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Lemma 4.3. Let \( M \) be a left \( R \) -module, let \( E = {\operatorname{End}}_{R}^{\mathrm{{op}}}\left( M\right) \), let \( n > 0 \) , and let \( F = {\operatorname{End}}_{R}^{\mathrm{{op}}}\left( {M}^{n}\right) \) . If \( \xi : M \rightarrow M \) is an E-endomorphism, then \( {\xi }^{n} : {M}^{n} \rightarrow {M}^{...
Proof. Let \( {\iota }_{i} : M \rightarrow {M}^{n} \) and \( {\pi }_{j} : {M}^{n} \rightarrow M \) be the injections and projections. Let \( \eta \in F \) . As in the proof of 1.4, every \( \eta : {M}^{n} \rightarrow {M}^{n} \) is determined by a matrix \( \left( {\eta }_{ij}\right) \), where \( {\eta }_{ij} = {\pi }_{...
No
Corollary 4.4. If \( D \) is a division ring, then the center of \( {M}_{n}\left( D\right) \) consists of all scalar matrices whose diagonal entry is in the center of \( D \) .
Proof. We prove this for \( {D}^{\mathrm{{op}}} \) . Let \( I \) denote the identity matrix. Let \( V \) be a left \( D \) -module with a basis \( {e}_{1},\ldots ,{e}_{n} \), so that \( {\operatorname{End}}_{D}\left( V\right) \cong {M}_{n}\left( {D}^{\mathrm{{op}}}\right) \) . Let \( E = {\operatorname{End}}_{D}^{\math...
Yes
Corollary 4.5 (Burnside [1905]). Let \( K \) be an algebraically closed field, let \( V \) be a finite-dimensional vector space over \( K \), and let \( R \) be a subring of \( {\operatorname{End}}_{K}\left( V\right) \) that contains all scalar transformations \( v \mapsto {av} \) with \( a \in K \) . If \( V \) is a s...
Proof. Identify \( a \in K \) with the scalar transformation \( {aI} : v \mapsto {av} \), so that \( K \) becomes a subfield of \( {\operatorname{End}}_{K}\left( V\right) \) that is contained in the center of \( {\operatorname{End}}_{K}\left( V\right) \) (in fact, \( K \) is the center of \( {\operatorname{End}}_{K}\le...
Yes
Theorem 4.6 (Jacobson Density Theorem). A ring \( R \) is left primitive if and only if it is isomorphic to a dense subring of \( {\operatorname{End}}_{D}\left( V\right) \) for some division ring \( D \) and right \( D \) -module \( V \) .
Proof. If \( S \) is a faithful simple left \( R \) -module, then \( D = {\operatorname{End}}_{R}^{\mathrm{{op}}}\left( S\right) \) is a division ring by 1.2, the canonical homomorphism \( \Phi : R \rightarrow {\operatorname{End}}_{D}\left( S\right) \) is injective since \( S \) is faithful, and \( \Phi \left( R\right)...
Yes
Theorem 4.7. Let \( R \) be a left primitive ring and let \( S \) be a faithful simple left \( R \) -module, so that \( D = {\operatorname{End}}_{R}^{\text{op }}\left( S\right) \) is a division ring.\n\n(1) If \( R \) is left Artinian, then \( n = {\dim }_{D}S \) is finite and \( R \cong {M}_{n}\left( D\right) \) .
Proof. As in the proof of Theorem 4.6, the canonical homomorphism \( \Phi \) : \( R \rightarrow {\operatorname{End}}_{D}\left( S\right) \) is injective, and \( \Phi \left( R\right) \) is dense in \( {\operatorname{End}}_{D}\left( S\right) \) .\n\n(1). If a basis of \( S \) contains an infinite sequence \( {e}_{1},\ldot...
Yes
In a ring \( R, J\left( R\right) \) is the intersection of all the annihilators of simple left \( R \) -modules; hence \( J\left( R\right) \) is a two-sided ideal of \( R \) .
Proof. If \( L \) is a maximal left ideal of \( R \), then \( S = {}_{R}R/L \) is a simple left \( R \) -module and \( \operatorname{Ann}\left( S\right) \subseteq L \) . Hence the intersection of all \( \operatorname{Ann}\left( S\right) \) is contained in \( J\left( R\right) \) . Conversely, let \( r \in J\left( R\righ...
Yes
Lemma 5.2. If \( x \in R \), then \( x \in J\left( R\right) \) if and only if \( 1 + {tx} \) has a left inverse for every \( t \in R \) .
Proof. If \( x \notin J\left( R\right) \), then \( x \notin L \) for some maximal left ideal \( L, L + {Rx} = R \) , \( 1 = \ell + {rx} \) for some \( \ell \in L \) and \( r \in R \), and \( 1 - {rx} \in L \) has no left inverse, since all its left multiples are in \( L \) . Conversely, if some \( 1 + {tx} \) has no le...
Yes
Proposition 5.3. In a ring \( R, J\left( R\right) \) is the largest two-sided ideal \( I \) of \( R \) such that \( 1 + x \) is a unit of \( R \) for all \( x \in I \) ; hence \( J\left( R\right) = J\left( {R}^{\mathrm{{op}}}\right) \) .
Proof. If \( x \in J\left( R\right) \), then, by 5.2, \( 1 + x \) has a left inverse \( y \), whence \( y = 1 - {yx} \) and \( y \) has a left inverse \( z \) . Since \( y \) already has a right inverse \( 1 + x \), it follows that \( 1 + x = z \), and \( y \) is a two-sided inverse of \( 1 + x \) . Thus \( J\left( R\r...
Yes
Proposition 5.6. In a ring \( R, J\left( R\right) \) contains all nilpotent left or right ideals of \( R \) . If \( R \) is commutative, then \( J\left( R\right) \) contains all nilpotent elements of \( R \) .
Proof. Let \( N \) be a nilpotent left ideal and let \( S \) be a simple left \( R \) -module. If \( {NS} \neq 0 \), then \( {NS} = S \) and \( S = {NS} = {N}^{2}S = \cdots = {N}^{n}S = 0 \), a contradiction; therefore \( {NS} = 0 \) and \( N \subseteq \operatorname{Ann}\left( S\right) \) . Hence \( N \subseteq J\left(...
Yes
Proposition 5.7 (Nakayama’s Lemma). Let \( M \) be a finitely generated left \( R \) -module. If \( J\left( R\right) M = M \), then \( M = 0 \) .
Proof. Assume \( M \neq 0 \) . Since \( M \) is finitely generated, the union of a chain \( {\left( {N}_{i}\right) }_{i \in I} \) of proper submodules of \( M \) is a proper submodule of \( M \) : otherwise, the finitely many generators of \( M \) all belong to some \( {N}_{i} \), and then \( {N}_{i} = M \) . By Zorn’s...
Yes
Proposition 5.8 (Nakayama’s Lemma). Let \( N \) be a submodule of a finitely generated left \( R \) -module \( M \) . If \( N + J\left( R\right) M = M \), then \( N = M \) .
Proof. First, \( M/N \) is finitely generated. If \( N + J\left( R\right) M = M \), then \( J\left( R\right) \left( {M/N}\right) = M/N \) and \( M/N = 0 \), by 5.7. \( ▱ \)
Yes
Theorem 6.1. A ring \( R \) is semisimple if and only if \( R \) is left Artinian and \( J\left( R\right) = 0 \), if and only if \( R \) is left Artinian and has no nonzero nilpotent ideal.
Proof. The last two conditions are equivalent, by 6.2 below.\n\nBy 3.5,3.3, a semisimple ring \( R \) is left Artinian, and is isomorphic to a direct product \( {M}_{{n}_{1}}\left( {D}_{1}\right) \times \cdots \times {M}_{{n}_{s}}\left( {D}_{s}\right) \) of finitely many matrix rings over division rings \( {D}_{1},\ldo...
No
Lemma 6.2. If \( R \) is left Artinian, then \( J\left( R\right) \) is nilpotent, and is the greatest nilpotent left ideal of \( R \) and the greatest nilpotent right ideal of \( R \) .
Proof. Let \( J = J\left( R\right) \) . Since \( R \) is left Artinian, the descending sequence \( J \supseteq {J}^{2} \supseteq \cdots \supseteq {J}^{n} \supseteq {J}^{n + 1} \supseteq \cdots \) terminates at some \( {J}^{m}\left( {{J}^{n} = {J}^{m}}\right. \) for all \( n \geqq m) \) . Suppose that \( {J}^{m} \neq 0 ...
