HW3

1. Let $M,N,M_i(i\in I)$ and $N_j(j\in J)$ be $R$-modules. Prove the following isomorphisms of abelian groups.

• (1) ${\mathrm{Hom}}_R\left({⨁}_{i\in I}{M}_i,N\right)\simeq {\prod }_{i\in I}{\mathrm{Hom}}_R\left(M_i,N\right)$;
• (2) ${\mathrm{Hom}}_R\left(M,{\prod }_{j\in J}{N}_j\right)\simeq {\prod }_{j\in J}{\mathrm{Hom}}_R\left(M,N_j\right)$.

\begin{proof} (1) Define

$$\begin{array}{rl}\phi :{\mathrm{Hom}}_R\left(\underset{i\in I}{⨁}{M}_i,N\right)& \to \prod _{i\in I}{\mathrm{Hom}}_R\left(M_i,N\right),\\ f_0& \mapsto (f_0{|}_{M_i}{)}_{i\in I}.\end{array}$$

It is easy to verify $\phi$ is a homomorphism of abelian groups, and $\phi$ is injective. For any $(f_i{)}_{i\in I}\in {\prod }_{i\in I}{\mathrm{Hom}}_R\left(M_i,N\right)$, define $f_0:{\oplus }_{i\in I}{M}_i\to N$ as $f_0((m_i{)}_{i\in I})={\sum }_{i\in I}{f}_i(m_i)$, then $\phi (f_0)=(f_i{)}_{i\in I}$ and so $\phi$ is surjective. Remark that ${\sum }_{i\in I}{f}_i(m_i)$ is a finite sum of elements in $N$ and so it is well-defined. Therefore, $\phi$ is an isomorphism and so ${\mathrm{Hom}}_R\left({⨁}_{i\in I}{M}_i,N\right)\simeq {\prod }_{i\in I}{\mathrm{Hom}}_R\left(M_i,N\right)$.

(2) Define

\Psi:\operatorname{Hom}_R\left(M, \prod_{j\in J}N_j\right) &\to \prod_{j \in J} \operatorname{Hom}_R\left(M, N_j\right),\\ g &\mapsto \Psi(g):m\mapsto (g(m)|_{N_j})_{j\in J} \end{aligned}

and notice that $\Psi$ is a homomorphism and is injective. For any $({\ell }_j{)}_{j\in J}\in {\prod }_{j\in J}{\mathrm{Hom}}_R(M,N_j)$, define $\ell \in {\mathrm{Hom}}_R(M,{\prod }_{j\in J}{N}_j)$ as $\ell (m)=({\ell }_j(m){)}_{j\in J}\in {\prod }_{j\in J}{N}_j$. Then $\Psi (\ell )=({\ell }_j{)}_{j\in J}$ and so $\psi$ is an isomorphism. \end{proof}

2. (Five Lemma) Suppose that the following diagram of homomorphisms of $R$-modules

is commutative and has exact rows. Prove that

• (1) if ${\phi }_1$ is surjective, and ${\phi }_2$ and ${\phi }_4$ are injective, then ${\phi }_3$ is injective;
• (2) if ${\phi }_5$ is injective, and ${\phi }_2$ and ${\phi }_4$ are surjective, then ${\phi }_3$ is surjective.

\begin{proof} (1) Assume that $m\in M_3$ such that ${\phi }_3(m_3)=0$, then ${\phi }_4(f_3(m))=g_3({\phi }_3(m))=0$ and $f_3(m)=0$. It deduces that $m\in \ker f_3=\mathrm{im}{f}_2$. There exists $n\in M_2$ such that $f_2(n)=m$. Since $g_2({\phi }_2(n))=0$, there is ${\phi }_2(n)\in \ker g_2=\mathrm{im}{g}_1$ and so we can find $\ell \in N_1$ such that $g_1(\ell )={\phi }_2(n)$. As ${\phi }_1$ is surjective, there is ${\ell }_0\in M_1$ such that ${\phi }_1({\ell }_0)=\ell$. Remark that ${\phi }_2(f_1({\ell }_0))={\phi }_2(n)$ yields that $f_1({\ell }_0)=n$ by ${\phi }_2$ injective. Then $n\in \mathrm{im}{f}_1=\ker f_2$ and so $f_2(n)=m=0$. Therefore, ${\phi }_3$ is injective.

