On a paper I'm reading, the author (Fulton) asks to apply the Snake Lemma to a square diagram:
how is that possible? I tried to construct a standard snake lemma diagram starting from this one, but didn't succeed. In this contest $R$ is a discrete valuation ring, and $A,B,C$ are matrices with entries in $R$ with determinant not $0$ which of course satisfy $C=AB$ (from the diagram).
Complete the square to the following diagram with exact rows:
\begin{array}{c} & & R^n\phantom C & \xrightarrow{B} & R^n\phantom A & \to & \operatorname{coker}B & \to & 0 \\ & & \downarrow C & & \downarrow A & & \downarrow \\ 0 & \to & R^n\phantom B & \xrightarrow{id} & R^n\phantom A & \to & 0 &\to &0 \end{array}
Now the snake lemma is applicable, and $\ker(\operatorname{coker}B\to 0) = \operatorname{coker}B$ and $\operatorname{coker}(\operatorname{coker}B\to 0) = 0$. This gives you the exact sequence $$ \ker C\to\ker A\to\operatorname{coker}B\to\operatorname{coker}C\to\operatorname{coker}A\to0, $$ and as $R$ is a DVR and $A$ has nonzero determinant, $\det A$ is a (nonzero) non-zero-divisor in $R$, and hence $\ker A = 0$.