Yes
Proposition 6.3. If \( R \) is left Artinian, then a left \( R \) -module \( M \) is semisimple if and only if \( J\left( R\right) M = 0 \) .
Proof. Let \( J = J\left( R\right) \) . We have \( {JS} = 0 \) for every simple \( R \) -module \( S \), since \( J \subseteq \operatorname{Ann}\left( S\right) \) ; hence \( {JM} = 0 \) whenever \( M \) is semisimple. Conversely, assume that \( {JM} = 0 \) . Then \( M \) is a left \( R/J \) -module, in which \( \left( ...
Yes
Theorem 6.4 (Hopkins-Levitzki). If \( R \) is left Artinian, then for a left \( R \) -module \( M \) the following properties are equivalent: (i) \( M \) is Noetherian; (ii) \( M \) is Artinian; (iii) \( M \) is of finite length.
Proof. If \( M \) is semisimple, then (i),(ii), and (iii) are equivalent, since a Noetherian or Artinian module cannot be the direct sum of infinitely many simple submodules.\n\nIn general, let \( J = J\left( R\right) \). Let \( M \) be Noetherian (or Artinian). By \( {6.2},{J}^{n} = 0 \) for some \( n > 0 \), which yi...
Yes
Proposition 7.1. Every finite-dimensional representation is a direct sum of irreducible representations.
Proof. This is shown by induction on the dimension of \( V \) : if \( \rho \) is not irreducible, then \( \rho = {\rho }_{1} \oplus {\rho }_{2} \), where \( V = {V}_{1} \oplus {V}_{2} \) and \( {V}_{1},{V}_{2} \) have lower dimension than \( V \) . \( ▱ \)
Yes
Proposition 7.2. Every multiplicative homomorphism of a group \( G \) into a \( K \) -algebra \( A \) extends uniquely to an algebra homomorphism of \( K\left\lbrack G\right\rbrack \) into \( A \) .
Proof. Let \( \varphi : G \rightarrow A \) be a multiplicative homomorphism (meaning \( \varphi \left( {gh}\right) = \varphi \left( g\right) \varphi \left( h\right) \) for all \( g, h \in G \), and \( \varphi \left( 1\right) = 1 \) ). Since \( G \) is a basis of \( K\left\lbrack G\right\rbrack ,\varphi \) extends uniqu...
Yes
There is a one-to-one correspondence between representations of a group \( G \) over a field \( K \) and \( K\left\lbrack G\right\rbrack \) -modules. Two representations are equivalent if and only if the corresponding modules are isomorphic.
Proof. By 7.2, a representation \( \rho : G \rightarrow {GL}\left( V\right) \subseteq {\operatorname{End}}_{K}\left( V\right) \) extends uniquely to an algebra homomorphism \( \rho : K\left\lbrack G\right\rbrack \rightarrow {\operatorname{End}}_{K}\left( V\right) \) that makes \( V \) a \( K\left\lbrack G\right\rbrack ...
Yes
Theorem 7.4 (Maschke [1898]). Let \( G \) be a finite group and let \( K \) be a field. If \( K \) has characteristic 0, or if \( K \) has characteristic \( p \neq 0 \) and \( p \) does not divide the order of \( G \), then \( K\left\lbrack G\right\rbrack \) is semisimple.
Proof. We show that every submodule \( W \) of a \( G \) -module \( V \) is a direct summand of \( V \) . We already have \( V = W \oplus {W}^{\prime } \) (as a vector space) for some subspace \( {W}^{\prime } \) . The projection \( \pi : V \rightarrow W \) is a linear transformation and is the identity on \( W \) . De...
Yes
Corollary 7.5. Let \( G \) be a finite group and let \( K \) be a field whose characteristic does not divide the order of \( G \) .\n\n(1) Up to isomorphism, there are only finitely many simple \( G \) -modules \( {S}_{1},\ldots ,{S}_{s} \) , and they all have finite dimension over \( K \) .\n\n(2) Every G-module is is...
Proof. That there are only finitely many simple \( G \) -modules follows from 3.6. By 1.3, every simple \( G \) -module is isomorphic to a minimal left ideal of \( K\left\lbrack G\right\rbrack \) , and has finite dimension over \( K \), like \( K\left\lbrack G\right\rbrack \) . Then (2) follows from 3.7, and (3) follow...
No
Proposition 7.6. Let \( G \) be a finite group and let \( K \) be an algebraically closed field whose characteristic does not divide the order of \( G \) . Let the nonisomorphic simple \( G \) -modules have dimensions \( {d}_{1},\ldots ,{d}_{s} \) over \( K \) . The simple components of \( K\left\lbrack G\right\rbrack ...
Proof. By 7.4, \( K\left\lbrack G\right\rbrack \) is semisimple. The simple components \( {R}_{1},\ldots ,{R}_{r} \) of \( K\left\lbrack G\right\rbrack \) are simple left Artinian rings \( {R}_{i} \cong {M}_{{n}_{i}}\left( {D}_{i}\right) \), where \( {D}_{1},\ldots ,{D}_{r} \) are division rings, as well as two-sided i...
Yes
Lemma 7.7. Let \( D \) be a division ring that has finite dimension over a central subfield \( K \) . If \( K \) is algebraically closed, then \( D = K \) .
Proof. If \( \alpha \in D \), then \( K \) and \( \alpha \) generate a commutative subring \( K\left\lbrack \alpha \right\rbrack \) of \( D \) , \( \alpha \) is algebraic over \( K \) since \( K\left\lbrack \alpha \right\rbrack \) has finite dimension over \( K \), and \( \alpha \in K \) since \( K \) is algebraically ...
Yes
Theorem 7.9. Let \( G \) be a finite group with \( s \) conjugacy classes and let \( K \) be an algebraically closed field whose characteristic does not divide the order of \( G \) . Up to equivalence, \( G \) has \( s \) distinct irreducible representations over \( K \) .
Proof. We look at the center \( Z\left( {K\left\lbrack G\right\rbrack }\right) \) of \( K\left\lbrack G\right\rbrack \) . By 7.6, \( K\left\lbrack G\right\rbrack \cong {R}_{1} \times \cdots \times \) \( {R}_{s} \), where \( s \) is now the number of distinct irreducible representations of \( G \) and \( {R}_{i} \cong {...
No
Proposition 7.10. Let \( G \) be a finite group and let \( K \) be an algebraically closed field whose characteristic does not divide the order of \( G \) . Let \( \rho : G \rightarrow {\operatorname{End}}_{K}\left( S\right) \) be an irreducible representation and let \( c \) be the sum of a conjugacy class. Then \( \r...
Proof. First, \( S \) is a simple \( K\left\lbrack G\right\rbrack \) -module. Let \( E = {\operatorname{End}}_{K\left\lbrack G\right\rbrack }^{\text{op }}\left( S\right) \) . Since \( K \) is central in \( K\left\lbrack G\right\rbrack \), the scalar linear transformations \( ╏ : x \mapsto {\lambda x} \) are \( K\left\l...
Yes
Proposition 8.2. If \( K \) is an algebraically closed field whose characteristic does not divide \( \left| G\right| \), then the regular character is \( {\chi }_{r} = \mathop{\sum }\limits_{i}{d}_{i}{\chi }_{i} \) .
Proof. By 7.6, \( {R}_{i} \cong {M}_{{d}_{i}}\left( K\right) \cong {S}_{i}^{{d}_{i}} \) ; hence \( K\left\lbrack G\right\rbrack \cong {S}_{1}^{{d}_{1}} \oplus \cdots \oplus {S}_{s}^{{d}_{s}} \), as a \( G \) -module, and \( {\chi }_{r} = \mathop{\sum }\limits_{i}{d}_{i}{\chi }_{i} \) . \( ▱ \)
Yes
Lemma 8.3. If the characteristic of \( K \) does not divide \( \left| G\right| \), then:\n\n(1) \( {\chi }_{i}\left( x\right) = 0 \) when \( x \in {R}_{j} \) and \( j \neq i \) ;\n\n(2) \( {\chi }_{i}\left( {e}_{j}\right) = 0 \) if \( j \neq i \) and \( {\chi }_{i}\left( {e}_{i}\right) = {\chi }_{i}\left( 1\right) = {d...
