Homework 5

Let's do a classic diagram chase! First let's suppose $h_1$ and $h_3$ are injective (and both rows are exact). We wish to show $h_2:Y\to Y^{\prime }$ is injective, so suppose $y\in Y$ is such that $h_2(y)=0$. We want to prove $y=0$ and we'll do so by "chasing" $y$ around the diagram.

First, by the commutativity of the right square we have $h_3(g(y))=g^{\prime }(h_2(y))=g^{\prime }(0)=0$. In other words, if we send $y$ down $h_2$ (where it's sent to 0) and then across $g^{\prime }$ (which must send 0 to 0) then we land at 0; by commutativity it follows that when we send $y$ across $g$ and down $h_3$ we must also get 0. This means that $g(y)$ is in the kernel of $h_3$ and since $h_3$ was assumed to be injective, we must therefore have that $g(y)=0$. But this implies $y$ is in the kernel of $g$, and since the top row is exact (and so in particular exact at $Y$) it now follows that $y$ is in the image of $f$. Thus, there must exist some $x\in X$ with $f(x)=y$.

Now follow $x$ around the left commutative square. If we send $x$ across (via $f$) and then down (via $h_2$), it maps to $y$ and then to 0. Thus, when we instead send $x$ down (via $h_1$) and across (via $f^{\prime })$ it must map to 0. This means $h_1(x)$ is in the kernel of $f^{\prime }$. But the bottom row is exact (in particular at $X^{\prime }$) and so $f^{\prime }$ is injective, hence we must have $h_1(x)=0$. Since $h_1$ is injective this means $x=0$. But then $y=f(x)=f(0)=0$. Thus, $h_2$ is injective.

Now let's try the surjectivity argument. Suppoes $h_1$ and $h_3$ are surjective (and both rows are exact). We wish to show $h_2:Y\to Y^{\prime }$ is surjective, so fix any element $y^{\prime }\in Y$. We must prove there is an element $y\in Y$ with $h_2(y)=y^{\prime }$.

Let $z^{\prime }=g^{\prime }(y^{\prime })$. Since $h_3$ is surjective there is some $z\in Z$ with $h_3(z)=z^{\prime }$. Since the top row is exact (in particular at $Z$), the map $g$ is surjective and so there is some $y_1\in Y$ with $g(y_1)=z$. Now note that $y$ must map down and over to $z^{\prime }$, but this is not quite enough to conclude $y_1$ maps down to $y^{\prime }$. In other words, $h_2(y_1)$ and $y^{\prime }$ are both elements that map (via $g^{\prime }$) to the same element $z^{\prime }$, but that's not enough to conclude $h_2(y_1)=y^{\prime }$.

So, let $\tilde{y}=y^{\prime }-h_2(y_1)$. Then $g^{\prime }(\tilde{y})=g(y^{\prime }-h_2(y_1))=g(y^{\prime })-g(h_2(y_1))=z^{\prime }-z^{\prime }=0$, so $\tilde{y}$ is in the kernel of $g^{\prime }.$ Since the bottom row is exact at $Y^{\prime }$, this implies $\tilde{y}$ is in the image of $f^{\prime }$ and hence there is some $x^{\prime }\in X^{\prime }$ with $f^{\prime }(x^{\prime })=\tilde{y}$. Now $h_1$ is surjective so there is some $x\in X$ with $h_1(x)=x^{\prime }$. Finally, we let $y=f(x)+y_1$. Then observe that

$h_2(y)=h_2(f(x))+h_2(y_1)=\tilde{y}+h_2(y_1)=y^{\prime }.$

Thus, we've proven $h_2$ is surjective.

1. You could use this characterization of projective modules, in which case this property is basically immediate. However, it is not hard to give a direct argument, so let's see that.

   First suppose $M_1$ and $M_2$ are projective. To prove $M_1\oplus M_2$ is projective, let $g:K\to L$ be any surjection and $\varphi :M_1\oplus M_2\to L$ be any morphism. Then consider the morphism $\varphi \circ i_1:M_1\to L$. Since $M_1$ is projective, this morphism factors through $g$, i.e., there is some morphism ${\psi }_1:M_1\to K$ such that $\varphi \circ i_1=g\circ {\psi }_1$. This is pictured below:

   We can repeat this same argument for the projective module $M_2$, obtaining a morphism ${\psi }_2:M_2\to K$ such that $\varphi \circ i_2=g\circ {\psi }_2$. By a universal property for the direct sum $M_1\oplus M_2$, we then get a unique module morphism $\psi :M_1\oplus M_2\to K$ with $\psi \circ i_1={\psi }_1$ and $\psi \circ i_2={\psi }_2$:

   By construction, we have

   $(g\circ \psi )\circ i_1=g\circ (\psi \circ i_1)=g\circ {\psi }_1=\varphi \circ i_1,$

   and similarly

   $(g\circ \psi )\circ i_2=g\circ (\psi \circ i_2)=g\circ {\psi }_2=\varphi \circ i_2.$

   By our universal property for $M_1\oplus M_2$, it follows that $g\circ \psi =\varphi$. Thus, we've proven $M_1\oplus M_2$ is projective.

