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Theorem 6.5 (Principal Axis Theorem). Suppose that \( \mathbb{K} \) has characteristic different from 2.\n\n(a) If \( \langle \cdot , \cdot \rangle \) is a symmetric bilinear form on a finite-dimensional vector space \( V \) , then there exists an ordered basis of \( V \) in which the matrix of \( \langle \cdot \) , \(... | Proof VIA MATRICES. If \( A \) is an \( n \) -by- \( n \) symmetric matrix, we seek a nonsingular \( M \) with \( {M}^{t}{AM} \) diagonal. We induct on the size of \( A \), the base case of the induction being \( n = 1 \), where there is nothing to prove. Assume the result to be known for size \( n - 1 \), and write th... | Yes |
Theorem 6.9 (Sylvester’s Law). The signature of an \( n \) -by- \( n \) Hermitian matrix is well defined. | The proof is the same as for Theorem 6.6 except for adjustments in notation. | No |
If \( E \) and \( F \) are vector spaces over \( \mathbb{K} \), then a tensor product of \( E \) and \( F \) exists and is unique up to canonical isomorphism in this sense: if \( \left( {{V}_{1},{\iota }_{1}}\right) \) and \( \left( {{V}_{2},{\iota }_{2}}\right) \) are tensor products, then there exists a unique linear... | Proof of UNIQUENESS. Let \( \left( {{V}_{1},{\iota }_{1}}\right) \) and \( \left( {{V}_{2},{\iota }_{2}}\right) \) be tensor products. Set up the diagrams in Figure 6.2, and use the universal mapping property to obtain linear maps \( {B}_{2} : {V}_{1} \rightarrow {V}_{2} \) and \( {B}_{1} : {V}_{2} \rightarrow {V}_{1} ... | Yes |
Corollary 6.11. If \( E, F \), and \( V \) are vector spaces over \( \mathbb{K} \), then the vector space \( {\operatorname{Hom}}_{\mathbb{K}}\left( {E{ \otimes }_{\mathbb{K}}F, V}\right) \) is canonically isomorphic (via restriction to pure tensors) to the vector space of all \( V \) -valued bilinear functions on \( E... | Proof. Restriction is a linear mapping from \( {\operatorname{Hom}}_{\mathbb{K}}\left( {E{ \otimes }_{\mathbb{K}}F, V}\right) \) to the vector space of all \( V \) -valued bilinear functions on \( E \times F \), and it is one-one since the image of \( E \times F \) in \( E{ \otimes }_{\mathbb{K}}F \) spans \( E{ \otime... | Yes |
Corollary 6.12. If \( E \) and \( F \) are vector spaces over \( \mathbb{K} \), then the vector space of all bilinear forms on \( E \times F \) is canonically isomorphic to \( {\left( E{ \otimes }_{\mathbb{K}}F\right) }^{\prime } \), the dual of the vector space \( E{ \otimes }_{\mathbb{K}}F \) . | Proof. This is the special case of Corollary 6.11 in which \( V = \mathbb{K} \) . | Yes |
Corollary 6.13. If \( E, F \), and \( V \) are vector spaces over \( \mathbb{K} \), then there is a canonical \( \mathbb{K} \) linear isomorphism \( \Phi \) of left side to right side in\n\n\[{\operatorname{Hom}}_{\mathbb{K}}\left( {E{ \otimes }_{\mathbb{K}}F, V}\right) \cong {\operatorname{Hom}}_{\mathbb{K}}\left( {E,... | Proof. The map \( \Phi \) is well defined and \( \mathbb{K} \) linear, and it carries the left side to the right side. For \( \psi \) in the right side, define \( \Psi \left( \psi \right) \left( {e, f}\right) = \psi \left( e\right) \left( f\right) \) . Then \( \Psi \left( \psi \right) \) is a bilinear map from \( E \ti... | Yes |
Proposition 6.15. Let \( E, F, V,{E}_{1},{F}_{1} \), and \( {V}_{1} \) be vector spaces over \( \mathbb{K} \), and let \( {L}_{{E}_{1}} : {E}_{1} \rightarrow E,{L}_{{F}_{1}} : {F}_{1} \rightarrow F \), and \( {L}_{V} : V \rightarrow {V}_{1} \) be \( \mathbb{K} \) linear maps. Then the isomorphism \( \Phi \) of Corollar... | Proof. For \( \varphi \) in \( {\operatorname{Hom}}_{\mathbb{K}}\left( {E{ \otimes }_{\mathbb{K}}F, V}\right) ,{e}_{1} \) in \( {E}_{1} \), and \( {f}_{1} \) in \( {F}_{1} \), we have\n\n\[ \left( {\operatorname{Hom}\left( {{L}_{{E}_{1}},\operatorname{Hom}\left( {{L}_{{F}_{1}},{L}_{V}}\right) }\right) \circ \Phi }\righ... | Yes |
Proposition 6.16. Let \( \mathcal{D} \) be a category, and suppose that \( U \) and \( {U}^{\prime } \) are objects in \( \mathcal{D} \) with the following property: to each object \( V \) in \( \mathcal{D} \) corresponds a one-one onto function \[ {T}_{V} : \operatorname{Morph}\left( {U, V}\right) \rightarrow \operato... | Proof. Let \( \varphi \) be the element \( {T}_{{U}^{\prime }}^{-1}\left( {1}_{{U}^{\prime }}\right) \) of \( \operatorname{Morph}\left( {U,{U}^{\prime }}\right) \), and let \( \psi \) be the element \( {T}_{U}\left( {1}_{U}\right) \) of \( \operatorname{Morph}\left( {{U}^{\prime }, U}\right) \) . To prove the proposit... | Yes |
Corollary 6.17. Let \( \mathcal{C} \) and \( \mathcal{D} \) be categories, and let \( F : \mathcal{C} \rightarrow \mathcal{D} \) and \( G : \mathcal{C} \rightarrow \mathcal{D} \) be covariant functors. Suppose that to each pair of objects \( \left( {A, V}\right) \) in \( \mathcal{C} \times \mathcal{D} \) corresponds a ... | Proof. By Proposition 6.16 and the hypotheses, the member \( {T}_{A, G\left( A\right) }^{-1}\left( {1}_{G\left( A\right) }\right) \) of \( {\operatorname{Morph}}_{\mathcal{D}}\left( {F\left( A\right), G\left( A\right) }\right) \) is an isomorphism. We are to prove that the system \( \left\{ {T}_{A, G\left( A\right) }\r... | Yes |
(a) If \( \left( {{E}^{\mathbb{L}},\iota }\right) \) is formed by extension of scalars, then \( \left( {{E}^{\mathbb{L}},\iota }\right) \) has the following universal mapping property: whenever \( U \) is a vector space over \( \mathbb{L} \) and \( \varphi : E \rightarrow U \) is a \( \mathbb{K} \) linear map, there ex... | In (a), for the uniqueness of \( \Phi \), we must have \( \Phi \left( {e \otimes c}\right) = {c\Phi }\left( {e \otimes 1}\right) = \) \( c\left( {\Phi \iota }\right) \left( e\right) = {c\varphi }\left( e\right) \) . Hence \( \Phi \) is determined by \( \varphi \) on pure tensors in \( E{ \otimes }_{\mathbb{K}}\mathbb{L... | Yes |
Proposition 6.19. If \( \mathbb{K} \) is a field and \( A, B, C \) are vector spaces over \( \mathbb{K} \), then\n\n(a) \( \left( {A{ \otimes }_{\mathbb{K}}B}\right) { \otimes }_{\mathbb{K}}C \) and \( A{ \otimes }_{\mathbb{K}}\left( {B{ \otimes }_{\mathbb{K}}C}\right) \) are triple tensor products.\n\n(b) there exists... | Proof. In (a), consider \( \left( {A{ \otimes }_{\mathbb{K}}B}\right) { \otimes }_{\mathbb{K}}C \) . Let \( t : A \times B \times C \rightarrow U \) be 3-linear. For \( c \in C \), define \( {t}_{c} : A \times B \rightarrow U \) by \( {t}_{c}\left( {a, b}\right) = t\left( {a, b, c}\right) \) . Then \( {t}_{c} \) is bil... | Yes |
Proposition 6.20. Let \( A, B, C,{A}_{1},{B}_{1} \), and \( {C}_{1} \) be vector spaces over a field \( \mathbb{K} \), and let \( {L}_{A} : A \rightarrow {A}_{1},{L}_{B} : B \rightarrow {B}_{1} \), and \( {L}_{C} : C \rightarrow {C}_{1} \) be linear maps. Then the isomorphism \( \Phi \) of Proposition \( {6.19}\mathrm{... | Proof. We have\n\n\[ \left( {\left( {{L}_{A} \otimes \left( {{L}_{B} \otimes {L}_{C}}\right) }\right) \circ \Phi }\right) \left( {\left( {a \otimes b}\right) \otimes c}\right) \]\n\n\[ = \left( {{L}_{A} \otimes \left( {{L}_{B} \otimes {L}_{C}}\right) }\right) \left( {a \otimes \left( {b \otimes c}\right) }\right) \]\n\... | Yes |
