Algebra Qual Study Problems: 6

Theorem: The following are equivalent on an R-module P-
(i) 0 \rightarrow A \rightarrow B \rightarrow P\rightarrow 0 is split exact.
(ii) P is a direct summand of a free module
(iii) P is projective.
Proof-
( 3\Rightarrow 1) Assume that P is projective and let 0\rightarrow A \xrightarrow{f} B \xrightarrow{g} P\rightarrow 0 be a short exact sequence. By the exactness we have that the map g is surjective, and for P we have the identity map from P to P. Since P is projective there exists a map h:P\rightarrow B such that gh=1_P. By one of the equivalent properties we know that this gives that the sequence splits.
(1\Rightarrow 2) Assume that every short exact sequence 0 \rightarrow A \rightarrow B \rightarrow P\rightarrow 0 splits. Every module is the homomorphic image of a free module, so there exists some free module F and some surjection f: F\rightarrow P. Then we can construct a short exact sequence, 0 \rightarrow Ker(f) \xrightarrow{i} F \xrightarrow{f} P\rightarrow 0 where the map i:Ker(f) \rightarrow F is the inclusion map. Then by assumption the sequence splits so that F = Ker(f) \bigoplus P.
(2\Rightarrow 3) Suppose that P is a direct summand of a free module F so that F \cong K \bigoplus P for some R module K. Then we have the canonical projection and inclusion maps, \pi:F \rightarrow P and i: P\rightarrow F. Let g:B\rightarrow C \rightarrow 0 and \phi :P\rightarrow C. Since F is free and any free module is projective, there exists a map \psi :F \rightarrow B such that g\psi = \phi \pi. Then consider \psi i :P \rightarrow B. We have that g ( \psi i ) = (g \psi ) i = (\phi \pi ) i = \phi (\pi i ) = \phi 1_P = \phi. Therefore the diagram commutes so that P is projective.

2009 Qual #3: Let R be a ring. Let M be a finitely generated R-module and let N be a projective R -module. Prove that if f:M\rightarrow N is a surjective homomorphism, then the Ker(f) is finitely generated.
Proof-
First consider the sequence 0\rightarrow Ker(f) \xrightarrow{i} M \xrightarrow{f} N \rightarrow 0. This is an exact sequence because f is surjective and i :Ker(f) \rightarrow M is the canonical inclusion map which is injective. Since N is projective we know that the sequence splits so that M \cong Ker(f) \bigoplus N. Since M is finitely generated it must follow then that Ker(f) is also finitely generated.

2010 Qual #7: Suppose that P_1,P_2 are R-modules. Prove that P_1 \bigoplus P_2 is projective iff P_1,P_2 are projective.
Proof-
(\Rightarrow ) Suppose that P_1 \bigoplus P_2 is projective. Then there eixsts a free module F and an R module K such that F = K\bigoplus (P_1 \bigoplus P_2) = (K \bigoplus P_1 )\bigoplus P_2 = (K \bigoplus P_2 )\bigoplus P_1. Then by the same reason we have that P_1, P_2 are projective.
(\Leftarrow ) Suppose that P_1,P_2 are projective. Then there exists free modules F_1,F_2 and R-modules K_1,K_2 such that F_1 = P_1 \bigoplus K_1 and F_2 = P_2\bigoplus K_2. Then we have that F_1\bigoplus F_2 =  (P_1 \bigoplus K_1)\bigoplus (P_2\bigoplus K_2) = (K_1 \bigoplus K_2)\bigoplus (P_1\bigoplus P_2). Since the direct sum of two free modules is still free this gives that P_1 \bigoplus P_2 is projective.

Lemma 4.3.9(general case): A R-module D is a divisible R module iff D is an injective R-module.
Proof-
(\Leftarrow ). Let D be an injective R-module. Take y\in D and r\in R\backslash \{0_R\}. We want to find some x\in D such that y = rx. Consider the principle ideal (r) = rR. Then we have 0\rightarrow (r) \xrightarrow{i}R is an exact sequence where i:(r) \rightarrow R is the inclusion map. Define f:(r) \rightarrow D by f(r) = y. Then since D is injective there exists a map g:R\rightarrow D such that gi=f. So we have that g(i(r)) = f(r) \Rightarrow g(r) = rg(1) = f(r) = y. Therefore we choose x= g(1) and we are done.

