Short exact sequences of finitely generated modules over a PID.

Question

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?

Answer 1

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 $B\cong C$.