Proposition 6.22. The pair \( \left( {T\left( E\right) ,\iota }\right) \) has the following universal mapping property: whenever \( l : E \rightarrow A \) is a linear map from \( E \) into an associative algebra with identity, then there exists a unique associative algebra homomorphism \( L : T\left( E\right) \rightarr... | Proof. Uniqueness is clear, since \( E \) and 1 generate \( T\left( E\right) \) as an algebra. For existence we define \( {L}^{\left( n\right) } \) on \( {T}^{n}\left( E\right) \) to be the linear extension of the \( n \) -multilinear map\n\n\[ \left( {{v}_{1},{v}_{2},\ldots ,{v}_{n}}\right) \mapsto l\left( {v}_{1}\rig... | Yes |
Let \( E \) be a vector space over the field \( \mathbb{K} \). (a) Let \( \iota \) be the \( n \)-multilinear function \( \iota \left( {{v}_{1},\ldots ,{v}_{n}}\right) = {v}_{1}\cdots {v}_{n} \) of \( E \times \cdots \times E \) into \( {S}^{n}\left( E\right) \). Then \( \left( {{S}^{n}\left( E\right) ,\iota }\right) \... | Proof. In both cases uniqueness is trivial. For existence we use the universal mapping properties of \( {T}^{n}\left( E\right) \) and \( T\left( E\right) \) to produce \( \widetilde{L} \) on \( {T}^{n}\left( E\right) \) or \( T\left( E\right) \). If we can show that \( \widetilde{L} \) annihilates the appropriate subsp... | Yes |
Corollary 6.24. If \( E \) and \( F \) are vector spaces over the field \( \mathbb{K} \), then the vector space \( {\operatorname{Hom}}_{\mathbb{K}}\left( {{S}^{n}\left( E\right), F}\right) \) is canonically isomorphic (via restriction to pure tensors) to the vector space of all \( F \) -valued symmetric \( n \) -multi... | Proof. Restriction is linear and one-one. It is onto by Proposition 6.23a. | No |
Corollary 6.25. If \( E \) is a vector space over the field \( \mathbb{K} \), then the dual \( {\left( {S}^{n}\left( E\right) \right) }^{\prime } \) of \( {S}^{n}\left( E\right) \) is canonically isomorphic (via restriction to pure tensors) to the vector space of symmetric \( n \) -multilinear forms on \( E \times \cdo... | Proof. This is a special case of Corollary 6.24. | No |
Proposition 6.26. Let \( E \) be a vector space over the field \( \mathbb{K} \), let \( {\left\{ {u}_{i}\right\} }_{i \in A} \) be a basis of \( E \), and suppose that a simple ordering has been imposed on the index set \( A \) . Then the set of all monomials \( {u}_{{i}_{1}}^{{j}_{1}}\cdots {u}_{{i}_{k}}^{{j}_{k}} \) ... | Proof. Since \( S\left( E\right) \) is commutative and since \( n \) -fold products of elements \( \iota \left( {u}_{i}\right) \) in \( {T}^{1}\left( E\right) \) span \( {T}^{n}\left( E\right) \), the indicated set of monomials spans \( {S}^{n}\left( E\right) \) . Let us see that the set is linearly independent. Take a... | Yes |
Proposition 6.28. If \( E \) is a finite-dimensional vector space over the field \( \mathbb{K} \), then the system of evaluation homomorphisms \( P\left( E\right) \rightarrow \mathbb{K} \) on polynomials given by \( p \mapsto \{ p\left( e\right) {\} }_{e \in E} \) is an algebra homomorphism of \( P\left( E\right) \) on... | Proof. Certainly \( p \mapsto \{ p\left( e\right) {\} }_{e \in E} \) is an algebra homomorphism of \( P\left( E\right) \) into the algebra of \( \mathbb{K} \) -valued polynomial functions on \( E \), and it carries the identity to the constant function 1 . We have seen that the image of \( {P}^{1}\left( E\right) \) is ... | Yes |
Proposition 6.29. Let the field \( \mathbb{K} \) have characteristic 0, and let \( E \) be a vector space over \( \mathbb{K} \) . Then the symmetrizer operator \( \sigma \) satisfies \( {\sigma }^{2} = \sigma \) . The kernel of \( \sigma \) on \( {T}^{n}\left( E\right) \) is exactly \( {T}^{n}\left( E\right) \cap I \),... | Proof. We have\n\n\[ \n{\sigma }^{2}\left( {{v}_{1} \otimes \cdots \otimes {v}_{n}}\right) = \frac{1}{{\left( n!\right) }^{2}}\mathop{\sum }\limits_{{\rho ,\tau \in {\mathfrak{S}}_{n}}}{v}_{{\rho \tau }\left( 1\right) } \otimes \cdots \otimes {v}_{{\rho \tau }\left( n\right) }\n\]\n\n![e96ae642-3395-4f58-8eae-2546a737f... | Yes |
Proposition 6.30. Let \( E \) be a vector space over the field \( \mathbb{K} \) .\n\n(a) Let \( \iota \) be the \( n \) -multilinear function \( \iota \left( {{v}_{1},\ldots ,{v}_{n}}\right) = {v}_{1} \land \cdots \land {v}_{n} \) of \( E \times \cdots \times E \) into \( \mathop{\bigwedge }\limits^{n}\left( E\right) \... | Proof. The proof is completely analogous to the proof of Proposition 6.23. \( ▱ \) | Yes |
Corollary 6.31. If \( E \) and \( F \) are vector spaces over the field \( \mathbb{K} \), then the vector space \( {\operatorname{Hom}}_{\mathbb{K}}\left( {\mathop{\bigwedge }\limits^{n}\left( E\right), F}\right) \) is canonically isomorphic (via restriction to pure tensors) to the vector space of all \( F \) -valued a... | Proof. Restriction is linear and one-one. It is onto by Proposition 6.30a. | No |
Corollary 6.32. If \( E \) is a vector space over the field \( \mathbb{K} \), then the dual \( {\left( \mathop{\bigwedge }\limits^{n}\left( E\right) \right) }^{\prime } \) of \( \mathop{\bigwedge }\limits^{n}\left( E\right) \) is canonically isomorphic (via restriction to pure tensors) to the vector space of alternatin... | Proof. This is a special case of Corollary 6.31. | Yes |
Proposition 6.33. Let \( E \) be a vector space over the field \( \mathbb{K} \), let \( {\left\{ {u}_{i}\right\} }_{i \in A} \) be a basis of \( E \), and suppose that a simple ordering has been imposed on the index set \( A \) . Then the set of all monomials \( {u}_{{i}_{1}} \land \cdots \land {u}_{{i}_{n}} \) with \(... | Proof. Since multiplication in \( \bigwedge \left( E\right) \) satisfies \( a \land b = {\left( -1\right) }^{mn}b \land a \) for \( a \in \mathop{\bigwedge }\limits^{m}\left( E\right) \) and \( b \in \mathop{\bigwedge }\limits^{n}\left( E\right) \) and since monomials span \( {T}^{n}\left( E\right) \), the indicated se... | Yes |
Corollary 6.34. Let \( E \) be a finite-dimensional vector space over \( \mathbb{K} \) of dimension \( N \) . Then\n\n(a) \( \dim \mathop{\bigwedge }\limits^{n}\left( E\right) = \left( \begin{matrix} N \\ n \end{matrix}\right) \) for \( 0 \leq n \leq N \) and \( = 0 \) for \( n > N \) ,\n\n(b) \( \mathop{\bigwedge }\li... | Proof. Part (a) is an immediate consequence of Proposition 6.33, and (b) is proved in the same way as Corollary 6.27b, using Proposition 6.30a as a tool. The \ | No |
Proposition 6.35. Let the field \( \mathbb{K} \) have characteristic 0, and let \( E \) be a vector space over \( \mathbb{K} \). Then the antisymmetrizer operator \( {\sigma }^{\prime } \) satisfies \( {\sigma }^{\prime 2} = {\sigma }^{\prime } \). The kernel of \( {\sigma }^{\prime } \) on \( {T}^{n}\left( E\right) \)... | Proof. We have\n\n\[ \n{\sigma }^{\prime 2}\left( {{v}_{1} \otimes \cdots \otimes {v}_{n}}\right) = \frac{1}{{\left( n!\right) }^{2}}\mathop{\sum }\limits_{{\rho ,\tau \in {\mathfrak{S}}_{n}}}\left( {\operatorname{sgn}{\rho \tau }}\right) {v}_{{\rho \tau }\left( 1\right) } \otimes \cdots \otimes {v}_{{\rho \tau }\left(... | Yes |