2005 Qual #4: Decide whether they are injective \mathbb{Z}-modules.
a) \mathbb{Q}
Yes because it is a divisible \mathbb{Z}-module
b) \mathbb{Z}_{12}
No, it is not divisible because 2x\neq 1 \forall x\in \mathbb{Z} \backslash \{0\}, since 2 is not a unit. Thus it is also not injective.

Final #5b (Hungerford 4.4.3): Let \pi : \mathbb{Z} \rightarrow \mathbb{Z}_2 be the canonical epimorphism. The induced map \bar{\pi} : Hom(\mathbb{Z}_2 ,\mathbb{Z} ) \rightarrow Hom(\mathbb{Z}_2 ,\mathbb{Z}_2 ) is the zero map. Since Hom(\mathbb{Z}_2 ,\mathbb{Z}_2 ) \neq 0 we have that \bar{\pi} is not an epimorphism.
(Give an example of a commutative ring R, R-modules M, M’, N and f\in Hom_R(M,M') such that f is surjective, but \bar{f} : Hom_R(N,M) \rightarrow Hom_R(N,M') is not surjective.)
Proof-
Let f \in Hom(\mathbb{Z}_2 ,\mathbb{Z} ). Suppose that f(1) = k \neq 0. Then 0 = f(0) = f(1+1) = f(1) + f(1) = 2k \Rightarrow k =0. This is a contradiction, so that f must be the zero map and so Hom(\mathbb{Z}_2 ,\mathbb{Z}) = 0. Thus \bar{\pi} : Hom(\mathbb{Z}_2 ,\mathbb{Z} ) = 0 \rightarrow Hom(\mathbb{Z}_2 ,\mathbb{Z}_2 ) must be the zero map. However we have that Hom(\mathbb{Z}_2 ,\mathbb{Z}_2 ) \neq 0 because the 1_{\mathbb{Z}_2} \in Hom(\mathbb{Z}_2 ,\mathbb{Z}_2 ). Thus \bar{\pi} cannot be an epimorphism.

Final #6 (Hungerford 4.4.11(a)): Let P be a finitely generated projective left module. Prove that P* is a projective right R module.
Proof-
Let P be a finitely generated projective left R module. Then there exists some free module left module F and an epimorphism f: F \rightarrow P since every module is the homomorphic image of a free R module. Since P is finitely generated, F can be chosen to be finitely generated. Also, there is some left R module K such that F = P \bigoplus K, since P is projective. Then we get that F* = (P\bigoplus K )* \cong P* \bigoplus K*. Since F is a free left module which is finitely generated we get that F* is a free right R module. Thus P* is a direct summand of a free module and so is projective.

Final #5a (Hungerford 4.5.7): The usual injection map f:\mathbb{Z}_2 \rightarrow \mathbb{Z}_4 is a monomorphism of abelian groups. Show that 1\otimes f : \mathbb{Z}_2 \bigotimes \mathbb{Z}_2 \rightarrow \mathbb{Z}_2 \bigotimes \mathbb{Z}_4 is the zero map.
Proof-
Consider what the map does to each generator: 0\otimes 0\mapsto 0 and 1\otimes 1 \mapsto 1\otimes 2 =1 \cdot 2 \otimes 1 = 2 \otimes 1 = 0 \otimes 1 = 0. Thus it is indeed the zero map. Note that \mathbb{Z}_2 \bigotimes \mathbb{Z}_2 \cong \mathbb{Z}_2 and \mathbb{Z}_2 \bigotimes \mathbb{Z}_4 \cong \mathbb{Z}_2 by exercise 2b. Thus we have the zero map between two nonzero spaces so that it is not a monomorphism. This also tells us that tensoring two injective maps need not result in an injective map.

Qual 2013 #3a, Final #4: a) Give an example of a divisible \mathbb{Z}-module. Prove that if N is a torsion right R module, M is a divisible left R module, then N \bigotimes_R M = 0.
Proof-
Consider the element n\otimes m \in N\bigotimes M. Since N is torsion there is some r\in R such that n \cdot r = 0. M is divisible so there exists some m'\in M such that m = r m'. So then n\otimes m = n \otimes rm' = nr \otimes m' = 0 \otimes m' = 0. Thus an arbitrary generator is zero so that N\bigotimes_R = 0.

b) Let R be a commutative ring and let S be a multiplicative set. Prove that if the natural homomorphism of R-modules, f: M\rightarrow S^{-1} M by m\mapsto \frac{sm}{s} is an isomorphism, then M is divisible with respect to S. Conversely if M is divisible with respect to S then the map f is surjective.
Proof-
Suppose that f is an isomorphism. Let m\in M and s\in S. Since f is an isomorphism we have that f^{-1} is an isomorphism. m =  f^{-1}(\frac{sm}{s} )= sf^{-1} ( \frac{m}{s}) and f^{-1}(\frac{m}{s} ) \in M so that M is divisible with respect to S.
Now suppose that M is divisible with respect to S. Let \frac{m}{s} \in S^{-1}M, then there exists some m'\in M such that m = sm'. So then \frac{m}{s} = \frac{sm'}{s} = f(m'). Thus f is surjective.

