Let $A$ be a commutative ring with the unit element (not necessarily being an integral domain) and $M$ be a finitely generated $A$-module. The following proposition is well-known:
If $A$ is an integral domain and $M$ is torsion-free, then there is an injective $A$-module morphism $M\to A^{n}$ with some $n\in\mathbb{Z}_{\ge0}$.
My question is: Is the above proposition still true even if we do not assume $A$ is an integral domain?
The essentially same question is asked here
but OP there gave too strong definition for "torsion-freeness" when $A$ is not an integral domain. I think if we use the following definitions, then the problem is not so immediate as in 1:
- We say $a\in A$ is a zero-divisor if there is a non-zero $b\in A$ such that $ab=0$.
- We say $a\in A$ is regular if $a$ is not a zero-divisor.
- We define the torsion submodule $T(M)$ of $M$ by
$$
T(M) := \{x\in M\mid\text{there is a regular $a\in A$ such that $ax=0$}\}.
$$ - We say $M$ is torsion-free if $T(M)=0$.
Best Answer
Take the ring to be $A=k[x,y]/(x^2,xy,y^2)$ and $M=A/(x)$. Then $M$ is torsion free in your definition (since the only regular elements in the ring are units). If $M\subset A^n$ for some. $n$, one easily checks that $M$ is contained in $(x,y)^n\subset A^n$. But, the annihilator of $(x,y)^n$ is $(x,y)$, while the annihilator of $M$ is $(x)$. This contradiction shows that $M$ is not contained in a free module.