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On a paper I'm reading, the author (Fulton) asks to apply the Snake Lemma to a square diagram:

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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).

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  • $\begingroup$ What is the paper? Does the author indicate the result you are supposed to prove by applying the Snake Lemma? $\endgroup$ – André 3000 May 19 '17 at 3:31
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    $\begingroup$ @Quasicoherent sure, we are supposed to prove that $0 \to coker B \to coker C \to coker A \to 0$ is exact. (The paper is "Eigenvalues, Invariant Factors, Highest Weights and Schubert Calculus"). $\endgroup$ – Maffred May 19 '17 at 3:41
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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 $\ker A = 0$.

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