Proof. If \( x \in {R}_{j} \) and \( j \neq i \), then \( {\rho }_{i}\left( x\right) \left( s\right) = {xs} = 0 \) for all \( s \in {S}_{i} \) , \( {\rho }_{i}\left( x\right) = 0 \), and \( {\chi }_{i}\left( x\right) = \operatorname{Tr}{\rho }_{i}\left( x\right) = 0 \) . In particular, \( {\chi }_{i}\left( {e}_{j}\righ...
Yes
Theorem 8.4. If \( G \) is finite and \( K \) has characteristic 0 :\n\n(1) the irreducible characters of \( G \) are linearly independent over \( K \) ;\n\n(2) every character of \( G \) can be written uniquely as a linear combination of irreducible characters with nonnegative integer coefficients;\n\n(3) two finite-d...
Proof. If \( {a}_{i} \in K \) and \( \mathop{\sum }\limits_{i}{a}_{i}{\chi }_{i} = 0 \), then \( {d}_{i}{a}_{i} = \mathop{\sum }\limits_{j}{a}_{j}{\chi }_{j}\left( {e}_{i}\right) = 0 \) for all \( i \) ; since \( K \) has characteristic 0 this implies \( {a}_{i} = 0 \) for all \( i \) .\n\nLet \( \chi \) be the charact...
Yes
Theorem 8.5. If \( G \) is a finite group and \( K \) is an algebraically closed field whose characteristic does not divide \( \left| G\right| \), then \( {d}_{i} \neq 0 \) in \( K \) and\n\n\[ \mathop{\sum }\limits_{{g \in G}}{\chi }_{i}\left( g\right) {\chi }_{j}\left( {g}^{-1}\right) = \left\{ \begin{array}{ll} \lef...
Proof. Let \( {e}_{ig} \) be the \( g \) coordinate of \( {e}_{i} \) in \( K\left\lbrack G\right\rbrack \), so that \( {e}_{i} = \mathop{\sum }\limits_{{g \in G}}{e}_{ig}g \) . Since \( {\chi }_{r}\left( 1\right) = \left| G\right| \) and \( {\chi }_{r}\left( g\right) = 0 \) for all \( g \neq 1 \) we have\n\n\[ {\chi }_...
Yes
Proposition 9.1. Let \( G \) be a finite group and let \( \chi \) be the character of a complex representation \( \rho : G \rightarrow {GL}\left( V\right) \) of dimension \( d \) . For every \( g \in G \) :\n\n(1) \( \rho \left( g\right) \) is diagonalizable;\n\n(2) \( \chi \left( g\right) \) is a sum of \( d \) roots ...
Proof. (1). Let \( H = \langle g\rangle \) . By 7.8 the representation \( {\rho }_{\mid H} : H \rightarrow {GL}\left( V\right) \) is a direct sum of representations of dimension 1. Hence \( V \), as an \( H \) -module, is a direct sum of submodules of dimension 1 over \( \mathbb{C} \), and has a basis \( {e}_{1},\ldots...
Yes
Lemma 9.2. \( {c}_{ij} \) is integral over \( \mathbb{Z} \) .
Proof. First, \( {c}_{j}{c}_{k} = \left( {\mathop{\sum }\limits_{{{h}^{\prime } \in {C}_{j}}}{h}^{\prime }}\right) \left( {\mathop{\sum }\limits_{{{h}^{\prime \prime } \in {C}_{k}}}{h}^{\prime \prime }}\right) = \mathop{\sum }\limits_{{g \in G}}{n}_{g}g \), where \( {n}_{g} \) is the number of ordered pairs \( \left( {...
Yes
Proposition 9.3. If \( {d}_{i} \) and \( \left| {C}_{j}\right| \) are relatively prime, then either \( \left| {{\chi }_{i}\left( g\right) }\right| = {d}_{i} \) for all \( g \in {C}_{j} \), or \( {\chi }_{i}\left( g\right) = 0 \) for all \( g \in {C}_{j} \) .
Proof. Assume \( \left| {{\chi }_{i}\left( g\right) }\right| < {d}_{i} \), where \( g \in {C}_{j} \) . Let \( \alpha = {\chi }_{i}\left( g\right) /{d}_{i} = {c}_{ij}/\left| {C}_{j}\right| \) . Then \( \left| \alpha \right| < 1 \) . Also, \( u{d}_{i} + v\left| {C}_{j}\right| = 1 \) for some \( u, v \in \mathbb{Z} \) ; h...
Yes
Theorem 9.4 (Burnside). Let \( p \) and \( q \) be prime numbers. Every group of order \( {p}^{m}{q}^{n} \) is solvable.
Proof. It is enough to show that simple groups of order \( {p}^{m}{q}^{n} \) are abelian.\n\nAssume that \( G \) is a simple nonabelian group of order \( {p}^{m}{q}^{n} \) . Since \( p \) -groups are solvable we may assume that \( p \neq q \) and that \( m, n > 0 \) . Number \( {\chi }_{1},\ldots ,{\chi }_{s} \) so tha...
Yes
Lemma 1.1. If \( 0 \rightarrow A\overset{\mu }{ \rightarrow }B\overset{\varphi }{ \rightarrow }C \) is exact, then every homomorphism \( \psi \) such that \( \varphi \circ \psi = 0 \) factors uniquely through \( \mu \):
Proof. We have \( \operatorname{Im}\psi \subseteq \operatorname{Ker}\varphi = \operatorname{Im}\mu \) . Since \( \mu \) is injective there is a unique mapping \( \chi : M \rightarrow A \) such that \( \psi \left( x\right) = \mu \left( {\chi \left( x\right) }\right) \) for all \( x \in M \) . Then \( \chi \) is a module...
Yes
Lemma 1.2. If \( A\overset{\varphi }{ \rightarrow }B\overset{\sigma }{ \rightarrow }C \rightarrow 0 \) is exact, then every homomorphism \( \psi \) such that \( \psi \circ \varphi = 0 \) factors uniquely through \( \sigma \) :
Proof. This follows from Theorem VIII.2.5, since Ker \( \sigma = \operatorname{Im}\varphi \subseteq \operatorname{Ker}\psi \) . \( ▱ \)
Yes
Lemma 1.3 (Short Five Lemma). In a commutative diagram with exact rows,\n\n![5e708ed9-3d6d-4f59-a748-eaac13dfd780_406_2.jpg](images/5e708ed9-3d6d-4f59-a748-eaac13dfd780_406_2.jpg)\n\nif \( \alpha \) and \( \gamma \) are isomorphisms, then so is \( \beta \) .
Proof. Assume that \( \beta \left( b\right) = 0 \) . Then \( \gamma \left( {\rho \left( b\right) }\right) = {\rho }^{\prime }\left( {\beta \left( b\right) }\right) = 0 \) and \( \rho \left( b\right) = 0 \) . By exactness, \( b = \mu \left( a\right) \) for some \( a \in A \) . Then \( {\mu }^{\prime }\left( {\alpha \lef...
Yes
Proposition 1.5. For a short exact sequence \( 0 \rightarrow A\overset{\mu }{ \rightarrow }B\overset{\rho }{ \rightarrow }C \rightarrow 0 \) the following conditions are equivalent:\n\n(1) \( \mu \) splits \( \left( {\sigma \circ \mu = {1}_{A}}\right. \) for some homomorphism \( \sigma : B \rightarrow A \) );\n\n(2) \(...
Proof. (1) implies (3). Assume that \( \sigma \circ \mu = {1}_{A} \) . There is a homomorphism \( \theta : B \rightarrow A \oplus C \) such that \( {\pi }^{\prime } \circ \theta = \sigma \) and \( \pi \circ \theta = \rho \) (where \( {\pi }^{\prime } \) : \( A \oplus C \rightarrow A \) is the projection), namely \( {\t...
Yes
Proposition 2.1. For every pair of homomorphisms \( \alpha : A \rightarrow C \) and \( \beta \) : \( B \rightarrow C \) of left \( R \) -modules, there exists a pullback \( \alpha \circ {\beta }^{\prime } = \beta \circ {\alpha }^{\prime } \), and it is unique up to isomorphism.
Proof. Uniqueness follows from the universal property. Let \( \alpha \circ {\beta }^{\prime } = \beta \circ {\alpha }^{\prime } \) and \( \alpha \circ {\beta }^{\prime \prime } = \beta \circ {\alpha }^{\prime \prime } \) be pullbacks. There exist homomorphisms \( \theta \) and \( \zeta \) such that \( {\beta }^{\prime ...
