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I'm interested in knowing a concrete description of what elements of Tor modules $\mathrm{Tor}^i_R(M,N)$ "are". As it stands I have no real intuition for, say, maps between Tor modules induced by functoriality.

I have a couple of pipe dreams -- one is to have a description in the same vein as the well-known description of Ext in terms of extensions (and the Baer sum). Another is to have a description in special cases, e.g. $\mathrm{Tor}^i(R/I,M)$ over a local ring $R$. (is there something along the lines of the description of $\mathrm{Ext}$ in terms of regular sequences?)

Of course there are a handful of nice cases, like $\mathrm{Tor}^1(R/I,R/J) = I \cap J / IJ$ and $\mathrm{Tor}^1(R/I,M) = I \otimes M$ when $I \subseteq \mathrm{Ann}(M)$. Or, somewhat similarly, if

$$0 \to K \to F \to M \to 0$$

is a short exact sequence with $F = R^n$ a free module, then $\mathrm{Tor}^1(R/I,M) = (K \cap IF)/IK$. So this gives a semi-concrete description as "$I$-linear relations between generators modulo $I$-multiples of relations" or something.

I will also accept explanations as to why such a description doesn't seem to exist, or why the best we can do is try to understand things in terms of free resolutions and "generators and relations and relations between relations...".

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  • $\begingroup$ You're familiar with the construction of $\text{Tor}^*(-, B)$ as left-derived functors of $-\otimes B$, right? The only other significance of $\text{Tor}^*$ (at least along the lines of the extension of group extensions you mentioned for $\text{Ext}$) I can think of is Serre's formula for intersection numbers, but I don't enough about that to talk about it usefully. $\endgroup$ – anomaly Oct 30 '14 at 4:12
  • $\begingroup$ @anomaly Yes, I'm fine with derived functors, at least in the abstract. But I don't have any idea what the actual elements of a Tor module represent (aside from generalities that would apply equally well to any derived functor). $\endgroup$ – Jake Levinson Oct 30 '14 at 4:43
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    $\begingroup$ How about this: Take an $k$-algebra $A$ with $A_0 = k$ and $A_n = 0$ for $n < 0$. Then $\text{Tor}^1_A(k, k)$ has a basis corresponding to a minimal system of generators of $A$, and $\text{Tor}^2_A(k, k)$ has a basis corresponding to a minimal system of relations for $A$. Unfortunately, I know I have a great reference for the results above but I can't remember exactly what it is. (A book of Serre's, I think? Maybe Eisenbud?) $\endgroup$ – anomaly Oct 30 '14 at 5:17
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    $\begingroup$ @anomaly Perhaps you're thinking of Serre's "Local Algebra". It would be nice if someone could explain the intuition behind why the higher $\text{Tor}$ functors are necessary to correctly define the intersection multiplicity, a supremely geometric thing. I have an intuition, kind of, but nothing as concrete as what the OP probably wants. $\endgroup$ – Alex Youcis Oct 30 '14 at 22:26
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    $\begingroup$ @anomaly Thanks for the example! $\endgroup$ – Jake Levinson Oct 31 '14 at 23:34
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I think in general this is tough, maybe because you can define $Tor$ without ever even saying the word element! Part of the beauty of the thing is that $Tor_1$ can be an obstruction to the preservation of many constructions (all those built out of sequences kernels, cokernels, direct sums and direct products). But you can come up with many physical interpretations this way.

For instance: say I'm in $k[x,y]$ and I want to intersect $I =(x^2, xy)$ and $J = (3x +5, y)$. Is intersecting these two ideals first and extending to $k[x,y,z]$ the same as taking $I (k[x,y,z]) \cap J (k[x,y,z])$? Yes, and you could probably prove it directly for any $I,J$. But you could also write down this construction in terms of kernels and cokernels-- and then just invoke the fact that $k[x,y,z]$ is flat over $k[x,y]$. There's no reason to do it in such a categorical way in this case-- but what if you take $k[x,y]/(x^2-y)$ instead of $k[x,y,z]$? How far are you off now by intersecting before extending instead of after? The answer is roughly "the size of a couple of Tor$_1$'s".

