## Every module is contained in an injective module

Let $R$ be a ring with 1 and let $M$ be a left unital $R$-module.

1. Show that $M$ is contained in an injective $\mathbb{Z}$-module.
2. Show that $\mathsf{Hom}_R(R,M) \subseteq \mathsf{Hom}_\mathbb{Z}(R,M) \subseteq \mathsf{Hom}_\mathbb{Z}(R,Q)$.
3. Conclude that $M$ is contained in an injective module.

We know that every abelian group is contained in a divisible abelian group- so as a $\mathbb{Z}$-module, $M$ is contained in an injective $\mathbb{Z}$-module $Q$. Let $\iota : M \rightarrow Q$ be the inclusion map.

Now $R$ is an $(R,R)$-bimodule and a $(\mathbb{Z},R)$-bimodule, $M$ a left $R$-module, and $Q$ is a left $\mathbb{Z}$-module. Using the action described in this prior exercise, $\mathsf{Hom}_R(R,M)$ and $\mathsf{Hom}_\mathbb{Z}(R,M)$ are left $R$-modules.

Moreover, as we argue, $\mathsf{Hom}_\mathbb{Z}(R,Q)$ is a unital left $R$-module via the action $(r \cdot \varphi)(s) = \varphi(rs)$. Indeed, $(r\cdot\varphi)(s_1+s_2) = \varphi((s_1+s_2)r)$ $= \varphi(s_1r) + \varphi(s_2r)$ $= (r\varphi)(s_1) + (r\varphi)(s_2)$, so that $r \cdot \varphi$ is a $\mathbb{Z}$-module homomorphism. Now $((r_1+r_2)\cdot\varphi)(s) = \varphi(s(r_1+r_2))$ $= \varphi(sr_1)\varphi(sr_2)$ $= (r_1\varphi)(s) + (r_2\varphi)(s)$ $= (r_1\varphi + r_2\varphi)(s)$, so that $(r_1+r_2)\varphi = r_1\varphi + r_2\varphi$; $((r_1r_2)\varphi)(s) = \varphi(sr_1r_2)$ $= (r_2)\varphi(sr_1)$ $= (r_1(r_2\varphi))(s)$, so that $(r_1r_2)\varphi = r_1(r_2\varphi)$; $(r(\varphi_1+\varphi_2))(s) = (\varphi_1+\varphi_2)(sr)$ $= \varphi_1(sr) + \varphi_2(sr)$ $= (r\varphi_1)(s) + (r\varphi_2)(s)$ $= (r\varphi_1 + r\varphi_2)(s)$, so that $r(\varphi_1 + r\varphi_2) = r\varphi_1 + r\varphi_2$, and $(1\varphi)(s) = \varphi(s)$, so that $1 \varphi = \varphi$.

Now every $R$-module homomorphism $R \rightarrow M$ is also an abelian group (hence $\mathbb{Z}$-module) homomorphism; this gives an injective map $\varphi : \mathsf{Hom}_R(R,M) \rightarrow \mathsf{Hom}_\mathbb{Z}(R,M)$. Note that for all $R$-module homomorphisms $\alpha,\beta : R \rightarrow M$ and $r \in R$, $\varphi(\alpha + r\beta)(a) = (\alpha+r\beta)(a)$ $= \alpha(a) + (r\beta)(a)$ $= \alpha(a) + \beta(ar)$ $= \varphi(\alpha)(a) + \varphi(\beta)(ar)$ $= \varphi(\alpha)(a) + (r\varphi(\beta))(a)$ $= (\varphi(\alpha) + r\varphi(\beta))(a)$, so that $\varphi(\alpha + r\beta) = \varphi(\alpha) + r\varphi(\beta)$. In particular, $\varphi$ is an $R$-module homomorphism.

We also have $(\iota \circ) : \mathsf{Hom}_\mathbb{Z}(R,M) \rightarrow \mathsf{Hom}_\mathbb{Z}(R,Q)$. Thus map is clearly well defined. Moreover, if $\alpha,\beta \in \mathsf{Hom}_\mathbb{Z}(R,M)$ and $r,a \in R$, we have $(\iota \circ)(\alpha + r\beta)(a) = ((\iota \circ \alpha) + (\iota \circ r\beta))(a)$ $= (\iota \circ \alpha)(a) + (\iota + r\beta)(a)$ $= (\iota \circ \alpha)(a) + \iota((r\beta)(a)$ $= (\iota \circ \alpha)(a) + \iota(\beta(ar))$ $= (\iota \circ \alpha)(a) + (\iota \circ \beta)(ar)$ $(\iota \circ \alpha) + (r(\iota \circ \beta))(a)$ $= ((\iota \circ)(\alpha) + r(\iota \circ)(\beta))(a)$. So $(\iota \circ)(\alpha + r\beta) = (\iota \circ)(\alpha) + r(\iota \circ)(\beta)$, and thus $(\iota \circ)$ is an $R$-module homomorphism. Now if $\iota \circ \alpha = \iota \circ \beta$, then $\alpha = \beta$ since $\iota$ is injective; thus $(\iota \circ)$ is injective.

Thus we have an $R$-module injection $\mathsf{Hom}_R(R,M) \rightarrow \mathsf{Hom}_\mathbb{Z}(R,Q)$. Now we saw in this previous exercise that (under this action) $M$ and $\mathsf{Hom}_R(R,M)$ are isomorphic as left $R$-modules. By this previous exercise, $\mathsf{Hom}_\mathbb{Z}(R,Q)$ is injective as a left $R$-module.

So $M$ is isomorphically embedded in an injective $R$-module.

• mcoulont  On September 23, 2011 at 12:30 pm

But Q is not a R-module (which is needed in the previous exercise), is it ?

• nbloomf  On September 23, 2011 at 12:52 pm

Indeed…

If I understood the problem correctly, I think I fixed it by showing directly that $\mathsf{Hom}_\mathbb{Z}(R,Q)$ is a left unital $R$-module.

I wrote this in about 15 minutes between classes, so it is likely that I introduced an error or missed the point entirely. Let me know if you find a problem.

Thanks!

• mcoulont  On September 23, 2011 at 1:04 pm

HomZ(R,Q) is a Z-module but at exercise 15 it seems that the hypothesis that Q is a R-module is necessary.

But I may also miss the point, too.

Bon appétit.

• mcoulont  On September 23, 2011 at 1:05 pm

HomZ(R,Q) is a R-module ; sorry.

• mcoulont  On September 23, 2011 at 1:07 pm

And I’m sorry for the redundancy at the second sentence, too.

• nbloomf  On September 23, 2011 at 3:41 pm

No problem. Thanks for pointing it out.