17
$\begingroup$

Let's work in the category $R$-${\sf mod}$, for concreteness. I know that one can see the modules ${\rm Ext}^n_R(M,N)$ as modules of equivalence classes of $n$-extensions of $M$ by $N$ (Yoneda extensions), namely, exact sequences of the form $$0 \to N \to E_1 \to \cdots \to E_n \to M \to 0,$$with certain operations (more precisely, if one denotes such collection by ${\rm E}^n(M,N)$, there are natural isomorphisms ${\rm E}^n(M,N)\cong {\rm Ext}_R^n(M,N)$ for each $n$).

Is there anything similar for ${\rm Tor}^R_n(M,N)$?

I expect the answer to be highly non-trivial, for the following analogy: this business about $n$-extensions effectively gives us a way to describe the elements of ${\rm Ext}^n_R(M,N)$, but why should we expect any simple explanation for ${\rm Tor}$ if we cannot even describe the elements of ${\rm Tor}_0^R(M,N) = M\otimes_RN$ in general?

Apologies if by any chance this is a repeated question, a quick search on the website didn't show up anything here.

$\endgroup$

migrated from math.stackexchange.com Mar 26 at 22:03

This question came from our site for people studying math at any level and professionals in related fields.

  • $\begingroup$ I will mention that on MathOverflow the tag abstract-algebra is deprecated and each question is supposed to have a top-level tag. So if the question stays on MO, it should probably be retagged in accordance with this sites' rules. (There are some differences between the tags on MO and Mathematics.) $\endgroup$ – Martin Sleziak Mar 27 at 4:32
3
$\begingroup$

I don't know an answer for general rings/modules but the best approach is probably to write Tor in terms of Ext when you want a similar interpretation for Tor with exact sequences. For Artin algebras $R$ with duality $D$ (which include for example all quiver algebras $KQ/I$) and finitely generated modules $Y$ and $Z$ one then has $$Tor_n^R(Y,Z)=DExt_R^n(Y,D(Z)).$$ So in this case Tor for $Y$ and $Z$ has an interpretation as a dual space of extensions between $Y$ and $D(Z)$. I think the same should work for general rings/modules with a duality having good properties.

$\endgroup$
  • 1
    $\begingroup$ You certainly need some kind of finiteness condition on either $Y$ or $Z$ for this to be true $\endgroup$ – Denis Nardin Mar 27 at 5:47
  • $\begingroup$ @DenisNardin I added that the modules should be finitely generated for safety. $\endgroup$ – Mare Mar 27 at 7:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.