2
$\begingroup$

$0 \rightarrow M'\xrightarrow[]{f_1} M \xrightarrow[]{f_2} M'' $ is exact if and only if $ 0 \rightarrow Hom (N,M') \xrightarrow[]{\phi} Hom(N,M) \xrightarrow[]{\chi} Hom(N,M'')$ is exact for all module $N$ over the same ring. Assuming the first sequence to be exact I have shown that $\phi$ is injective and $Im \phi$ is a subset of $ker\chi$ but I could not show the reverse inclusion i.e. $ker \chi $is a subset of $Im \phi$. I tried to show it by a contradiction but did not succeed. Also the converse part of the proof seems hard to me.

$\endgroup$
3
  • $\begingroup$ What is $N{{}}$? $\endgroup$ Jan 22, 2018 at 7:14
  • $\begingroup$ N is any module over the same ring $\endgroup$ Jan 22, 2018 at 7:17
  • $\begingroup$ Sir, I have made correction .Please check and respond $\endgroup$ Jan 22, 2018 at 7:29

1 Answer 1

2
$\begingroup$

Showing that $\operatorname{Im} \phi \subseteq \operatorname{ker} \chi$ is a good start, you don't have to do too much more work to get that the sets are equal.

Suppose $\chi (\theta) = 0$ for some $\theta \colon N \to M,$ then this tells us that $g \circ \theta = 0$, in particular $g(\theta(x)) = 0$ for all $x \in N$. This tells you that $\theta(x) \in \operatorname{ker} g = \operatorname{Im} f$ for all $x \in N$. You can then observe that since $f$ is injective, it is in fact an isomorphism onto its image. Then, since $\theta$ maps into $\operatorname{Im} f$ we have that $f^{-1} \circ \theta$ is defined (where of course here $f^{-1} \colon \operatorname{Im} f \to M'$), and we can see that $$\theta = f \circ f^{-1} \circ \theta = \phi( f^{-1} \circ \theta) \in \operatorname{Im} \phi.$$

As for the converse, you should note that you know (if $0 \to M' \to M \to M''$ is an exact sequence of $R$-modules) $$0 \to \operatorname{Hom}(N,M') \to \operatorname{Hom}(N,M) \to \operatorname{Hom}(N,M'')$$ is exact for all $R$-modules $N$. So you just want to pick a particular $N$ which will spit out the answer immediately.

Hint: There is an $R$-module $N$ such that $\operatorname{Hom}_R(N,M) \cong M$ for all $R$-modules $M$.

I hope this helps, if not do let me know!

$\endgroup$
4
  • $\begingroup$ what should be the choice of N? Should it satisfy $Hom (N,M')=M'$also $\endgroup$ Jan 23, 2018 at 17:09
  • 1
    $\begingroup$ But $\chi \circ \phi =0$ implies$ Im \phi \subset ker \chi.$Is not it? I just want to prove the reverse inclusion. $\endgroup$ Jan 23, 2018 at 17:42
  • $\begingroup$ @jaggu Whoops, my bad! Hopefully I've fixed that now. As for the module $N$, as I said above it gives you that $\operatorname{Hom}_R (N,M) \cong M$ for all $R$-modules $M$, so it will also hold true for $M'$ and $M''$. Another hint: It is perhaps the first or second example of an $R$-module one would ever think of. $\endgroup$ Jan 24, 2018 at 0:09
  • $\begingroup$ If the ring $R$ doesn't contain the identity, can the inverse be true? $\endgroup$
    – Mod.esty
    Nov 2, 2020 at 3:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .