TITLE: Exact sequence of $n$th powers of abelian groups
QUESTION [8 upvotes]: Let $A,B,C$ be finitely generated abelian groups. Assume that there is an exact sequence $$0 \to C \to A^n \to B^n \to 0,$$where $A^n = A \oplus \dotsc \oplus A$ as usual. It is not assumed that $A^n \to B^n$ is induced by some $A \to B$.

Assume that $C$ can be generated by $<n$ elements, does it follow $C=0$?

I have proven this for $n=2$ here and numerical experiments suggest that it also holds for $n=3$. Of course the same question can be asked for finitely generated modules over a PID.

REPLY [5 votes]: Here's a partial reduction; I write it here since I was too sloppy in the comments.
Lemma Let $R$ be a noetherian (commutative) ring. Let $$0\to M\to N\to Q\to 0$$ be a short exact sequence of f.g. modules with $M\neq 0$. Then there exists an artinian local ring quotient $R'$ of $R$ such that the induced surjective map $N\otimes_R R'\to Q\otimes_R R'$ is non-injective.
Indeed, let $m\in N$ be a nonzero element in the kernel of $N\to Q$. Since any f.g. module over a noetherian ring is residually of finite length ($\star$), there exists an ideal $I$ of $R$ such that $R'=R/I$ is of finite length and $m\notin IN$. This proves the claim at least with $R'$ artinian (maybe not local), but local follows since $R'$ is a direct product of finitely many local rings.
This reduces the question from noetherian rings to its local artinian quotients (this, in the case of $R=\mathbf{Z}$, reduces to finite $p$-groups). 
($\star$) Let $W$ be a maximal submodule of $N$ not containing $m\neq 0$. It's enough to prove that $N/W$ has finite length. Indeed, in the quotient $N/W$, the intersection of nonzero submodules is nonzero. But if $N/W$ has infinite length, it has some non-maximal associated prime $P$, but then in $R/P$ the intersection of nonzero ideals is zero (e.g., using powers of a single ideal), contradicting that $R/P$ is isomorphic to a submodule of $N/W$.

For people not familiar to commutative algebra, here's the particular case of abelian groups.
Lemma'  Let $$0\to M\to N\to Q\to 0$$ be a short exact sequence of f.g. abelian groups with $M\neq 0$. Then there exists a prime power $n\ge 1$ such that the induced surjective map $N/nN\to Q/nQ$ is non-injective.
Indeed, let $m\in N$ be a nonzero element in the kernel of $N\to Q$. There exists $n\ge 1$ such that $m\notin nN$. This proves the lemma at least with $n\ge 1$ (maybe not prime power), and then passing to some $p$-power is immediate by the Chinese remainder theorem.