Yes
Proposition 2.2. Given homomorphisms \( \alpha : A \rightarrow C \) and \( \beta : B \rightarrow C \) of left \( R \) -modules, let \( P = \{ \left( {a, b}\right) \in A \oplus B \mid \alpha \left( a\right) = \beta \left( b\right) \} ,{\alpha }^{\prime } : P \rightarrow B \) , \( \left( {a, b}\right) \mapsto b \), and \...
Proof. First, \( \alpha \circ {\beta }^{\prime } = \beta \circ {\alpha }^{\prime } \), by the choice of \( P \) . Assume \( \alpha \circ \varphi = \beta \circ \psi \) , where \( \varphi : M \rightarrow A \) and \( \psi : M \rightarrow B \) are module homomorphisms. Then \( \left( {\varphi \left( m\right) ,\psi \left( m...
Yes
Proposition 2.5. For every pair of homomorphisms \( \alpha : C \rightarrow A \) and \( \beta \) : \( C \rightarrow B \) of left \( R \) -modules, there exists a pushout \( {\beta }^{\prime } \circ \alpha = {\alpha }^{\prime } \circ \beta \), and it is unique up to isomorphism.
Uniqueness follows from the universal property, and existence from a construction:
No
Proposition 2.6. Given homomorphisms \( \alpha : C \rightarrow A \) and \( \beta : C \rightarrow B \) of left \( R \) -modules, let \( K = \{ \left( {\alpha \left( c\right) , - \beta \left( c\right) }\right) \in A \oplus B \mid c \in C\}, P = \left( {A \oplus B}\right) /K \) , \( {\alpha }^{\prime } = \pi \circ \kappa ...
Proof. First, \( \operatorname{Ker}\pi = K = \operatorname{Im}\left( {\iota \circ \alpha - \kappa \circ \beta }\right) \) ; hence \( {\beta }^{\prime } \circ \alpha = \pi \circ \iota \circ \alpha = \) \( \pi \circ \kappa \circ \beta = {\alpha }^{\prime } \circ \beta \) . Assume that \( \varphi \circ \alpha = \chi \circ...
Yes
Proposition 3.1. Every free module is projective.
Proof. Let \( \varphi : F \rightarrow N \) and \( \rho : M \rightarrow N \) be homomorphisms and let \( {\left( {e}_{i}\right) }_{i \in I} \) be a basis of \( F \) . If \( \rho \) is surjective, there is for every \( i \in I \) some \( {m}_{i} \in M \) such that \( \varphi \left( {e}_{i}\right) = \rho \left( {m}_{i}\ri...
Yes
Proposition 3.2. A left \( R \) -module \( P \) is projective if and only if every epimorphism \( M \rightarrow P \) splits, if and only if every short exact sequence \( 0 \rightarrow A \rightarrow \) \( B \rightarrow P \rightarrow 0 \) splits.
Proof. By 1.5, \( 0 \rightarrow A \rightarrow B \rightarrow P \rightarrow 0 \) splits if and only if \( B \rightarrow P \) splits. Assume that every epimorphism \( M \rightarrow P \) splits. Let \( \varphi : P \rightarrow N \) and \( \rho : M \rightarrow N \) be homomorphisms, with \( \rho \) surjective. In the pullbac...
No
Corollary 3.3. A ring \( R \) is semisimple if and only if every short exact sequence of left \( R \) -modules splits, if and only if every left \( R \) -module is projective.
Proof. A left \( R \) -module \( B \) is semisimple if and only if every submodule of \( B \) is a direct summand, if and only if every short exact sequence \( 0 \rightarrow A \rightarrow B \rightarrow \) \( C \rightarrow 0 \) splits. \( ▱ \)
No
Corollary 3.6. A module is projective if and only if it is isomorphic to a direct summand of a free module.
Proof. For every module \( P \) there is by VIII.4.6 an epimorphism \( \rho : F \rightarrow P \) where \( F \) is free; if \( P \) is projective, then the exact sequence \( 0 \rightarrow \operatorname{Ker}\varphi \rightarrow \) \( F \rightarrow P \rightarrow 0 \) splits, and \( P \) is isomorphic to a direct summand of...
Yes
Corollary 3.7. If \( R \) is a PID, then an \( R \) -module is projective if and only if it is free.
Proof. Every submodule of a free \( R \) -module is free, by Theorem VIII.6.1. \( ▱ \)
No
Proposition 4.1. For a left \( R \) -module \( J \) the following conditions are equivalent:\n\n(1) \( J \) is injective;\n\n(2) every monomorphism \( J \rightarrow M \) splits;\n\n(3) every short exact sequence \( 0 \rightarrow J \rightarrow B \rightarrow C \rightarrow 0 \) splits;\n\n(4) \( J \) is a direct summand o...
Proof. We prove that (2) implies (1); the other implications are clear. Assume (2). Let \( \varphi : M \rightarrow J \) and \( \mu : M \rightarrow N \) be homomorphisms, with \( \mu \) injective. In the pushout \( {\varphi }^{\prime } \circ \mu = {\mu }^{\prime } \circ \varphi ,{\mu }^{\prime } \) is injective, by 2.7....
No
Proposition 4.5 (Baer’s Criterion). For a left R-module \( J \) the following conditions are equivalent:\n\n(1) \( J \) is injective;\n\n(2) every module homomorphism of a left ideal of \( R \) into \( J \) can be extended to \( {}_{R}R \) ;\n\n(3) for every module homomorphism \( \varphi \) of a left ideal \( L \) of ...
Proof. (1) implies (2), and readers will happily show that (2) and (3) are equivalent. We show that (2) implies (1): every module homomorphism \( \varphi \) : \( M \rightarrow J \) extends through every monomorphism \( \mu : M \rightarrow N \) . Then \( M \) is isomorphic to a submodule of \( N \) ; we may assume that ...
No
Proposition 4.6. If \( R \) is a domain, then every injective \( R \) -module is divisible.
Proof. Let \( J \) be injective. Let \( a \in J \) and \( 0 \neq r \in R \) . Since \( R \) is a domain, every element of \( {Rr} \) can be written in the form \( {tr} \) for some unique \( t \in R \) . Hence \( \varphi : {tr} \mapsto {ta} \) is a module homomorphism of \( {Rr} \) into \( J \) . By 4.5 there exists \( ...
Yes
Proposition 4.7. If \( R \) is a PID, then an \( R \) -module is injective if and only if it is divisible.
Proof. Let \( M \) be a divisible \( R \) -module. Let \( {Rr} \) be any (left) ideal of \( R \) and let \( \varphi : {Rr} \rightarrow M \) be a module homomorphism. If \( r = 0 \), then \( \varphi \left( s\right) = {s0} \) for all \( s \in {Rr} \) . Otherwise \( \varphi \left( r\right) = {rm} \) for some \( m \in M \)...
No
Proposition 4.8. The group \( {\mathbb{Z}}_{{p}^{\infty }} \) is the union of cyclic subgroups \( {C}_{1} \subseteq {C}_{2} \subseteq \) \( \cdots \subseteq {C}_{n} \subseteq \cdots \) of orders \( p,{p}^{2},\ldots ,{p}^{n},\ldots \) and is a divisible abelian group.
Proof. First we find a model of \( {\mathbb{Z}}_{{p}^{\infty }} \) . Let \( U \) be the multiplicative group of all complex numbers of modulus 1 . Let \( {\alpha }_{n} = {e}^{{2i\pi }/{p}^{n}} \in U \) . Then \( {\alpha }_{1}^{p} = 1 \) and \( {\alpha }_{n}^{p} = {\alpha }_{n - 1} \) for all \( n > 1 \) . Hence there i...
Yes
Proposition 4.10. Every abelian group can be embedded into a divisible abelian group.
Proof. For every abelian group \( A \) there is an epimorphism \( F \rightarrow A \) where \( F \) is a free abelian group. Now, \( F \) is a direct sum of copies of \( \mathbb{Z} \) and can be embedded into a direct sum \( D \) of copies of \( \mathbb{Q} \) . Then \( D \) is divisible, like \( \mathbb{Q} \) . By 2.7, ...
Yes
Theorem 4.12. A ring \( R \) is left Noetherian if and only if every direct sum of injective left \( R \) -modules is injective.