If \( S \) is a set and \( W\left( {S}^{\prime }\right) \) is the corresponding set of words built from \( {S}^{\prime } = S \cup {S}^{-1} \), then the product operation defined on \( W\left( {S}^{\prime }\right) \) descends in a well-defined fashion to the set \( F\left( S\right) \) of equivalence classes of members o... | Let us denote equivalence classes by brackets. We want to define multiplication in \( F\left( S\right) \) by \( \left\lbrack {w}_{1}\right\rbrack \left\lbrack {w}_{2}\right\rbrack = \left\lbrack {{w}_{1}{w}_{2}}\right\rbrack \) . To see that this formula makes sense in \( F\left( S\right) \), let \( {x}_{1},{x}_{2} \),... | Yes |
Proposition 7.2. Let \( S \) be a set, \( F \) be a group, and \( {\iota }^{\prime } : S \rightarrow F \) be a function. Suppose that the pair \( \left( {F,{\iota }^{\prime }}\right) \) has the following universal mapping property: whenever \( G \) is a group and \( \varphi : S \rightarrow G \) is a function, then ther... | Proof. We apply the universal mapping property of \( \left( {F\left( S\right) ,\iota }\right) \), as stated in Theorem 7.1, to the group \( G = F \) and the function \( \varphi = {\iota }^{\prime } \), obtaining a group homomorphism \( \Phi : F\left( S\right) \rightarrow F \) such that \( {\iota }^{\prime } = \Phi \cir... | Yes |
Proposition 7.4. If \( G \) is a group, then the commutator subgroup is normal, and \( G/{G}^{\prime } \) is abelian. If \( \varphi : G \rightarrow H \) is any homomorphism of \( G \) into an abelian group \( H \), then \( \ker \varphi \supseteq {G}^{\prime } \) . | PROOF. The computation\n\n\[ \n{axy}{x}^{-1}{y}^{-1}{a}^{-1} = \left( {{ax}{a}^{-1}}\right) \left( {{ay}{a}^{-1}}\right) {\left( ax{a}^{-1}\right) }^{-1}{\left( ay{a}^{-1}\right) }^{-1} \]\n\nshows that \( {G}^{\prime } \) is normal. If \( \psi : G \rightarrow G/{G}^{\prime } \) is the quotient homomorphism, then \( \p... | Yes |
Corollary 7.5. If \( F \) is the free group on a set \( S \) and if \( {F}^{\prime } \) is the commutator subgroup of \( F \), then \( F/{F}^{\prime } \) is isomorphic to the free abelian group \( {\bigoplus }_{s \in S}{\mathbb{Z}}_{s} \) . | Proof. Let \( H = {\bigoplus }_{s \in S}{\mathbb{Z}}_{s} \), and let \( \varphi : S \rightarrow H \) be the function with \( \varphi \left( s\right) = {1}_{s} \) , i.e., \( \varphi \left( s\right) \) is to be the member of \( H \) that is 1 in the \( {s}^{\text{th }} \) coordinate and is 0 elsewhere. Application of the... | Yes |
If \( {F}_{1} \) and \( {F}_{2} \) are isomorphic free groups on sets \( {S}_{1} \) and \( {S}_{2} \) , respectively, then \( {S}_{1} \) and \( {S}_{2} \) have the same cardinality. | Corollary 7.5 shows that an isomorphism of \( {F}_{1} \) with \( {F}_{2} \) induces an isomorphism of the free abelian groups \( {\bigoplus }_{s \in {S}_{1}}{\mathbb{Z}}_{{s}_{1}} \) and \( {\bigoplus }_{s \in {S}_{2}}{\mathbb{Z}}_{{s}_{2}} \) . The rank of a free abelian group is a well-defined cardinal, and the resul... | Yes |
Proposition 7.7. Each group \( G \) is the homomorphic image of a free group. | Proof. Let \( S \) be a set of generators for \( G \) ; for example, \( S \) can be taken to be \( G \) itself. Let \( \varphi : S \rightarrow G \) be the inclusion of the set of generators into \( G \) , and let \( \widetilde{\varphi } : F\left( S\right) \rightarrow G \) be the group homomorphism of Theorem 7.1 such t... | Yes |
Proposition 7.8. Let \( G = \langle S;R\rangle \) be a group given by generators and relations, let \( {G}^{\prime } \) be a second group, let \( \varphi \) be a one-one function \( \varphi \) from \( S \) onto a set of generators for \( {G}^{\prime } \), and let \( \Phi : F\left( S\right) \rightarrow {G}^{\prime } \) ... | Proof. The proposition follows immediately from the universal mapping property in Theorem 7.1 in combination with Proposition 4.11. | No |
Proposition 7.9. Let \( S \) be a set, and let \( R = \left\{ {{st}{s}^{-1}{t}^{-1} \mid s \in S, t \in S}\right\} \) . Then the smallest normal subgroup of the free group \( F\left( S\right) \) containing \( R \) is the commutator subgroup \( F{\left( S\right) }^{\prime } \), and therefore \( \langle S;R\rangle \) is ... | Proof. The members of \( R \) are in \( F{\left( S\right) }^{\prime } \), the product of two members of \( F{\left( S\right) }^{\prime } \) is in \( F{\left( S\right) }^{\prime } \), and any conjugate of a member of \( F{\left( S\right) }^{\prime } \) is in \( F{\left( S\right) }^{\prime } \) . Therefore the smallest n... | Yes |
Theorem 7.10 (Nielsen-Schreier Theorem). Every subgroup of a free group is a free group. | Let the given free group be \( F \), let the subgroup be \( H \), and form the right cosets \( {Hg} \) in \( F \) . Let \( C \) be a set of representatives for these cosets, with 1 chosen as the representative of the identity coset; we shall impose further conditions on \( C \) shortly.\n\nWe define a function \( \rho ... | No |
Let \( S \) be the set of generators of \( F \), and let \( {S}^{\prime } = S \cup {S}^{-1} \) . Every element of \( H \) is a product of elements of the form \( {gb\rho }{\left( gb\right) }^{-1} \) with \( g \) in \( C \) and \( b \) in \( {S}^{\prime } \) . Furthermore the element \( {g}^{\prime } = \rho \left( {gb}\... | Proof. Any \( h \) in \( F \) can be written as a product \( h = {b}_{1}\cdots {b}_{n} \) with each \( {b}_{j} \) in \( {S}^{\prime } \) . Define \( {r}_{0} = 1 \) and \( {r}_{k} = \rho \left( {{b}_{1}\cdots {b}_{k}}\right) \) for \( 1 \leq k \leq n \) . Then\n\n\[ h{r}_{n}^{-1} = \left( {{r}_{0}{b}_{1}{r}_{1}^{-1}}\ri... | Yes |
Lemma 7.12. With \( F \) free it is possible to choose the set \( C \) of coset representatives in such a way that all of its members have expansions in terms of \( {S}^{\prime } \) as \( g = {b}_{1}\cdots {b}_{n} \) in which\n\n(a) \( g = {b}_{1}{b}_{2}\cdots {b}_{n} \) is a reduced word as written,\n\n(b) \( {b}_{1}{... | Proof. If \( {S}^{\prime } \) is finite or countably infinite, we enumerate it. In the uncountable case (which is of less practical interest), we introduce a well ordering in \( {S}^{\prime } \) by means of Zermelo's Well-Ordering Theorem as in Section A5 of the appendix.\n\nThe ordering of \( {S}^{\prime } \) will be ... | Yes |
Theorem 7.13. If \( S \) is a nonempty set of groups \( {G}_{s} \) and \( W\left( \left\{ {G}_{s}\right\} \right) \) is the set of all words from the groups \( {G}_{s} \), then the product operation defined on \( W\left( \left\{ {G}_{s}\right\} \right) \) descends in a well-defined fashion to the set \( { * }_{s \in S}... | Proof. Let us write \( \sim \) for the equivalence relation on words, and let us denote equivalence classes by brackets. We want to define multiplication in \( { * }_{s \in S}{G}_{s} \) by \( \left\lbrack {w}_{1}\right\rbrack \left\lbrack {w}_{2}\right\rbrack = \left\lbrack {{w}_{1}{w}_{2}}\right\rbrack \) . To see tha... | Yes |
