\right) + {M}_{1}... | Proof. By the Second Isomorphism Theorem (Theorem 8.4),\n\n\[ \left( {{M}_{1} \cap {M}_{2}}\right) /\left( {\left( {\left( {{M}_{1} \cap {M}_{2}^{\prime }}\right) + {M}_{1}^{\prime }}\right) \cap \left( {{M}_{1} \cap {M}_{2}}\right) }\right) \]\n\n\[ \cong \left( {\left( {{M}_{1} \cap {M}_{2}}\right) + \left( {{M}_{1} ... | Yes |
Theorem 10.6 (Schreier). Any two finite filtrations of a module \( M \) in \( \mathcal{C} \) have equivalent refinements. | PROOF. Let the two finite filtrations be\n\n\[ M = {M}_{0} \supseteq {M}_{1} \supseteq \cdots \supseteq {M}_{m} = 0 \]\n\nand\n\n\[ M = {N}_{0} \supseteq {N}_{1} \supseteq \cdots \supseteq {N}_{n} = 0, \]\n\nand define\n\n\[ {M}_{ij} = \left( {{M}_{i} \cap {N}_{j}}\right) + {M}_{i + 1}\;\text{ for }0 \leq i \leq m - 1\... | Yes |
Corollary 10.7 (Jordan-Hölder Theorem). If \( M \) is a unital left \( R \) module with a composition series, then\n\n(a) any finite filtration of \( M \) in which all consecutive quotients are nonzero can be refined to a composition series, and\n\n(b) any two composition series of \( M \) are equivalent. | Proof. We apply Theorem 10.6 to a given filtration and a known composition series. After discarding redundant terms from each refinement (those that lead to 0 as a consecutive quotient), we arrive at a refinement of our given finite filtration that is equivalent to the known composition series. Hence the refinement is ... | Yes |
Proposition 10.8. If \( R \) is a ring with identity and \( M \) is a unital left \( R \) module, then the following conditions on \( R \) submodules of \( M \) are equivalent:\n\n(a) (ascending chain condition) every strictly ascending chain of \( R \) submodules \( {M}_{1} \subsetneqq {M}_{2} \subsetneqq \cdots \) te... | Proof. To see that (a) implies (b), let \( \mathcal{S} \) be a nonempty collection of \( R \) submodules of \( M \) . Take \( {M}_{1} \) in \( \mathcal{S} \) . If \( {M}_{1} \) is not maximal, choose \( {M}_{2} \) in \( \mathcal{S} \) properly containing \( {M}_{1} \) . If \( {M}_{2} \) is not maximal, choose \( {M}_{3... | Yes |
Proposition 10.9. If \( R \) is a ring with identity and \( M \) is a unital left \( R \) module, then the following conditions on \( R \) submodules of \( M \) are equivalent:\n\n(a) (descending chain condition) every strictly descending chain of \( R \) submodules \( {M}_{1} \supsetneqq {M}_{2} \supsetneqq \cdots \) ... | Proof. To see that (a) implies (b), let \( \mathcal{S} \) be a nonempty collection of \( R \) submodules of \( M \) . Take \( {M}_{1} \) in \( \mathcal{S} \) . If \( {M}_{1} \) is not minimal, choose \( {M}_{2} \) in \( \mathcal{S} \) properly contained in \( {M}_{1} \) . If \( {M}_{2} \) is not minimal, choose \( {M}_... | Yes |
Proposition 10.10. Let \( R \) be a ring with identity, let \( M \) be a unital left \( R \) module, and let \( N \) be an \( R \) submodule of \( M \). Then\n\n(a) \( M \) satisfies the ascending chain condition if and only if \( N \) and \( M/N \) satisfy the ascending chain condition,\n\n(b) \( M \) satisfies the de... | Proof. We prove (a), and the proof of (b) is completely similar. Suppose \( M \) satisfies the ascending chain condition and hence also the maximum condition by Proposition 10.8. The \( R \) submodules of \( N \) are in particular \( R \) submodules of \( M \) and hence satisfy the maximum condition. The \( R \) submod... | Yes |