Proof. Assume that every direct sum of injective left \( R \) -modules is injective, and let \( {L}_{1} \subseteq {L}_{2} \subseteq \cdots \subseteq {L}_{n} \subseteq \cdots \) be an ascending sequence of left ideals of \( R \) . Then \( L = \mathop{\bigcup }\limits_{{n > 0}}{L}_{n} \) is a left ideal of \( R \) . By 4...
Yes
Proposition 5.2. If \( \mu \) is an essential monomorphism, and \( \varphi \circ \mu \) is injective, then \( \varphi \) is injective.
Proof. If \( \varphi \circ \mu \) is injective, then \( \operatorname{Ker}\varphi \cap \operatorname{Im}\mu = 0 \) ; hence \( \operatorname{Ker}\varphi = 0 \) .
Yes
Proposition 5.3. If \( \mu : A \rightarrow B \) and \( v : B \rightarrow C \) are monomorphisms, then \( v \circ \mu \) is essential if and only if \( \mu \) and \( v \) are essential.
This follows from Proposition 5.1; the details make nifty exercises.
No
Proposition 5.4. A left \( R \) -module \( J \) is injective if and only if \( J \) has no proper essential extension \( J \subsetneqq M \), if and only if every essential monomorphism \( J \rightarrow M \) is an isomorphism.
Proof. Let \( J \) be injective. If \( J \subseteq M \), then \( J \) is a direct summand of \( M \) , \( M = J \oplus N \) ; then \( N \cap J = 0 \) ; if \( J \) is essential in \( M \), then \( N = 0 \) and \( J = M \) . If in turn \( J \) has no proper essential extension, and \( \mu : J \rightarrow M \) is an essen...
Yes
Proposition 5.5. Let \( \mu : M \rightarrow N \) and \( v : M \rightarrow J \) be monomorphisms. If \( \mu \) is essential and \( J \) is injective, then \( v = \kappa \circ \mu \) for some monomorphism \( \kappa : N \rightarrow J \) .
Proof. Since \( J \) is injective, there exists a homomorphism \( \kappa : N \rightarrow J \) such that \( v = \kappa \circ \mu \), which is injective by 5.2. \( ▱ \)
No
Theorem 5.6. Every left R-module \( M \) is an essential submodule of an injective \( R \) -module, which is unique up to isomorphism.
Proof. By Theorem 4.11, \( M \) is a submodule of an injective module \( K \) . Let \( \mathcal{S} \) be the set of all submodules \( M \subseteq S \subseteq K \) of \( K \) in which \( M \) is essential (for instance, \( M \) itself). If \( {\left( {S}_{i}\right) }_{i \in I} \) is a chain in \( \mathcal{S} \), then \(...
Yes
A ring \( R \) is left hereditary if and only if every left ideal of \( R \) is projective, and then every submodule of a free left R-module is isomorphic to a direct sum of left ideals of \( R \) .
Proof. If \( R \) is left hereditary, then every left ideal of \( R \) is projective (as an \( R \) -module), since \( {}_{R}R \) is projective. Conversely, assume that every left ideal of \( R \) is projective. We show that every submodule of a free module is isomorphic to a direct sum of left ideals of \( R \) ; then...
Yes
Lemma 6.2. (1) A module \( P \) is projective if and only if every module homomorphism \( P \rightarrow B \) factors through every epimorphism \( J \rightarrow B \) in which \( J \) is injective.
Proof. We prove (1) and let readers reverse arrows to prove (2). Assume that every module homomorphism \( P \rightarrow B \) factors through every epimorphism \( J \rightarrow B \) in which \( J \) is injective. Let \( \varphi : P \rightarrow B \) be a homomorphism and let \( \sigma : A \rightarrow B \) be any epimorph...
No
Proposition 6.3. A ring \( R \) is left hereditary if and only if every quotient of an injective left \( R \) -module is injective.
Proof. Assume that \( R \) is left hereditary. Let \( J \) be an injective module and let \( \sigma : J \rightarrow B \) be an epimorphism. To prove that \( B \) is injective we use 6.2 and show that every homomorphism \( \varphi : A \rightarrow B \) factors through every monomorphism \( \mu : A \rightarrow P \) in whi...
Yes
Proposition 6.4. A nonzero fractional ideal of a domain \( R \) is invertible if and only if it is projective as an \( R \)-module.
Proof. Let \( \mathfrak{A} \neq 0 \) be a fractional ideal of \( R \) . Let \( \mathfrak{{AB}} = R \) for some fractional ideal \( \mathfrak{B} \) . As in the proof of VII.5.2, \( 1 = {a}_{1}{b}_{1} + \cdots + {a}_{n}{b}_{n} \) for some \( {a}_{1},\ldots ,{a}_{n} \in \) \( \mathfrak{A} \) and \( {b}_{1},\ldots ,{b}_{n}...
Yes
Proposition 6.6. A domain \( R \) is a Dedekind domain if and only if every divisible \( R \) -module is injective.
Proof. Injective modules are divisible, by 4.6, and quotients of divisible modules are divisible. If divisible \( R \) -modules are injective, then quotients of injective \( R \) -module are injective, \( R \) is hereditary by 6.3, and \( R \) is Dedekind by 6.5 .\n\nConversely, let \( R \) be a Dedekind domain and let...
Yes
Proposition 1.1. If \( \varphi : A \rightarrow B \) is a homomorphism of left \( R \) -modules, then \( {\varphi }_{ * } = {\operatorname{Hom}}_{R}\left( {M,\varphi }\right) : {\operatorname{Hom}}_{R}\left( {M, A}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {M, B}\right) \) is a homomorphism of abelian groups. Mo...
(1) if \( \varphi \) is the identity on \( A \), then \( {\varphi }_{ * } \) is the identity on \( {\operatorname{Hom}}_{R}\left( {M, A}\right) \) ;\n\n(2) \( {\left( \psi \circ \varphi \right) }_{ * } = {\psi }_{ * } \circ {\varphi }_{ * } \) whenever \( \varphi : A \rightarrow B \) and \( \psi : B \rightarrow C \) ;\...
Yes
Proposition 1.3. For every \( \varphi : A \rightarrow B \) and \( \psi : M \rightarrow N \) the following square commutes:\n\n\[ \n{\operatorname{Hom}}_{R}\left( {M, A}\right) \xrightarrow[]{{\operatorname{Hom}}_{R}\left( {M,\varphi }\right) }{\operatorname{Hom}}_{R}\left( {M, B}\right) \]\n\n\[ \n{\operatorname{Hom}}_...
Proof. For every \( \alpha : N \rightarrow A,{\varphi }_{ * }\left( {{\psi }^{ * }\left( \alpha \right) }\right) = \varphi \circ \alpha \circ \psi = {\psi }^{ * }\left( {{\varphi }_{ * }\left( \alpha \right) }\right) \) . \( ▱ \)
Yes
Proposition 1.4. If \( M \) is an \( R \) -S-bimodule and \( A \) is an \( R \) -T-bimodule, then \( {\operatorname{Hom}}_{R}\left( {M, A}\right) \) is an \( S \) - \( T \) -bimodule, in which\n\n\[ \left( {s\alpha }\right) \left( x\right) = \alpha \left( {xs}\right) \text{ and }\left( {\alpha t}\right) \left( x\right)...
Proof. In the above, \( {s\alpha } \) and \( {\alpha t} \) are homomorphisms of left \( R \) -modules, since \( M \) and \( A \) are bimodules. Moreover, \( s\left( {\alpha + \beta }\right) = {s\alpha } + {s\beta } \), and\n\n\[ s\left( {{s}^{\prime }\alpha }\right) \left( x\right) = \left( {{s}^{\prime }\alpha }\right...
Yes
Proposition 1.6. If \( M \) is an \( R \) -S-bimodule and \( \varphi \) is a homomorphism of \( R \) -T-bimodules, then \( {\operatorname{Hom}}_{R}\left( {M,\varphi }\right) \) is a homomorphism of \( S \) - \( T \) -bimodules. If \( A \) is an \( R \) - \( T \) -bimodule and \( \psi \) is a homomorphism of \( R \) - \...
The proof is an exercise. In Proposition 1.6, \( {\operatorname{Hom}}_{R}\left( {M, - }\right) \) is now a functor from \( R \) - \( T \) -bimodules to \( S \) - \( T \) -bimodules, and \( {\operatorname{Hom}}_{R}\left( {-, A}\right) \) is a contravariant functor from \( R \) - \( S \) -bimodules to \( S \) - \( T \) -...
