Is Tor always torsion? - MathOverflow

Is Tor always torsion?

2010-07-13T15:15:46Z

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).

Answer by norondion for Is Tor always torsion?

2010-07-13T15:44:02Z

Some thoughts:

Since Tor commutes with colimits, one can reduce to the finitely generated case.
By choosing projective resolution, we can reduce to $\mathrm{Tor}_1^R$.
If $M = R/r$ is cyclic, we have $\mathrm{Tor}_1^R(R/r, N) = {}_rN$ the $r$-torsion in $N$.

Answer by Sasha for Is Tor always torsion?

2010-07-13T16:08:23Z

$Tor$ 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.

Answer by David Speyer for Is Tor always torsion?

2010-07-13T16:09:49Z

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.