(2) For any $n\in N_3$, we aim to find $m\in M_3$ such that ${\phi }_3(m)=n$. Since ${\phi }_4$ is surjective, there is $m_4\in M_4$ such that ${\phi }_4(m_4)=g_3(n)$. Since $g_4(g_3(n))=0$, we have ${\phi }_5\circ f_4(m_4)=g_4\circ {\phi }_4(m_4)=0$. Then $f_4(m_4)=0$ by ${\phi }_5$ injective, and so $m_4\in \ker f_4=\mathrm{im}{f}_3$. Take $m_3\in M_3$ such that $f_3(m_3)=m_4$, then $g_3(n)={\phi }_4(m_4)={\phi }_4\circ f_3(m_3)=g_3\circ {\phi }_3(m_3)$ and so ${\phi }_3(m_3)-n\in \ker g_3=\mathrm{im}{g}_2$. Let $n_2\in N_2$ be an element such that $g_2(n_2)={\phi }_3(m_3)-n$, and let $m_2$ be an element such that ${\phi }_2(m_2)=n_2$. Then ${\phi }_3\circ f_2(m_2)=g_2(n_2)={\phi }_3(m_3)-n$ and so $n={\phi }_3(m_3-f_2(m_2))$. That is, $m_3-f_2(m_2)$ is what we desired. Now we finish the proof. \end{proof}

3. (Snake Lemma) Suppose that the following diagram of homomorphisms of $R$ modules

is commutative and has exact rows. Prove that

$$0\to \ker {\phi }^{\prime }\stackrel{{\alpha }^{\prime }}{\to }\ker \phi \stackrel{{\beta }^{\prime }}{\to }\ker {\phi }^{\prime \prime }\stackrel{\delta }{\to }\mathrm{coker}{\phi }^{\prime }\stackrel{{\mu }^{\prime }}{\to }\mathrm{coker}\phi \stackrel{{\nu }^{\prime }}{\to }\mathrm{coker}{\phi }^{\prime \prime }\to 0$$

is a long exact sequence, where ${\alpha }^{\prime },{\beta }^{\prime },{\mu }^{\prime },{\nu }^{\prime }$ are homomorphisms induced by $\alpha ,\beta ,\mu ,\nu$, respectively. (Hint: for any $x\in \ker {\phi }^{\prime \prime }$, take $y\in M$ such that $\beta (y)=x$, and take $z\in N^{\prime }$ such that $\mu (z)=\phi (y)$. Define $\delta (x)=\bar{z}=z+\mathrm{im}{\phi }^{\prime })$

\begin{proof} We redefine the notations as follows.

Notice that $\ker f^{\prime }\to \ker f\to \ker f^{\prime \prime }$ are restriction of $M^{\prime }\to M\to M^{\prime \prime }$. We can check that $g$ and $h$ induce maps $\mathrm{coker}{f}^{\prime }\to \mathrm{coker}f$ and $\mathrm{coker}f\to \mathrm{coker}{f}^{\prime \prime }$. So it suffices to construct $d:\ker f^{\prime \prime }\to \mathrm{coker}{f}^{\prime }$. For any $x\in M^{\prime \prime }$, take $\hat{x}\in M$ and so $f(\hat{x})\in \ker h$. Then define $\check{x}\in N^{\prime }$ such that $g(\check{x})=\hat{x}$ and $\overline{\check{x}}\in \mathrm{coker}{f}^{\prime }$ is what we desire. It is easy to check this map is well-defined.

