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Theorem 3.6. Let \( \mathrm{R} \) be a commutative ring with identity and \( \mathrm{B} \) an \( \mathrm{R} \) -module satisfying the ascending chain condition on submodules. Then every submodule \( \mathrm{A}\left( { \neq \mathrm{B}}\right) \) has a reduced primary decomposition. In particular, every submodule \( \mat... | PROOF OF 3.6. Let \( \mathcal{S} \) be the set of all submodules of \( B \) that do not have a primary decomposition. Clearly no primary submodule is in \( S \) . We must show that \( S \) is actually empty. If \( \mathcal{S} \) is nonempty, then \( \mathcal{S} \) contains a maximal element \( C \) by Theorem 1.4. Sinc... | Yes |
Proposition 4.1. (I. S. Cohen). A commutative ring \( \mathbf{R} \) with identity is Noetherian if and only if every prime ideal of \( \mathbf{R} \) is finitely generated. | SKETCH OF PROOF. ( \( \Leftarrow \) ) Let \( \mathcal{S} \) be the set of all ideals of \( R \) which are not finitely generated. If \( \mathcal{S} \) is nonempty, then use Zorn’s Lemma to find a maximal element \( P \) of \( \mathbb{S} \) . \( P \) is prime by Proposition 2.4 and hence finitely generated by hypothesis... | No |
Lemma 4.2. Let \( \mathrm{B} \) be a finitely generated module over a commutative ring \( \mathrm{R} \) with identity and let \( \mathbf{I} \) be the annihilator of \( \mathbf{B} \) in \( \mathbf{R} \). Then \( \mathbf{B} \) satisfies the ascending [resp. descending] chain condition on submodules if and only if \( \mat... | SKETCH OF PROOF. Let \( B \) be generated by \( {b}_{1},\ldots ,{b}_{n} \) and assume \( B \) satisfies the ascending chain condition. Then \( B = R{b}_{1} + \cdots + R{b}_{n} \) by Theorem IV.1.5. Consequently, \( I = {I}_{1} \cap {I}_{2} \cap \cdots \cap {I}_{n} \), where \( {I}_{j} \) is the annihilator of the submo... | No |
Lemma 4.3. Let \( \mathrm{P} \) be a prime ideal in a commutative ring \( \mathrm{R} \) with identity. If \( \mathrm{C} \) is a P-primary submodule of the Noetherian R-module A, then there exists a positive integer \( \mathrm{m} \) such that \( {\mathrm{P}}^{\mathrm{m}}\mathrm{A} \subset \mathrm{C} \) . | PROOF. Let \( I \) be the annihilator of \( A \) in \( R \) and consider the ring \( \bar{R} = R/I \) . Denote the coset \( r + {I\varepsilon }\bar{R} \) by \( \bar{r} \) . Clearly \( I \subset \{ r \in R \mid {rA} \subset C\} \subset P \), whence \( \bar{P} = P/I \) is an ideal of \( \bar{R}.A \) and \( C \) are each ... | Yes |
Theorem 4.4. (Krull Intersection Theorem). Let \( \mathrm{R} \) be a commutative ring with identity, I an ideal of \( \mathrm{R} \) and \( \mathrm{A} \) a Noetherian \( \mathrm{R} \) -module. If \( \mathrm{B} = \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{\mathrm{I}}^{\mathrm{n}}\mathrm{A} \), then \( \mathrm{{IB}} = \... | PROOF OF 4.4. If \( {IB} = A \), then \( A = {IB} \subset B \), whence \( B = A = {IB} \) . If \( {IB} \neq A \), then by Theorem \( {3.6IB} \) has a primary decomposition:\n\n\[ \n{IB} = {A}_{1} \cap {A}_{2} \cap \cdots \cap {A}_{s} \n\]\n\nwhere each \( {A}_{i} \) is a \( {P}_{i} \) -primary submodule of \( A \) for ... | Yes |
Lemma 4.5. (Nakayama) If \( \mathrm{J} \) is an ideal in a commutative ring \( \mathrm{R} \) with identity, then the following conditions are equivalent.\n\n(i) \( \mathrm{J} \) is contained in every maximal ideal of \( \mathrm{R} \) ;\n\n(ii) \( {1}_{\mathrm{R}} - \mathrm{j} \) is a unit for every \( \mathrm{j}\vareps... | PROOF OF 4.5. (i) \( \Rightarrow \) (ii) if \( {j\varepsilon J} \) and \( {1}_{R} - j \) is not a unit, then the ideal \( \left( {{1}_{R} - j}\right) \) is not \( R \) itself (Theorem III.3.2) and therefore is contained in a maximal ideal \( M \neq R \) (Theorem III.2.18). But \( {1}_{R} - j \in M \) and \( {j\varepsil... | Yes |
Proposition 4.6. Let \( \mathrm{J} \) be an ideal in a commutative ring \( \mathrm{R} \) with identity. Then \( \mathrm{J} \) is contained in every maximal ideal of \( \mathrm{R} \) if and only if for every \( \mathrm{R} \) -module \( \mathrm{A} \) satisfying the ascending chain condition on submodules, \( \mathop{\big... | PROOF. \( \left( \Rightarrow \right) \) If \( B = \mathop{\bigcap }\limits_{n}{J}^{n}A \), then \( {JB} = B \) by Theorem 4.4. Since \( B \) is finitely generated by Theorem \( {1.9}, B = 0 \) by Nakayama’s Lemma 4.5.\n\n\( \left( \Leftarrow \right) \) We may assume \( R \neq 0 \) . If \( M \) is any maximal ideal of \... | Yes |
Corollary 4.7. If \( \mathrm{R} \) is a Noetherian local ring with maximal ideal \( \mathrm{M} \), then \( \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{M}^{n} = 0 \) . | PROOF. If \( J = M \) and \( A = R \), then \( {J}^{n}A = {M}^{n} \) ; apply Proposition 4.6. | No |
Theorem 5.3. Let \( \mathrm{S} \) be an extension ring of \( \mathrm{R} \) and \( \mathrm{s} \in \mathrm{S} \) . Then the following conditions are equivalent.\n\n(i) \( \mathrm{s} \) is integral over \( \mathrm{R} \) ;\n\n(ii) \( \mathrm{R}\left\lbrack \mathrm{s}\right\rbrack \) is a finitely generated \( \mathrm{R} \)... | SKETCH OF PROOF. (i) \( \Rightarrow \) (ii) Suppose \( s \) is a root of the monic polynomial \( {f\varepsilon R}\left\lbrack x\right\rbrack \) of degree \( n \) . We claim that \( {1}_{R} = {s}^{0}, s,{s}^{2},\ldots ,{s}^{n - 1} \) generate \( R\left\lbrack s\right\rbrack \) as an \( R \) -module. As observed above, e... | No |
Corollary 5.4. If \( \mathrm{S} \) is a ring extension of \( \mathrm{R} \) and \( \mathrm{S} \) is finitely generated as an \( \mathrm{R} \) -module, then \( \mathrm{S} \) is an integral extension of \( \mathrm{R} \) . | PROOF. For any \( s \in S \) let \( S = T \) in part (iii) of Theorem 5.3. Then \( s \) is integral over \( R \) by Theorem 5.3(i). | No |
Theorem 5.5. If \( \mathrm{S} \) is an extension ring of \( \mathrm{R} \) and \( {\mathrm{s}}_{1},\ldots ,{\mathrm{s}}_{\mathrm{t}} \in \mathrm{S} \) are integral over \( \mathrm{R} \) , then \( \mathrm{R}\left\lbrack {{s}_{1},\ldots ,{\mathrm{s}}_{\mathrm{t}}}\right\rbrack \) is a finitely generated \( \mathrm{R} \) -... | PROOF. We have a tower of extension rings:\n\n\[ R \subset R\left\lbrack {s}_{1}\right\rbrack \subset R\left\lbrack {{s}_{1},{s}_{2}}\right\rbrack \subset \cdots \subset R\left\lbrack {{s}_{1},\ldots ,{s}_{t}}\right\rbrack . \]\n\nFor each \( i,{s}_{i} \) is integral over \( R \) and hence integral over \( R\left\lbrac... | Yes |
Theorem 5.6. If \( \mathrm{T} \) is an integral extension ring of \( \mathrm{S} \) and \( \mathrm{S} \) is an integral extension ring of \( \mathrm{R} \), then \( \mathrm{T} \) is an integral extension ring of \( \mathrm{R} \). | PROOF. \( T \) is obviously an extension ring of \( R \) . If \( t \in T \), then \( t \) is integral over \( S \) and therefore the root of some monic polynomial \( {f\varepsilon S}\left\lbrack x\right\rbrack \), say \( f = \mathop{\sum }\limits_{{i = 0}}^{n}{s}_{i}{x}^{i} \) . Since \( f \) is also a polynomial over ... | Yes |