Proposition 10.11. If \( R \) is a ring with identity and \( M \) is a unital left \( R \) module, then \( M \) has a composition series if and only if \( M \) satisfies both the ascending chain condition and the descending chain condition. | Proof. If \( M \) has a composition series of length \( n \), then the Jordan-Hölder Theorem (Corollary 10.7a) shows that every finite filtration of \( M \) with nonzero consecutive quotients has length \( \leq n \) , and hence \( M \) satisfies both chain conditions.\n\nConversely suppose that \( M \) satisfies both c... | Yes |
Proposition 10.12. Let \( S \) be a nonempty set, let \( {M}_{s} \) and \( {N}_{s} \) be unital left \( R \) modules for each \( s \in S \), and let \( M \) and \( N \) be unital left \( R \) modules. Then there are isomorphisms of abelian groups\n\n(a) \( {\operatorname{Hom}}_{R}\left( {{\bigoplus }_{s \in S}{M}_{s}, ... | Proof. For (a), let \( {e}_{s} : {M}_{s} \rightarrow {\bigoplus }_{t}{M}_{t} \) be the \( {s}^{\text{th }} \) inclusion, and let \( {p}_{s} : {\bigoplus }_{t}{M}_{t} \rightarrow {M}_{s} \) be the \( {s}^{\text{th }} \) projection; the latter is defined as the restriction of the projection associated with the direct pro... | Yes |
Corollary 10.13. Let \( V \) be a unital left \( R \) module, and let \( S \) be the ring \( S = {\operatorname{End}}_{R}\left( V\right) \). For integers \( m \geq 1 \) and \( n \geq 1 \), there is a canonical isomorphism of abelian groups\n\n\[{\operatorname{Hom}}_{R}\left( {{V}^{n},{V}^{m}}\right) \cong {M}_{mn}\left... | Proof. Let \( {e}_{j} : V \rightarrow {V}^{n} = {\bigoplus }_{k = 1}^{n}V = \mathop{\prod }\limits_{{k = 1}}^{n}V \) be the \( {j}^{\text{th }} \) inclusion for whatever \( n \) is under discussion, and let \( {p}_{i} : {V}^{m} \rightarrow V \) be the \( {i}^{\text{th }} \) projection for whatever \( m \) is under disc... | Yes |
Proposition 10.14. The mapping \( \varphi \mapsto \varphi \left( 1\right) \) is a ring isomorphism \( {\operatorname{End}}_{R}\left( R\right) \cong \) \( {R}^{o} \) of \( {\operatorname{End}}_{R}\left( R\right) \) onto the opposite ring \( {R}^{o} \) of \( R \) . | Proof. The mapping \( \varphi \mapsto \varphi \left( 1\right) \) certainly respects addition. If \( \varphi \) maps to \( \varphi \left( 1\right) \) and \( \tau \) maps to \( \tau \left( 1\right) \), then \( {\varphi \tau } \) maps to \( \left( {\varphi \tau }\right) \left( 1\right) = \varphi \left( {\tau \left( 1\righ... | Yes |
Corollary 10.15. For any integer \( n \geq 1,{\operatorname{End}}_{R}\left( {R}^{n}\right) \) is ring isomorphic to \( {M}_{n}\left( {R}^{o}\right) \) . | Proof. Corollary 10.13 shows that \( {\operatorname{End}}_{R}\left( {R}^{n}\right) \) is isomorphic to \( {M}_{n}\left( {{\operatorname{End}}_{R}\left( R\right) }\right) \) , and Proposition 10.14 shows that the latter ring is isomorphic to \( {M}_{n}\left( {R}^{o}\right) \) . | Yes |
Proposition 10.16. For any unital left \( R \) module \( M \), there is a canonical \( R \) isomorphism\n\n\[{\operatorname{Hom}}_{R}\left( {R, M}\right) \cong M\]\n\nand this isomorphism is natural in the variable \( M \) . | Proof. The map \( \Phi \) from left to right is given by \( \Phi \left( \sigma \right) = \sigma \left( 1\right) \), and the inverse will be seen to be given by \( {\Phi }^{\prime }\left( m\right) = {\tau }_{m} \) with \( {\tau }_{m}\left( r\right) = {rm} \) . The computation \( \Phi \left( {r\sigma }\right) = \left( {r... | Yes |