It remains to check the sequence is exact. For convenience, define

$$0\to \ker f^{\prime }\stackrel{G^{\prime }}{\to }\ker f\stackrel{H^{\prime }}{\to }\ker f^{\prime \prime }\stackrel{d}{\to }\mathrm{coker}{f}^{\prime }\stackrel{g^{\prime }}{\to }\mathrm{coker}f\stackrel{h^{\prime }}{\to }\mathrm{coker}{f}^{\prime \prime }\to 0.$$

It is easy to verify the sequence is exact at $\ker f^{\prime },\ker f,\mathrm{coker}f$ and $\mathrm{coker}{f}^{\prime \prime }$ by their definition. Now we verify the sequence is exact at the two places $\ker f^{\prime \prime }$ and $\mathrm{coker}{f}^{\prime }$. For any $x\in \ker d⩽M^{\prime \prime }$, there exists $y\in M$ such that $H(y)=x$. By the definition of $d$, we know $f(y)=g(d(x))=0$ and so $y\in \ker f$. It deduces that $x\in \mathrm{im}(H^{\prime })$. Conversely, for any $x\in \mathrm{im}(H^{\prime })$, there exists $y\in \ker f$ such that $x=H^{\prime }(y)$. By the definition of $d$, there is $g(d(x))=f(y)=0$ and $d(x)=0$. Thus $x\in \ker d$. Now we have proved $\ker d=\mathrm{im}{H}^{\prime }$.

For any $x\in \ker g^{\prime }$, there is $g^{\prime }(x)=0$ and so $g(\hat{x})\in \mathrm{im}f$ for any given preimage $\hat{x}\in N^{\prime }$. Then there is $y\in M$ such that $f(y)=g(\hat{x})$. By the definition of $d$, we know $d(H(y))=x$. In addition, since $f^{\prime \prime }(H(y))=h(f(y))=h(g(\hat{x}))=0$, $H(y)\in \ker f^{\prime \prime }$. It yields that $x\in \mathrm{im}d$. Conversely, for any $x\in \mathrm{im}d$, there exists $y$ such that $d(y)=x$ and $z\in M$ such that $H(z)=y$ and $f(z)=g(\hat{x})$ for any given $x\in N^{\prime }$. By the definition of $d$, there is $g^{\prime }(x)=\overline{f(z)}\in \mathrm{coker}f$ and so $g^{\prime }(x)=0$, $x\in \ker g^{\prime }$. Now we have proved $\mathrm{im}d=\ker g^{\prime }$. \end{proof}

4. Let $J$ be an $R$-module. Show that the following are equivalent.

• (1) $J$ is an injective module.
• (2) For any $R$-module monomorphism $\phi :J\to M$, there exists an $R$-module homomorphism $\psi :M\to J$ such that $\psi \phi ={\mathrm{id}}_J$.
• (3) If $J$ is a submodule of some $R$-module $M$, then $J$ is a direct summand of $M$.

\begin{proof} (1)→(2). If $J$ is an injective module, then the functor $\mathrm{Hom}(-,J)$ is an exact functor. Note that

$$J\stackrel{\phi }{\to }M\stackrel{\pi }{\to }M/\mathrm{im}\phi \to 0$$

is exact, so we have the following exact sequence

$$0\to \mathrm{Hom}(M/\mathrm{im}\phi ,J)\to \mathrm{Hom}(M,J)\stackrel{{\phi }^{\ast }}{\to }\mathrm{Hom}(J,J)\to 0.$$

Since ${\phi }^{\ast }$ is surjective and ${\mathrm{id}}_J\in \mathrm{Hom}(J,J)$, there exists $\psi \in \mathrm{Hom}(M,J)$ such that ${\phi }^{\ast }(\psi )={\mathrm{id}}_J$, that is, $\psi \phi ={\mathrm{id}}_J$.

(2)→(3). Since $J\subseteq M$, we define a monomorphism $\phi =\mathrm{id}:J\hookrightarrow M$. Then there exists $\psi :M\to J$ such that $\psi \phi ={\mathrm{id}}_J$ and so $\psi$ is surjective. Note that $\ker \psi :=N$ is a submodule and $\psi {|}_J={\mathrm{id}}_J$. For any $m\in M$, $m=m-\psi (m)+\psi (m)$ with $m-\psi (m)\in N$ and $\psi (m)\in J$. It follows that $M=N+J$. Furthermore, $N\cap J=\{0\}$ yields that $M=N\oplus J$, that is, $J$ is a direct summand of $M$.

