# Proposition 2.10 Atiyah MacDonald

I have been self studying from Atiyah and Macdonald's Intoduction to Commutative Algebra and am having difficulty with the following proposition:

Proposition 2.10 Let $$\require{AMScd} \begin{CD} 0 @>>> M^{'} @>{u}>> M @>{v}>> M^{''} @>>> 0 \\ @. @V{f^{'}}VV @V{f}VV @VV{f^{''}}V \\ 0 @>>> N^{'} @>{u^{'}}>> N @>{v^{'}}>> N^{''} @>>> 0 \end{CD}$$ be a commutative diagram of $$A$$-modules and homomorphisms, with the rows exact. Then there exists an exact sequence $$0\longrightarrow \mathrm{Ker}(f')\stackrel{\bar{u}}\longrightarrow \mathrm{Ker}(f) \stackrel{\bar{v}}\longrightarrow \mathrm{Ker}(f^{''})\stackrel{d}\longrightarrow \mathrm{Coker}(f^{'})\stackrel{\bar{u}^{'}}\longrightarrow \mathrm{Coker}(f)\stackrel{\bar{v}^{'}}\longrightarrow \mathrm{Coker}(f'') \longrightarrow 0$$ in which $$\bar{u}, \bar{v}$$ are restrictions of $$u, v$$, and $$\bar{u}^{'}, \bar{v}^{'}$$ are induced by $$u^{'}, v^{'}$$.

Here, $$d: \mathrm{Ker}(f'') \to \mathrm{Coker}(f')$$ is defined by $$x'' \mapsto y+\mathrm{Im}(f')$$, where $$y$$ is such that $$v(x)=x''$$ with $$f(x)=u'(y)$$.

I showed that $$d$$ is well-defined. However, I am struggling to show exactness, particularly for $$\mathrm{Ker(f'')}$$. I showed that $$\mathrm{Im}(\bar{v}) \subset \mathrm{Ker}(d)$$, but I don't know where to proceed in showing $$\mathrm{Ker}(d) \subset \mathrm{Im}(\bar{v})$$.

I know that if $$x'' \in \mathrm{Ker}(d)$$, we have that there exists $$x\in M$$ such that $$x''=v(x)$$ where $$f(x)=u'(n')$$, $$n'\in \mathrm{Im}(f')$$. Also, $$f''(x'')=0$$, since $$x''\in \mathrm{Ker}(f'')$$. To show $$x'' \in \mathrm{Im}(\bar{v})$$, we need to find $$a\in M$$ such that $$v(a)=x''$$ and $$f(a)=0$$.

I simply don't know what to do. I tried using $$a=x$$ but I got stuck. $$v$$ is surjective, so there are certainly points of $$M$$ that map to $$x''$$, but I fail to see how to find them in $$\mathrm{Ker}(f)$$. I also don't see where to use $$f''(x'')=0$$. If anyone could guide me in the right direction, it would be much appreciated.

• This is a good question; thanks for sharing your progress so far :) It's worth nothing that result is very well-known: people call it the "Snake Lemma". Famously, the movie "It's My Turn" features a scene where a professor (correctly!) proves that the connecting homomorphism $d$ is well-defined. Worth a watch if you haven't seen it before! youtube.com/watch?v=etbcKWEKnvg. Mar 2 at 2:51

Suppose $$x'' \in \ker(d)$$. As you note, we then know that here is some $$x \in M$$ and $$y \in N'$$ such that $$u'(y) = f(x)$$, $$v(x) = x''$$, and $$y \in \operatorname{img}(f')$$. We want to find some $$a \in M$$ such that $$v(a) = x''$$ and $$f(a) = 0$$.

Whatever $$a$$ ends up being, we will have $$v(a-x) = v(a) - v(x) = x'' - x'' = 0$$, so we will have $$a-x \in \ker(v) = \operatorname{img}(u)$$. So we will try to construct $$a$$ as $$x + b$$ for some clever choice of $$b \in \operatorname{img}(u)$$.

What property will we need $$b$$ to satisfy? Well, we just need that $$f(a) = 0$$, and we know $$f(x) = u'(y)$$, so we will need $$f(b) = -u'(y)$$.

If we want $$b \in \operatorname{img}(u)$$, we really will need to find some $$z \in M'$$, and then set $$b = u(z)$$. Then our desired condition reads $$f(u(z)) = -u'(y)$$. My commutativity of the diagram, this is the same as $$u'(f'(z)) = -u'(y)$$. So wouldn't it be nice if we could choose $$f'(z) = -y$$?

We can! We already know $$y \in \operatorname{img}(f')$$ (this is the part of our assumption coming from the fact that $$d(x'') = 0$$). So that does it :)

Putting this all together:

Since $$y \in \operatorname{img}(f')$$, pick some $$z \in M'$$ such that $$f'(z) = y$$. Then set $$a = x'' - u(z)$$. We have $$v(a) = v(x'') - v(u(z)) = x - 0 = x$$ and $$f(a) = f(x'') - f(u(z)) = u'(y) - u'(f'(z)) = u'(y) - u'(y) = 0,$$ so $$x \in \operatorname{img}(\overline{v})$$, as desired. $$\square$$

• Thank you for the clear and thoughtful response! Makes sense to me now. Mar 2 at 3:29