Proposition 7.14. Let \( S \) be a nonempty set of groups \( {G}_{s} \) . Suppose that \( {G}^{\prime } \) is a group and that \( {i}_{s}^{\prime } : {G}_{s} \rightarrow {G}^{\prime } \) for \( s \in S \) is a system of group homomorphisms with the following universal mapping property: whenever \( H \) is a group and \... | Proof. Put \( G = { * }_{s \in S}{G}_{s} \) . In the universal mapping property of Theorem 7.13, let \( H = {G}^{\prime } \) and \( {\varphi }_{s} = {i}_{s}^{\prime } \), and let \( \Phi : G \rightarrow {G}^{\prime } \) be the homomorphism \( \varphi \) produced by that theorem. Then \( \Phi \) satisfies \( \Phi \circ ... | Yes |
Proposition 7.15. (solution of the word problem for free products). If \( S \) is a nonempty set of groups \( {G}_{s} \) and \( W\left( \left\{ {G}_{s}\right\} \right) \) is the set of all words from the groups \( {G}_{s} \), then each word in \( W\left( \left\{ {G}_{s}\right\} \right) \) is equivalent to one and only ... | Proof of Proposition 7.15. Both operations-eliminating factors of the identity and multiplying consecutive factors in each word when they come from the same group-reduce the length of a word. Since the length has to remain \( \geq 0 \), the process of successively carrying out these two operations as much as possible h... | Yes |
Proposition 7.18 (Schur’s Lemma). If \( \left( {{R}_{1},{V}_{1}}\right) \) and \( \left( {{R}_{2},{V}_{2}}\right) \) are irreducible representations of the finite group \( G \) on finite-dimensional complex vector spaces and if \( A : {V}_{1} \rightarrow {V}_{2} \) is an intertwining operator, then \( A \) is invertibl... | Proof. The equality \( {R}_{2}\left( g\right) A{v}_{1} = A{R}_{1}\left( g\right) v \) shows that \( \ker A \) and image \( A \) are invariant subspaces. By the assumed irreducibility, \( \ker A \) equals 0 or \( {V}_{1} \), and image \( A \) equals 0 or \( {V}_{2} \) . The first statement follows. When \( \left( {{R}_{... | No |
Every irreducible finite-dimensional representation of a finite abelian group \( G \) is 1-dimensional. | Proof. If \( \left( {R, V}\right) \) is given, then the linear map \( A = R\left( g\right) \) satisfies \( {AR}\left( g\right) = \) \( R\left( {xg}\right) = R\left( {gx}\right) = R\left( g\right) A \) for all \( x \) in \( G \) . By Schur’s Lemma (Proposition 7.18), \( A = R\left( g\right) \) is scalar. Since \( g \) i... | Yes |
Proposition 7.20. If \( \left( {R, V}\right) \) is a finite-dimensional representation of the finite group \( G \) and if an inner product is introduced in \( V \) that makes the representation unitary, then the orthogonal complement of an invariant subspace is invariant. | Proof. Let \( U \) be an invariant subspace. If \( u \) is in \( U \) and \( {u}^{ \bot } \) is in \( {U}^{ \bot } \), then \( \left( {R\left( g\right) {u}^{ \bot }, u}\right) = \left( {R{\left( g\right) }^{-1}R\left( g\right) {u}^{ \bot }, R{\left( g\right) }^{-1}u}\right) = \left( {{u}^{ \bot }, R{\left( g\right) }^{... | Yes |
Corollary 7.21. Any finite-dimensional representation of the finite group \( G \) is a direct sum of irreducible representations. | Proof. This is immediate by induction on the dimension. For dimension 0 , the representation is the empty direct sum of irreducible representations. If the decomposition is known for dimension \( < n \) and if \( U \) is an invariant subspace under \( R \) of smallest possible dimension \( \geq 1 \), then \( U \) is ir... | Yes |
Proposition 7.22 (Schur orthogonality). For finite-dimensional representations of a finite group \( G \) in which inner products have been introduced to make the representations unitary,\n\n(a) if \( \\left( {{R}_{1},{V}_{1}}\\right) \) and \( \\left( {{R}_{2},{V}_{2}}\\right) \) are inequivalent and irreducible, then\... | Proof. For (a), let \( l : {V}_{2} \\rightarrow {V}_{1} \) be any linear map, and form the linear map\n\n\[ \nL = \\mathop{\\sum }\\limits_{{x \\in G}}{R}_{1}\\left( x\\right) l{R}_{2}\\left( {x}^{-1}\\right) .\n\]\n\nMultiplying on the left by \( {R}_{1}\\left( g\\right) \) and on the right by \( {R}_{2}\\left( {g}^{-... | Yes |
Corollary 7.24. Let \( \\left\\{ \\left( {{R}_{\\alpha },{U}_{\\alpha }}\\right) \\right\\} \) be a complete set of inequivalent irreducible finite-dimensional representations of the finite group \( G \), and let \( {d}_{\\alpha } = \\dim {U}_{\\alpha } \). In each \( {U}_{\\alpha } \), introduce an inner product makin... | Proof. This follows form (a), (c), and (e) in Theorem 7.23, together with Theorem 3.11 and the remarks made before the statement of Theorem 7.23. | Yes |
Corollary 7.25. Let \( \\left\\{ \\left( {{R}_{\\alpha },{U}_{\\alpha }}\\right) \\right\\} \) be a complete set of inequivalent irreducible finite-dimensional representations of the finite group \( G \), and let \( {d}_{\\alpha } = \\dim {U}_{\\alpha } \). Then \( \\mathop{\\sum }\\limits_{\\alpha }{d}_{\\alpha }^{2} ... | Proof. This follows by counting the number of members listed in the orthonormal basis of \( C\\left( {G,\\mathbb{C}}\\right) \) given in Corollary 7.24. | Yes |
Proposition 7.26. Let \( R,{R}_{1} \), and \( {R}_{2} \) be irreducible finite-dimensional representations of a finite group \( G \) . Then their characters satisfy\n\n(a) \( \mathop{\sum }\limits_{{x \in G}}{\left| {\chi }_{R}\left( x\right) \right| }^{2} = \left| G\right| \),\n\n(b) \( \mathop{\sum }\limits_{{x \in G... | Proof. These follow from Schur orthogonality (Proposition 7.22): For (a), let \( R \) act on the vector space \( V \), let \( d = \dim V \), introduce an inner product with respect to which \( R \) is unitary, and let \( \left\{ {v}_{i}\right\} \) be an orthonormal basis of \( V \) . Then Proposition 7.22b gives\n\n\[ ... | Yes |
Theorem 7.27 (Fourier inversion formula for class functions). For the finite group \( G \), let \( \left\{ \left( {{R}_{\alpha },{U}_{\alpha }}\right) \right\} \) be a complete set of inequivalent irreducible finite-dimensional representations of \( G \) . If \( f \) is a class function on \( G \), then\n\n\[ f\left( x... | Proof. Using the result and notation of Corollary 7.24, we have\n\n\[ f\left( x\right) = {\left| G\right| }^{-1}\mathop{\sum }\limits_{\alpha }{d}_{\alpha }\mathop{\sum }\limits_{{i, j}}\left( {\mathop{\sum }\limits_{{y \in G}}f\left( y\right) \overline{\left( {R}_{\alpha }\left( y\right) {v}_{i}^{\left( \alpha \right)... | Yes |
Corollary 7.28. If \( G \) is a finite group, then the number of irreducible finite-dimensional representations of \( G \), up to equivalence, equals the number of conjugacy classes of \( G \) . | Proof. Theorem 7.27 shows that the irreducible characters span the vector space of class functions. Proposition 7.26b shows that the irreducible characters are orthogonal and hence are linearly independent. Thus the number of irreducible characters equals the dimension of the space of class functions, which equals the ... | Yes |