Theorem 5.7. Let \( \mathrm{S} \) be an extension ring of \( \mathrm{R} \) and let \( \widehat{\mathrm{R}} \) be the set of all elements of \( \mathrm{S} \) that are integral over \( \mathrm{R} \) . Then \( \widehat{\mathrm{R}} \) is an integral extension ring of \( \mathrm{R} \) which contains every subring of \( \mat... | PROOF. If \( s,{t\varepsilon }\widehat{R} \), then \( s,{t\varepsilon R}\left\lbrack {s, t}\right\rbrack \), whence \( t - {s\varepsilon R}\left\lbrack {s, t}\right\rbrack \) and \( t\bar{s}{\varepsilon R}\left\lbrack {s, t}\right\rbrack \) . Since \( s \) and \( t \) are integral over \( R \), so is the ring \( R\left... | Yes |
Theorem 5.8. Let \( \mathrm{T} \) be a multiplicative subset of an integral domain \( \mathrm{R} \) such that \( 0 \in \mathrm{T} \) . If \( \mathrm{R} \) is integrally closed, then \( {\mathrm{T}}^{-1}\mathrm{R} \) is an integrally closed integral domain. | SKETCH OF PROOF. \( {T}^{-1}R \) is an integral domain (Theorem III.4.3(ii)) and \( R \) may be identified with a subring of \( {T}^{-1}R \) (Theorem III.4.4(ii)). Extending this identification, the quotient field \( Q\left( R\right) \) of \( R \) may be considered as a subfield of the quotient field \( Q\left( {{T}^{-... | No |
Theorem 5.9. (Lying-over Theorem) Let \( \mathrm{S} \) be an integral extension ring of \( \mathrm{R} \) and \( \mathrm{P}a \) prime ideal of \( \mathrm{R} \) . Then there exists a prime ideal \( \mathrm{Q} \) in \( \mathrm{S} \) which lies over \( \mathrm{P} \) (that is, \( \mathrm{Q} \cap \mathrm{R} = \mathrm{P} \) )... | PROOF. Since \( P \) is prime, \( R - P \) is a multiplicative subset of \( R \) (Theorem 2.1) and hence a multiplicative subset of \( S \) . Clearly \( 0 \in R - P \) . By Theorem 2.2 there is an ideal \( Q \) of \( S \) that is maximal in the set of all ideals \( I \) of \( S \) such that \( I \cap \left( {R - P}\rig... | Yes |
Corollary 5.10. (Going-up Theorem) Let \( \mathrm{S} \) be an integral extension ring of \( \mathrm{R} \) and \( {\mathrm{P}}_{1} \) , \( \mathrm{P} \) prime ideals in \( \mathrm{R} \) such that \( {\mathrm{P}}_{1} \subset \mathrm{P} \) . If \( {\mathrm{Q}}_{1} \) is a prime ideal of \( \mathrm{S} \) lying over \( {\ma... | SKETCH OF PROOF. As in the proof of Theorem 5.9, \( R - P \) is a multiplicative set in \( S \) . Since \( {Q}_{1} \cap R = {P}_{1} \subset P \), we have \( {Q}_{1} \cap \left( {R - P}\right) = \varnothing \) . By Theorem 2.2 there is a prime ideal \( Q \) of \( S \) that contains \( {Q}_{1} \) and is maximal in the se... | Yes |
Theorem 5.11. Let \( \mathrm{S} \) be an integral extension ring of \( \mathrm{R} \) and \( \mathrm{P} \) a prime ideal in \( \mathrm{R} \). If \( \mathrm{Q} \) and \( {\mathrm{Q}}^{\prime } \) are prime ideals in \( \mathrm{S} \) such that \( \mathrm{Q} \subset {\mathrm{Q}}^{\prime } \) and both \( \mathrm{Q} \) and \... | PROOF. It suffices to prove the following statement: if \( Q \) is a prime ideal in \( S \) such that \( Q \cap R = P \), then \( Q \) is maximal in the set \( \mathcal{S} \) of all ideals \( I \) in \( S \) with the property \( I \cap \left( {R - P}\right) = \varnothing \).\n\nIf \( Q \) is not maximal in \( \mathcal{... | No |
Theorem 5.12. Let \( \mathrm{S} \) be an integral extension ring of \( \mathrm{R} \) and let \( \mathrm{Q} \) be a prime ideal in \( \mathrm{S} \) which lies over a prime ideal \( \mathrm{P} \) in \( \mathrm{R} \) . Then \( \mathrm{Q} \) is maximal in \( \mathrm{S} \) if and only if \( \mathrm{P} \) is maximal in \( \m... | PROOF. Suppose \( Q \) is maximal in \( S \) . By Theorem III.2.18 there is a maximal ideal \( M \) of \( R \) that contains \( P.M \) is prime by Theorem III.2.19. By Corollary 5.10 there is a prime ideal \( {Q}^{\prime } \) in \( S \) such that \( Q \subset {Q}^{\prime } \) and \( {Q}^{\prime } \) lies over \( M \) .... | Yes |
Theorem 6.3. If \( \mathrm{R} \) is an integral domain with quotient field \( \mathbf{K} \), then the set of all fractional ideals of \( \mathrm{R} \) forms a commutative monoid, with identity \( \mathrm{R} \) and multiplication given by \( \mathbf{{IJ}} = \left\{ {\mathop{\sum }\limits_{{i = 1}}^{n}{\mathrm{a}}_{\math... | PROOF. Exercise; note that if \( I \) and \( J \) are ideals in \( R \), then \( {IJ} \) is the usual product of ideals. | No |
Lemma 6.4. Let \( \mathrm{I},{\mathrm{I}}_{1},{\mathrm{I}}_{2},\ldots ,{\mathrm{I}}_{\mathrm{n}} \) be ideals in an integral domain \( \mathrm{R} \). (i) The ideal \( {\mathrm{I}}_{1}{\mathrm{I}}_{2}\cdots {\mathrm{I}}_{\mathrm{n}} \) is invertible if and only if each \( {\mathrm{I}}_{\mathrm{j}} \) is invertible. | PROOF. (i) If \( J \) is a fractional ideal such that \( J\left( {{I}_{1}\cdots {I}_{n}}\right) = R \), then for each \( j = 1,2,\ldots, n,{I}_{j}\left( {J{I}_{1}\cdots {I}_{j - 1}{I}_{j + 1}\cdots {I}_{n}}\right) = R \), whence \( {I}_{j} \) is invertible. Conversely, if each \( {I}_{j} \) is invertible, then \( \left... | Yes |
Theorem 6.5. If \( \mathrm{R} \) is a Dedekind domain, then every nonzero prime ideal of \( \mathrm{R} \) is invertible and maximal. | PROOF. We show first that every invertible prime ideal \( P \) is maximal. If \( a \in R - P \), we must show that the ideal \( P + {Ra} \) generated by \( P \) and \( a \) is \( R \) . If \( P + {Ra} \neq R \), then since \( R \) is Dedekind, there exist prime ideals \( {P}_{i} \) and \( {Q}_{j} \) such that \( P + {R... | Yes |
Lemma 6.4(ii) implies \( n = {2m} \) and (after reindexing) \( \pi \left( {P}_{i}\right) = \pi \left( {Q}_{2i}\right) = \pi \left( {Q}_{{2i} - 1}\right) \) for \( i = 1,2,\ldots, m \) . Since Ker \( \pi = P \subset {P}_{i} \) and \( P \subset {Q}_{i} \) for all \( i, i \) . | \[ {P}_{i} = {\pi }^{-1}\left( {\pi \left( {P}_{i}\right) }\right) = {\pi }^{-1}\left( {\pi \left( {Q}_{2i}\right) }\right) = {Q}_{2i} \] and similarly \( {P}_{i} = {Q}_{{2i} - 1} \) for \( i = 1,2,\ldots, m \) . Consequently, \( P + R{a}^{2} = {\left( P + Ra\right) }^{2} \) and \( P \subset P + R{a}^{2} \subset {\left... | Yes |
Lemma 6.6. If \( \mathbf{I} \) is a fractional ideal of an integral domain \( \mathbf{R} \) with quotient field \( \mathbf{K} \) and \( \mathrm{f}\varepsilon {\operatorname{Hom}}_{\mathrm{R}}\left( {\mathrm{I},\mathrm{R}}\right) \), then for all \( \mathrm{a},\mathrm{b}\varepsilon \mathrm{I} : \mathrm{{af}}\left( \math... | PROOF. Now \( a = r/s \) and \( b = v/t\left( {r, s, v, t \in R;s, t \neq 0}\right) \) so \( {sa} = r \) and \( {tb} = v \) . Hence \( {sab} = {rb} \in I \) and \( {tab} = {va} \in I \) . Thus \( {sf}\left( {tab}\right) = f\left( {stab}\right) = {tf}\left( {sab}\right) \) in \( \mathrm{R} \) . Therefore, \( {af}\left( ... | Yes |
Lemma 6.7. Every invertible fractional ideal of an integral domain \( \mathbf{R} \) with quotient field \( \mathrm{K} \) is a finitely generated \( \mathrm{R} \) -module. | PROOF. Since \( {I}^{-1}I = R \), there exist \( {a}_{i} \in {I}^{-1},{b}_{i} \in I \) such that \( {1}_{R} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{b}_{i} \) . If \( c \in I \), then \( c = \mathop{\sum }\limits_{{i = 1}}^{n}\left( {c{a}_{i}}\right) {b}_{i} \) . Furthermore each \( c{a}_{i} \in R \) since \( {a}_{... | Yes |