Proposition 10.17. If \( R \) and \( S \) are two rings with identity, if \( P \) is a unital \( \left( {R, S}\right) \) bimodule, and if \( M \) is any unital left \( R \) module, then the abelian group \( {\operatorname{Hom}}_{R}\left( {P, M}\right) \) becomes a unital left \( S \) module under the definition \( \lef... | Proof. To see that \( {s\varphi } \) is an \( R \) homomorphism, we compute that \( \left( {s\varphi }\right) \left( {rp}\right) = \) \( \varphi \left( {\left( {rp}\right) s}\right) = \varphi \left( {r\left( {ps}\right) }\right) = r\left( {\varphi \left( {ps}\right) }\right) = r\left( {\left( {s\varphi }\right) \left( ... | Yes |
Theorem 10.18. Let \( R \) be a ring with identity. If \( M \) is a unital right \( R \) module and \( N \) is a unital left \( R \) module, then there exists a tensor product \( \left( {M{ \otimes }_{R}N,\iota }\right) \) of \( M \) and \( N \) over \( R \), and it is unique in the following sense: if \( \left( {{V}_{... | Proof. Form the free abelian group \( G \) with a \( \mathbb{Z} \) basis parametrized by the elements of \( M \times N \) . We write \( e\left( {m, n}\right) \) for the basis element in \( G \) corresponding to the element \( \left( {m, n}\right) \) of \( M \times N \), and we regard \( e \) as a one-one function from ... | Yes |
Corollary 10.19. Let \( R, S \), and \( T \) be rings with identity, and suppose that \( M \) is a unital right \( R \) module and \( N \) is a unital left \( R \) module. Under the additional hypothesis that\n\n(a) \( M \) is a unital \( \left( {S, R}\right) \) bimodule, then \( M{ \otimes }_{R}N \) is a unital left \... | Proof. In (a), let left multiplication by \( s \in S \) within \( M \) be given by \( {\varphi }_{s} : M \rightarrow \) \( M \) with \( {\varphi }_{s}\left( m\right) = {sm} \) . Then multiplication by \( s \) in \( S \) within \( M{ \otimes }_{R}N \) is given by \( {\varphi }_{s} \otimes 1 \) . The covariant-functor pr... | Yes |
Proposition 10.20. Let \( R \) be a ring with identity, let \( M \) be a unital right \( R \) module, and let \( N \) be a unital left \( R \) module. Let \( {R}^{o} \) be the opposite ring of \( R \) , let \( {M}^{o} \) be \( M \) regarded as a left \( {R}^{o} \) module, and let \( {N}^{o} \) be \( N \) regarded as a ... | Proof. The map \( \left( {m, n}\right) \mapsto {n}^{o} \otimes {m}^{o} \) is additive in each variable and carries \( \left( {m,{rn}}\right) \) to \( {\left( rn\right) }^{o} \otimes {m}^{o} = {n}^{o}{r}^{o} \otimes {m}^{o} = {n}^{o} \otimes {r}^{o}{m}^{o} = {n}^{o} \otimes {\left( mr\right) }^{o} \) . This expression i... | Yes |
Proposition 10.21. Let \( R \) be a ring with identity, let \( S \) be a nonempty set, let \( {M}_{s} \) be a unital right \( R \) module for each \( s \in S \), and let \( N \) be a unital left \( R \) module. Then\n\n\[ \left( {{\bigoplus }_{s \in S}{M}_{s}}\right) { \otimes }_{R}N \cong {\bigoplus }_{s \in S}\left( ... | Proof. The map \( \left( {{\left\{ {m}_{s}\right\} }_{s}, n}\right) \mapsto {\left\{ {m}_{s} \otimes n\right\} }_{s} \) is \( R \) bilinear from \( \left( {{\bigoplus }_{s \in S}{M}_{s}}\right) \times N \) into \( {\bigoplus }_{s \in S}\left( {{M}_{s}{ \otimes }_{R}N}\right) \), and its additive extension \( \Phi \) is... | Yes |
