Extension of linearly independent set to a basis of a free module over a quotient of a Dedekind domain

Question

Let $A$ be a Dedekind domain and a non-zero ideal $N$; we consider the commutative ring $R = A/N$.

Let $S$ be a linearly independent subset of $R^m$. Can $S$ necessarily be extended to a basis of $R^m$?

Answer 1

Yes (and the assumption that $R$ is finite is unnecessary; you just need $N$ to be a nonzero ideal). Factoring $N$ into prime ideals as $\prod P_i^{d_i}$, the Chinese remainder theorem gives $R\cong \prod A/P_i^{d_i}$. Working on each coordinate separately, we may thus assume that $N=P^d$ for some prime $P$. We can then localize at $P$ and assume $A$ is a DVR with maximal ideal $P$ generated by a uniformizer $p$ and $R=A/P^d$.

Now I claim that the linearly independent set $S$ remains linearly independent over $R/P$ in the quotient $(R/P)^m$. Indeed, a linear relation mod $P$ would give scalars $c_s\in R$ such that $\sum_{s\in S} c_ss \in P$ but $c_s\not\in P$ for some $s$. This just means that $\sum c_s s$ is divisible by $p$, and thus is annihilated by $p^{d-1}$. That is, $\sum p^{d-1}c_ss=0$. But this is a nontrivial relation among the elements of $S$, since some $c_s$ is not in $P$ and thus $p^{d-1}c_s$ is nonzero for that $s$.

We can thus extend $S$ to a set $T$ whose image mod $P$ is a basis for the vector space $(R/P)^m$ over $R/P$. By Nakayama, $T$ generates $R^m$, and thus is a basis since it has $m$ elements (it induces a surjective homomorphism $R^m\to R^m$ which is then automatically an isomorphism; see Surjective endomorphisms of finitely generated modules are isomorphisms though there are simpler arguments you can give in this case since $R$ is Noetherian and even Artinian).