Theorem 1.3. A left module \( \mathrm{A} \) over a ring \( \mathrm{R} \) is simple if and only if \( \mathrm{A} \) is isomorphic to \( \mathrm{R}/\mathrm{I} \) for some regular maximal left ideal \( \mathrm{I} \). | PROOF OF 1.3. The discussion preceding Definition 1.2 shows that if \( A \) is simple, then \( A = {Ra} \cong R/I \) where the maximal left ideal \( I \) is the kernel of \( \theta \) . Since \( A = {Ra}, a = {ea} \) for some \( {e\varepsilon R} \) . Consequently, for any \( {r\varepsilon R},{ra} = {rea} \) or \( \left... | Yes |
Theorem 1.4. Let \( \mathrm{B} \) be a subset of a left module \( \mathrm{A} \) over a ring \( \mathrm{R} \) . Then \( \mathcal{Q}\left( \mathrm{B}\right) = \{ \mathrm{r}\varepsilon \mathrm{R} \mid \mathrm{{rb}} = 0 \) for all \( \mathrm{b}\varepsilon \mathrm{B}\} \) is a left ideal of \( \mathrm{R} \) . If \( \mathrm{... | SKETCH OF PROOF OF 1.4. It is easy to verify that \( \mathcal{Q}\left( B\right) \) is a left ideal. Let \( B \) be a submodule. If \( r \in R \) and \( s \in \mathcal{Q}\left( B\right) \), then for every \( b \in B\left( {sr}\right) b = s\left( {rb}\right) = 0 \) since \( {rb} \in B \) . Consequently, \( {sr} \in \math... | No |
Proposition 1.6. A simple ring \( \mathrm{R} \) with identity is primitive. | PROOF. \( R \) contains a maximal left ideal \( I \) by Theorem III.2.18. Since \( R \) has an identity \( I \) is regular, whence \( R/I \) is a simple \( R \) -module by Theorem 1.3. Since \( \mathcal{Q}\left( {R/I}\right) \) is an ideal of \( R \) that does not contain \( {1}_{R}, Q\left( {R/I}\right) = 0 \) by simp... | No |
Proposition 1.7. A commutative ring \( \mathrm{R} \) is primitive if and only if \( \mathrm{R} \) is a field. | PROOF. A field is primitive by Proposition 1.6. Conversely, let \( A \) be a faithful simple left \( R \) -module. Then \( A \cong R/I \) for some regular maximal left ideal \( I \) of \( R \) . Since \( R \) is commutative, \( I \) is in fact an ideal and \( I \subset \mathcal{Q}\left( {R/I}\right) = \mathcal{Q}\left(... | Yes |
Theorem 1.9. Let \( \mathrm{R} \) be a dense ring of endomorphisms of a vector space \( \mathrm{V} \) over a division ring \( \mathrm{D} \) . Then \( \mathrm{R} \) is left [resp. right] Artinian if and only if \( {\mathrm{{dim}}}_{\mathrm{D}}\mathrm{V} \) is finite, in which case \( \mathrm{R} = {\operatorname{Hom}}_{\... | PROOF. If \( R \) is left Artinian and \( {\dim }_{D}V \) is infinite, then there exists an infinite linearly independent subset \( \left\{ {{u}_{1},{u}_{2},\ldots }\right\} \) of \( V \) . By Exercise IV.1.7 \( V \) is a left \( {\operatorname{Hom}}_{D}\left( {V, V}\right) \) -module and hence a left \( R \) -module. ... | Yes |
Lemma 1.10. (Schur) Let \( \mathrm{A} \) be a simple module over a ring \( \mathrm{R} \) and let \( \mathrm{B} \) be any R-module.\n\n(i) Every nonzero R-module homomorphism \( \mathrm{f} : \mathrm{A} \rightarrow \mathrm{B} \) is a monomorphism;\n\n(ii) every nonzero \( \mathrm{R} \) -module homomorphism \( \mathrm{g} ... | PROOF. (i) Ker \( f \) is a submodule of \( A \) and Ker \( f \neq A \) since \( f \neq 0 \) . Therefore \( \operatorname{Ker}f = 0 \) by simplicity. (ii) \( \operatorname{Im}g \) is a nonzero submodule of \( A \) since \( g \neq 0 \), whence \( \operatorname{Im}g = A \) by simplicity. (iii) If \( h \in D \) and \( h \... | Yes |
Lemma 1.11. Let \( \mathrm{A} \) be a simple module over a ring \( \mathrm{R} \) . Consider \( \mathrm{A} \) as a vector space over the division ring \( \mathrm{D} = {\operatorname{Hom}}_{\mathrm{R}}\left( {\mathrm{A},\mathrm{A}}\right) \) . If \( \mathrm{V} \) is a finite dimensional \( \mathrm{D} \) -subspace of the ... | PROOF. The proof is by induction on \( n = {\dim }_{D}V \) . If \( n = 0 \), then \( V = 0 \) and \( a \neq 0 \) . Since \( A \) is simple, \( A = {Ra} \) by Remark (iii) after Definition 1.1. Consequently, there exists \( {r\varepsilon R} \) such that \( {ra} = a \neq 0 \) and \( {rV} = {r0} = 0 \) . Suppose \( {\dim ... | Yes |
Theorem 1.12. (Jacobson Density Theorem) Let \( \mathrm{R} \) be a primitive ring and \( \mathrm{A} \) a faithful simple \( \mathrm{R} \) -module. Consider \( \mathrm{A} \) as a vector space over the division ring \( {\operatorname{Hom}}_{\mathrm{R}}\left( {\mathrm{A},\mathrm{A}}\right) = \mathrm{D} \) . Then \( \mathr... | PROOF OF 1.12. For each \( {r\varepsilon R} \) the map \( {\alpha }_{r} : A \rightarrow A \) given by \( {\alpha }_{r}\left( a\right) = {ra} \) is easily seen to be a \( D \) -endomorphism of \( A \) : that is, \( {\alpha }_{r}\varepsilon {\operatorname{Hom}}_{D}\left( {A, A}\right) \) . Furthermore for all \( r,{s\var... | Yes |
Corollary 1.13. If \( \mathrm{R} \) is a primitive ring, then for some division ring \( \mathrm{D} \) either \( \mathrm{R} \) is isomorphic to the endomorphism ring of a finite dimensional vector space over \( \mathbf{D} \) or for every positive integer \( \mathrm{m} \) there is a subring \( {\mathrm{R}}_{\mathrm{m}} \... | SKETCH OF PROOF OF 1.13. In the notation of Theorem 1.12, \[ \alpha : R \rightarrow {\operatorname{Hom}}_{D}\left( {A, A}\right) \] is a monomorphism such that \( R = \operatorname{Im}\alpha \) and \( \operatorname{Im}\alpha \) is dense in \( {\operatorname{Hom}}_{D}\left( {A, A}\right) \) . If \( {\dim }_{D}A = n \) i... | Yes |
Theorem 1.14. (Wedderburn-Artin) The following conditions on a left Artinian ring \( \mathbf{R} \) are equivalent.\n\n(i) \( \mathrm{R} \) is simple;\n\n(ii) \( \mathrm{R} \) is primitive;\n\n(iii) \( \mathrm{R} \) is isomorphic to the endomorphism ring of a nonzero finite dimensional vector space \( \mathbf{V} \) over... | PROOF. (i) \( \Rightarrow \) (ii) We first observe that \( I = \{ r \in R \mid {Rr} = 0\} \) is an ideal of \( R \) , whence \( I = R \) or \( I = 0 \) . Since \( {R}^{2} \neq 0 \), we must have \( I = 0 \) . Since \( R \) is left Artinian the set of all nonzero left ideals of \( R \) contains a minimal left ideal \( J... | No |
Lemma 1.15. Let \( \mathrm{V} \) be a finite dimensional vector space over a division ring \( \mathrm{D} \) . If \( \mathrm{A} \) and \( \mathrm{B} \) are simple faithful modules over the endomorphism ring \( \mathrm{R} = {\operatorname{Hom}}_{\mathrm{D}}\left( {\mathrm{V},\mathrm{V}}\right) \), then A and B are isomor... | PROOF. By Theorems VII.1.4, VIII.1.4 and Corollary VIII.1.12, the ring \( R \) contains a (nonzero) minimal left ideal \( I \) . Since \( A \) is faithful, there exists \( {a\varepsilon A} \) such that \( {Ia} \neq 0 \) . Thus \( {Ia} \) is a nonzero submodule of \( A \) (Exercise IV.1.3), whence \( {Ia} = A \) by simp... | No |
Lemma 1.16. Let \( \mathrm{V} \) be a nonzero vector space over a division ring \( \mathrm{D} \) and let \( \mathrm{R} \) be the endomorphism ring \( {\operatorname{Hom}}_{\mathrm{D}}\left( {\mathrm{V},\mathrm{V}}\right) \) . If \( \mathrm{g} : \mathrm{V} \rightarrow \mathrm{V} \) is a homomorphism of additive groups s... | PROOF. Let \( u \) be a nonzero element of \( V \) . We claim that \( u \) and \( g\left( u\right) \) are linearly dependent over \( D \) . If \( {\dim }_{D}V = 1 \), this is trivial. Suppose \( {\dim }_{D}V \geq 2 \) and \( \{ u, g\left( u\right) \} \) is linearly independent. Since \( R \) is dense in itself (Example... | Yes |
