# Is Tor always torsion?

Question: Is the following statement true?

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.

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.

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 structure theorem).

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By the way, is my question in some sense ill-posed, if $R$ has zero divisors? $R$'s being an integral domain seems crucial at least for Sasha's and David's approaches. –  Rasmus Bentmann Jul 13 '10 at 18:20
Rasmus, I think you are right. If R is not a domain, sometimes "torsion" means being killed by a nonzero divisor. If that's the case, then Tor(M,N) may not be torsion. As an example, take $R=k[x,y]/(x^2)$ and $M=R/(x)$. Then $Tor_i(M,M)$ is either $0$ or $M$, which is not torsion in that sense. –  Hailong Dao Jul 13 '10 at 18:46

$Tor$ 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.

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Do you have a reference or explanation for "Tor commutes with extension of scalars"? –  Rasmus Bentmann Jul 13 '10 at 16:26
Tor does not commute with extension of scalars, unless the extension of scalars is flat. The localization which you need inthis case is flat, though. –  Mariano Suárez-Alvarez Jul 13 '10 at 16:47
In the derived sense, it commutes always. Just because the tensor product commutes with extension of scalars: $(M\otimes_R R')\otimes_{{R'}} (N\otimes_R R') \cong (M\otimes_R N)\otimes_R R'$. –  Sasha Jul 13 '10 at 20:17
But your formulas are not about derived tensor products but their homologies... –  Mariano Suárez-Alvarez Jul 14 '10 at 7:19
Tensor product commutes with extension of scalars (which in fact is a tensor product itself), hence their derived functors also commute --- $(M\otimes_R^L R')\otimes_{{R'}}^L (N\otimes_R^L R') \cong (M\otimes_R^L N)\otimes_R^L R'$. In the case of $R' = K$, as it was mentioned by Mariano, $R' = K$ is flat over $R$, so the derived product can be replaced with the usual one in the corresponding places. This gives $(M\otimes_R K)\otimes_K^L (N\otimes_R K) \cong (M\otimes_R^L N)\otimes_R K$. Taking the cohomology, we get $Tor_n^K(M\otimes_R K,N\otimes_R K) \cong Tor_n^R(M,N)\otimes_R K$. –  Sasha Jul 14 '10 at 14:55

Some thoughts:

1. Since Tor commutes with colimits, one can reduce to the finitely generated case.

2. By choosing projective resolution, we can reduce to $\mathrm{Tor}_1^R$.

3. If $M = R/r$ is cyclic, we have $\mathrm{Tor}_1^R(R/r, N) = {}_rN$ the $r$-torsion in $N$.

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Could you please elaborate on 2.? –  Rasmus Bentmann Jul 13 '10 at 15:54
Isn't 2 just connected with Tor and Ext dropping in dimension if you replace the correct of its argument modules with a projective cover of it? So you could do things like $Tor_n(M,N) = Tor_{n-1}(M,\Sigma N)$ (likely to be wrong in details - I have no books around me now!) –  Mikael Vejdemo-Johansson Jul 13 '10 at 15:59

Yes. Set $K = \mathrm{Frac} \ R$.

Lemma: 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 $k>0$.

Proof: 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

Take resolutions $A_{\bullet} \to M$ and $B_{\bullet} \to N$ by free $R$-modules. Then $\mathrm{Tor}_{\bullet}(M,N)$ is the homology of the complex formed by collappsing the double complex $A_{\bullet} \otimes_R B_{\bullet}$. Note that $\left( A_{\bullet} \otimes_R B_{\bullet} \right) \otimes_R K \cong (A_{\bullet} \otimes_R K) \otimes_K (B_{\bullet} \otimes_R K)$.

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 $\begin{smallmatrix} K & \to & K \\ \downarrow & & \downarrow \\ K & \to & K \end{smallmatrix}$, vertical strips $\begin{smallmatrix} K \\ \downarrow \\ K \end{smallmatrix}$, horizontal strips $\begin{smallmatrix} K & \to & K \end{smallmatrix}$ and, in position $(0,0)$, some isolated copies of $K$.

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.

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$A_\bullet\otimes K$ is a resolution of $M\otimes K$ as a $K$-module, and the same with $B_\bullet\otimes K$, so the complex $(A\otimes B)\otimes K$ computes $\Tor^K(M\otimes K,N\otimes K)$. The vanishing in positive degrees follows from $K$'s being a field. –  Mariano Suárez-Alvarez Jul 13 '10 at 16:13
«computes $\mathrm{Tor}^K(M\otimes K,N\otimes K)$» –  Mariano Suárez-Alvarez Jul 13 '10 at 16:14
You're right; that's a cleaner way to finish. –  David Speyer Jul 13 '10 at 16:19
(Not that it avoids AoC, I guess) In fact, the cleanest would be to remark that $\mathrm{Tor}$ commutes with localization (which is, essentially, what you did), so that $\mathrm{Tor}^R(M,N)_{(0)}=\mathrm{Tor}^K(M\otimes K,N\otimes K)$, which is zero in positive degrees. –  Mariano Suárez-Alvarez Jul 13 '10 at 16:28