Example of a submodule $N\subseteq M$ (over some commutative ring) such that the morphism $N\otimes M\to M\otimes M$ is not injective.

Question

I would like to find an example of a submodule $N\subseteq M$ (over some commutative ring) such that the morphism $N\otimes M\to M\otimes M$ is not injective. Notice that the ambient module is the same as the one we're tensoring with (this is the point of the question).

Answer 1

Let $R=k[\![x,y]\!]$ with $\mathfrak{m}=(x,y)$. Take $M:=\mathfrak{m}$ and let $N:=\mathfrak{m}^2=(x^2,xy,y^2)$. Then $N \subseteq M$ and there is a short exact sequence $$0 \to \mathfrak{m}^2 \to \mathfrak{m} \to \mathfrak{m}/\mathfrak{m}^2 \to 0$$

The module on the right is isomorphic to $k^{\oplus 2}$. We note by dimension shifting that $\operatorname{Tor}^R_1(\mathfrak{m},\mathfrak{m}) \cong \operatorname{Tor}^R_3(k,k)$. Since $R$ has global dimension $2$ (or one can compute the minimal free resolution of $k$ and argue directly), we thus have $\operatorname{Tor}^R_1(\mathfrak{m},\mathfrak{m})=0$. Applying $- \otimes_R \mathfrak{m}$ and considering the long exact sequence in $\operatorname{Tor}$, we get an exact sequence $$0 \to \operatorname{Tor}^R_1(k^{\oplus 2},\mathfrak{m}) \to \mathfrak{m}^2 \otimes_R \mathfrak{m} \to \mathfrak{m} \otimes_R \mathfrak{m} \to k^{\oplus 2} \otimes_R \mathfrak{m}.$$ By dimension shifting, $\operatorname{Tor}^R_1(k^{\oplus 2},m) \cong \operatorname{Tor}^R_2(k,k)^{\oplus 2}$. But $R$ is a local ring of global dimension $2$, so $\operatorname{Tor}^R_2(k,k) \ne 0$. In fact, one can show directly by computing the minimal free resolution of $k$, that $\operatorname{Tor}^R_2(k,k) \cong k$. Thus the induced map $\mathfrak{m}^2 \otimes_R \mathfrak{m} \to \mathfrak{m} \otimes_R \mathfrak{m}$ is not injective.