Lemma 2.4. If \( \mathrm{I}\left( { \neq \mathrm{R}}\right) \) is a regular left ideal of a ring \( \mathrm{R} \), then \( \mathrm{I} \) is contained in a maximal left ideal which is regular. | SKETCH OF PROOF. Since \( I \) is regular, there exists \( e \in R \) such that \( r - {re\varepsilon I} \) for all \( {r\varepsilon R} \) . Thus any left ideal \( J \) containing \( I \) is also regular (with the same element \( e \in R) \) . If \( I \subset J \) and \( {e\varepsilon J} \), then \( r - {re\varepsilon ... | No |
Lemma 2.5. Let \( \mathrm{R} \) be a ring and let \( \mathrm{K} \) be the intersection of all regular maximal left ideals of \( \mathbf{R} \) . Then \( \mathbf{K} \) is a left quasi-regular left ideal of \( \mathbf{R} \) . | PROOF. \( K \) is obviously a left ideal. If \( a \in K \) let \( T = \{ r + {ra} \mid r \in R\} \) . If \( T = R \) , then there exists \( {r\varepsilon R} \) such that \( r + {ra} = - a \) . Consequently \( r + a + {ra} = 0 \) and hence \( a \) is left quasi-regular. Thus it suffices to show that \( T = R \) .\n\nVer... | Yes |
Lemma 2.6. Let \( \mathrm{R} \) be a ring that has a simple left \( \mathrm{R} \) -module. If \( \mathrm{I} \) is a left quasi-regular left ideal of \( \mathrm{R} \), then \( \mathrm{I} \) is contained in the intersection of all the left annihilators of simple left \( \mathrm{R} \) -modules. | PROOF. If \( I ⊄ \cap \mathcal{Q}\left( A\right) \), where the intersection is taken over all simple left \( R \) -modules \( A \), then \( {IB} \neq 0 \) for some simple left \( R \) -module \( B \), whence \( {Ib} \neq 0 \) for some nonzero \( {b\varepsilon B} \) . Since \( I \) is a left ideal, \( {Ib} \) is a nonze... | Yes |
Lemma 2.7. An ideal \( \mathrm{P} \) of a ring \( \mathrm{R} \) is left primitive if and only if \( \mathrm{P} \) is the left annihilator of a simple left \( \mathrm{R} \) -module. | PROOF. If \( P \) is a left primitive ideal, let \( A \) be a simple faithful \( R/P \) -module. Verify that \( A \) is an \( R \) -module, with \( {ra}\left( {r \in R, a \in A}\right) \) defined to be \( \left( {r + P}\right) a \) . Then \( {RA} = \left( {R/P}\right) A \neq 0 \) and every \( R \) -submodule of \( A \)... | Yes |
Lemma 2.8. Let \( \mathrm{I} \) be a left ideal of a ring \( \mathrm{R} \) . If \( \mathrm{I} \) is left quasi-regular, then \( \mathrm{I} \) is right quasi-regular. | PROOF. If \( I \) is left quasi-regular and \( {a\varepsilon I} \), then there exists \( {r\varepsilon R} \) such that \( r \circ a = r + a + {ra} = 0 \) . Since \( r = - a - {ra\varepsilon I} \), there exists \( s \in R \) such that \( s \circ r = s + r + {sr} = 0 \), whence \( s \) is right quasi-regular. The operati... | Yes |
Theorem 2.10. Let \( \mathrm{R} \) be a ring.\n\n(i) If \( \mathrm{R} \) is primitive, then \( \mathrm{R} \) is semisimple.\n\n(ii) If \( \mathrm{R} \) is simple and semisimple, then \( \mathrm{R} \) is primitive.\n\n(iii) If \( \mathrm{R} \) is simple, then \( \mathrm{R} \) is either a primitive semisimple or a radica... | PROOF. (i) \( R \) has a faithful simple left \( R \) -module \( A \), whence \( J\left( R\right) \subset \mathcal{Q}\left( A\right) = 0 \) .\n\n(ii) \( R \neq 0 \) by simplicity. There must exist a simple left \( R \) -module \( A \) ; (otherwise by Theorem 2.3 (i) \( J\left( R\right) = R \neq 0 \), contradicting semi... | Yes |
Theorem 2.12. If \( \mathrm{R} \) is a ring, then every nil right or left ideal is contained in the radical \( \mathrm{J}\left( \mathrm{R}\right) \) . | PROOF OF 2.12. If \( {a}^{n} = 0 \), let \( r = - a + {a}^{2} - {a}^{3} + \cdots + {\left( -1\right) }^{n - 1}{a}^{n - 1} \) . Verify that \( r + a + {ra} = 0 = a + r + {ar} \), whence \( a \) is both left and right quasi-regular. Therefore every nil left [right] ideal is left [right] quasi-regular and hence is contain... | Yes |
Proposition 2.13. If \( \mathrm{R} \) is a left [resp. right] Artinian ring, then the radical \( \mathrm{J}\left( \mathrm{R}\right) \) is a nilpotent ideal. Consequently every nil left or right ideal of \( \mathrm{R} \) is nilpotent and \( \mathrm{J}\left( \mathrm{R}\right) \) is the unique maximal nilpotent left (or r... | PROOF OF 2.13. Let \( J = J\left( R\right) \) and consider the chain of (left) ideals \( J \supset {J}^{2} \supset {J}^{3} \supset \cdots \) . By hypothesis there exists \( k \) such that \( {J}^{i} = {J}^{k} \) for all \( i \geq k \) . We claim that \( {J}^{k} = 0 \) . If \( {J}^{k} \neq 0 \), then the set \( S \) of ... | Yes |
Lemma 2.15. Let \( \mathbf{R} \) be a ring and \( \mathbf{a}\varepsilon \mathbf{R} \) .\n\n(i) If \( - {\mathrm{a}}^{2} \) is left quasi-regular, then so is \( \mathrm{a} \) . | PROOF. (i) If \( r + \left( {-{a}^{2}}\right) + r\left( {-{a}^{2}}\right) = 0 \), let \( s = r - a - {ra} \) . Verify that \( s + a + {sa} = 0 \), whence \( a \) is left quasi-regular. | Yes |
Theorem 2.16. (i) If an ideal \( I \) of a ring \( \mathrm{R} \) is itself considered as a ring, then \( \mathrm{J}\left( \mathrm{I}\right) = \mathrm{I} \cap \mathrm{J}\left( \mathrm{R}\right) \). | PROOF. (i) \( I \cap J\left( R\right) \) is clearly an ideal of \( I \). If \( a \in I \cap J\left( R\right) \), then \( a \) is left quasi-regular in \( R \), whence \( r + a + {ra} = 0 \) for some \( {r\varepsilon R} \). But \( r = - a - {ra\varepsilon I} \). Thus every element of \( I \cap J\left( R\right) \) is lef... | Yes |
Theorem 2.17. If \( \left\{ {{\mathrm{R}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) is a family of rings, then \( \mathrm{J}\left( {\mathop{\prod }\limits_{{i\varepsilon I}}{R}_{i}}\right) = \mathop{\prod }\limits_{{i\varepsilon I}}J\left( {R}_{i}\right) \) . | SKETCH OF PROOF. Verify that an element \( \left\{ {a}_{i}\right\} \varepsilon \prod {R}_{i} \) is left quasi-regular in \( \prod {R}_{i} \) if and only if \( {a}_{i} \) is left quasi-regular in \( {R}_{i} \) for each \( i \) . Consequently \( \prod J\left( {R}_{i}\right) \) is a left quasi-regular ideal of \( \prod {R... | No |
Proposition 3.2. A nonzero ring \( \mathrm{R} \) is semisimple if and only if \( \mathrm{R} \) is isomorphic to a subdirect product of primitive rings. | SKETCH OF PROOF OF 3.2. Suppose \( R \) is nonzero semisimple and let \( O \) be the set of all left primitive ideals of \( R \) . Then for each \( {P\varepsilon }\mathcal{O}, R/P \) is a primitive ring (Definition 2.1). By Theorem 2.3 (iii), \( 0 = J\left( R\right) = \mathop{\bigcap }\limits_{{{Pe}{0}^{ \circ }}}P \) ... | Yes |
Corollary 3.4. (i) A semisimple left Artinian ring has an identity. | SKETCH OF PROOF OF 3.4. (i) Theorem 3.3. | No |