This kind of leaves the higher Tor's a mystery. As anomaly alluded to, when you're working over a local or graded ring $R$ with residue field $k$, there is a unique minimal free resolution $F_i$ of any module $M$ (essentially because of Nakayama's lemma). Then $Tor_i(k, M) = k \otimes F_i$, so that Tor really is just a sequence of vector spaces of the same rank the free resolution.

To see why this can be neat in more geometric setting: take $R = k[x,y,z]$ thought of as the homogeneous coordinate ring of $\mathbb P^2$. We're considering graded modules over this ring, and consequently $Tor_i(M,k)$ is a graded vector space. Let $X$ be a set of six points in $\mathbb P^2$. Sometimes all the points will lie on a conic, but usually not. And via a minimal free resolution, we can check whether or not this is true in terms of the grading on $Tor_1(R/I_X, k)$: If the degree 2 part is nonzero then yes, and otherwise no. Further, if the degree two part is two dimensional then a full pencil of conics contains $X$. This is a bit tautological (we're just considering the degree of the minimal generators of $I_X$). For $Tor_2(R/I_X, k)$ we're playing the "relations between relations" game again, but it's still pretty physical: say that $X$ miraculously lies on $2$ conics $\mathbb V(f_1), \mathbb V(f_2)$. Then are there linear forms $l_1, l_2$ such that $l_1 f_1 = l_2 f_2$? We might expect not, this a lot like assuming that the conics share a common factor-- but maybe it does happen for certain points . Anyways, you can detect whether the conics are close in this sense by looking at the degree $3$ component of $Tor_2(R/I_X, k)$, and so on. Going on like this, anything to do with the betti numbers in the minimal free resolution is translated into a statement about Tors.

It would be cool to have a better interpretation than special cases like these, but I don't know anything good.

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  • $\begingroup$ Thanks for the example! I had heard of minimal free resolutions, but didn't know the connection to Tor. I like that example a lot, since it actually gives a basis for the module, not just a computation of its size (and/or whether it vanishes). It also illustrates the rigidity nicely, since the construction clearly shows that once one Tor module vanishes, the higher ones will too. $\endgroup$ – Jake Levinson Oct 31 '14 at 23:41
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    $\begingroup$ You can find a lot of examples of this sort in the literature on syzygies, by the way; Eisenbud has a whole book. $\endgroup$ – Hoot Nov 1 '14 at 21:40
  • $\begingroup$ Yeah, the Geometry of Syzygies is a very good read for these minimal free resolutions/betti numbers $\endgroup$ – Phil Tosteson Nov 2 '14 at 17:12
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I'm really not an expert here, but I guess that the Serre intersection formula indicates that the Tor functors resp. their alternating lengths allow us to bring two subvarieties into general position instead of just taking the naive intersection formula. In the language of derived algebraic geometry the formula simplifies a lot (MO/12236) and indicates that Tor functors are just relicts of the past in which tensor products were not derived and one had to deal with the non-exactness of the tensor product.

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  • $\begingroup$ I'm not sure I agree that the derived formula is 'simpler', since the derived tensor product functor still encapsulates all the Tor modules. For that matter, the same reasoning with Hom instead of $\otimes$ indicates (reasonably enough) that Ext functors are also 'relics of the past', but that doesn't prevent us from having a concrete description of them! $\endgroup$ – Jake Levinson Oct 31 '14 at 23:44
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    $\begingroup$ @JakeLevinson: Sure, but often it is much more convenient to work with $\otimes^L$ instead of all functors $\mathrm{Tor}^i$ separately. Also, the fact that $\otimes^L$ encapsulates $\mathrm{Tor}^i$ doesn't mean that you have think about $\mathrm{Tor}^i$ when working with $\otimes^L$. Generally speaking, a complex is better and more flexible than its sequence of homology groups. And similarly a space is better and more flexible than its sequence of homotopy groups. $\endgroup$ – Martin Brandenburg Nov 1 '14 at 8:29
  • $\begingroup$ Fair enough. Thanks for the extra thoughts. $\endgroup$ – Jake Levinson Nov 1 '14 at 16:01

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