(3)→(1). By Theorem 3.18 in Jacobson’s Basic Algebra II (p. 160), each module is a submodule of an injective module, so we can assume that $J$ is a submodule of injective module $Q$ and $Q=J\oplus R$. Define the inclusion map $\tau :J\hookrightarrow Q$ and the projective map $\gamma :Q\to J$.

To show $J$ is an injective module, it suffices to show if $0\to L\stackrel{\phi }{\to }M\stackrel{\psi }{\to }N\to 0$ is exact, then $0\to \mathrm{Hom}(P,J)\to \mathrm{Hom}(L,J)\to \mathrm{Hom}(N,J)\to 0$ is exact. Since $\mathrm{Hom}(-,J)$ is left exact, it is enough to show $\mathrm{Hom}(L,J)\to \mathrm{Hom}(N,J)$ is surjective. Notice that

$$0\to \mathrm{Hom}(N,Q)\stackrel{{\psi }^{\ast }}{\to }\mathrm{Hom}(M,Q)\stackrel{{\phi }^{\ast }}{\to }\mathrm{Hom}(L,Q)\to 0$$

is exact as $Q$ is an injective module. For any $f\in \mathrm{Hom}(L,J)$, there is $\tau \circ f\in \mathrm{Hom}(L,Q)$ and so there exists $g\in \mathrm{Hom}(M,Q)$ such that ${\phi }^{\ast }g=\tau \circ f$. Note that $\gamma \circ \tau ={\mathrm{id}}_J$, then we have

$${\phi }^{\ast }(\gamma \circ g)=\gamma \circ g\circ \phi =\gamma \circ \tau \circ f=f$$

which deduces that $f\in \mathrm{im}{\phi }^{\ast }$. Now we finish the proof.

• By ghm: Pasted image 20250402152040.png
• By zsw: Pasted image 20250402152151.png and Pasted image 20250402152213.png
• In midterm, I give a proof about it. See here. \end{proof}

5. Suppose that $M_i(i\in I)$ are $R$-modules. Prove that ${⨁}_{i\in I}{M}_i$ is a flat $R$-module if and only if each $M_i,i\in I$ is a flat $R$-module.

\begin{proof} For any exact sequence $0\to N\stackrel{\phi }{\to }K\stackrel{\psi }{\to }L\to 0$, it suffices to show that the sequence

$$0\to \left({\oplus }_{i\in I}{M}_i\right)\otimes N\stackrel{\mathrm{id}\otimes \phi }{\to }\left({\oplus }_{i\in I}{M}_i\right)\otimes K\stackrel{\mathrm{id}\otimes \psi }{\to }\left({\oplus }_{i\in I}{M}_i\right)\otimes L\to 0$$

is exact iff $0\to M_i\otimes N\to M_i\otimes K\to M_i\otimes L\to 0$ is exact for all $i\in I$.

Note that $\left({\oplus }_{i\in I}{M}_i\right)\otimes N\stackrel{{\theta }_N}{\simeq }{\oplus }_{i\in I}{M}_i\otimes N$, and we can define ${\theta }_K,{\theta }_L$ similarly. Since $\mathrm{id}\otimes \phi$ is injective iff

$${\phi }^{\ast }={\theta }_K\circ (\mathrm{id}\otimes \phi )\circ {\theta }_N^{-1}:{\oplus }_{i\in I}(M_i\otimes N)\to {\oplus }_{i\in I}(M_i\otimes K)$$

is injective iff ${\phi }^{\ast }{|}_{M_i\otimes N}$ is injective for all $i$ iff $M_i\otimes N\to M_i\otimes K$ is injective for all $i$, we know $0\to \left({\oplus }_{i\in I}{M}_i\right)\otimes N\stackrel{\mathrm{id}\otimes \phi }{\to }\left({\oplus }_{i\in I}{M}_i\right)\otimes K$ is exact iff $0\to M_i\otimes N\to M_i\otimes K$ is exact for all $i$. Similarly we can prove $\left({\oplus }_{i\in I}{M}_i\right)\otimes K\stackrel{\mathrm{id}\otimes \psi }{\to }\left({\oplus }_{i\in I}{M}_i\right)\otimes L\to 0$ is exact iff $M_i\otimes K\to M_i\otimes L\to 0$ is exact for all $i$. \end{proof}