No
Proposition 2.1 (Left Exactness). If \( 0 \rightarrow A\overset{\mu }{ \rightarrow }B\overset{\rho }{ \rightarrow }C \) is exact, then\n\n\( 0 \rightarrow {\operatorname{Hom}}_{R}\left( {M, A}\right) \xrightarrow[]{{\operatorname{Hom}}_{R}\left( {M,\mu }\right) }{\operatorname{Hom}}_{R}\left( {M, B}\right) \xrightarrow...
Proof. If \( {\mu }_{ * }\left( \alpha \right) = 0 \), then \( \mu \circ \alpha = 0,\operatorname{Im}\alpha \subseteq \operatorname{Ker}\mu = 0 \), and \( \alpha = 0 \) . Similarly, \( {\rho }_{ * }\left( {{\mu }_{ * }\left( \alpha \right) }\right) = \rho \circ \mu \circ \alpha = 0 \) . Conversely, if \( {\rho }_{ * }\...
Yes
Proposition 2.5. There is an isomorphism\n\n\[ \n{\operatorname{Hom}}_{R}\left( {M,\mathop{\prod }\limits_{{i \in I}}{A}_{i}}\right) \cong \mathop{\prod }\limits_{{i \in I}}{\operatorname{Hom}}_{R}\left( {M,{A}_{i}}\right) \]\n\nwhich is natural in \( M \) and in \( {\left( {A}_{i}\right) }_{i \in I} \) .
Proof. The projections \( {\pi }_{i} : \mathop{\prod }\limits_{{i \in I}}{A}_{i} \rightarrow {A}_{i} \) induce homomorphisms \( {\operatorname{Hom}}_{R}\left( {M,{\pi }_{i}}\right) : {\operatorname{Hom}}_{R}\left( {M,\mathop{\prod }\limits_{{i \in I}}{A}_{i}}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {M,{A}_{i}...
Yes
Lemma 2.7. Let \( A \) be an abelian group and let \( M \) be a left \( R \) -module. If \( \varphi \in {\operatorname{Hom}}_{\mathbb{Z}}\left( {M, A}\right) \), then the mapping \( \xi \) that sends \( x \in M \) to \( \xi \left( x\right) : r \mapsto \) \( \varphi \left( {rx}\right) \) is a module homomorphism of \( M...
Proof. \( {\operatorname{Hom}}_{\mathbb{Z}}\left( {{R}_{R}, A}\right) \) is a left \( R \) -module, in which \( \left( {r\alpha }\right) s = \alpha \left( {sr}\right) \) for all \( r, s \in R \) . If \( x \in M \), then \( \xi \left( x\right) : r \mapsto \varphi \left( {rx}\right) \) is in \( {\operatorname{Hom}}_{\mat...
Yes
Proposition 2.8. If \( A \) is a divisible abelian group, then \( {\operatorname{Hom}}_{\mathbb{Z}}\left( {{R}_{R}, A}\right) \) is an injective left \( R \) -module.
Proof. Let \( \psi : M \rightarrow N \) be any monomorphism of left \( R \) -modules. Since \( D \) is injective as a \( \mathbb{Z} \) -module, \( {\psi }^{ * } : {\operatorname{Hom}}_{\mathbb{Z}}\left( {N, A}\right) \rightarrow {\operatorname{Hom}}_{\mathbb{Z}}\left( {M, A}\right) \) is surjective, by 2.4. Since the d...
Yes
Proposition 3.6. Let \( \mathcal{A} = \left( {A,\alpha }\right) \) be a direct system of left \( R \) -modules over a directed preordered set \( I \) . A cone \( \varphi : \mathcal{A} \rightarrow M \) is a limit cone of \( \mathcal{A} \) if and only if (i) \( M = \mathop{\bigcup }\limits_{{i \in I}}\operatorname{Im}{\v...
Proof. The limit cone \( \lambda : \mathcal{A} \rightarrow L \) constructed above, \( {\lambda }_{i}\left( x\right) = \operatorname{cls}\left( {x, i}\right) \), has properties (i) \( L = \mathop{\bigcup }\limits_{{i \in I}}{\lambda }_{i}\left( {A}_{i}\right) \) and (iii) \( {\lambda }_{i}\left( x\right) = {\lambda }_{j...
Yes
Proposition 3.7. Let \( \mathcal{A} = \left( {A,\alpha }\right) \) and \( \mathcal{B} = \left( {B,\beta }\right) \) be direct systems of left \( R \) -modules over the same directed preordered set \( I \), with limit cones \( \lambda : \mathcal{A} \rightarrow L \) and \( \mu : \mathcal{B} \rightarrow M \) . Every homom...
Proof. In the statement, \( \mu \circ \varphi : \mathcal{A} \rightarrow M \) is a cone and factors uniquely through \( \lambda \) . The last parts of the statement follow from this uniqueness. \( ▱ \)
No
Proposition 3.8. Let \( \varphi : \mathcal{A} \rightarrow \mathcal{B} \) and \( \psi : \mathcal{B} \rightarrow \mathcal{C} \) be homomorphisms of direct systems of left R-modules over the same directed preordered set I. If \( {A}_{i}\overset{{\varphi }_{i}}{ \rightarrow }{B}_{i}\overset{{\psi }_{i}}{ \rightarrow }{C}_{...
Proof. Let \( \lambda : \mathcal{A} \rightarrow L,\mu : \mathcal{B} \rightarrow M \), and \( v : \mathcal{C} \rightarrow N \) be limit cones; let \( \bar{\varphi } = \underline{\lim }{\varphi }_{i},\bar{\psi } = \underline{\lim }{\psi }_{i} \), so that the diagram below commutes for every \( i \in I \) : ![5e708ed9-3d6...
Yes
Proposition 4.3. Every inverse system of left R-modules has an inverse limit.
Proof. Let \( \mathcal{A} = \left( {A,\alpha }\right) \) be an inverse system of left \( R \) -modules over a directed preordered set \( I \) . We retrieve an inverse limit of \( \mathcal{A} \) from the direct product \( P = \mathop{\prod }\limits_{{i \in I}}{A}_{i} \) and its projections \( {\pi }_{i} : P \rightarrow ...
Yes
Proposition 4.4. Let \( \mathcal{A} = \left( {A,\alpha }\right) \) be an inverse system of left \( R \) -modules over a directed preordered set \( I \) . A cone \( \varphi : M \rightarrow \mathcal{A} \) is a limit cone of \( \mathcal{A} \) if and only if (i) \( \mathop{\bigcap }\limits_{{i \in I}}\operatorname{Ker}{\va...
Proof. The limit cone \( \lambda : L \rightarrow \mathcal{A} \) constructed in the proof of 4.3 has properties (i) and (ii): if \( x \in L \) and \( {\lambda }_{i}\left( x\right) = 0 \) for all \( i \), then \( x = 0 \) ; if \( {x}_{i} \in {A}_{i} \) and \( {\alpha }_{ij}\left( {x}_{i}\right) = {x}_{j} \) whenever \( i...
Yes
Proposition 4.6. Let \( \varphi : \mathcal{A} \rightarrow \mathcal{B} \) and \( \psi : \mathcal{B} \rightarrow \mathcal{C} \) be homomorphism of inverse systems of left \( R \) -modules over the same directed preordered set \( I \) . If \( 0 \rightarrow {A}_{i}\overset{{\varphi }_{i}}{ \rightarrow }{B}_{i}\overset{{\ps...
Proof. Let \( \lambda : \mathcal{A} \rightarrow L,\mu : \mathcal{B} \rightarrow M \), and \( v : \mathcal{C} \rightarrow N \) be limit cones; let \( \bar{\varphi } = \underline{\lim }{\varphi }_{i},\bar{\psi } = \underline{\lim }{\psi }_{i} \), so that the diagram below commutes for every \( i \in I \) :\n\n![5e708ed9-...
Yes
Proposition 5.2. Let \( R \) be a ring, let \( A \) be a right \( R \) -module, let \( B \) be a left \( R \) -module, and let \( C \) be an abelian group. For a mapping \( \beta : A \times B \rightarrow C \) the following conditions are equivalent:\n\n(1) for all \( a,{a}^{\prime } \in A, b,{b}^{\prime } \in B \), and...
