Let $J$ be an ideal in a Noetherian local domain $(R, \mathfrak m,k )$ such that the natural map $J \otimes_R J \to J^2$ sending $a \otimes b \to ab$ is an isomorphism.
Then is it true that $J$ is flat as an $R$-module i.e. is it true that $\operatorname {Tor}_1^R (J,k)=0$ ?
Looking at the induced long exact sequence of $\operatorname{Tor}$ after tensor the short exact sequence $0\to J \to R \to R/J\to 0$ with $-\otimes J$, we can see that $\operatorname {Tor}_1^R (J,R/J)=0$ .
So in particular my claim is true if $J=\mathfrak m$. But I'm not sure what happens otherwise …
Please help.
Best Answer
$\require{AMScd}$
Yes. Note that every finitely generated flat module over a Noetherian local ring is free. As $J$ is an ideal in a domain, your question reduces to asking why $J$ is principal when the natural map $J \otimes_R J \to J^2$ is an isomorphism. Let $f:J \otimes_R J \to J^2$ be the natural map defined on elementary tensors by $f(a \otimes b)=ab$.
Suppose $J$ is not principal and let $x_1,\dots,x_n$ be a minimal generating set for $J$. We claim that one of the elements $x_i \otimes x_j-x_j \otimes x_i$ must be nonzero in $J \otimes_R J$. As $f(x_i \otimes x_j-x_j \otimes x_i)=x_ix_j-x_jx_i=0$, this will conclude the proof. To see that one of these elements must be nonzero, suppose, for the sake of contradiction, that they are all $0$. Given any $R$-module $M$, we have the natural antisymmetrization map $\phi_M:\bigwedge^2_R(M) \to M \otimes_R M$ given on elementary wedges by $\phi(a \wedge b)=a \otimes b-b \otimes a$. To say that an $x_i \otimes x_j-x_j \otimes x_i=0$ is to say that $x_i \wedge x_j \in \ker \phi_J$. As $\bigwedge^2_R(J)$ is generated by the $x_i \wedge x_j$, our assumption forces $\phi_J$ to be the zero map. However, $\phi_{J \otimes_R k}$ is naturally identified with $\phi_J \otimes \operatorname{id}_k$ via the following commutative digram:
\begin{CD} \bigwedge\nolimits^2_R(J \otimes_R k) @>\phi_{J \otimes_R k}>> (J \otimes_R k) \otimes_R (J \otimes_R k)\\ @VhVV @VVgV\\ \bigwedge\nolimits^2_R(J) \otimes_R k @>\phi_J \otimes \operatorname{id}_k>> (J \otimes_R J) \otimes_R k \end{CD}
where $h$ is the isomorphism defined via $h((a \otimes 1) \wedge (b \otimes 1))=(a \wedge b) \otimes 1$, and where $g$ is the isomorphism defined via $g((a \otimes 1) \otimes (b \otimes 1))=(a \otimes b) \otimes 1$.
Thus $\phi_{J \otimes_R k}$ is the zero map. But $\phi_{J \otimes_R k}$ is always injective, since $J \otimes_R k$ is a $k$-vector space. Thus $\bigwedge^2_R(J \otimes_R k)=0$, and therefore $\dim_k(J \otimes_R k) \le 1$. By Nakayama's lemma, $J$ is principal, a contradiction.