   Conversely, suppose $M_1\oplus M_2$ is projective. To prove $M_1$ is projective, let $g:K\to L$ be any surjection and $\varphi :M_1\to L$ be any morphism. Let ${\pi }_1:M_1\oplus M_2\to M_1$ be the projection (available via the isomorphism $M_1\oplus M_2\simeq M_1\times M_2$ and the canonical projection $M_1\times M_2\to M_1$). Since $M_1\oplus M_2$ is projective, the morphism $\varphi \circ {\pi }_1:M_1\oplus M_2\to L$ must factor through $g$, i.e., there is some $\psi :M_1\oplus M_2\to K$ such that $g\circ \psi =\varphi \circ {\pi }_1$.

   Now let ${\psi }_1=\psi \circ i_1$, where $i_1:M_1\to M_1\oplus M_2$ is the canonical injection. Then observe that

   $\begin{array}{rl}g\circ {\psi }_1& =g\circ (\psi \circ i_1)\\ & =(g\circ \psi )\circ i_1\\ & =(\varphi \circ \pi )\circ i_1\\ & =\varphi \circ ({\pi }_1\circ i_1)\\ & =\varphi \circ 1_{M_1}\\ & =\varphi .\end{array}$

   Thus, $M_1$ is indeed projective. The same argument, mutatis mutandis, proves $M_2$ is also projective.

2. This argument is basically the dual of the projective argument. First suppose $M_1$ and $M_2$ are both injective. To prove $M_1\oplus M_2$ is injective, let $g:L\to K$ be any injection and $\varphi :L\to M_1\oplus M_2$ be any morphism. Then consider the morphism ${\pi }_1\circ \varphi :L\to M_1$. Since $M_1$ is injective, this morphism factors through $g$, i.e., there is some morphism ${\psi }_1:K\to M_1$ such that ${\pi }_1\circ \varphi ={\psi }_1\circ g$. This is pictured below:

   We can repeat this same argument for the projective module $M_2$, obtaining a morphism ${\psi }_2:K\to M_2$ such that ${\pi }_2\circ \varphi ={\psi }_2\circ g$. By a universal property for the direct product $M_1\times M_2\simeq M_1\oplus M_2$, we then get a unique module morphism $\psi :K\to M_1\oplus M_2$ with ${\pi }_1\circ \psi ={\psi }_1$ and ${\pi }_2\circ \psi ={\psi }_2$:

   By construction, we have

   ${\pi }_1\circ (\psi \circ g)=({\pi }_1\circ \psi )\circ g={\psi }_1\circ g={\pi }_1\circ \varphi ,$

   and similarly

   ${\pi }_2\circ (\psi \circ g)=({\pi }_2\circ \psi )\circ g={\psi }_2\circ g={\pi }_2\circ \varphi .$

   By our universal property for $M_1\times M_2\simeq M_1\oplus M_2$, it follows that $\psi \circ g=\varphi$. Thus, we've proven $M_1\oplus M_2$ is injective.

   Conversely, suppose $M_1\oplus M_2$ is injective. To prove $M_1$ is injective, let $g:L\to K$ be any injection and $\varphi :L\to M_1$ be any morphism. Since $M_1\oplus M_2$ is injective, the morphism $i_1\circ \varphi :L\to M_1\oplus M_2$ must factor through $g$, i.e., there is some $\psi :K\to M_1\oplus M_2$ such that $\psi \circ g=i_1\circ \varphi$.

   Now let ${\psi }_1={\pi }_1\circ \psi$. Then observe that

   $\begin{array}{rl}\psi \circ g& =({\pi }_1\circ \psi )\circ g\\ & ={\pi }_1\circ (\psi \circ g)\\ & ={\pi }_1\circ (i_1\circ \varphi )\\ & =({\pi }_1\circ i_1)\circ \varphi \\ & =1_{M_1}\circ \varphi \\ & =\varphi .\end{array}$

   Thus, $M_1$ is indeed injective. The same argument, mutatis mutandis, proves $M_2$ is also injective.