Proposition 10.22. Let \( R \) and \( S \) be rings with identity, let \( M \) be a unital right \( R \) module, let \( N \) be a unital \( \left( {R, S}\right) \) bimodule, and let \( P \) be a unital left \( S \) module. Then\n\n\[ \left( {M{ \otimes }_{R}N}\right) { \otimes }_{S}P \cong M{ \otimes }_{R}\left( {N{ \o... | Proof. For fixed \( p \), the map \( \left( {m, n, p}\right) \mapsto m \otimes \left( {n \otimes p}\right) \) is \( R \) bilinear. In fact, the map is certainly additive in \( m \) and in \( n \) . For the transformation law with an element \( r \) of \( R \), the calculation is \( \left( {{mr}, n, p}\right) \mapsto {m... | Yes |
Proposition 10.23. Let \( R \) and \( S \) be rings with identity, let \( M \) be a unital left \( R \) module, let \( N \) be a unital \( \left( {S, R}\right) \) bimodule, and let \( P \) be a unital left \( S \) module. Then \[ {\operatorname{Hom}}_{S}\left( {N{ \otimes }_{R}M, P}\right) \cong {\operatorname{Hom}}_{R... | Proof. The homomorphism \( \Phi \) is well defined. We construct its inverse. If \( \psi \) is in \( {\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{S}\left( {N, P}\right) }\right) \), then the map \( \left( {n, m}\right) \mapsto \psi \left( m\right) \left( n\right) \) sends \( \left( {{nr}, m}\right) \) to \( ... | Yes |
Proposition 10.25. Let \( R \) be a ring with identity, let\n\n\[ 0 \rightarrow M\overset{\varphi }{ \rightarrow }N\overset{\psi }{ \rightarrow }P \rightarrow 0 \]\n\nbe a short exact sequence in the category \( {\mathcal{C}}_{R} \), let \( E \) be a module in \( {\mathcal{C}}_{R} \), and let \( {E}^{\prime } \) be a m... | Proof. For the first sequence in \( {\mathcal{C}}_{\mathbb{Z}} \), we are to show that \( 1 \otimes \psi \) is onto \( {E}^{\prime }{ \otimes }_{R}P \) and that every member of the kernel of \( 1 \otimes \psi \) is in the image of \( 1 \otimes \varphi \) . (Recall that \( \ker \left( {1 \otimes \psi }\right) \supseteq ... | Yes |
Lemma 1. Suppose that the vectors \( {x}_{1},\ldots ,{x}_{n} \) span a linear space \( X \) and that the vectors \( {y}_{1},\ldots ,{y}_{j} \) in \( X \) are linearly independent. Then\n\n\[ j \leq n\text{.} \] | Proof. Since \( {x}_{1},\ldots ,{x}_{n} \) span \( X \), every vector in \( X \) can be written as a linear combination of \( {x}_{1},\ldots ,{x}_{n} \) . In particular, \( {y}_{1} \) :\n\n\[ {y}_{1} = {k}_{1}{x}_{1} + \cdots + {k}_{n}{x}_{n} \]\n\nSince \( {y}_{1} \neq 0 \) (see Exercise 10), not all \( k \) are equal... | No |
Lemma 2. A linear space \( X \) which is spanned by a finite set of vectors \( {x}_{1},\ldots ,{x}_{n} \) has a basis. | Proof. If \( {x}_{1},\ldots ,{x}_{n} \) are linearly dependent, there is a nontrivial relation between them; from this one of the \( {x}_{i} \) can be expressed as a linear combination of the rest. So we can drop that \( {x}_{i} \). Repeat this step until the remaining \( {x}_{j} \) are linear independent: they still s... | Yes |
Theorem 3. All bases for a finite-dimensional linear space \( X \) contain the same number of vectors. This number is called the dimension of \( X \) and is denoted as | Proof. Let \( {x}_{1},\ldots ,{x}_{n} \) be one basis, and let \( {y}_{1},\ldots ,{y}_{m} \) be another. By Lemma 1 and the definition of basis we conclude that \( m \leq n \), and also \( n \leq m \) . So we conclude that \( n \) and \( m \) are equal. | Yes |