Corollary 3.5. If \( \mathrm{I} \) is an ideal in a semisimple left Artinian ring \( \mathrm{R} \), then \( \mathrm{I} = \mathrm{{Re}} \), where \( \mathrm{e} \) is an idempotent which is in the center of \( \mathrm{R} \) . | SKETCH OF PROOF. By Theorem 3.3 \( R \) is a (ring) direct product of simple ideals, \( R = {I}_{1} \times \cdots \times {I}_{n} \) . For each \( j, I \cap {I}_{j} \) is either 0 or \( {I}_{j} \) by simplicity. After reindexing if necessary we may assume that \( I \cap {I}_{j} = {I}_{j} \) for \( j = 1,2,\ldots, t \) a... | No |
Theorem 3.7. The following conditions on a nonzero ring \( \mathbf{R} \) with identity are equivalent.\n\n(i) \( \mathrm{R} \) is semisimple left Artinian;\n\n(ii) every unitary left \( \mathrm{R} \) -module is projective;\n\n(iii) every unitary left \( \mathrm{R} \) -module is injective;\n\n(iv) every short exact sequ... | SKETCH OF PROOF OF 3.7. (ii) \( \Leftrightarrow \) (iii) \( \Leftrightarrow \) (iv) is Exercise IV.3.1. To complete the proof we shall prove the implications (iv) \( \Leftrightarrow \) (v) and (v) \( \Rightarrow \) (vii) \( \Rightarrow \) (vi) \( \Rightarrow \) (i) \( \Rightarrow \) (viii) \( \Rightarrow \) (v).\n\n(iv... | No |
Proposition 3.8. Let \( \mathrm{R} \) be a semisimple left Artinian ring.\n\n(i) \( \mathrm{R} = {\mathrm{I}}_{1} \times \cdots \times {\mathrm{I}}_{\mathrm{n}} \) where each \( {\mathrm{I}}_{\mathrm{j}} \) is a simple ideal of \( \mathrm{R} \).\n\n(ii) If \( \mathrm{J} \) is any simple ideal of \( \mathrm{R} \), then ... | PROOF OF 3.8. (i) is true by Theorem 3.3. (ii) If \( J \) is a simple ideal of \( R \), then \( {RJ} \neq 0 \), whence \( {I}_{k}J \neq 0 \) for some \( k \). Since \( {I}_{k}J \) is a nonzero ideal that is contained in both \( {I}_{k} \) and \( J \), the simplicity of \( {I}_{k} \) and \( J \) implies \( {I}_{k} = {I}... | Yes |
Proposition 3.9. Let \( \\mathrm{A} \) be a semisimple module over a ring \( \\mathrm{R} \) . If there are direct sum decompositions\n\n\[ \n\\mathrm{A} = {\\mathrm{B}}_{1} \\oplus \\cdots \\oplus {\\mathrm{B}}_{\\mathrm{m}}\\;\\text{ and }\\;\\mathrm{A} = {\\mathrm{C}}_{1} \\oplus \\cdots \\oplus {\\mathrm{C}}_{\\math... | PROOF OF 3.9. The series\n\n\[ \nA = {B}_{1} \\oplus \\cdots \\oplus {B}_{m} \\supset {B}_{2} \\oplus \\cdots \\oplus {B}_{m} \\supset \\cdots \\supset {B}_{m} \\supset 0\n\]\n\n is a composition series for \( A \) with simple factors \( {B}_{1},{B}_{2},\\ldots ,{B}_{m} \) (see p. 375). Similarly \( A = {C}_{1} \\oplus... | Yes |
Proposition 4.3. \( \mathrm{K} \) is a prime ideal of a ring \( \mathrm{R} \) if and only if \( \mathrm{R}/\mathrm{K} \) is a prime ring. | SKETCH OF PROOF OF 4.3. If \( R/K \) is prime, let \( \pi : R \rightarrow R/K \) be the canonical epimorphism. If \( I \) and \( J \) are ideals of \( R \) such that \( {IJ} \subset K \), then \( \pi \left( I\right) ,\pi \left( J\right) \) are ideals of \( R/K \) (Exercise III.2.13(b)) such that \( \pi \left( I\right) ... | No |
Proposition 4.4. A ring \( \mathrm{R} \) is semiprime if and only if \( \mathrm{R} \) is isomorphic to a subdirect product of prime rings. | SKETCH OF PROOF. Proposition 4.4 is simply Proposition 3.2 with the words \ | No |
Corollary 4.9. R is a semiprime [resp. prime] left Goldie ring if and only if \( \mathbf{R} \) has a quotient ring \( \mathrm{Q}\left( \mathrm{R}\right) \) such that \( \mathrm{Q}\left( \mathrm{R}\right) \cong {\operatorname{Mat}}_{{n}_{1}}{\mathrm{D}}_{1} \times \cdots \times {\operatorname{Mat}}_{{\mathrm{n}}_{k}}{\m... | ## PROOF. Theorems 1.14, 3.3, and 4.8. | No |
Theorem 5.2. Let \( \mathrm{A} \) be a \( \mathrm{K} \) -algebra.\n\n(i) A subset \( \mathbf{I} \) of \( \mathbf{A} \) is a regular maximal left algebra ideal if and only if \( \mathbf{I} \) is a regular maximal left ideal of the ring \( \mathrm{A} \) .\n\n(ii) The Jacobson radical of the ring A coincides with the Jaco... | PROOF OF 5.2. (i) If \( I \) is a regular maximal left ideal of the ring \( A \), it suffices to show that \( {kI} \subset I \) for all \( k \in K \) . Suppose \( {kI} ⊄ I \) for some \( k \in K \) . Since \( r\left( {kI}\right) = k\left( {rI}\right) \) by Definition 5.1(i), \( I + {kI} \) is a left ideal of \( A \) th... | Yes |
Theorem 5.3. Let A be a K-algebra. Every simple algebra A-module is a simple module over the ring A. Every simple module M over the ring A can be given a unique \( \mathrm{K} \) -module structure in such a way that \( \mathrm{M} \) is a simple algebra A-module. | PROOF. Let \( N \) be a simple algebra \( A \) -module, whence \( {AN} \neq 0 \) . If \( {N}_{1} \) is a submodule of \( N \), then \( A{N}_{1} \) is an algebra submodule of \( N \), whence \( A{N}_{1} = N \) or \( A{N}_{1} = 0 \) . If \( A{N}_{1} = N \), then \( {N}_{1} = N \) . If \( A{N}_{1} = 0 \), then \( {N}_{1} ... | Yes |
Theorem 5.4. A is a semisimple left Artinian K-algebra if and only if there is an isomorphism of \( \mathrm{K} \) -algebras\n\n\[ \mathrm{A} \cong {\operatorname{Mat}}_{{\mathrm{n}}_{1}}{\mathrm{D}}_{1} \times {\operatorname{Mat}}_{{\mathrm{n}}_{2}}{\mathrm{D}}_{2} \times \cdots \times {\operatorname{Mat}}_{{\mathrm{n}... | SKETCH OF PROOF OF 5.4. Use Theorems 5.2 and 5.3 and Exercises 3 and 4 to carry over the proof of the Wedderburn-Artin Theorem 3.3 to \( K \) -algebras. | No |
Lemma 5.6. If \( \mathrm{D} \) is an algebraic division algebra over an algebraically closed field \( \mathbf{K} \) , then \( \mathrm{D} = \mathrm{K} \) . | PROOF. \( K \) is contained in the center of \( D \) by the convention adopted above. If \( a \in D \), then \( f\left( a\right) = 0 \) for some \( f \in K\left\lbrack x\right\rbrack \) . Since \( K \) is algebraically closed \( f\left( x\right) = k\left( {x - {k}_{1}}\right) \left( {x - {k}_{2}}\right) \cdots \left( {... | Yes |
Theorem 5.7. Let \( \mathbf{A} \) be a finite dimensional semisimple algebra over an algebraically closed field \( \mathrm{K} \) . Then there are positive integers \( {\mathrm{n}}_{1},\ldots ,{\mathrm{n}}_{\mathrm{t}} \) and an isomorphism of K-algebras\n\n\[ \mathrm{A} \cong {\operatorname{Mat}}_{{\mathrm{n}}_{1}}\mat... | PROOF. By Theorem 5.4 (and the subsequent Remark) \( A \cong {\operatorname{Mat}}_{{n}_{1}}{D}_{1} \times \) \( {\operatorname{Mat}}_{{n}_{2}}{D}_{2} \times \cdots \times {\operatorname{Mat}}_{{n}_{t}}{D}_{t} \) where each \( {D}_{i} \) is a division algebra over \( K \) . Each \( {D}_{i} \) is necessarily finite dimen... | Yes |
Corollary 5.9. Let \( \mathrm{K}\left( \mathrm{G}\right) \) be the group algebra of a finite group \( \mathrm{G} \) over an algebraically closed field \( \mathrm{K} \) . If char \( \mathrm{K} = 0 \) or char \( \mathrm{K} = \mathrm{p} \) and \( \mathrm{p} \nmid \left| \mathrm{G}\right| \), then there exist positive inte... | PROOF. Since \( G \) is finite, \( K\left( G\right) \) is a finite dimensional \( K \) -algebra and hence left Artinian (Exercise 2). Apply Theorem 5.7 and Proposition 5.8. | No |
