Quotient of a torsion free module is torsion free

Question

If $R$ is an integral domain and $M$ is a torsion-free $R$ module. Then is $M/IM$ a torsion-free $R/I$ module, where $I\neq 0$ is an ideal of $R$?

I think there is an obvious counterexample to it, but I cannot find it.

Answer 1

Let's try to write down the naive proof attempt, see where it goes wrong, and extract a counterexample from it :

Let $x\in M/IM, r\in R/I$ such that $rx = 0$. Then, pulling back to $M$ and $R$, we hvae $rx \in IM$.

The point is that this does not imply $x\in IM$ or $r\in I$.

For instance, pick $I$ which is not a prime ideal, so for instance $xy\in I,x\notin I, y\notin I$ and $M=R$. Then in $R/I$, you get $xy = 0$, but $x\not = 0$ in $R/I$, $y\not = 0$ in $M/IM$ ($=R/I$).

e.g. $\mathbb Z/6$ is not torsion-free as a $\mathbb Z/6$-module, because $2\times 3= 0$

($R$ is torsion-free as an $R$-module if and only if it's an integral domain)

Here's an example where $r$ is not a zero divisor, in fact an example where $R/I$ is integral : let $R=\mathbb Z[X]$ and $M = \bigoplus_\mathbb N \mathbb Z$, where $X$ acts by shifting copies of $\mathbb Z$ and multiplying by $2$, that is, $n$ in position $i$ is sent by $X$ to $2n$ in position $i+1$.

This is torsion free : indeed, if $P=a_n X^n +$ lower degree terms is a nonzero polynomial acting on some element whose maximal nonzero coordinate is $x$ in position $r$, then $P\cdot x$ has $2^n a_n x$ in position $r+n$ (everyone else is sent to lower positions), which is nonzero.

However, with $I=(X), R/I \cong \mathbb Z$ and $M/IM$ has torsion, for instance $1$ in degree $2$ is $2$-torsion.