Lemma 7.30. The set \( \mathcal{O} \) of algebraic integers is a ring, and \( \mathcal{O} \cap \mathbb{Q} = \mathbb{Z} \) . | Proof. Suppose that \( x \) and \( y \) are complex numbers satisfying the polynomial equations \( {x}^{m} + {a}_{m - 1}{x}^{m - 1} + \cdots + {a}_{1}x + {a}_{0} = 0 \) and \( {y}^{n} + {b}_{n - 1}{y}^{n - 1} + \cdots + {b}_{1}y + {b}_{0} = \) 0, each with integer coefficients. Form the subset of \( \mathbb{C} \) given... | Yes |
Proposition 7.36 (Schreier). Let two groups \( N \) and \( G \) be given, along with a family of automorphisms \( x \mapsto {x}^{u} \) of \( N \) parametrized by \( u \) in \( G \), as well as a function \( a : G \times G \rightarrow N \) such that\n\n(a) \( {\left( {x}^{v}\right) }^{u} = a\left( {u, v}\right) {x}^{uv}... | Proof. Reverting to the earlier notation, let us write \( x\bar{u} \) in place of \( \left( {x, u}\right) \) for elements of \( E \) . Associativity of multiplication follows from the computation\n\n\[ \left( {x\bar{u}y\bar{v}}\right) \left( {z\bar{w}}\right) = \left( {x{y}^{u}a\left( {u, v}\right) \overline{uv}}\right... | Yes |
Proposition 7.37. Suppose that \( {E}_{1} \) and \( {E}_{2} \) are group extensions of \( N \) by \( G \) with respective inclusions \( {i}_{1} : N \rightarrow {E}_{1} \) and \( {i}_{2} : N \rightarrow {E}_{2} \) and with respective quotient homomorphisms \( {\varphi }_{1} : {E}_{1} \rightarrow G \) and \( {\varphi }_{... | Proof. For the direct part, suppose that \( \Phi \) exists. For each \( u \) in \( G \), select \( \bar{u} \) in \( {E}_{1} \) with \( {\varphi }_{1}\left( \bar{u}\right) = u \) . Then we can form the extension data \( \left\{ {x \mapsto {x}^{u}}\right\} \) and \( \{ a\left( {u, v}\right) \} \) for \( {E}_{1} \) relati... | Yes |
Proposition 7.38. Let \( G \) and \( N \) be groups with \( N \) abelian, and suppose that \( \tau : G \rightarrow \) Aut \( N \) is a homomorphism. Then the set of equivalence classes of group extensions of \( N \) by \( G \) corresponding to the action \( \tau : G \rightarrow \) Aut \( N \) is parametrized by the quo... | The extension \( E \) corresponding to the 0 factor set is of special interest. In this case the multiplication law for the coset representatives is \( \bar{u}\bar{v} = \overline{uv} \) since the member \( a\left( {u, v}\right) = 0 \) of \( N \) is to be interpreted multiplicatively in this product formula. Consequentl... | Yes |
Proposition 7.40. Let \( G \) and \( N \) be groups with \( N \) abelian, and suppose that \( \tau : G \rightarrow \) Aut \( N \) is a homomorphism. Then the set of equivalence classes of group extensions of \( N \) by \( G \) corresponding to the action \( \tau : G \rightarrow \) Aut \( N \) is parametrized by \( {H}^... | Since group extensions have such a nice interpretation in terms of cohomology groups \( {H}^{2} \), it is reasonable to look for a nice interpretation for \( {H}^{1} \) as well. Indeed, \( {H}^{1} \) has an interpretation in terms of uniqueness up to inner isomorphisms for semidirect-product decompositions. We continue... | Yes |
Theorem 8.4 (Second Isomorphism Theorem). Let \( R \) be a ring, let \( M \) be a left \( R \) module, and let \( {N}_{1} \) and \( {N}_{2} \) be \( R \) submodules of \( M \). Then \( {N}_{1} \cap {N}_{2} \) is an \( R \) submodule of \( {N}_{1} \), the set \( {N}_{1} + {N}_{2} \) of sums is an \( R \) submodule of \(... | \[ {N}_{1}/\left( {{N}_{1} \cap {N}_{2}}\right) \cong \left( {{N}_{1} + {N}_{2}}\right) /{N}_{2} \] | Yes |
Proposition 8.5. Let \( R \) be a ring, let \( M = {\bigoplus }_{s \in S}{M}_{s} \) be a direct sum of left \( R \) modules, and for each \( s \) in \( S \), let \( {N}_{s} \) be a left \( R \) submodule of \( {M}_{s} \) . Then the natural map of \( {\bigoplus }_{s \in S}{M}_{s} \) to the direct sum of quotients descen... | Proof. Let \( \varphi : {\bigoplus }_{s \in S}{M}_{s} \rightarrow {\bigoplus }_{s \in S}\left( {{M}_{s}/{N}_{s}}\right) \) be the \( R \) homomorphism defined by \( \varphi \left( {\left\{ {m}_{s}\right\} }_{s \in S}\right) = {\left\{ {m}_{s} + {N}_{s}\right\} }_{s \in S} \) . The mapping \( \varphi \) is onto \( {\big... | Yes |
Proposition 8.6. Let \( R \) be a nonzero integral domain, let \( F \) be its field of fractions, and let \( \eta \) be the canonical embedding of \( R \) into \( F \) . Whenever \( \varphi \) is a one-one ring homomorphism of \( R \) into a field \( {F}^{\prime } \) carrying 1 to 1, then there exists a unique ring hom... | Proof. If \( \left( {a, b}\right) \) with \( b \neq 0 \) is a pair in \( \widetilde{F} \), we define \( \Phi \left( {a, b}\right) = \varphi \left( a\right) \varphi {\left( b\right) }^{-1} \) . This is well defined since \( b \neq 0 \) and since \( \varphi \) , being one-one, cannot have \( \varphi \left( b\right) = 0 \... | Yes |
Proposition 8.7. An ideal \( I \) in the commutative ring \( R \) is prime if and only if \( R/I \) is an integral domain. | Proof. If a proper ideal \( I \) fails to be prime, choose \( {ab} \) in \( I \) with \( a \notin I \) and \( b \notin I \) . Then \( a + I \) and \( b + I \) are nonzero in \( R/I \) and have product \( 0 + I \) . So \( R/I \) is nonzero and has a zero divisor; by definition, \( R/I \) fails to be an integral domain.\... | Yes |
Proposition 8.8. In a commutative ring \( R \) with identity, any proper ideal is contained in a maximal ideal. | Proof. This follows from Zorn's Lemma (Section A5 of the appendix). Specifically let \( I \) be the given proper ideal, and form the set \( S \) of all proper ideals that contain \( I \) . This set is nonempty, containing \( I \) as a member, and we order it by inclusion upward. If we have a chain in \( S \), then the ... | Yes |
Lemma 8.9. If \( R \) is a nonzero commutative ring with identity, then \( R \) is a field if and only if the only proper ideal in \( R \) is 0 . | Proof. If \( R \) is a field and \( I \) is a nonzero ideal in \( R \), let \( a \neq 0 \) be in \( I \) . Then \( 1 = a{a}^{-1} \) is in \( I \), and consequently \( I = R \) . Conversely if the only ideals in \( R \) are 0 and \( R \), let \( a \neq 0 \) be given in \( R \), and form the ideal \( I = {aR} \) . Since ... | Yes |
Proposition 8.10. If \( R \) is a commutative ring with identity, then an ideal \( I \) is maximal if and only if \( R/I \) is a field. | Proof. We consider \( R \) and \( R/I \) as unital \( R \) modules, the ideals for each of \( R \) and \( R/I \) being the \( R \) submodules. The quotient ring homomorphism \( R \rightarrow R/I \) is an \( R \) homomorphism. By the First Isomorphism Theorem for modules (Theorem 8.3), there is a one-one correspondence ... | No |
Corollary 8.11. If \( R \) is a commutative ring with identity, then every maximal ideal is prime. | Proof. If \( I \) is maximal, then \( R/I \) is a field by Proposition 8.10. Hence \( R/I \) is an integral domain, and \( I \) must be prime by Proposition 8.7. | Yes |
