# Torsion-free tensor powers

Let $R$ be an integral domain. If $M$ is an $R$-module such that every tensor power of $M$ over $R$ is $R$-torsion-free, then is $M$ necessarily flat as an $R$-module? If not, then does this implication hold for $R$-algebras $M$, or at least for $R$-algebras $M$ between $R[X]$ and $K[X]$, where $K$ is the quotient field of $R$?

A while ago David Speyer showed in his nice answer below that the answer is yes if $M$ is finitely generated, but the particular modules I'm interested in ($R$-algebras between $R[X]$ and $K[X]$) are not finitely generated.

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It was hard for me to understand the content your question, so let me rephrase it (perhaps also for others): Every flat module is torsionfree. Over a Prüfer domain, every torsionfree module is flat. Now in general assume that even all powers of $M$ are torsionfree and nonzero; does this suffice to conclude that $M$ is flat? If not (which I suspect), what is a counterexample? –  Martin Brandenburg Aug 18 '11 at 9:38

There are no examples with $M$ finitely generated. We first reduce to the case that $M$ is local. Suppose that $M^{\otimes n}$ is torsion-free for all $n$. Then any localization of $M^{\otimes n}$ is torsion-free. (See, for example, Exercise 14.5.I in Ravi's notes. I had a nagging suspicion that there was a noetherian hypothesis needed for this, but Ravi is pretty careful about that and he doesn't give one.) So, if we have proved the local case, then we know that every $M_{\mathfrak{p}}$ is flat. Flatness can be checked locally; see Ravi's proposition 25.2.3.

$\def\mm{\mathfrak{m}}$ So we now assume that $R$ is local, with $\mm$ the maximal ideal and $k = R/\mm$. Suppose that $M$ is not flat. Let $V = M \otimes k$ and let $n = \dim_k V$. We will show that $M^{\otimes n}$ has torsion.

Proof: Let $f_i$ be a basis of $V$ and let $e_i$ in $M$ be a preimage of $f_i$. By Nakayama's lemma, the map $R^{\oplus n} \to M$ sending $(x_1, \ldots, x_n)$ to $\sum x_i e_i$ is surjective so, if $M$ is not flat, it must have a kernel. In other words, there must be some $(x_1, \ldots, x_n)$ in $\mm^n$, not all $0$, such that $\sum x_i e_i=0$. Without loss of generality, let $x_n$ be nonzero.

Set $$\Delta := \sum_{\sigma \in S_n} \epsilon(\sigma) e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)} \in M^{\otimes n}.$$ Here $\epsilon(\sigma)$ is the sign of the permutation $\sigma$. I claim that $\Delta$ is nonzero but $x_n \Delta =0$.

Proof that $\Delta$ is nonzero: By the associativity of tensor product, $M^{\otimes n} \otimes k \cong V^{\otimes n}$. The image of $\Delta$ in $V^{\otimes n}$ is nonzero, so $\Delta$ is nonzero.

Proof that $x_n \Delta=0$ is zero: Note that $$x_n e_1 \otimes \cdots \otimes e_n = e_1 \otimes \cdots \otimes e_{n-1} \otimes \left(- x_1 e_1 -x_2 e_2 -\cdots - x_{n-1} e_{n-1} \right).$$ Similarly expand each of the $n!$ terms. You get an antisymmetric expression of degree $n$ in $e_1$, ..., $e_{n-1}$, so it must be zero.

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Very nice, +1. It is perhaps worth noting a counter-example in the non-domain case (when torsion-free means that no non-zero element is killed by a regular element). We can take $R= k[x,y]/(xy)$ and $M=R/(x)$. Then $M$ is torsion-free and $M^{\otimes n}=M$ for all $n$. –  Hailong Dao Aug 18 '11 at 16:31
Nice, indeed! I agree with Martin Brandenburg that there should be an example, which you show must be non finitely generated. –  Jesse Elliott Aug 18 '11 at 19:17

It is hard for me not to mention the following splendid result by Auslander-Lichtenbaum:

If $R$ is regular local with Krull dimension $d$, then for any finitely generated module $M$, $M^{\otimes n}$ is torsion-free for some $n\geq d$ if and only if $M$ is flat.

Auslander only proved it for unfamified regular local rings (see the references here) but the key ingredient for the proof, namely certain Tor-rigidity property for finitely generated modules was later completed by Lichtenbaum.

Very recently, this was generalized to some extend to non-finitely generated case in this paper.

I believe Auslander's theorem extends to the case when $R$ is an isolated hypersurface singularity over a char. $0$ field and $d =\dim R$ is even (because the key ingredients are now available for that case).

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Thank you for posting that. These are nice results I wasn't aware of. The recent paper you posted extends the result to modules $M$ of essentially finite type over an essentially smooth algebra $R$ over a field, so any counterexample must not satisfy those hypotheses. –  Jesse Elliott Aug 18 '11 at 19:33

Here is a vague idea for how one might prove this in general. $M$ is flat if and only if, for all finitely generated ideals $\langle x_1, x_2, \ldots, x_n \rangle$ of $R$, we have $\mathrm{Tor}_1(M, R/\langle x_1, \ldots, x_n \rangle) = 0$. $N$ is torsion free if and only if, for all nonzero $x$ in $R$, we have $\mathrm{Tor}_1(N, R/x)=0$. Can we somehow get a relation between $\mathrm{Tor}_1(M, R/\langle x_1, \ldots, x_n \rangle)$ and $\mathrm{Tor}_1(M^{\otimes n}, R/(x_1 \cdots x_n))$?

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I'm not sure how to find such a relation. Are you implying that you think torsion-freeness of the tensor powers of $M$ implies $M$ is flat, even if $M$ is not finitely generated? –  Jesse Elliott Aug 23 '11 at 7:14