# Short exact sequence of modules over Artinian local ring

Let $$(R,\mathfrak m, k)$$ be an Artinian local ring. So for every non-zero finitely generated $$R$$-module $$M$$, we have $$\mathfrak m\in \mathrm{Ass}(M)$$, hence we have an exact sequence $$0\to k\to M$$, so in particular, $$0\to k\to R$$.

My question is:

If $$0\to A\to B\to C\to 0$$ is an exact sequence of finitely generated $$R$$-modules such that $$A,C$$ are torsion-less, then is $$B$$ also torsion-less ?

Here, a finitely generated module $$M$$ is called torsion-less iff it embeds into a free module of finite rank, or equivalently, if the canonical map $$M\to M^{**}$$ is injective.

Let $$R=(\mathbb{Z}/4\mathbb{Z})[x]\,/\,(2x,x^2).$$ Then $$R$$ is finite, so Artinian, and contains a unique maximal ideal $$(2,x)$$, so local.

Consider the exact sequence of finitely generated $$R$$ modules:$$0\to\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}/4\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\to 0,$$ where $$x$$ acts as $$0$$ on all three modules. Clearly the modules at both ends above embed in $$R$$, with the generators of the modules mapping to $$x\in R$$.

If the middle module embedded in $$R^n$$ for some integer $$n$$, then each generator of the middle module would map to an element of $$R^n$$ with co-ordinates all lying in the annihilator of $$x$$, and at least one co-ordinate having additive order $$4$$.

However no element of $$R$$ has order $$4$$ and lies in the annihilator of $$x$$. We conclude that the middle module is not torsion-less.