Proposition 8.12. In \( R = \mathbb{Z} \) or \( R = \mathbb{K}\left\lbrack X\right\rbrack \) with \( \mathbb{K} \) a field, every nonzero prime ideal is maximal. | Proof. Examples 1 and 2 at the beginning of this section show that every nonzero prime ideal is of the form \( I = {pR} \) with \( p \) prime. If such an \( I \) is given and if \( J \) is any ideal strictly containing \( I \), choose \( a \) in \( J \) with \( a \) not in \( I \) . Since \( a \) is not in \( I = {pR} ... | Yes |
Proposition 8.13. In an integral domain \( R \) in which (UFD1) holds, the condition (UFD2) is equivalent to the condition\n\n(UFD2') every irreducible element is prime. | Proof. Suppose that (UFD2) holds, that \( p \) is an irreducible element, and that \( p \) divides \( {ab} \) . We are to show that \( p \) divides \( a \) or \( p \) divides \( b \) . We may assume that \( {ab} \neq 0 \) . Write \( {ab} = {pc} \), and let \( a = \mathop{\prod }\limits_{i}{p}_{i}, b = \mathop{\prod }\l... | No |
Proposition 8.14. A nonzero element \( p \) in an integral domain \( R \) is prime if and only if the ideal \( \left( p\right) \) in \( R \) is prime. | Proof. Suppose that the element \( p \) is prime. Then the ideal \( \left( p\right) \) is not \( R \) ; in fact, otherwise 1 would have to be of the form \( 1 = {rp} \) for some \( r \in R, r \) would be a multiplicative inverse of \( p \), and \( p \) would be a unit. Now suppose that a product \( {ab} \) is in the id... | Yes |
Corollary 8.16. In a principal ideal domain, every nonzero prime ideal is maximal. | Proof. Let \( \left( p\right) \) be a nonzero prime ideal. Proposition 8.14 shows that \( p \) is prime, and prime elements are automatically irreducible. The proof of the uniqueness part of Theorem 8.15 then deduces in the context of a principal ideal domain that \( \left( p\right) \) is maximal. | Yes |
Theorem 8.17. Every Euclidean domain is a principal ideal domain. | Proof. Let \( I \) be an ideal in \( R \) . We are to show that \( I \) is principal. Without loss of generality, we may assume that \( I \neq 0 \) . Choose \( b \neq 0 \) in \( I \) with \( \delta \left( b\right) \) as small as possible. Certainly \( I \supseteq \left( b\right) \) . If \( a \neq 0 \) is in \( I \), wr... | Yes |
Theorem 8.18 (Gauss’s Lemma). If \( R \) is a unique factorization domain, then the product of primitive polynomials is primitive. | Proof #1. Arguing by contradiction, let \( A\left( X\right) = {a}_{m}{X}^{m} + \cdots + {a}_{0} \) and \( B\left( X\right) = {b}_{n}{X}^{n} + \cdots + {b}_{0} \) be primitive polynomials such that every coefficient of \( A\left( X\right) B\left( X\right) \) is divisible by some prime \( p \) . Since \( A\left( X\right)... | Yes |
Proposition 8.19. Let \( R \) be a unique factorization domain, and let \( F \) be its field of fractions. If \( A\left( X\right) \) is any nonzero polynomial in \( F\left\lbrack X\right\rbrack \), then there exist \( \alpha \) in \( F \) and \( {A}_{0}\left( X\right) \) in \( R\left\lbrack X\right\rbrack \) such that ... | Proof. Let \( A\left( X\right) = {c}_{n}{X}^{n} + \cdots + {c}_{1}X + {c}_{0} \) with each \( {c}_{k} \) in \( F \) . We can write each \( {c}_{k} \) as \( {a}_{k}{b}_{k}^{-1} \) with \( {a}_{k} \) and \( {b}_{k} \) in \( R \) and \( {b}_{k} \neq 0 \) . We clear fractions. That is, we let \( \beta = \mathop{\prod }\lim... | Yes |
Let \( R \) be a unique factorization domain, and let \( F \) be its field of fractions.\n\n(a) Let \( A\left( X\right) \) and \( B\left( X\right) \) be nonzero polynomials in \( R\left\lbrack X\right\rbrack \), and suppose that \( B\left( X\right) \) is primitive. If \( B\left( X\right) \) divides \( A\left( X\right) ... | Proof. In (a), write \( A\left( X\right) = B\left( X\right) Q\left( X\right) \) in \( F\left( X\right) \), and let \( Q\left( X\right) = \rho {Q}_{0}\left( X\right) \) be a decomposition of \( Q\left( X\right) \) as in Proposition 8.19. Since \( c{\left( A\right) }^{-1}A\left( X\right) \) is primitive, the correspondin... | Yes |
Corollary 8.22 (Eisenstein’s irreducibility criterion). Let \( R \) be a unique factorization domain, let \( F \) be its field of fractions, and let \( p \) be a prime in \( R \) . If \( A\left( X\right) = {a}_{N}{X}^{N} + \cdots + {a}_{1}X + {a}_{0} \) is a polynomial of degree \( \geq 1 \) in \( R\left\lbrack X\right... | Proof. Without loss of generality, we may replace \( A\left( X\right) \) by \( c{\left( A\right) }^{-1}A\left( X\right) \) and thereby assume that \( A\left( X\right) \) is primitive. Corollary \( {8.20}\mathrm{\;b} \) shows that it is enough to prove irreducibility in \( R\left\lbrack X\right\rbrack \) . Assuming the ... | Yes |
Lemma 8.23. Let \( R \) be a commutative ring with identity, and let \( \varphi : M \rightarrow N \) be a homomorphism of unital \( R \) modules. If \( \ker \varphi \) and image \( \varphi \) are finitely generated, then \( M \) is finitely generated. | Proof. Let \( \left\{ {{x}_{1},\ldots ,{x}_{m}}\right\} \) and \( \left\{ {{y}_{1},\ldots ,{y}_{n}}\right\} \) be respective finite sets of generators for \( \ker \varphi \) and image \( \varphi \) . For \( 1 \leq j \leq n \), choose \( {x}_{j}^{\prime } \) in \( M \) with \( \varphi \left( {x}_{j}^{\prime }\right) = {... | Yes |
Proposition 8.24. If \( R \) is a principal ideal domain, then any \( R \) submodule of a finitely generated unital \( R \) module is finitely generated. Moreover, any \( R \) submodule of a singly generated unital \( R \) module is singly generated. | Proof. Let \( M \) be unital and finitely generated with a set \( \left\{ {{m}_{1},\ldots ,{m}_{n}}\right\} \) of \( n \) generators, and define \( {M}_{k} = R{m}_{1} + \cdots + R{m}_{k} \) for \( 1 \leq k \leq n \) . Then \( {M}_{n} = M \) since \( M \) is unital. We shall prove by induction on \( k \) that every \( R... | Yes |
Theorem 8.27 (Chinese Remainder Theorem). Let \( R \) be a commutative ring with identity, and let \( {I}_{1},\ldots ,{I}_{n} \) be ideals in \( R \) such that \( {I}_{i} + {I}_{j} = R \) whenever \( i \neq j \). (a) If elements \( {x}_{1},\ldots ,{x}_{n} \) of \( R \) are given, then there exists \( x \) in \( R \) su... | Proof. For existence in (a) when \( n = 1 \), we take \( x = {x}_{1} \). For existence when \( n = 2 \), the assumption \( {I}_{1} + {I}_{2} = R \) implies that there exist \( {a}_{1} \in {I}_{1} \) and \( {a}_{2} \in {I}_{2} \) with \( {a}_{1} + {a}_{2} = 1 \). Given \( {x}_{1} \) and \( {x}_{2} \), we put \( x = {x}_... | Yes |
Corollary 8.28. Let \( R \) be a principal ideal domain, and let \( a = \varepsilon {p}_{1}^{{k}_{1}}\cdots {p}_{n}^{{k}_{n}} \) be a factorization of a nonzero nonunit element \( a \) into the product of a unit and powers of nonassociate primes. Then there is a ring isomorphism\n\n\[ R/\left( a\right) \cong R/\left( {... | Proof. Let \( {I}_{j} = \left( {p}_{j}^{{k}_{j}}\right) \) in Theorem 8.27. For \( i \neq j \), we have \( \operatorname{GCD}\left( {{p}_{i}^{{k}_{i}},{p}_{j}^{{k}_{j}}}\right) = \) 1. Since \( R \) is a principal ideal domain, there exist \( a \) and \( b \) in \( R \) with \( a{p}_{i}^{{k}_{i}} + b{p}_{j}^{{k}_{j}} =... | Yes |