Theorem 4. Every linearly independent set of vectors \( {y}_{1},\ldots ,{y}_{j} \) in a finite-dimensional linear space \( X \) can be completed to a basis of \( X \) . | Proof. If \( {y}_{1},\ldots ,{y}_{j} \) do not span \( X \), there is some \( {x}_{1} \) that cannot be expressed as a linear combination of \( {y}_{1},\ldots ,{y}_{j} \) . Adjoin this \( {x}_{1} \) to the \( y \) ’s. Repeat this step until the \( y \) ’s span \( X \) . This will happen in less than \( n \) steps, \( n... | Yes |
Theorem 5. (a) Every subspace \( Y \) of a finite-dimensional linear space \( X \) is finite dimensional. | Proof. We can construct a basis in \( Y \) by starting with any nonzero vector \( {y}_{1} \), and then adding another vector \( {y}_{2} \) and another, as long as they are linearly independent. According to Lemma 1, there can be no more of these \( {y}_{i} \) than the dimension of \( X \) . A maximal set of linearly in... | Yes |
Theorem 6. \( Y \) is a subspace of a finite-dimensional linear space \( X \) ; then\n\n\[ \n\dim Y + \dim \left( {X/Y}\right) = \dim X \n\] | Proof. Let \( {y}_{1},\ldots ,{y}_{j} \) be a basis for \( Y, j = \dim Y \) . According to Theorem 4, this set can be completed to form a basis for \( X \) by adjoining \( {x}_{j + 1},\ldots ,{x}_{n}, n = \dim X \) . We claim that\n\n\[ \n\left\{ {x}_{j + 1}\right\} ,\ldots ,\left\{ {x}_{n}\right\} \n\]\n\n\( {\left( {... | Yes |
Theorem 7. Suppose \( X \) is a finite-dimensional linear space, \( U \) and \( V \) two subspaces of \( X \) such that \( X \) is the sum of \( U \) and \( V \) :\n\n\[ \nX = U + V \n\]\n\nDenote by \( W \) the intersection of \( U \) and \( V \) :\n\n\[ \nW = U \cap V. \n\]\n\nThen\n\n\[ \n\dim X = \dim U + \dim V - ... | Proof. When the intersection \( W \) of \( U \) and \( V \) is the trivial space \( \{ 0\} ,\dim W = 0 \) , and (14) is relation \( {\left( {11}\right) }^{\prime } \) of Theorem 5 . We show now how to use the notion of quotient space to reduce the general case to the simple case \( \dim W = 0 \) .\n\nDefine \( {U}_{0} ... | Yes |
Example 1. \( X = \{ \) continuous functions \( f\left( s\right) ,0 \leq s \leq 1\} \) . Then for any point \( {s}_{1} \) in \( \left\lbrack {0,1}\right\rbrack \) , | \[ l\left( f\right) = f\left( {s}_{1}\right) \] is a linear function. So is \[ l\left( f\right) = \mathop{\sum }\limits_{1}^{n}{k}_{j}f\left( {s}_{j}\right) \] where \( {s}_{j} \) is an arbitrary collection of points in \( \left\lbrack {0,1}\right\rbrack ,{k}_{j} \) arbitrary scalars. So is \[ l\left( f\right) = {\int ... | Yes |
Theorem 1. Let \( X \) be a linear space of dimension \( n \) . The elements \( x \) of \( X \) can be represented as arrays of \( n \) scalars:\n\n\[ x = \left( {{c}_{1},\ldots ,{c}_{n}}\right) \]\n\n(3)\n\nAddition and multiplication by a scalar is defined componentwise. Let \( {a}_{1},\ldots ,{a}_{n} \) be any array... | Proof. That \( l\left( x\right) \) defined by (4) is a linear function of \( x \) is obvious. The converse is not much harder. Let \( l \) be any linear function defined on \( X \) . Define \( {x}_{j} \) to be the vector whose \( j \) th component is 1, with all other components zero. Then \( x \) defined by (3) can be... | Yes |