Proof. Addition on \( {\operatorname{Hom}}_{\mathbb{Z}}\left( {B, C}\right) \) is pointwise and that \( r \in R \) acts on \( \varphi \in {\operatorname{Hom}}_{\mathbb{Z}}\left( {B, C}\right) \) by \( \left( {\varphi r}\right) \left( b\right) = \varphi \left( {rb}\right) \) for all \( b \in B \) ; hence (1) states that...
Yes
Proposition 5.3. For every right \( R \) -module \( A \) and left \( R \) -module \( B, A{ \otimes }_{R}B \) and its tensor map exist, and they are unique up to isomorphism.
Proof. Uniqueness follows from the universal property. Existence is proved by constructing a tensor product. Let \( T = F/K \), where \( F \) is the free abelian group on the set \( A \times B \), and \( K \) is the subgroup of \( F \) generated by all \( (a + \) \( \left. {{a}^{\prime }, b}\right) - \left( {a, b}\righ...
Yes
Every element of \( A{ \otimes }_{R}B \) is a finite sum \( \mathop{\sum }\limits_{i}\left( {{a}_{i} \otimes {b}_{i}}\right) \), where \( {a}_{i} \in A \) and \( {b}_{i} \in B \) .
Let \( T = F/K \) as above. Every element of \( F \) is a finite linear combination \( \mathop{\sum }\limits_{i}{n}_{i}\left( {{a}_{i},{b}_{i}}\right) \) with integer coefficients. Hence every element of \( T = \pi \left( F\right) \) is a finite linear combination \( \mathop{\sum }\limits_{i}{n}_{i}\left( {{a}_{i} \oti...
Yes
Proposition 5.5. If \( \varphi : A \rightarrow {A}^{\prime } \) is a homomorphism of right \( R \) -modules, and \( \psi : B \rightarrow {B}^{\prime } \) is a homomorphism of left \( R \) -modules, then there is a unique homomorphism \( \varphi \otimes \psi : A{ \otimes }_{R}B \rightarrow {A}^{\prime }{ \otimes }_{R}{B...
Proof. Let \( \tau : A \times B \rightarrow A{ \otimes }_{R}B \) and \( {\tau }^{\prime } : {A}^{\prime } \times {B}^{\prime } \rightarrow {A}^{\prime }{ \otimes }_{R}{B}^{\prime } \) be the tensor maps. Since \( {\tau }^{\prime } \) is a bihomomorphism, \( {\tau }^{\prime } \circ \left( {\varphi \times \psi }\right) :...
Yes
Proposition 5.7. Let \( A \) be a right \( R \) -module and let \( B \) be a left \( R \) -module.\n\n(1) if \( A \) is an \( S \) - \( R \) -bimodule, then \( A{ \otimes }_{R}B \) is a left \( S \) -module, in which \( s\left( {a \otimes b}\right) = \) \( {sa} \otimes b \) for all \( a \in A, b \in B \), and \( s \in ...
Proof. We prove (1), (2), and (3), and leave the other parts to eager readers. Let \( A \) be an \( S \) - \( R \) -bimodule. If \( s \in S \), then \( {\alpha }_{s} : a \mapsto {sa} \) is a right \( R \) -module endomorphism of \( A \) . By 5.5, there is a unique endomorphism \( {\bar{\alpha }}_{s} = {\alpha }_{s} \ot...
No
Proposition 5.10. Let \( M \) be a left \( R \) -module and let \( \rho : R \rightarrow S \) be a homomorphism of rings [with identity].\n\n(1) \( S{ \otimes }_{R}M \) is a left \( S \) -module, and \( \iota : x \mapsto 1 \otimes x \) is a homomorphism of left \( R \) -modules of \( M \) into \( S{ \otimes }_{R}M \) ;
Proof. (1). \( S \) is an \( S \) - \( R \) -bimodule, in which \( s \cdot r = {s\rho }\left( r\right) \) for all \( s \in S \) and \( r \in R \) . By 5.7, \( S{ \otimes }_{R}M \) is a left \( S \) -module. Hence \( S{ \otimes }_{R}M \) is also a left \( R \) -module, in which \( r\left( {s \otimes x}\right) = \rho \le...
Yes
Proposition 6.1. For every right \( R \) -module \( A \) and left \( R \) -module \( B \) there is a commutativity isomorphism \( B{ \otimes }_{{R}^{\text{op }}}A \cong A{ \otimes }_{R}B \), which sends \( b \otimes a \) to \( a \otimes b \) and is natural in \( A \) and \( B \) . If \( A \) and \( B \) are bimodules, ...
Proof. If \( C \) is an abelian group, then \( \beta \) is a bihomomorphism of \( {A}_{R} \times {}_{R}B \) into \( C \) if and only if \( {\beta }^{\mathrm{{op}}} : \left( {b, a}\right) \mapsto \beta \left( {a, b}\right) \) is a bihomomorphism of \( {B}_{{R}^{\mathrm{{op}}}} \times {}_{{R}^{\mathrm{{op}}}}A \) into \(...
No
For every right R-module \( A, R \) -S-bimodule \( B \), and left \( S \) -module \( C \), there is an associativity isomorphism \( \left( {A{ \otimes }_{R}B}\right) { \otimes }_{S}C \cong A{ \otimes }_{R}\left( {B{ \otimes }_{S}C}\right) \) , which sends \( \left( {a \otimes b}\right) \otimes c \) to \( a \otimes \lef...
We prove the bimodule case and let \( A \) be a \( Q - R \) -bimodule and \( C \) be an \( S \) - \( T \) -bimodule, so that \( \left( {A{ \otimes }_{R}B}\right) { \otimes }_{S}C \) and \( A{ \otimes }_{R}\left( {B{ \otimes }_{S}C}\right) \) are \( Q \) - \( T \) -bimodules; the first part of the statement is the case ...
Yes
Proposition 6.5. For every \( Q \) -R-bimodule \( A, R \) -S-bimodule \( B \), and \( S \) - \( T \) -bimodule \( C \), a tensor product \( A{ \otimes }_{R}B{ \otimes }_{S}C \) and its tensor map exist, and they are unique up to isomorphism.
To Proposition 6.2 can be added natural isomorphisms \( \left( {A{ \otimes }_{R}B}\right) { \otimes }_{S}C \cong \) \( A{ \otimes }_{R}B{ \otimes }_{S}C \cong A{ \otimes }_{R}\left( {B{ \otimes }_{S}C}\right) \), which send \( \left( {a \otimes b}\right) \otimes c \) to \( a \otimes b \otimes c \) and to \( a \otimes \...
No
Proposition 6.6 (Adjoint Associativity). Let \( A \) be a right \( R \) -module, let \( B \) be a left \( R \) -module, and let \( C \) be an abelian group. There are adjoint associativity isomorphisms\n\n\[ \Theta : {\operatorname{Hom}}_{\mathbb{Z}}\left( {A{ \otimes }_{R}B, C}\right) \cong {\operatorname{Hom}}_{R}\le...
Proof. The set Bihom \( \left( {A \times B, C}\right) \) of all bihomomorphisms of \( A \times B \) into \( C \) is an abelian group under pointwise addition. The universal property of the tensor map \( \tau : \left( {a, b}\right) \mapsto a \otimes b \) provides a bijection \( \varphi \mapsto \varphi \circ \tau \) of \...
Yes
Proposition 6.7 (Right Exactness). For every right R-module \( M \) and left \( R \)-module \( N \), the functors \( M{ \otimes }_{R} - \) and \( - { \otimes }_{R}N \) are right exact: if \( A \rightarrow B \rightarrow C \rightarrow 0 \) is exact, then so is\n\n\[ M{ \otimes }_{R}A \rightarrow M{ \otimes }_{R}B \righta...
Proof. We prove the first half of the statement; then 6.1 yields the second half. Let \( A\overset{\varphi }{ \rightarrow }B\overset{\psi }{ \rightarrow }C \rightarrow 0 \) be exact; let \( \bar{\varphi } = {1}_{M} \otimes \varphi \) and \( \bar{\psi } = {1}_{M} \otimes \psi \).\n\n![5e708ed9-3d6d-4f59-a748-eaac13dfd78...
