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.

  • 2
    $\begingroup$ 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. $\endgroup$ Mar 2 at 2:51

1 Answer 1


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. $\square$

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

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