Lemma 6.4. Let \( \mathrm{A} \) be an algebra with identity over a field \( \mathrm{K} \) and \( \mathrm{F} \) a field containing \( \mathrm{K} \) ; then \( \mathrm{A}{\bigotimes }_{\mathrm{K}}\mathrm{F} \) is an \( \mathrm{F} \) -algebra such that \( {\dim }_{\mathrm{K}}\mathrm{A} = {\dim }_{\mathrm{F}}\left( {\mathrm... | SKETCH OF PROOF. Since \( F \) is commutative and a \( K - F \) bimodule, \( A{\bigotimes }_{K}F \) is a vector space over \( F \) with \( b\left( {a \otimes {b}_{1}}\right) = \left( {a \otimes {b}_{1}}\right) b = a \otimes {b}_{1}b\left( {{a\varepsilon A};b,{b}_{1}{\varepsilon F}}\right. \) ; see Theorem IV.5.5 and th... | No |
Lemma 6.5. Let \( \mathrm{D} \) be a division algebra over a field \( \mathrm{K} \) and \( \mathrm{A} \) a finite dimensional \( \mathrm{K} \) -algebra with identity. Then \( \mathrm{D}{\bigotimes }_{\mathrm{K}}\mathrm{A} \) is a left Artinian \( \mathrm{K} \) -algebra. | SKETCH OF PROOF. \( D{ \otimes }_{K}A \) is a vector space over \( D \) with the action of \( {d\varepsilon D} \) on a generator \( {d}_{1} \otimes a \) of \( D{ \otimes }_{K}A \) given by \( d\left( {{d}_{1} \otimes a}\right) = d{d}_{1} \otimes a = \left( {d \otimes {1}_{A}}\right) \left( {{d}_{1} \otimes a}\right) \)... | No |
Theorem 6.6. Let \( \mathrm{D} \) be a division ring with center \( \mathrm{K} \) and maximal subfield \( \mathrm{F} \) . Then \( {\dim }_{\mathrm{K}}\mathrm{D} \) is finite if and only if \( {\dim }_{\mathrm{K}}\mathrm{F} \) is finite, in which case \( {\dim }_{\mathrm{F}}\mathrm{D} = {\dim }_{\mathrm{K}}\mathrm{F} \)... | PROOF. If \( {\dim }_{K}F \) is infinite, so is \( {\dim }_{K}D \) . If \( {\dim }_{K}F \) is finite, then \( D{\bigotimes }_{K}F \) is a left Artinian \( K \) -algebra by Lemma 6.5. Thus \( D{ \otimes }_{K}F \) is isomorphic to a dense left Artinian subalgebra of \( {\operatorname{Hom}}_{F}\left( {D, D}\right) \) by T... | Yes |
Corollary 6.8. (Frobenius) Let \( \mathbf{D} \) be an algebraic division algebra over the field \( \mathbf{R} \) of real numbers. Then \( \mathbf{D} \) is isomorphic to either \( \mathbf{R} \) or the field \( \mathbf{C} \) of complex numbers or the division algebra \( \mathrm{T} \) of real quaternions. | SKETCH OF PROOF. Let \( K \) be the center of \( D \) and \( F \) a maximal subfield. We have \( \mathbf{R} \subset K \subset F \subset D \), with \( F \) an algebraic field extension of \( \mathbf{R} \) . Consequently \( {\dim }_{K}F \leq {\dim }_{R}F \leq 2 \) by Corollary V.3.20. By Theorem 6.6 \( {\dim }_{F}D = {\d... | No |
Corollary 6.9. (Wedderburn) Every finite division ring \( \mathrm{D} \) is a field. | REMARK. An elementary proof of this fact, via cyclotomic polynomials, is given in Exercise V.8.10.\n\nPROOF OF 6.9. Let \( K \) be the center of \( D \) and \( F \) any maximal subfield. By Theorem \( {6.6}{\dim }_{K}D = {n}^{2} \), where \( {\dim }_{K}F = n \) . Thus every maximal subfield is a finite field of order \... | No |
Lemma 6.10. If \( \mathrm{G} \) is a finite (multiplicative) group and \( \mathrm{H} \) is a proper subgroup, then\n\n\[ \mathop{\bigcup }\limits_{{x \in G}}{xH}{x}^{-1} \subseteq G \] | PROOF. The number of distinct conjugates of \( H \) is \( \left\lbrack {G : N}\right\rbrack \), where \( N \) is the normalizer of \( H \) in \( G \) (Corollary II.4.4). Since \( H < N < G \) and \( H \neq G,\left\lbrack {G : N}\right\rbrack \leq \) \( \left\lbrack {G : H}\right\rbrack \) and \( \left\lbrack {G : H}\ri... | Yes |
Lemma 1.5. Let \( \mathrm{T} : \mathcal{C} \rightarrow \mathcal{S} \) be a covariant functor from a category \( \mathcal{C} \) to the category S of sets and let \( \mathrm{A} \) be an object of \( \mathcal{C} \). (i) If \( \alpha : {\mathrm{h}}_{\mathrm{A}} \rightarrow \mathrm{T} \) is a natural transformation from the... | PROOF. (i) Let \( C \) be an object of \( \mathcal{C} \) and \( g \in {\hom }_{\mathcal{C}}\left( {A, C}\right) \) . By hypothesis the\n\ndiagram\n\n\[ \n{h}_{A}\left( A\right) = {\hom }_{\mathcal{C}}\left( {A, A}\right) \rightarrow T\left( A\right) \n\]\n\n\[ \n{h}_{A}\left( g\right) \downarrow T\left( g\right) \n\]\n... | Yes |
Theorem 1.6. Let \( \mathrm{T} : \mathcal{C} \rightarrow \mathcal{S} \) be a covariant functor from a category \( \mathcal{C} \) to the category S of sets. There is a one-to-one correspondence between the class \( \mathrm{X} \) of all representations of \( \mathrm{T} \) and the class \( \mathrm{Y} \) of all universal e... | PROOF OF 1.6. Let \( \left( {A,\alpha }\right) \) be a representation of \( T \) and let \( {\alpha }_{A}\left( {1}_{A}\right) = u \) e \( T\left( A\right) \) . Suppose \( \left( {B, s}\right) \) is an object of \( {\mathcal{C}}_{T} \) . By hypothesis \( {\alpha }_{B} : {h}_{A}\left( B\right) = {\hom }_{\mathcal{C}}\le... | Yes |
Corollary 1.7. Let \( \mathrm{T} : \mathcal{C} \rightarrow \mathcal{S} \) be a covariant functor from a category \( \mathcal{C} \) to the category S of sets. If \( \left( {\mathrm{A},\alpha }\right) \) and \( \left( {\mathrm{B},\beta }\right) \) are representations of \( \mathrm{T} \), then there is a unique equivalenc... | PROOF. Let \( u = {\alpha }_{A}\left( {1}_{A}\right) \) and \( v = {\beta }_{B}\left( {1}_{B}\right) \) . By Theorem 1.6 \( \left( {A, u}\right) \) and \( \left( {B, v}\right) \) are universal elements of \( T \), whence by Lemma I.7.10 there is a unique equivalence \( f : A \rightarrow B \) in \( \mathcal{C} \) such t... | Yes |
Corollary 1.8. (Yoneda) Let \( \mathrm{T} : \mathcal{C} \rightarrow \mathcal{S} \) be a covariant functor from a category \( \mathcal{C} \) to the category \( \mathcal{S} \) of sets and let \( \mathrm{A} \) be an object of \( \mathcal{C} \) . Then there is a one-to-one correspondence between the set \( \mathrm{T}\left(... | SKETCH OF PROOF. Define a function \( \psi = {\psi }_{A} : \operatorname{Nat}\left( {{h}_{A}, T}\right) \rightarrow T\left( A\right) \) by\n\n\[ \alpha \mapsto {\alpha }_{A}\left( {1}_{A}\right) \in T\left( A\right) \]\n\nand a function \( \phi : T\left( A\right) \rightarrow \operatorname{Nat}\left( {{h}_{A}, T}\right)... | No |
Theorem 1.9. Let \( \mathcal{C} \) and \( \mathcal{D} \) be categories and \( \mathbf{T} \) a functor from the product category \( \operatorname{\mathcal{C} } \times \operatorname{\mathcal{D} } \) to the category \( \operatorname{\mathbb{S}} \) of sets, contravariant in the first variable and covariant in the second, s... | PROOF OF 1.9. The object function of the functor \( S \) is defined by \( S\left( C\right) = {A}_{C} \) for each object \( C \) of \( \mathcal{C} \) . The morphism function of \( S \) is defined as follows. For each object \( C \) of \( \& {\alpha }^{C}{}_{AC} : {\hom }_{\mathfrak{D}}\left( {{A}_{C},{A}_{C}}\right) \ri... | Yes |
Proposition 2.2. A covariant functor \( \mathrm{T} : \mathfrak{D} \rightarrow \mathcal{C} \) has a left adjoint if and only if for each object \( \mathrm{C} \) in \( \mathcal{C} \) the functor home \( \left( {\mathrm{C},\mathrm{T}\left( -\right) }\right) : \mathfrak{D} \rightarrow \mathcal{S} \) is representable. | PROOF. If \( S : \mathcal{C} \rightarrow \mathfrak{D} \) is a left adjoint of \( T \), then there is for each object \( C \) of \( \mathcal{C} \) and \( D \) of \( \mathfrak{D} \) a bijection\n\n\[ \n{\alpha }_{C, D} : {\hom }_{\mathfrak{D}}\left( {S\left( C\right), D}\right) \rightarrow {\hom }_{\mathfrak{C}}\left( {C... | Yes |
