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

- Show that is contained in an injective -module.
- Show that .
- Conclude that is contained in an injective module.

We know that every abelian group is contained in a divisible abelian group- so as a -module, is contained in an injective -module . Let be the inclusion map.

Now is an -bimodule and a -bimodule, a left -module, and is a left -module. Using the action described in this prior exercise, and are left -modules.

Moreover, as we argue, is a unital left -module via the action . Indeed, , so that is a -module homomorphism. Now , so that ; , so that ; , so that , and , so that .

Now every -module homomorphism is also an abelian group (hence -module) homomorphism; this gives an injective map . Note that for all -module homomorphisms and , , so that . In particular, is an -module homomorphism.

We also have . Thus map is clearly well defined. Moreover, if and , we have . So , and thus is an -module homomorphism. Now if , then since is injective; thus is injective.

Thus we have an -module injection . Now we saw in this previous exercise that (under this action) and are isomorphic as left -modules. By this previous exercise, is injective as a left -module.

So is isomorphically embedded in an injective -module.

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

Indeed…

If I understood the problem correctly, I think I fixed it by showing directly that is a left unital -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!

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.

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

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

No problem. Thanks for pointing it out.