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.

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  • 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


      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.


  • 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.

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