Tensoring a module with an ideal

Question

say $M$ is an $A$-module without any other presumed properties. When is the map $\phi: I\otimes_A M\longrightarrow IM$, defined by: $\phi(i\otimes_A m)=im$ not injective?

Thanks in advance!

Answer 1

It can be shown that the map you ask about is injective for all ideals $I$ of $A$ if and only if $M$ is flat over $A$. (In fact it is enough to consider finitely generated ideals. See Tag 00HD.)

For a particular ideal $I$, injectivity of the map $I\otimes_AM\to M$ can be reformulated in terms of $\mathrm{Tor}$. (I am not sure how useful this reformulation is to you, as it is pretty formal.) Since $A$ is a free (hence flat) $A$-module, $\mathrm{Tor}_1^A(A,M)=0$, so there is an exact sequence

$$0\to\mathrm{Tor}_1^A(A/I,M)\to I\otimes_AM\to M\to M/IM\to 0.$$

(Look at the long exact $\mathrm{Tor}$ sequence obtained by applying $-\otimes_AM$ to the short exact sequence $0\to I\to A\to A/I\to 0$.) So the kernel of the map you ask about can be identified with $\mathrm{Tor}_1^A(A/I,M)$, and the map $I\otimes_AM\to M$ is (not) injective if and only if this $A$-module is (not) zero.

EDIT: To elaborate on Bernard's comment, if $I$ is a pure ideal of $A$, which means that $A/I$ is a flat $A$-module, then $I\otimes_AM\to M$ will be injective for every $A$-module $M$, regardless of whether or not $M$ is flat over $A$. A finitely generated ideal is pure if and only if it is generated by an idempotent (see 05KK). So any ring of the form $A_1\times A_2$ with $A_1$ and $A_2$ nonzero will have proper, nonzero pure ideals.

Answer 2

As the general question was answered in a comment already (non-flatness), let me give an example. Consider the $\mathbb{Z}$-module $M = \mathbb{Z}/m\mathbb{Z}$ and the ideal $I = n\mathbb{Z} \subset \mathbb{Z}$. Then we get $$IM=n\mathbb{Z}\cdot \mathbb{Z}/m\mathbb{Z}=\gcd(n,m)\mathbb{Z}/m\mathbb{Z}$$ and since $I\cong \mathbb{Z}$ we also get $$I\otimes_\mathbb{Z}M\cong\mathbb{Z}\otimes_\mathbb{Z}M\cong M=\mathbb{Z}/m\mathbb{Z}.$$ Now $IM$ has $\frac{m}{\gcd(n,m)}$ many elements, while $I\otimes_\mathbb{Z}M$ has $m$ elements. That means if we choose $m$ and $n$ properly (in a way such that $\gcd(n,m) > 1$), there cannot be an injective map $I\otimes_\mathbb{Z} M\longrightarrow IM$.