If \( R \) is a principal ideal domain, then any finitely generated unital \( R \) module \( M \) is the direct sum of a nonunique free \( R \) submodule \( {\bigoplus }_{i = 1}^{s}R \) of a well-defined finite rank \( s \geq 0 \) and the \( R \) submodule \( T \) of all members \( m \) of \( M \) such that \( {rm} = 0... | Proof. Theorem 8.25c gives \( M = F \oplus {\bigoplus }_{j = 1}^{n}R{a}_{j} \), where \( F \) is a free \( R \) submodule of some finite rank \( s \) and the \( {a}_{j} \) ’s are nonzero members of \( M \) that are each annihilated by some nonzero member of \( R \) . The set \( T \) of all \( m \) with \( {rm} = 0 \) f... | Yes |
Proposition 8.30. If \( R \) is a commutative ring with identity and \( M \) is a unital \( R \) 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 \cdot... | Proof. To see that (a) implies (b), let \( \mathcal{C} \) be a nonempty collection of \( R \) submodules of \( M \) . Take \( {M}_{1} \) in \( \mathcal{C} \) . If \( {M}_{1} \) is not maximal, choose \( {M}_{2} \) in \( \mathcal{C} \) properly containing \( {M}_{1} \) . If \( {M}_{2} \) is not maximal, choose \( {M}_{3... | Yes |
Corollary 8.31. If \( R \) is a commutative ring with identity, then the following conditions on \( R \) are equivalent:\n\n(a) ascending chain condition for ideals: every strictly ascending chain of ideals in \( R \) is finite,\n\n(b) maximum condition for ideals of \( R \) : every nonempty collection of ideals in \( ... | The corollary follows immediately from Proposition 8.30. A commutative ring with identity satisfying the equivalent conditions of Corollary 8.31 is said to be a Noetherian commutative ring. | No |
Proposition 8.33. In a Noetherian integral domain, every nonzero nonunit is a product of irreducible elements. | Proof. In other words, the assertion is that the condition (UFD1) of Section 4 is satisfied. The proof of the assertion is essentially the same as the proof of (UFD1) for Theorem 8.15. | No |
Proposition 8.34. If \( R \) is a Noetherian commutative ring, then any \( R \) submodule of a finitely generated unital \( R \) module is finitely generated. | Proof. Let \( M \) be a unital finitely generated \( R \) module with a set \( \left\{ {{m}_{1},\ldots ,{m}_{n}}\right\} \) of \( n \) generators, and define \( {M}_{k} = R{m}_{1} + \cdots + R{m}_{k} \) for \( 1 \leq k \leq n \) . Then \( {M}_{n} = M \) since \( M \) is unital. We shall prove by induction on \( k \) th... | Yes |
Proposition 8.35. Let \( R \) be an integral domain, \( F \) be its field of fractions, and \( K \) be any field containing \( F \) . Then the following conditions on an element \( x \) of \( K \) are equivalent:\n\n(a) \( x \) is a root of a monic polynomial in \( R\left\lbrack X\right\rbrack \) ,\n\n(b) the subring \... | Proof. If (a) holds, we can write \( {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} = 0 \) for suitable coefficients in \( R \) . Solving for \( {x}^{n} \) and substituting, we see that the subring \( R\left\lbrack x\right\rbrack \), which equals \( R + {Rx} + R{x}^{2} + \cdots \), is actually given by ... | Yes |
Lemma 8.36. If \( A, B \), and \( C \) are integral domains with \( A \subseteq B \subseteq C \) such that \( C \) is a finitely generated \( B \) module and \( B \) is a finitely generated \( A \) module, then \( C \) is a finitely generated \( A \) module. | Proof. Let \( C \) be generated over \( B \) by \( {c}_{1},\ldots ,{c}_{r} \), and let \( B \) be generated over \( A \) by \( {b}_{1},\ldots ,{b}_{s} \) . Then \( C \) is generated over \( A \) by the \( {sr} \) elements \( {b}_{j}{c}_{i} \) for \( 1 \leq i \leq r \) and \( 1 \leq j \leq s \) . | Yes |
Proposition 8.37. Let \( R \) be an integral domain, \( F \) be its field of fractions, and \( K \) be any field containing \( F \) . If \( {x}_{1},\ldots ,{x}_{r} \) are members of \( K \) integral over \( R \) , then the subring \( R\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \) of \( K \) generated by \( R \... | Proof. We induct on \( r \) . Since \( {x}_{1} \) is assumed integral over \( R \), the case \( r = 1 \) follows from Proposition \( {8.35}\mathrm{\;b} \) . For the inductive step, suppose that \( R\left\lbrack {{x}_{1},\ldots ,{x}_{s}}\right\rbrack \) is a finitely generated \( R \) module. Since \( {x}_{s + 1} \) is ... | Yes |
Let \( R \) be an integral domain, \( F \) be its field of fractions, and \( K \) be any field containing \( F \) . Then the integral closure of \( R \) in \( K \) is a subring of \( K \). | Proof. Let \( x \) and \( y \) be integral over \( R \) . Then \( R\left\lbrack {x, y}\right\rbrack \) is a finitely generated \( R \) module by Proposition 8.37. We have \( \left( {x \pm y}\right) R\left\lbrack {x, y}\right\rbrack \subseteq R\left\lbrack {x, y}\right\rbrack \) and \( \left( {xy}\right) R\left\lbrack {... | Yes |
Corollary 8.39. Let \( A, B \), and \( C \) be integral domains with \( A \subseteq B \subseteq C \) . If every member of \( B \) is integral over \( A \) and if every member of \( C \) is integral over \( B \), then every member of \( C \) is integral over \( A \) . | Proof. Let \( K \) be the field of fractions of \( C \), and regard \( C \) as a subring of \( K \) . If \( x \) is in \( C \), then \( x \) is a root of some monic polynomial with coefficients in \( B \), say \( {x}^{n} + {b}_{n - 1}{x}^{n - 1} + \cdots + {b}_{0} = 0 \) . By Proposition 8.37 the subring \( D = \) \( A... | Yes |
Corollary 8.40. Let \( R \) be an integral domain, \( F \) be its field of fractions, and \( K \) be any field containing \( F \) . Then the integral closure \( T \) of \( R \) in \( K \) is integrally closed. | Proof. Corollary 8.38 shows that \( T \) is a subring of \( K \) . Let \( C \) be the integral closure of \( T \) in \( K \) . We apply Corollary 8.39 to the integral domains \( R \subseteq T \subseteq C \) . The corollary says that every member of \( C \) is integral over \( R \), and hence \( C \subseteq T \) . That ... | Yes |
Proposition 8.41. If \( R \) is a unique factorization domain, then \( R \) is integrally closed. | Proof. Suppose that \( {y}^{-1}x \) is a member of the field of fractions \( F \) of \( R \), with \( x \) and \( y \) in \( R \) and \( y \neq 0 \), and suppose that \( {y}^{-1}x \) satisfies the equation\n\n\[{\left( {y}^{-1}x\right) }^{n} + {a}_{n - 1}{\left( {y}^{-1}x\right) }^{n - 1} + \cdots + {a}_{1}\left( {{y}^... | Yes |
Proposition 8.42. Let \( R \) be an integral domain, \( F \) be its field of fractions, and \( K \) be any field containing \( F \) . If \( {\dim }_{F}K < \infty \), then any \( x \) in \( K \) has the property that there is some \( c \neq 0 \) in \( R \) such that \( {cx} \) is integral over \( R \) . | Proof. Since \( {\dim }_{F}K < \infty \), the elements \( 1, x,{x}^{2},\ldots \) of \( K \) are linearly dependent over \( F \) . Therefore \( {a}_{n}{x}^{n} + \cdots + {a}_{1}x + {a}_{0} = 0 \) for a suitable \( n \) and for suitable members of \( F \) with \( {a}_{n} \neq 0 \) . Clearing fractions, we may assume that... | Yes |