Hungerford 4.5.3b: \mathbb{Q} \bigotimes \mathbb{Q} \cong \mathbb{Q}.
Proof-
Consider the short exact sequence 0\rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q} / \mathbb{Z} \rightarrow 0. Since \mathbb{Q} is flat we have that 0 \rightarrow \mathbb{Q} \bigotimes \mathbb{Z} \rightarrow \mathbb{Q} \bigotimes \mathbb{Q} \rightarrow \mathbb{Q} \bigotimes \mathbb{Q} / \mathbb{Z} \rightarrow 0 is also exact. Then since \mathbb{Q} / \mathbb{Z} is torsion and \mathbb{Q} is a divisible \mathbb{Z}- module we have that \mathbb{Q} \bigotimes \mathbb{Q} / \mathbb{Z} = 0. Thus \mathbb{Q} \bigotimes \mathbb{Z} \cong \mathbb{Q} \bigotimes \mathbb{Q}. And since \mathbb{Q} \cong \mathbb{Q} \bigotimes \mathbb{Z} we are done.

2002 Qual #4(ii): Prove or disprove, A \bigotimes B \cong A \bigotimes C \Rightarrow B\cong C.
Proof-
False \mathbb{Q} \cong \mathbb{Q} \bigotimes \mathbb{Z} \cong \mathbb{Q} \bigotimes \mathbb{Q}, but obviously the rationals are not isomorphic to the integers.

2013 Qual #5: R a ring, f\in Hom_R(M,M') and g\in Hom_R(N,N'). Is it always true that Ker(f\otimes g) = M \bigotimes Ker(g) + Ker(f) \bigotimes N.
Proof-
False, consider the maps 1:\mathbb{Z}_2 \rightarrow \mathbb{Z}_2 and f:\mathbb{Z}_2 \rightarrow  \mathbb{Z}_4. They are both injective however 1\otimes f is not injective so that Ker(1\otimes f) \neq 0 but M \bigotimes Ker(g) + Ker(f) \bigotimes N = M \bigotimes 0 + 0 \bigotimes N = 0.

Qual 2011 #3: Let R be a ring and let M, N be left R-modules. Construct a natural homomorphism of abelian groups M* \bigotimes_R N \rightarrow Hom_R(M,N). If R = K is a field and M, N are finite dimensional K vector spaces prove that the natural homomorphism is in fact an isomorphism of K vector spaces.
Proof-
Define the map \phi: M* \bigotimes_R N \rightarrow Hom_R(M,N) by sending the element f\otimes n \mapsto \psi where \psi (m) = f(m) n. Check that this is a homomorphism of abelian groups.
Then, assuming that R = K is a field we show that \phi is an isomorphism. First,
dim(M* \bigotimes N ) = dim(M*) dim N by Corollary 4.5.12
= dim(M) dim(N) by Thm 4.4.11(ii)
= dim(Hom_R(M,N)).
Thus it is sufficient to show that \phi is surjective by previous exercise. To show that \phi is surjective, let \{m_i\}_1^s be a basis of M and \{n_j\}_1^t be a basis of N. By theorem 4.4.11, for each m_i \in \{m_i\}_1^s we have that f_{m_i}: M \rightarrow R given by f_{m_i}(y) = \delta_{m_iy}. Then these maps form a basis for M* since M is finitely generated. Now, Hom_R(M,N) has a basis \{g_{ij}\}_{i = 1, j = 1}^{s,t} where g_{ij}(m_k) =\delta_{ik} n_j with g_{ij} sending m_i to n_j and all other m_k to zero. Then by definition of \phi we have \phi (f_{m_i} \otimes n_j) (m_k )= f_{m_i}(m_k) n_j = \delta_{m_i,m_k} n_j = \delta_{i,k} n_j. This tells us that \phi (f_{m_i} \otimes n_j) = g_{ij}. So every basis element g_{ij} \in Hom_R(M,N) gets hit by \phi (f_{m_i} \otimes n_j) \in M*\bigotimes N through \phi. Therefore \phi is onto so we are done.

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