Theorem 2. The dual \( {X}^{\prime } \) of a finite-dimensional linear space \( X \) is a finite-dimensional linear space, and\n\n\[ \dim {X}^{\prime } = \dim X \] | Since \( {X}^{\prime } \) is a linear space, it has its own dual \( {X}^{\prime \prime } \) consisting of all linear functions on \( {X}^{\prime } \) . For fixed \( x,\left( {l, x}\right) \) is such a linear function. By Theorem 1, all linear functions are of this form. This proves the following theorem. | No |
Theorem 4. Let \( Y \) be a subspace of a finite-dimensional space \( X,{Y}^{ \bot } \) its annihilator. Then\n\n\[ \dim {Y}^{ \bot } + \dim Y = \dim X \] | Proof. We shall establish a natural isomorphism between \( {Y}^{ \bot } \) and the dual \( {\left( X/Y\right) }^{\prime } \) of \( X/Y \) . Given \( l \) in \( {Y}^{ \bot } \) we define \( L \) in \( {\left( X/Y\right) }^{\prime } \) as follows: for any congruence class \( \{ x\} \) in \( X/Y \), we define\n\n\[ L\{ x\... | Yes |
Theorem 5. Under the identification (5) of \( {X}^{\prime \prime } \) and \( X \), for every subspace \( Y \) of a finite-dimensional space \( X \) ,\n\n\[ \n{Y}^{ \bot \bot } = Y\text{. } \n\] | Proof. It follows from definition (6) of the annihilator of \( Y \) that all \( y \) in \( Y \) belong to \( {Y}^{ \bot \bot } \), the annihilator of \( {Y}^{ \bot } \) . To show that \( Y \) is all of \( {Y}^{ \bot \bot } \), we make use of (7) applied to \( {X}^{\prime } \) and its subspace \( {Y}^{ \bot } \) :\n\n\[... | Yes |
Theorem 7. Let \( l \) be an interval on the real axis, \( {t}_{1},\ldots ,{t}_{n}n \) distinct points. Then there exist \( n \) numbers \( {m}_{1},\ldots ,{m}_{n} \) such that the quadrature formula,\n\n\[ \n{\int }_{l}p\left( t\right) {dt} = {m}_{1}p\left( {t}_{1}\right) + \cdots + {m}_{n}p\left( {t}_{n}\right) \n\]\... | Proof. Denote by \( X \) the space of all polynomials \( p\left( t\right) = {a}_{0} + {a}_{1}t + \cdots + {a}_{n - 1}{t}^{n - 1} \) of degree less than \( n \) . Since \( X \) is isomorphic to the space \( \left( {{a}_{0},{a}_{1},\ldots ,{a}_{n - 1}}\right) = \) \( {\mathbb{R}}^{n},\dim X = n \) . We define \( {l}_{j} ... | Yes |
Theorem 2. Let T: \( X \rightarrow U \) be a linear map; then\n\n\[ \dim {N}_{\mathrm{T}} + \dim {R}_{\mathrm{T}} = \dim X \] | Proof. Since \( \mathrm{T} \) maps \( {N}_{\mathrm{T}} \) into \( 0,\mathrm{\;T}{x}_{1} = \mathrm{T}{x}_{2} \) when \( {x}_{1} \) and \( {x}_{2} \) are equivalent \( {\;\operatorname{mod}\;{N}_{\mathrm{T}}} \) . So we can define \( \mathrm{T} \) acting on the quotient space \( X/{N}_{\mathrm{T}} \) by setting\n\n\[ \ma... | Yes |
Example 10. \( X = {\mathbb{R}}^{n}, U = {\mathbb{R}}^{m} \) , \( \mathrm{T} \) as in Example 7. | \[ {u}_{i} = \sum {t}_{ij}{x}_{j} \] (10) \( {U}^{\prime } \) is then also \( {\mathbb{R}}^{m},{X}^{\prime } = {\mathbb{R}}^{n} \), with \( \left( {l, u}\right) = \mathop{\sum }\limits_{1}^{m}{l}_{i}{u}_{i},\left( {m, x}\right) = \mathop{\sum }\limits_{1}^{n}{m}_{j}{x}_{j} \) . Then with \( u = \mathrm{T}x \), using (1... | Yes |
Theorem 5. The annihilator of the range of \( T \) is the nullspace of its transpose: | Proof. By the definition in Chapter 2 of annihilator, the annihilator of the range \( {R}_{\mathrm{T}} \) consists of those linear functions \( l \) defined on the target space \( \mathrm{U} \) for which\n\n\[ \left( {l, u}\right) = 0\;\text{ for all }u\text{ in }{R}_{\mathrm{T}}. \]\n\nSince \( u \) in \( {R}_{\mathrm... | Yes |
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