Proposition 8.43. Let \( R \) be an integral domain, \( F \) be its field of fractions, \( K \) be any field containing \( F \), and \( T \) be the integral closure of \( R \) in \( K \). If \( Q \) is a nonzero prime ideal of \( T \), then \( P = R \cap Q \) is a nonzero prime ideal of \( R \). | Proof. Let \( Q \) be a nonzero prime ideal of \( T \), and put \( P = R \cap Q \). The ideal \( P \) is proper since 1 is not in \( Q \) and cannot be in \( P \). It is prime since \( {xy} \in P \) implies that \( {xy} \) is in \( Q, x \) or \( y \) is in \( Q \), and \( x \) or \( y \) is in \( R \cap Q = P \). To se... | Yes |
Lemma 8.44. Let \( R \) and \( T \) be integral domains with \( R \subseteq T \) and with every element of \( T \) integral over \( R \) . If \( {T}^{\prime } \) is an integral domain and \( \varphi : T \rightarrow {T}^{\prime } \) is a homomorphism of rings onto \( {T}^{\prime } \), then every member of \( {T}^{\prime... | Proof. If \( t \) is in \( T \), then \( t \) satisfies some monic polynomial equation of the form \( {t}^{n} + {a}_{n - 1}{t}^{n - 1} + \cdots + {a}_{1}t + {a}_{0} = 0 \) with coefficients in \( R \) . Applying \( \varphi \) to this equation, we see that \( \varphi \left( t\right) \) satisfies a monic polynomial equat... | Yes |
Proposition 8.45. Let \( R \) be an integral domain, \( F \) be its field of fractions, \( K \) be any field containing \( F \), and \( T \) be the integral closure of \( R \) in \( K \). If every nonzero prime ideal of \( R \) is maximal, then every nonzero prime ideal of \( T \) is maximal. | Proof. Let \( Q \) be a nonzero prime ideal in \( T \), and let \( P = R \cap Q \). Since \( P \) is a nonzero prime ideal of \( R \) by Proposition 8.43, the hypotheses say that \( P \) is maximal in \( R \). We shall apply Lemma 8.44 to the quotient homomorphism \( T \rightarrow T/Q \). The lemma says that every elem... | Yes |
Proposition 8.46. Let \( R \) be a commutative ring with identity, let \( S \) be a multiplicative system in \( R \), let \( {S}^{-1}R \) be the localization of \( R \) at \( S \), and let \( \eta \) be the canonical homomorphism of \( R \) into \( {S}^{-1}R \) . Whenever \( \varphi \) is a ring homomorphism of \( R \)... | Proof. If \( \left( {r, s}\right) \) with \( s \in S \) is a pair in \( \widetilde{R} \), we define \( \Phi \left( {r, s}\right) = \varphi \left( r\right) \varphi {\left( s\right) }^{-1} \) . This is well defined since \( \varphi \left( s\right) \) is assumed to be a unit in \( T \) . Let us see that \( \Phi \) is cons... | Yes |
Proposition 8.47. Let \( R \) be a commutative ring with identity, and let \( {S}^{-1}R \) be a localization. If \( J \) is an ideal in \( {S}^{-1}R \), then \( {S}^{-1}\left( {R \cap J}\right) = J \) . Consequently the mapping \( I \mapsto {S}^{-1}I \) is a one-one mapping of the set of all ideals \( I \) in \( R \) o... | Proof. From the facts that \( R \cap J \subseteq J \) and \( J \) is an ideal in \( {S}^{-1}R \), we obtain \( {S}^{-1}\left( {R \cap J}\right) \subseteq {S}^{-1}J \subseteq J \) . For the reverse inclusion let \( x \) be in \( J \), and write \( x = {s}^{-1}r \) with \( r \) in \( R \) and \( s \) in \( S \) . Then \(... | Yes |
(a) If \( R \) is Noetherian, then \( {S}^{-1}R \) is Noetherian. | For (a), let \( \left\{ {J}_{\alpha }\right\} \) be a nonempty collection of ideals in \( {S}^{-1}R \) . Contraction of ideals is one-one by the first conclusion of Proposition 8.47, and it respects inclusions because it is given by the inverse image of a function. Since \( R \) is Noetherian, Corollary \( {8.31}\mathr... | Yes |
Proposition 8.49. A nonzero commutative ring \( R \) with identity is a local ring if and only if the nonunits of \( R \) form an ideal. | Proof. If the nonunits of \( R \) form an ideal, then that ideal is a unique maximal ideal since a proper ideal cannot contain a unit; hence \( R \) is local. Conversely suppose that \( R \) is local and that \( M \) is the unique maximal ideal. If \( x \) is any nonunit, then the principal ideal \( \left( x\right) \) ... | Yes |
Corollary 8.50. Let \( R \) be an integral domain, let \( P \) be a prime ideal of \( R \), let \( S \) be the set-theoretic complement of \( P \), and let \( {R}_{P} = {S}^{-1}R \) be the localization of \( R \) at \( P \). Then \( {R}_{P} \) is a local ring, its unique maximal ideal is \( M = {S}^{-1}P \), and \( P \... | Proof. The subset \( {S}^{-1}P \) of \( {S}^{-1}R \) is an ideal by Proposition 8.47, and Corollary 8.48d shows that it is proper. Every member of \( {S}^{-1}R \) that is not in \( {S}^{-1}P \) is of the form \( {s}^{\prime - 1}s \) with \( s \) and \( {s}^{\prime } \) in \( S \) and hence is a unit. Since no unit lies... | Yes |
Lemma 8.51 (Nakayama’s Lemma). Let \( R \) be a commutative ring with identity, let \( I \) be an ideal of \( R \) contained in all maximal ideals, and let \( M \) be a finitely generated unital \( R \) module. If \( {IM} = M \), then \( M = 0 \) . | Proof. We induct on the number of generators of \( M \) . If \( M \) is singly generated, say by a generator \( m \), then the hypothesis \( {IM} = M \) implies that \( {rm} = m \) for some \( r \) in \( I \) . Thus \( \left( {1 - r}\right) m = 0 \) . If \( 1 - r \) is a unit, then we can multiply by its inverse and ob... | Yes |
Proposition 8.52. Let \( R \) be a Noetherian commutative ring, and let \( I \) and \( P \) be ideals in \( R \) with \( P \) prime. If \( {IP} = I \), then \( I = 0 \) . | Proof. Let us localize with respect to the prime ideal \( P \) . If we write \( S \) for the set-theoretic complement of \( P \) in \( R \), then \( {R}_{P} = {S}^{-1}R \) is a local ring by Corollary 8.50, and its unique maximal ideal is \( {S}^{-1}P \) . Since \( \left( {{S}^{-1}I}\right) \left( {{S}^{-1}R}\right) = ... | Yes |
Proposition 8.53. Let \( R \) be an integral domain, \( F \) be its field of fractions, \( K \) be any field containing \( F \), and \( T \) be the integral closure of \( R \) in \( K \). If \( P \) is a maximal ideal in \( R \), then \( {PT} \neq T \), and there exists a maximal ideal \( Q \) of \( T \) with \( P = R ... | Proof. If \( {PT} \neq T \), then Proposition 8.8 supplies a maximal ideal \( Q \) of \( T \) with \( {PT} \subseteq Q \). Since 1 is not in \( Q \), we then have \( P \subseteq R \cap Q \subsetneqq R \). Consequently the maximality of \( P \) implies that \( P = R \cap Q \).\n\nArguing by contradiction, we now assume ... | Yes |
Lemma 8.56. In an integral domain, if \( P \) is a prime ideal such that \( P \supseteq {I}_{1}\cdots {I}_{n} \) for the product of the ideals \( {I}_{1},\ldots ,{I}_{n} \), then \( P \supseteq {I}_{j} \) for some \( j \) . | Proof. By induction it is enough to handle \( n = 2 \) . Thus suppose \( P \supseteq {I}_{1}{I}_{2} \) . We are to show that \( P \supseteq {I}_{1} \) or \( P \supseteq {I}_{2} \) . Arguing by contradiction, suppose on the contrary that \( x \in {I}_{1} \) and \( y \in {I}_{2} \) are elements with \( x \notin P \) and ... | Yes |
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