0
$\begingroup$

We know that for any $R$-module exist injective $R$-module $\overline{M}$ such that there is inclusion $i:M\rightarrow \overline{M}$, where we treat $M,\overline{M}$ as $\mathbb{Z}$-modules.

Show that $\widetilde{M}:=\mathrm{Hom}_{\mathbb{Z}}(R,\overline{M})$ is injective $R$-module.


$\widetilde{M}$ is injective module iff functor $\mathrm{Hom_{R}(\cdot,\widetilde{M}})$ is exact.

Let $0\rightarrow N'\rightarrow N\stackrel{\alpha}{\rightarrow} N''\rightarrow 0$ be exact sequence of $R$-modules. Then we obtain sequence $0\rightarrow \mathrm{Hom}_R(N',\widetilde{M})\rightarrow \mathrm{Hom}_R(N,\widetilde{M})\stackrel{\alpha^{*}}{\rightarrow}\mathrm{Hom}_R(N'',\widetilde{M})\rightarrow 0$

We know that this functor left-exact, so we should prove that $\mathrm{im}\alpha^*=\mathrm{Hom}_R(N'',\widetilde{M})$, but we know that $\mathrm{Hom}_R(N'',\widetilde{M})=\mathrm{Hom}_R\left(N'',\mathrm{Hom}_{\mathbb{Z}}(R,\overline{M})\right)\cong \mathrm{Hom}_{\mathbb{Z}}\left(N''\otimes_RR,\overline{M}\right)\cong \mathrm{Hom}_{\mathbb{Z}}(N'',\overline{M})$

So, I should prove that $\mathrm{im}\alpha^*=\mathrm{Hom}_{\mathbb{Z}}(N'',\overline{M})$.

How to do it?

$\endgroup$
  • $\begingroup$ you may do the same thing for the middle term and lift the surjection via the isomorphisms. $\endgroup$ – Math137 Jul 5 '15 at 22:09
1
$\begingroup$

You forgot the functor $\operatorname{Hom}_R(\text{--},M)$ is contravariant.

Let $0\longrightarrow N'\xrightarrow{\ u\ }N$. What has to be proved is $$\mathrm{Hom}_R(N,\widetilde{M})\rightarrow \mathrm{Hom}_R(N',\widetilde{M})$$ is surjective. For this, note first $$\operatorname{Hom}_R(N,\widetilde{M})=\operatorname{Hom}_R(N,\operatorname{Hom}_\mathbf Z(R,\overline{\!M\mkern-0.5mu})\simeq\operatorname{Hom}_\mathbf Z(N\otimes_R R,\overline{\!M\mkern-0.5mu})\simeq \operatorname{Hom}_\mathbf Z(N,\overline{\!M\mkern-0.5mu}).$$ Now consider the commutative diagram: $$\begin{matrix} \operatorname{Hom}_R(N,\widetilde{M})&\xrightarrow{\operatorname{Hom}_R(u,1)}&\operatorname{Hom}_R(N',\widetilde{M})\\ \downarrow &&\downarrow\\ \operatorname{Hom}_\mathbf Z(N,\overline{\!M\mkern-0.5mu})&\xrightarrow{\operatorname{Hom}_\mathbf Z(u,1)}& \operatorname{Hom}_\mathbf Z(N',\overline{\!M\mkern-0.5mu}) \end{matrix} $$ The two vertical maps are isomorphisms, and the bottom horizontal map is surjective since $\overline{\!M\mkern-0.5mu}$ is an injective $\mathbf Z$-module, hence the top horizontal map is surjective, as was to be proved.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.