3. To make our lives easier, let's first make an observation. Let $f:J\to K$ be any morphism and consider the result of tensoring with $(M_1\oplus M_2)$ over $R$:

   $(M_1\oplus M_2){\otimes }_{R}J\stackrel{1_{M_1\oplus M_2}\otimes f}{\to }(M_1\oplus M_2)\otimes K$

   Recall that tensor product commutes with direct sum, and so the above morphism is naturally isomorphic to the morphism

   $(M_1{\otimes }_{R}J)\oplus (M_2{\otimes }_{R}J)\stackrel{(1_{M_1}\otimes f)\oplus (1_{M_2}\otimes f)}{\to }(M_1{\otimes }_{R}K)\oplus (M_2{\otimes }_{R}K)$

   An immediate consequence of a universal property of the direct sum is that the kernel of the above morphism is (isomorphic to ) the direct sum of the kernels of the component morphisms:

   $\ker ((1_{M_1}\otimes f)\oplus (1_{M_2}\otimes f))\simeq \ker (1_{M_1}\otimes f)\oplus \ker (1_{M_2}\otimes f).$

   As a consequence, the morphism $(1_{M_1}\otimes f)\oplus (1_{M_2}\otimes f)$ is injective if and only if the morphisms $1_{M_1}\otimes f$ and $1_{M_2}\otimes f$ are both injective.

   From this observation, it is now immediate that $M_1\oplus M_2$ is flat if and only if $M_1$ and $M_2$ are both flat.

Let $A$ be a nonzero finite abelian group of order $n$, say. Then $na=0_A$ for every $a\in A$, and so every element in $A$ is torsion and $nA=0\ne A$.

Now suppose $A$ is projective. Then by this characterization of projective modules it follows that $A$ is (isomorphic to) a direct summand of a free $\mathbf{\text{Z}}$-module. In other words, there is an isomorphism of the form ${\mathbf{\text{Z}}}^k\simeq A\oplus B$. Since every element in $A$ is torsion, but there are no nonzero torsion elements in ${\mathbf{\text{Z}}}^k$, this is a contradiction.

Now suppose $A$ is injective. Then by this characterizations of injective modules over PIDs (which includes $\mathbf{\text{Z}}$), we must have $mA=A$ for every $m\in \mathbf{\text{Z}}$. But we've already noted that $nA=0\ne A$, so we've once again arrived at a contradiction.

1. Suppose $M$ and $N$ are free $R$-modules, say $M\simeq F(X)$ and $N\simeq F(Y)$ for some sets $X$ and $Y$. Recalling that $F(A)\simeq \underset{a\in A}{⨁}R$ for any set $A$, and that tensor product commutes with direct sum, observe that

   $\begin{array}{rl}M{\otimes }_{R}N& \simeq \left(\underset{x\in X}{⨁}R\right){\otimes }_R\left(\underset{y\in Y}{⨁}R\right)\\ & \simeq \underset{x\in X}{⨁}\left(R{\otimes }_R\left(\underset{y\in Y}{⨁}R\right)\right)\\ & \simeq \underset{x\in X}{⨁}(\underset{y\in Y}{⨁}(R{\otimes }_{R}R))\\ & \simeq \underset{(x,y)\in X\times Y}{⨁}R\\ & \simeq \underset{z\in X\times Y}{⨁}R\\ & \simeq F(X\times Y).\end{array}$

2. Suppose $M$ and $N$ are projective $R$-modules. By this characterization of projective modules, there exist isomorphisms of the form $F(X)\simeq M\oplus M^{\prime }$ and $F(Y)\simeq N\oplus N^{\prime }$ for some sets $X,Y$ and modules $M^{\prime },N^{\prime }$. Then observe that

   $\begin{array}{rl}F(X\bigsqcup Y)& \simeq F(X){\otimes }_{R}F(Y)\\ & \simeq (M\oplus M^{\prime }){\otimes }_R(N\oplus N^{\prime })\\ & \simeq (M{\otimes }_{R}N)\oplus (M{\otimes }_R{N}^{\prime })\oplus (M^{\prime }{\otimes }_{R}N)\oplus (M^{\prime }{\otimes }_R{N}^{\prime }).\end{array}$

   So, $M{\otimes }_{R}N$ is isomorphic to a direct summand of a free module and hence is projective.