Corollary 2.3. A covariant functor \( \mathrm{T} : \mathfrak{D} \rightarrow \mathcal{C} \) has a left adjoint if and only if there exists for each object \( \mathrm{C} \) of \( \mathcal{C} \) an object \( \mathrm{S}\left( \mathrm{C}\right) \) of \( \mathfrak{D} \) and a morphism \( {\mathrm{u}}_{\mathrm{C}} : \mathrm{C... | PROOF. Exercise; see Theorem 1.6. ∎ | No |
Corollary 2.4. Any two left adjoints of a covariant functor \( \mathrm{T} : \mathfrak{D} \rightarrow \mathcal{C} \) are naturally isomorphic. | PROOF. If \( {S}_{1} : \mathcal{C} \rightarrow \mathcal{D} \) and \( {S}_{2} : \mathcal{C} \rightarrow \mathcal{D} \) are left adjoints of \( T \), then there are natural isomorphisms\n\n\[ \alpha : {\hom }_{\mathfrak{D}}\left( {{S}_{1}\left( -\right) , - }\right) \rightarrow {\hom }_{\mathfrak{e}}\left( {-, T\left( -\... | Yes |
Proposition 3.2. Let \( \mathrm{f} : \mathrm{B} \rightarrow \mathrm{C} \) and \( \mathrm{g} : \mathrm{C} \rightarrow \mathrm{D} \) be morphisms of a category \( \mathrm{C} \) . (i) \( \mathrm{f} \) and \( \mathrm{g} \) monic \( \Rightarrow \mathrm{{gf}} \) monic; (ii) gf monic \( \Rightarrow \mathrm{f} \) monic; (iii) ... | PROOF. Exercise. | No |
Proposition 3.4. Let \( \mathcal{C} \) be a category which has a zero object 0 . Then for each pair \( \mathrm{C},\mathrm{D} \) of objects of \( \mathrm{C} \) there is a unique morphism \( {0}_{\mathrm{C},\mathrm{D}} : \mathrm{C} \rightarrow \mathrm{D} \) such that\n\n\[ \mathrm{f} \circ {0}_{\mathrm{C},\mathrm{D}} = {... | REMARK. \( {0}_{C, D} \) is called a zero morphism.\n\nPROOF OF 3.4. (Uniqueness) If \( \left\{ {0}_{C, D}^{\prime }\right\} \) and \( \left\{ {0}_{C, D}\right\} \) are two families of morphisms with the stated properties, then for each pair \( C, D \)\n\n\[ {0}_{C, D} = {0}_{D, D}^{\prime }{0}_{C, D} = {0}_{C, D}^{\pr... | Yes |
Proposition 3.6. Let \( \mathrm{f} : \mathrm{C} \rightarrow \mathrm{D} \) and \( \mathrm{g} : \mathrm{C} \rightarrow \mathrm{D} \) be morphisms of a category \( \mathrm{C} \). (i) If \( \mathrm{i} : \mathrm{B} \rightarrow \mathrm{C} \) is a difference kernel of \( \left( {\mathrm{f},\mathrm{g}}\right) \), then \( \math... | PROOF. (i) Let \( h, k : F \rightarrow B \) be morphisms such that \( {ih} = {ik} \). Then \( f\left( {ih}\right) = \left( {fi}\right) h = \left( {gi}\right) h = g\left( {ih}\right) \). Since \( i \) is a difference kernel of \( \left( {f, g}\right) \), there is a unique morphism \( t : F \rightarrow B \) such that \( ... | Yes |
Proposition 1. Let \( f : A \rightarrow B \) .\n\n(1) The map \( f \) is injective if and only if \( f \) has a left inverse.\n\n(2) The map \( f \) is surjective if and only if \( f \) has a right inverse.\n\n(3) The map \( f \) is a bijection if and only if there exists \( g : B \rightarrow A \) such that \( f \circ ... | Proof: Exercise. | No |
(1) If \( \\sim \) defines an equivalence relation on \( A \) then the set of equivalence classes of \( \\sim \) form a partition of \( A \) . | Proof: Omitted. | No |
Theorem 3. The operations of addition and multiplication on \( \mathbb{Z}/n\mathbb{Z} \) defined above are both well defined, that is, they do not depend on the choices of representatives for the classes involved. More precisely, if \( {a}_{1},{a}_{2} \in \mathbb{Z} \) and \( {b}_{1},{b}_{2} \in \mathbb{Z} \) with \( \... | Proof: Suppose \( {a}_{1} \equiv {b}_{1}\left( {\;\operatorname{mod}\;n}\right) \), i.e., \( {a}_{1} - {b}_{1} \) is divisible by \( n \) . Then \( {a}_{1} = {b}_{1} + {sn} \) for some integer \( s \) . Similarly, \( {a}_{2} \equiv {b}_{2}{\;(\operatorname{mod}\;n)} \) means \( {a}_{2} = {b}_{2} + {tn} \) for some inte... | Yes |
Proposition 4. \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } = \{ \bar{a} \in \mathbb{Z}/n\mathbb{Z} \mid \left( {a, n}\right) = 1\} \) . | It is easy to see that if any representative of \( \bar{a} \) is relatively prime to \( n \) then all representatives are relatively prime to \( n \) so that the set on the right in the proposition is well defined. | No |
Proposition 2. Let \( G \) be a group and let \( a, b \in G \) . The equations \( {ax} = b \) and \( {ya} = b \) have unique solutions for \( x, y \in G \) . In particular, the left and right cancellation laws hold in \( G \), i.e.,\n\n(1) if \( {au} = {av} \), then \( u = v \), and\n\n(2) if \( {ub} = {vb} \), then \(... | Proof: We can solve \( {ax} = b \) by multiplying both sides on the left by \( {a}^{-1} \) and simplifying to get \( x = {a}^{-1}b \) . The uniqueness of \( x \) follows because \( {a}^{-1} \) is unique. Similarly, if \( {ya} = b, y = b{a}^{-1} \) . If \( {au} = {av} \), multiply both sides on the left by \( {a}^{-1} \... | Yes |
Proposition 1. (The Subgroup Criterion) A subset \( H \) of a group \( G \) is a subgroup if and only if\n\n(1) \( H \neq \varnothing \), and\n\n(2) for all \( x, y \in H, x{y}^{-1} \in H \) . | Proof: If \( H \) is a subgroup of \( G \), then certainly (1) and (2) hold because \( H \) contains the identity of \( G \) and the inverse of each of its elements and because \( H \) is closed under multiplication.\n\nIt remains to show conversely that if \( H \) satisfies both (1) and (2), then \( H \leq G \) . Let ... | Yes |
Proposition 2. If \( H = \langle x\rangle \), then \( \left| H\right| = \left| x\right| \) (where if one side of this equality is infinite, so is the other). More specifically\n\n(1) if \( \left| H\right| = n < \infty \), then \( {x}^{n} = 1 \) and \( 1, x,{x}^{2},\ldots ,{x}^{n - 1} \) are all the distinct elements of... | Proof: Let \( \left| x\right| = n \) and first consider the case when \( n < \infty \) . The elements \( 1, x,{x}^{2},\ldots ,{x}^{n - 1} \) are distinct because if \( {x}^{a} = {x}^{b} \), with, say, \( 0 \leq a < b < n \), then \( {x}^{b - a} = {x}^{0} = 1 \), contrary to \( n \) being the smallest positive power of ... | Yes |
Proposition 3. Let \( G \) be an arbitrary group, \( x \in G \) and let \( m, n \in \mathbb{Z} \) . If \( {x}^{n} = 1 \) and \( {x}^{m} = 1 \), then \( {x}^{d} = 1 \), where \( d = \left( {m, n}\right) \) . In particular, if \( {x}^{m} = 1 \) for some \( m \in \mathbb{Z} \), then \( \left| x\right| \) divides \( m \) . | Proof: By the Euclidean Algorithm (see Section 0.2 (6)) there exist integers \( r \) and \( s \) such that \( d = {mr} + {ns} \), where \( d \) is the g.c.d. of \( m \) and \( n \) . Thus\n\n\[ \n{x}^{d} = {x}^{{mr} + {ns}} = {\left( {x}^{m}\right) }^{r}{\left( {x}^{n}\right) }^{s} = {1}^{r}{1}^{s} = 1.\n\]\n\nThis pro... | Yes |
Theorem 4. Any two cyclic groups of the same order are isomorphic. More specifically,\n\n(1) if \( n \in {\mathbb{Z}}^{ + } \) and \( \langle x\rangle \) and \( \langle y\rangle \) are both cyclic groups of order \( n \), then the map\n\n\[ \varphi : \langle x\rangle \rightarrow \langle y\rangle \]\n\n\[ {x}^{k} \mapst... | Proof: Suppose \( \langle x\rangle \) and \( \langle y\rangle \) are both cyclic groups of order \( n \) . Let \( \varphi : \langle x\rangle \rightarrow \langle y\rangle \) be defined by \( \varphi \left( {x}^{k}\right) = {y}^{k} \) ; we must first prove \( \varphi \) is well defined, that is,\n\n\[ \text{if}{x}^{r} = ... | Yes |
