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Let $R$ be a PID and $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow0$$ a short exact sequence of finitely generated $R$-modules. If $A\neq 0$, can we conclude that $B$ and $C$ are not isomorphic?

user26857
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Tom
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  • You can conclude the specific map $B \to C$ is not an isomorphism, which is usually the thing you want to know know about. You could do so even without the hypothesis that the modules are finitely generated or that $R$ is a PID; the argument is valid in any abelian category. –  Dec 09 '17 at 20:11
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    If $M\simeq M/N$ then $M\stackrel{\pi}\to M/N\simeq M$ (here $\pi$ stands for the canonical projection) is a onto endomorphism of $M$, so an isomorphism, and thus $N=0$. See https://math.stackexchange.com/questions/239364/surjective-endomorphisms-of-finitely-generated-modules-are-isomorphisms. (I've changed the notation in order to match with the linked thread.) – user26857 Dec 09 '17 at 20:18
  • Btw, you don't need $R$ a PID. – user26857 Dec 09 '17 at 20:26

1 Answers1

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Suppose $A$ is nonzero but $B\cong C$.

As the rank of $B$ is the rank of $A$ plus the rank of $C$, then for $B\cong C$ we need $A$ to have rank zero; that is be torsion. Then $A$ is a direct sum of modules of the shape $R/P^k$ with $P$ a prime ideal. Tensor with $R_P$, the localisation of $R$ at $P$ (which is flat) for some $P$ for which $R/P^k$ appears in the decomposition of $A$. We may thus assume that $R$ is a discrete valuation ring. We can regard $A$ as a submodule of the torsion of $B$. As $B$ splits as a direct sum of a free module and a torsion module then we may assume $B$ is torsion. Now we use the notion of module index: for a torsion $A\subseteq B$ with $B/A\cong\bigoplus R/P^{k_i}$ then $|B:A|=\prod P^{k_i}$. This is multiplicative: $|B:A||A:0|=|B:0|$. This means that $|B:A|=|C:0|\ne |B:0|$, so we get a contradiction to $B\cong C$.

Angina Seng
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