Torsion-free tensor powers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:44:20Z http://mathoverflow.net/feeds/question/73120 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73120/torsion-free-tensor-powers Torsion-free tensor powers Jesse Elliott 2011-08-18T03:28:14Z 2011-08-18T20:28:24Z <p>Does there exist an integral domain $R$ and an $R$-module $M$ that is not flat over $R$ such that every tensor power of $M$ over $R$ is nonzero and $R$-torsion-free? (If such a domain $R$ exists then it cannot be a Prufer domain.)</p> http://mathoverflow.net/questions/73120/torsion-free-tensor-powers/73161#73161 Answer by David Speyer for Torsion-free tensor powers David Speyer 2011-08-18T15:19:53Z 2011-08-18T16:28:26Z <p>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 <a href="http://math.stanford.edu/~vakil/0708-216/216winter0708.pdf" rel="nofollow">Ravi's notes</a>. 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 <code>$M_{\mathfrak{p}}$</code> is flat. Flatness can be checked locally; see Ravi's proposition 25.2.3.</p> <p>$\def\mm{\mathfrak{m}}$ So we now assume that $R$ is local, with $\mm$ the maximal ideal and $k = R/\mm$. Suppose that $M$ is <em>not</em> flat. Let $V = M \otimes k$ and let $n = \dim_k V$. We will show that $M^{\otimes n}$ has torsion.</p> <p><strong>Proof:</strong> 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.</p> <p>Set <code>$$\Delta := \sum_{\sigma \in S_n} \epsilon(\sigma) e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)} \in M^{\otimes n}.$$</code> Here $\epsilon(\sigma)$ is the sign of the permutation $\sigma$. I claim that $\Delta$ is nonzero but $x_n \Delta =0$.</p> <p><strong>Proof that $\Delta$ is nonzero:</strong> 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.</p> <p><strong>Proof that $x_n \Delta=0$ is zero:</strong> Note that <code>$$x_n e_1 \otimes \cdots \otimes e_n = e_1 \otimes \cdots \otimes e_{n-1} \otimes \left(- x_1 e_1 -x_2 e_2 -\cdots - x_{n-1} e_{n-1} \right).$$</code> 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.</p> http://mathoverflow.net/questions/73120/torsion-free-tensor-powers/73164#73164 Answer by Hailong Dao for Torsion-free tensor powers Hailong Dao 2011-08-18T16:13:43Z 2011-08-18T16:13:43Z <p>It is hard for me not to mention the following splendid result by Auslander-Lichtenbaum:</p> <blockquote> <p>If $R$ is regular local with Krull dimension $d$, then for any finitely generated module $M$, $M^{\otimes n}$ is torsion-free for <em>some</em> $n\geq d$ if and only if $M$ is flat. </p> </blockquote> <p>Auslander only proved it for unfamified regular local rings (see the references <a href="http://mathoverflow.net/questions/72205/a-missing-paper-by-auslander" rel="nofollow">here</a>) but the key ingredient for the proof, namely certain Tor-rigidity property for finitely generated modules was later completed by Lichtenbaum. </p> <p>Very recently, this was generalized to some extend to non-finitely generated case in <a href="http://front.math.ucdavis.edu/1002.3652" rel="nofollow">this paper</a>. </p> <p>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 <em>even</em> (because the key ingredients are now available for that case). </p> http://mathoverflow.net/questions/73120/torsion-free-tensor-powers/73185#73185 Answer by David Speyer for Torsion-free tensor powers David Speyer 2011-08-18T20:28:24Z 2011-08-18T20:28:24Z <p>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 $\mathrm{Tor}_1(M^{\otimes n}, R/(x_1 \cdots x_n))$?</p>