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.


1 Answer 1


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.


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