Yes
Proposition 6.9. There are natural isomorphisms \( M{ \otimes }_{R}\left( {{\bigoplus }_{i \in I}{B}_{i}}\right) \cong \) \( {\bigoplus }_{i \in I}\left( {M{ \otimes }_{R}{B}_{i}}\right) \) and \( \left( {{\bigoplus }_{i \in I}{A}_{i}}\right) { \otimes }_{R}N \cong {\bigoplus }_{i \in I}\left( {{A}_{i}{ \otimes }_{R}N}...
Proof. Let \( M \) be a right \( R \) -module and let \( {\left( {A}_{i}\right) }_{i \in I} \) be a family of left \( R \) -modules. The injection \( {\iota }_{i} : {A}_{i} \rightarrow {\bigoplus }_{i \in I}{A}_{i} \) induces a homomorphism \( {\bar{\iota }}_{i} = {1}_{M} \otimes {\iota }_{i} : M{ \otimes }_{R}{A}_{i} ...
Yes
Proposition 7.1. If \( F \) is a free left (or right) \( R \) -module with a finite basis \( {\left( {e}_{i}\right) }_{i \in I} \) , then \( {F}^{ * } \) is free with a finite basis \( {\left( {e}_{i}^{ * }\right) }_{i \in I} \), such that \( {e}_{i}^{ * }\left( {e}_{i}\right) = 1,{e}_{i}^{ * }\left( {e}_{j}\right) = 0...
Then \( {\left( {e}_{i}^{ * }\right) }_{i \in I} \) is the dual basis of the given basis \( {\left( {e}_{i}\right) }_{i \in I} \) . Proposition 7.1 does not extend to all free modules: for instance, \( {\left( {\bigoplus }_{i \in I}\mathbb{Z}\right) }^{ * } \cong \mathop{\prod }\limits_{{i \in I}}\mathbb{Z} \) is not f...
No
Proposition 7.4. If \( M \) is a finitely generated projective module, then the evaluation homomorphism \( M \rightarrow {M}^{* * } \) is an isomorphism.
Proof. If \( F \) is free, with a basis \( {\left( {e}_{i}\right) }_{i \in I} \), then applying 7.1 twice yields a basis \( {\left( {e}_{i}^{* * }\right) }_{i \in I} \) of \( {F}^{* * } \) such that \( {e}_{i}^{* * }\left( {e}_{i}^{ * }\right) = 1,{e}_{i}^{* * }\left( {e}_{j}^{ * }\right) = 0 \) for all \( j \neq i \) ...
Yes
Proposition 7.5. Let \( A \) and \( B \) be left \( R \) -modules. There is a homomorphism \( \zeta : {A}^{ * }{ \otimes }_{R}B \rightarrow {\operatorname{Hom}}_{R}\left( {A, B}\right) \), which is natural in \( A \) and \( B \), such that \( \zeta \left( {\alpha \otimes b}\right) \left( a\right) = \alpha \left( a\righ...
Proof. For every \( \alpha \in {A}^{ * } \) and \( b \in B,\beta \left( {\alpha, b}\right) : a \mapsto \alpha \left( a\right) b \) is a module homomorphism of \( A \) into \( B \) . We see that \( \beta : {A}^{ * } \times B \rightarrow {\operatorname{Hom}}_{R}\left( {A, B}\right) \) is a bihomomorphism. Hence \( \beta ...
No
Corollary 7.6. Let \( A \) be a finitely generated projective right \( R \) -module. There is an isomorphism \( A{ \otimes }_{R}B \cong {\operatorname{Hom}}_{R}\left( {{A}^{ * }, B}\right) \), which is natural in \( A \) and \( B \) .
Proof. By 7.4,7.5, \( A{ \otimes }_{R}B \cong {A}^{* * }{ \otimes }_{R}B \cong {\operatorname{Hom}}_{R}\left( {{A}^{ * }, B}\right) \).
No
Corollary 7.7. Let \( R \) be commutative and let \( A, B \) be finitely generated projective \( R \) -modules. There is an isomorphism \( {A}^{ * }{ \otimes }_{R}{B}^{ * } \cong {\left( A{ \otimes }_{R}B\right) }^{ * } \), which is natural in \( A \) and \( B \) .
Proof. By 7.5,6.6, \( {A}^{ * }{ \otimes }_{R}{B}^{ * } \cong {\operatorname{Hom}}_{R}\left( {A,{B}^{ * }}\right) = {\operatorname{Hom}}_{R}\left( {A,{\operatorname{Hom}}_{R}\left( {B, R}\right) }\right) \) \( \cong {\operatorname{Hom}}_{R}\left( {A{ \otimes }_{R}B, R}\right) = {\left( A{ \otimes }_{R}B\right) }^{ * }....
Yes
Proposition 8.1. Every projective module is flat.
Proof. Readers will verify that free modules are flat. Now, let \( P \) be a projective left \( R \) -module. There exist a free left \( R \) -module \( F \) and homomorphisms \( \pi : F \rightarrow \) \( P,\iota : P \rightarrow F \) such that \( \pi \circ \iota = {1}_{P} \) . Every monomorphism \( \mu : A \rightarrow ...
No
Proposition 8.6. An abelian group is flat (as a \( \mathbb{Z} \) -module) if and only if it is torsion-free.
Proof. Finitely generated torsion-free abelian groups are free, and are flat by 8.1. Now every torsion-free abelian group is the direct limit of its finitely generated subgroups, which are also torsion-free, and is flat by 8.4.\n\nOn the other hand, finite cyclic groups are not flat: if \( C \) is cyclic of order \( m ...
Yes
Proposition 8.8. Every module is a direct limit of finitely presented modules.
Proof. Let \( M \cong F/K \), where \( F \) is free with a basis \( X \) . We show that \( M \) is the direct limit of the finitely presented modules \( {F}_{Y}/S \), where \( {F}_{Y} \) is the free submodule of \( F \) generated by a finite subset \( Y \) of \( X \) and \( S \) is a finitely generated submodule of \( ...
Yes
Corollary 8.10. Every finitely presented flat module is projective.
Proof. The identity on such a module factors through a free module.
No
Theorem 8.11 (Lazard [1969]). For a left R-module \( M \) the following conditions are equivalent:\n\n(1) \( M \) is flat;\n\n(2) every homomorphism of a finitely presented free module into \( M \) factors through a finitely generated free module;\n\n(3) \( M \) is a direct limit of finitely generated free modules;\n\n...
Proof. (3) implies (4); (4) implies (1), by 8.1 and 8.4; (1) implies (2), by 8.9. Now assume (2). Let \( \pi : F \rightarrow M \) be an epimorphism, where \( F \) is free with a basis \( X \) . Choose \( \pi : F \rightarrow M \) so that there are for every \( m \in M \) infinitely many \( x \in X \) such that \( \pi \l...
Yes
Lemma 9.1. Let \( \mathfrak{a} \) be an ideal of a commutative ring \( R \), let \( M \) be an \( R \) -module, and let \( {M}_{1} \supseteq {M}_{2} \supseteq \cdots \) be an \( \mathfrak{a} \) -filtration on \( M \). (1) \( {R}^{ + } = R \oplus \mathfrak{a} \oplus {\mathfrak{a}}^{2} \oplus \cdots \) is a ring (the blo...
Proof. (1). The elements of \( {R}^{ + } \) are infinite sequences \( a = \left( {{a}_{0},{a}_{1},\ldots ,{a}_{i}}\right. \) , \( \ldots ) \) in which \( {a}_{0} \in R,{a}_{i} \in {\mathfrak{a}}^{i} \) for all \( i > 0 \), and \( {a}_{i} = 0 \) for almost all \( i \) . Addition in \( {R}^{ + } \) is componentwise; mult...
Yes
Lemma 9.2 (Artin-Rees). Let \( \mathfrak{a} \) be an ideal of a commutative Noetherian ring \( R \), and let \( M \) be a finitely generated \( R \) -module. If \( {M}_{1} \supseteq {M}_{2} \supseteq \cdots \) is an \( \mathfrak{a} \) -stable filtration on \( M \), then \( N \cap {M}_{1} \supseteq N \cap {M}_{2} \supse...
Proof. First, \( N \cap {M}_{1} \supseteq N \cap {M}_{2} \supseteq \cdots \) is an \( \mathfrak{a} \) -filtration on \( N \) ; \( M \) is a Noetherian \( R \) -module, by VIII.8.5; \( \bar{N} \) and all \( {M}_{i}, N \cap {M}_{i} \) are finitely generated; and \( {N}^{ + } \) is an \( {R}^{ + } \) -submodule of \( {M}^...
Yes