## 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.