# Homomorphic image of a module

abelian-categoriesabstract-algebracommutative-algebramodules

I have this pretty trivial question: if $$M\subseteq N$$ are $$A$$-modules, is it true in general that $$M$$ is the direct sum of $$N$$ and $$L:= N/M$$? Or, if $$0\to M\xrightarrow{\alpha} N\xrightarrow{\beta} L\to 0$$ is an exact sequence of $$A$$-modules, does $$N=M\oplus L$$?

I don't think so, because I never saw such a theorem; however in this case I don't understand why in Atiyah-MacDonald (proof of proposition 6.3) it is used that, given two $$A$$-modules $$H,H'\subseteq N$$, if $$\alpha^{-1}(H)=\alpha^{-1}(H')$$ and $$\beta(H)=\beta(H')$$, then $$H=H'$$. Thanks for any clarify

No, not every short exact sequence is split. For example, consider the exact sequence of abelian groups: $$0\to \mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/4\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}\to 0$$

Now suppose we have a short exact sequence $$0\rightarrow M\xrightarrow{\alpha} N\xrightarrow{\beta} L\rightarrow 0$$ and submodules $$H,H'\subseteq N$$ with $$\alpha^{-1}(H) = \alpha^{-1}(H')$$ and $$\beta(H) = \beta(H')$$. In your question, you suggest that this implies $$H = H'$$. In fact, this is also false.

Consider the exact sequence of abelian groups: $$0\to \mathbb{Z}\xrightarrow{\delta} \mathbb{Z}\oplus\mathbb{Z}\xrightarrow{\mu} \mathbb{Z}\to 0$$ where $$\delta(n) = (n,n)$$ and $$\mu(a,b) = a-b$$. Let $$H = \mathbb{Z}\oplus \{0\}$$, and let $$H' = \{0\}\oplus \mathbb{Z}$$. Then $$\mu(H) = \mu(H') = \mathbb{Z}$$, and $$\delta^{-1}(H) = \delta^{-1}(H') = \{0\}$$, but $$H\neq H'$$.

What is true is that if $$H'\subseteq H$$ and $$\beta(H) = \beta(H')$$ and $$\alpha^{-1}(H) = \alpha^{-1}(H')$$, then $$H = H'$$. This is what Atiyah and MacDonald are using in the proof of Proposition 6.3.

To see this, let $$x\in H$$. Since $$\beta(H) = \beta(H')$$, there is some $$x'\in H'$$ such that $$\beta(x) = \beta(x')$$. Thus $$0 = \beta(x) - \beta(x') = \beta(x-x')$$, so $$x-x'\in \ker(\beta) = \mathrm{im}(\alpha)$$. Let $$y\in M$$ with $$\alpha(y) = x-x'$$. Note also that $$x-x'\in H$$, so $$y\in \alpha^{-1}(H) = \alpha^{-1}(H')$$. Thus $$x-x' = \alpha(y)\in H'$$. So $$x = (x-x') + x' \in H'$$, as was to be shown.

For a fancier proof, or to put the result in context, note that we have a commutative diagram: $$\require{AMScd}$$ $$\begin{CD} 0 @>>> \alpha^{-1}(H') @>>> H'@>>> \beta(H') @>>> 0\\ @V{=}VV @V{=}VV @V{i}VV @V{=}VV @V{=}VV\\ 0 @>>> \alpha^{-1}(H) @>>> H @>>> \beta(H) @>>> 0 \end{CD}$$ The rows are exact and the vertical arrows, except possibly for the inclusion $$i\colon H\to H'$$, are isomorphisms. It follows from the Five Lemma (or its special case the Short Five Lemma) that the inclusion is an isomorphism, so $$H = H'$$. Of course, the proof of the Five Lemma includes reasoning like the direct proof I gave above.

Atiyah and MacDonald do not explicitly state the Five Lemma, but they give a more general statement (leaving most of the proof to the reader) in Proposition 2.10. In our case, the exact sequence from Proposition 2.10 is: $$0\to \ker(=)\to \ker(i)\to \ker(=)\to \mathrm{coker}(=)\to \mathrm{coker}(i)\to \mathrm{coker}(=)\to 0$$ Now $$\ker(=) = \mathrm{coker}(=) = 0$$, so $$\ker(i)$$ and $$\mathrm{coker}(i)$$ are both sandwiched between $$0$$s in an exact sequence. It follows that $$i$$ is an isomorphism.