Proposition 5. Let \( G \) be a group, let \( x \in G \) and let \( a \in \mathbb{Z} - \{ 0\} \) . (1) If \( \left| x\right| = \infty \), then \( \left| {x}^{a}\right| = \infty \) . | Proof: (1) By way of contradiction assume \( \left| x\right| = \infty \) but \( \left| {x}^{a}\right| = m < \infty \) . By definition of order \[ 1 = {\left( {x}^{a}\right) }^{m} = {x}^{am}. \] Also, \[ {x}^{-{am}} = {\left( {x}^{am}\right) }^{-1} = {1}^{-1} = 1. \] Now one of \( {am} \) or \( - {am} \) is positive (si... | Yes |
Proposition 6. Let \( H = \langle x\rangle \) . (2) Assume \( \left| x\right| = n < \infty \) . Then \( H = \left\langle {x}^{a}\right\rangle \) if and only if \( \left( {a, n}\right) = 1 \) . In particular, the number of generators of \( H \) is \( \varphi \left( n\right) \) (where \( \varphi \) is Euler’s \( \varphi ... | Proof: We leave (1) as an exercise. In (2) if \( \left| x\right| = n < \infty \), Proposition 2 says \( {x}^{a} \) generates a subgroup of \( H \) of order \( \left| {x}^{a}\right| \) . This subgroup equals all of \( H \) if and only if \( \left| {x}^{a}\right| = \left| x\right| \) . By Proposition 5,\n\n\[ \left| {x}^... | No |
Theorem 7. Let \( H = \langle x\rangle \) be a cyclic group.\n\n(1) Every subgroup of \( H \) is cyclic. More precisely, if \( K \leq H \), then either \( K = \{ 1\} \) or \( K = \left\langle {x}^{d}\right\rangle \), where \( d \) is the smallest positive integer such that \( {x}^{d} \in K \) . | Proof: (1) Let \( K \leq H \) . If \( K = \{ 1\} \), the proposition is true for this subgroup, so we assume \( K \neq \{ 1\} \) . Thus there exists some \( a \neq 0 \) such that \( {x}^{a} \in K \) . If \( a < 0 \) then since \( K \) is a group also \( {x}^{-a} = {\left( {x}^{a}\right) }^{-1} \in K \) . Hence \( K \) ... | Yes |
Proposition 8. If \( \mathcal{A} \) is any nonempty collection of subgroups of \( G \), then the intersection of all members of \( \mathcal{A} \) is also a subgroup of \( G \) . | Proof: This is an easy application of the subgroup criterion (see also Exercise 10, Section 1). Let\n\n\[ K = \mathop{\bigcap }\limits_{{H \in \mathcal{A}}}H \]\n\nSince each \( H \in \mathcal{A} \) is a subgroup, \( 1 \in H \), so \( 1 \in K \), that is, \( K \neq \varnothing \). If \( a, b \in K \), then \( a, b \in ... | No |
Proposition 9. \( \bar{A} = \langle A\rangle \) . | Proof: We first prove \( \bar{A} \) is a subgroup. Note that \( \bar{A} \neq \varnothing \) (even if \( A = \varnothing \) ). If \( a, b \in \bar{A} \) with \( a = {a}_{1}^{{\epsilon }_{1}}{a}_{2}^{{\epsilon }_{2}}\ldots {a}_{n}^{{\epsilon }_{n}} \) and \( b = {b}_{1}^{{\delta }_{1}}{b}_{2}^{{\delta }_{2}}\ldots {b}_{m... | Yes |
Proposition 1. Let \( G \) and \( H \) be groups and let \( \varphi : G \rightarrow H \) be a homomorphism.\n\n(1) \( \varphi \left( {1}_{G}\right) = {1}_{H} \), where \( {1}_{G} \) and \( {1}_{H} \) are the identities of \( G \) and \( H \), respectively.\n\n(2) \( \varphi \left( {g}^{-1}\right) = \varphi {\left( g\ri... | Proof: (1) Since \( \varphi \left( {1}_{G}\right) = \varphi \left( {{1}_{G}{1}_{G}}\right) = \varphi \left( {1}_{G}\right) \varphi \left( {1}_{G}\right) \), the cancellation laws show that (1) holds.\n\n(2) \( \varphi \left( {1}_{G}\right) = \varphi \left( {g{g}^{-1}}\right) = \varphi \left( g\right) \varphi \left( {g}... | No |
Proposition 2. Let \( \varphi : G \rightarrow H \) be a homomorphism of groups with kernel \( K \) . Let\n\n\( X \in G/K \) be the fiber above \( a \), i.e., \( X = {\varphi }^{-1}\left( a\right) \) . Then\n\n(1) For any \( u \in X,\;X = \{ {uk} \mid k \in K\} \)\n\n(2) For any \( u \in X,\;X = \{ {ku} \mid k \in K\} \... | Proof: We prove (1) and leave the proof of (2) as an exercise. Let \( u \in X \) so, by definition of \( X,\varphi \left( u\right) = a \) . Let\n\n\[ \n{uK} = \{ {uk} \mid k \in K\} .\n\]\n\nWe first prove \( {uK} \subseteq X \) . For any \( k \in K \) ,\n\n\[ \n\varphi \left( {uk}\right) = \varphi \left( u\right) \var... | No |
Proposition 4. Let \( N \) be any subgroup of the group \( G \). The set of left cosets of \( N \) in \( G \) form a partition of \( G \). Furthermore, for all \( u, v \in G,{uN} = {vN} \) if and only if \( {v}^{-1}u \in N \) and in particular, \( {uN} = {vN} \) if and only if \( u \) and \( v \) are representatives of... | Proof: First of all note that since \( N \) is a subgroup of \( G,1 \in N \). Thus \( g = g \cdot 1 \in {gN} \) for all \( g \in G \), i.e.,\n\n\[ G = \mathop{\bigcup }\limits_{{g \in G}}{gN} \]\n\nTo show that distinct left cosets have empty intersection, suppose \( {uN} \cap {vN} \neq \varnothing \). We show \( {uN} ... | Yes |
Proposition 5. Let \( G \) be a group and let \( N \) be a subgroup of \( G \). (1) The operation on the set of left cosets of \( N \) in \( G \) described by \[ {uN} \cdot {vN} = \left( {uv}\right) N \] is well defined if and only if \( {gn}{g}^{-1} \in N \) for all \( g \in G \) and all \( n \in N \). (2) If the abov... | Proof: (1) Assume first that this operation is well defined, that is, for all \( u, v \in G \), \[ \text{if}u,{u}_{1} \in {uN}\text{and}v,{v}_{1} \in {vN}\;\text{then}\;{uvN} = {u}_{1}{v}_{1}N\text{.} \] Let \( g \) be an arbitrary element of \( G \) and let \( n \) be an arbitrary element of \( N \). Letting \( u = 1,... | Yes |
Theorem 6. Let \( N \) be a subgroup of the group \( G \) . The following are equivalent:\n\n(1) \( N \trianglelefteq G \)\n\n(2) \( {N}_{G}\left( N\right) = G \) (recall \( {N}_{G}\left( N\right) \) is the normalizer in \( G \) of \( N \) )\n\n(3) \( {gN} = {Ng} \) for all \( g \in G \)\n\n(4) the operation on left co... | Proof: We have already done the hard equivalences; the others are left as exercises. | No |
Theorem 8. (Lagrange’s Theorem) If \( G \) is a finite group and \( H \) is a subgroup of \( G \) , then the order of \( H \) divides the order of \( G \) (i.e., \( \left| H\right| \left| \right| G \mid \) ) and the number of left cosets of \( H \) in \( G \) equals \( \frac{\left| G\right| }{\left| H\right| } \) . | Proof: Let \( \left| H\right| = n \) and let the number of left cosets of \( H \) in \( G \) equal \( k \) . By\n\nProposition 4 the set of left cosets of \( H \) in \( G \) partition \( G \) . By definition of a left coset the map:\n\n\[ H \rightarrow {gH}\;\text{ defined by }\;h \mapsto {gh} \]\n\nis a surjection fro... | Yes |
Corollary 9. If \( G \) is a finite group and \( x \in G \), then the order of \( x \) divides the order of \( G \) . In particular \( {x}^{\left| G\right| } = 1 \) for all \( x \) in \( G \) . | Proof: By Proposition 2.2, \( \left| x\right| = \left| {\langle x\rangle }\right| \) . The first part of the corollary follows from Lagrange’s Theorem applied to \( H = \langle x\rangle \) . The second statement is clear since now \( \left| G\right| \) is a multiple of the order of \( x \) . | Yes |
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