Is Tor always torsion? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T13:34:53Z http://mathoverflow.net/feeds/question/31714 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31714/is-tor-always-torsion Is Tor always torsion? Rasmus 2010-07-13T15:15:46Z 2010-07-13T16:12:47Z <p><strong>Question:</strong> Is the following statement true?</p> <blockquote> <p>Let $R$ be an associative, commutative, unital ring. Let $M$ and $N$ be $R$-modules. Let $n\geq 1$. Then $Tor_n^R(M,N)$ is torsion.</p> </blockquote> <p>By " $Tor_n^R(M,N)$ is torsion" I mean that every of its elements is a torsion element. Maybe I want to assume that $R$ is an integral domain.</p> <p>Remark: The above statement is true if $R$ is a principal ideal domain (then $Tor_n^R$ vanishes for $n\geq 2$) and $M$ and $N$ are finitely generated (then we can apply the <a href="http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain" rel="nofollow">structure theorem</a>).</p> http://mathoverflow.net/questions/31714/is-tor-always-torsion/31723#31723 Answer by norondion for Is Tor always torsion? norondion 2010-07-13T15:44:02Z 2010-07-13T15:44:02Z <p>Some thoughts:</p> <ol> <li><p>Since Tor commutes with colimits, one can reduce to the finitely generated case.</p></li> <li><p>By choosing projective resolution, we can reduce to $\mathrm{Tor}_1^R$.</p></li> <li><p>If $M = R/r$ is cyclic, we have $\mathrm{Tor}_1^R(R/r, N) = {}_rN$ the $r$-torsion in $N$.</p></li> </ol> http://mathoverflow.net/questions/31714/is-tor-always-torsion/31725#31725 Answer by Sasha for Is Tor always torsion? Sasha 2010-07-13T16:08:23Z 2010-07-13T16:08:23Z <p>$Tor$ commutes with extension of scalars, hence (if $R$ is an integral domain and $K$ is its field of fractions), we have $$Tor_n^R(M,N) \otimes_R K = Tor_n^K(M\otimes_R K,N\otimes_R K).$$ The right-hand-side vanishes for $n\ge 1$, because $K$ is a field. Hence $Tor$ vanishes after tensoring with $K$, which means that $Tor$ is torsion.</p> http://mathoverflow.net/questions/31714/is-tor-always-torsion/31727#31727 Answer by David Speyer for Is Tor always torsion? David Speyer 2010-07-13T16:09:49Z 2010-07-13T16:09:49Z <p>Yes. Set $K = \mathrm{Frac} \ R$.</p> <p><strong>Lemma:</strong> Let $\ldots \to C_2 \to C_1 \to C_0$ be a complex of $R$-modules. Suppose that $C^{\bullet} \otimes_R K$ is exact (but not necessarily surjective at $C_0$). Then $H_k(C_{\bullet})$ is $R$-torsion for $k>0$.</p> <p><strong>Proof:</strong> Let $v \in C_k$ with $dv=0$. So $d(v \otimes 1)=0$. By the exactness of $C^{\bullet} \otimes_R K$, there is $u \in C_{k+1} \otimes_R K$ with $du=v$. Write $u=\sum w_i \otimes (f_i/g_i)$, with $f_i/g_i \in K$ and $w_i \in C_{k+1}$. Set $g=\prod g_i$ and $w=\sum \left( \prod_{j \neq i} g_j \right) f_i w_i$. Then $dw=gv$, so $[v]$ is $g$-torsion in $H_k(C_{\bullet})$. QED</p> <p>Take resolutions <code>$A_{\bullet} \to M$</code> and <code>$B_{\bullet} \to N$</code> by free $R$-modules. Then <code>$\mathrm{Tor}_{\bullet}(M,N)$</code> is the homology of the complex formed by collappsing the double complex <code>$A_{\bullet} \otimes_R B_{\bullet}$</code>. Note that <code>$\left( A_{\bullet} \otimes_R B_{\bullet} \right) \otimes_R K \cong (A_{\bullet} \otimes_R K) \otimes_K (B_{\bullet} \otimes_R K)$</code>.</p> <p>Since $A^{\bullet}$ is exact, so is $A^{\bullet} \otimes_R K$. Thus $A_{\bullet} \otimes_R K$ breaks up as a direct sum of complexes of the form $\ldots \ldots 0 \to K \to K \to 0 \ldots$, and the complex $\ldots \to 0 \to K$, with the $K$ in the last position. (This uses the Axiom of Choice; I suspect you should be able to avoid it, but I don't see how right now.) The complex $B \otimes_R K$ breaks up into pieces of the same kind. So the double complex breaks up into squares <code>$\begin{smallmatrix} K &amp; \to &amp; K \\ \downarrow &amp; &amp; \downarrow \\ K &amp; \to &amp; K \end{smallmatrix}$</code>, vertical strips <code>$\begin{smallmatrix} K \\ \downarrow \\ K \end{smallmatrix}$</code>, horizontal strips <code>$\begin{smallmatrix} K &amp; \to &amp; K \end{smallmatrix}$</code> and, in position $(0,0)$, some isolated copies of $K$.</p> <p>Only summands of the last type contribute to the cohomology of the double complex, so the double complex obeys the hypotheses of the lemma and we are done.</p>