# Abstract Nonsense

## Injective Modules (Pt. III)

Point of Post: This is a continuation of this post.

Divisibility and Injectiveness

We would now like to discuss how the use of injective modules becomes easier if we insist on using only PIDs (such as $\mathbb{Z}$). What is going to end up happen is that while things aren’t quite as simple as projective modules over PIDs (they are just the free ones) PIDs allow us to give a fairly nice description of injective modules over $R$. Namely, if $R$ is an integral domain call a left $R$-module $M$ divisible if for any $m\in M$ and $r\in R$ there exists $m'\in M$ such that $rm'=m$. To start to understand why it makes sense that injective modules and divisible modules should be related suppose for a second that $M$ is a left $R$-module. Then, for any $x\in M$ and $r\in R$ we have a well-defined $R$-map $f:(r)\to M$ defined by $rr'\mapsto r'x$. This is well-defined precisely because $R$ is an integral domain. Now, if $M$ is injective we can extend this to a map $\widetilde{f}:R\to M$ and we then note that $x=f(r)=\widetilde{f}(r)=r\widetilde{f}(1)$ and so $r$ divides $x$. Thus, if $M$ is injective it is necessarily divisible.

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The key theorem is that the converse is true if we assume that $R$ is a PID. Why does this make sense? Well thanks to Baer’s criterion we really only have to worry about extending maps from ideals of $R$, but if we assume that $R$ is a PID we know that all of these ideals are singly generated. So, we have some map $f:(r)\to D$ where $D$ is a divisible $R$-module. Since $D$ is divisible we know we are able to find $x\in D$ such that $f(r)=rx$. Define then $\widetilde{f}(a)=ax$ where $a\in R$. Clearly this is an $R$-map and $\widetilde{f}(ra)=r\widetilde{f}(a)=rax=arx=af(r)=f(ar)$. Thus, we have proven:

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Theorem: Let $R$ be a PID. Then, an $R$-module $M$ is injective if and only if it’s divisible.

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It’s fairly clear that both quotients and direct sums of divisible $R$-modules are divisible and so we may conclude the following:

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Theorem: Let $R$ be a PID. The quotient of injective $R$-modules is injective. Moreover, the direct limit of injective $R$-modules is injective.

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Thus, we now know a lot of examples of injective abelian groups. Namely, if $k$ is any characteristic zero field (or any abelian group capable of supporting such a structure) then $k$ is an injective abelian group.

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Enough Injectives

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I’d like to finish by proving that categories of modules have “enough injectives” which means that they all modules sit inside an injective one. This will be useful later when we try to approximate arbitrary modules by injective ones leading to the notion of derived functors.

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So, we’ve already (secretly) proven that every abelian group sits inside an injective group. Indeed, we merely take any abelian group $A$ and find a surjection $\mathbb{Z}^{\oplus\lambda}\to A$ (this is possible because the category of abelian groups has enough projectives in the form of free modules!). We know then that if $K$ is the kernel of this map that $A\cong \mathbb{Z}^{\oplus\lambda}/K\subseteq\mathbb{Q}^{\oplus\lambda}/K$. Now, since $\mathbb{Q}^{\oplus\lambda}$ (being the direct sum of divisible(injective) modules is divisible(injective) and quotients of divisible(injective) modules are divisible(injective)) we are done.

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We next note the following fact:

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Theorem: Let $D$ be a divisible abelian group and $R$ a left $R$-module. Then, $\text{Hom}_\mathbb{Z}(R,D)$ with $(rf)(x)=f(xr)$ is an injective left $R$-module.

Proof: We must show that $\text{Hom}_R(\bullet,\text{Hom}_\mathbb{Z}(R,D))$ is an exact functor. But, by adjointness of Hom and tensor we have that $\text{Hom}_R(\bullet,\text{Hom}_\mathbb{Z}(R,D))\cong\text{Hom}_\mathbb{Z}(R\otimes_R\bullet,D)$ but this is clearly just isomorphic to $\text{Hom}_\mathbb{Z}(\bullet,D)$ (since $R\otimes\bullet$ is naturally isomorphic to the identity functor) and since $D$ is divisible we know that $\text{Hom}_\mathbb{Z}(\bullet,D)$ is exact. $\blacksquare$

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We can now prove our desired theorem:

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Theorem(Enough Injectives): Let $R$ be a ring and $M$ a left $R$-module. Then, there exists an injective left $R$-module $I$ and a monomorphism $M\to I$.

Proof: We start by embedding $M$ into $\text{Hom}_\mathbb{Z}(R,M)$ by $\varphi_m(r)=rm$. This is clearly an $R$-map and is injective by considering $\varphi_m(1)$.

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We know that we can find an embedding $i:M\to D$ of abelian groups where $D$ is some divisible abelian group. Then, since the functor $\text{Hom}_\mathbb{Z}(R,\bullet)$ is left exact we get an injection $i^\ast:\text{Hom}_\mathbb{Z}(R,M)\to\text{Hom}_\mathbb{Z}(R,D)$. If we can verify that $i^\ast\circ\varphi$ is an $R$-map then $i\circ\varphi:M\to\text{Hom}_\mathbb{Z}(R,D)$ will be the desired embedding of $M$ into an injective left $R$-module.

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To see that this is true let $m\in M$ and $r\in R$ be arbitrary. We need to show that $i\circ\varphi_{rm}=r(i\circ\varphi_m)$, but this is simple (I leave it to you). $\blacksquare$

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.