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I'm doing the exercises in "Introduction to commutive algebra" by Atiyah&MacDonald. In chapter two, exercises 24-26 assume knowledge of the Tor functor.

I have tried Googling the term, but I don't find any readable sources. Wikipedia's explanation use the the term "take the homology", which I don't understand (yet).

Are there any good explanations of what the Tor functor is available online not assuming any knowledge about homology?

The first exercise: "If $M$ is an $A$-module, TFAE:

1) $M$ is flat

2) $\operatorname{Tor}_n^A (M,N)=0$ for all $n>0$ and all $A$-modules $N$.

3) $\operatorname{Tor}_1^A (M,N)=0$ for all $A$-modules $N$."

Thanks in advance.

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The Tor functor is a homological construction, so I'm not sure I understand the question. A definition of the homology of a chain complex is available at en.wikipedia.org/wiki/Chain_complex . –  Qiaochu Yuan Jan 4 '11 at 13:22
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In the context of Atiyah+MacDonald, you can simply ignore those 3 exercises. –  Mariano Suárez-Alvarez Jan 4 '11 at 13:23
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A good text reference for this is the book by Dummit and Foote. But as @Mariano wrote, you can skip those exercises since understanding Tor is not essential to the rest of the book. –  Vitaly Lorman Jan 4 '11 at 16:14
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3 Answers

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The Tor functors are the derived functors of the tensor product. The starting observation is that if $0 \to M' \to M \to M'' \to 0$ is a ses of modules and $N$ is any module (let's work over a fixed commutative ring $R$), then $M' \otimes N \to M \otimes N \to M'' \otimes N \to 0$ is exact, but you don't necessarily have exactness at the first step. (This is what "the tensor product is right-exact" means.)

Now, whenever you have such a sequence that fails to be short exact at just one step, the general philosophy is that it should be the end of a long exact sequence. The long exact sequence should be like $$\dots \to ?_1 \to ?_2 \to ?_3 \to M' \otimes N \to M \otimes N \to M'' \otimes N \to 0$$ where the question marks indicate that we don't know what goes there yet.

The general approach is provided by the theory of derived functors. Derived functors allow one to construct, associated to a right (or left) exact functor $F$, a collection of functors $L_i F$ for $i \geq 0$ (I'm just going to handle the right-exact case henceforth), such that for each ses $0 \to M' \to M \to M'' \to 0$, the "almost ses" $$FM' \to FM \to FM'' \to 0$$ can be completed to a les $$\dots \to L_2F(M'') \to L_1F(M') \to L_1F(M) \to L_1F(M'') \to FM' \to FM \to FM'' \to 0.$$ This long exact sequence relies on "connecting homomorphisms" $L_nF(M'') \to L_{n-1}F(M')$, which are required to be functorial in the ses.

The exact construction of a derived functor uses projective resolutions of the object $M$ and a reasonable amount of diagram-chasing. I will refer you to books on homological algebra.

So suppose we consider the functor sending $M \to M \otimes N$. Its derived functors are denoted $\mathrm{Tor}_i(M,N)$. This means that the Tor functors provide the missing part of the les that expands out the "almost ses" of the tensor product above.

One reason the Tor functors are so useful is that they provide a very effective criterion for flatness. Recall that a module $N$ is flat if tensoring with it is exact. When you derive an exact functor, you just get the initial functor in dimension zero (actually, you always do for this), and then zero elsewhere. So if $N$ is flat, $\mathrm{Tor}_i(M,N) = 0$ for $i>0$. In fact, using the les, one can easily show that the implication is reversible, even if one only has $\mathrm{Tor}_1(M,N)=0$ for all $M$.

Let me give an example where things work out very nicely. If $R$ is a noetherian local ring with residue field $k$, and $M$ is a finite $R$-module, then it turns out $M$ is free if and only if $\mathrm{Tor}_1(k,M)=0$; in particular, freeness is the same thing for flatness (under these hypotheses). (For a proof, see the argument behind Theorem 3.2 in ch. 14 of http://www.people.fas.harvard.edu/~amathew/CRing.pdf.)

An application of this is provided by the observation (which can be deduced from this) that if the $\mathrm{Tor}$ functors of the pair $(k,k)$ vanish in high dimension, then the global dimension of the ring is finite. This is fairly interesting because the global dimension is a statement about all modules (and, strictly speaking, involving the functors $\mathrm{Ext}$ that derive Hom, not Tor), while the above reduction is to one involving only the residue field.

This is, in fact, the key observation behind one half the proof that a noetherian local ring is regular iff its global dimension is finite. This argument can be found in EGA 0-IV, sec. 17 (it doesn't presuppose anything beyond dimension theory, which you can read about in sec. 16).

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Thanks a lot for the long answer. I understand a lot more now. However, a question: the Wikipedia article about derived functors says the extension of the SES is extended "canonically". Does this mean that for any other such extension, it can be factored through the canonical one? (given the definition, I see now that the exercise I gave in the question is trivial) –  Fredrik Meyer Jan 4 '11 at 18:16
    
@Fredrik: Dear Fredrik, there is a sense in which derived functors are "universal." The notion of a "$\delta$-functor," (or connected sequence of functors $\{T_i\}$) due to Grothendieck in his Tohoku paper, axiomatizes the notion of a derived functor (namely, there is required to be a functorial long exact sequence associated to each les). It turns out that if a $\delta$-functor in positive dimensions is "coeffaceable" (i.e. there for each $M$ there is $P$ such that $T_i(P) \to T_i(M)$ is zero for $i>0$), then the $\delta$-functor is universal in the sense that, given two $\delta$-functors... –  Akhil Mathew Jan 4 '11 at 18:28
    
$\{T_i\}$ and $\{U_i\}$ such that one of them is coeffaceable (I forget which), and there is a morphism $T_0 \to U_0$, then there is a unique morphism of $\delta$-functors $T_i \to U_i$ commuting with the boundary maps. This is explained in Lang's Algebra as well. So it is not so much the sequence that is universal, but the $\delta$-functor, if I understand correctly. –  Akhil Mathew Jan 4 '11 at 18:29
    
best not to try to explain universal $\delta$-functors in the comments, I think. I recommend that interested readers look in any book on homological algebra instead. –  Pete L. Clark Jan 4 '11 at 20:31
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@Pete, @Akhil: in fact, I was going to answer the question by redirecting to a book! I admire your tenacity, Akhil, in writing this here :) –  Mariano Suárez-Alvarez Jan 4 '11 at 22:52
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You will be a lot more motivated to learn about Tor once you observe closely how horribly tensor product behaves.

Let us look at the simplest example possible. Consider the ring $R=\mathbb C[x,y]$ and the ideal $I=(x,y)$. These are about the most well-understood objects, right? What is the tensor product $I\otimes_RI$? This is quite nasty, it has torsions: the element $u = x\otimes y - y\otimes x$ is non-zero, but $xu=yu=0$!

Tor gives you a black box to understand this kind of things. Take the short exact sequence $0 \to I \to R \to R/I \to 0$ and tensor with $I$ we get:

$$0 \to \text{Tor}_1(R/I,I) \to I\otimes I \to I \to I/I^2 \to 0$$

from which you can extract:

$$0 \to \text{Tor}_1(R/I,I) \to I\otimes I \to I^2 \to 0$$

But $\text{Tor}_1(R/I,I) = \text{Tor}_2(R/I, R/I) = \mathbb C$ by standard homological algebra. So now everything fits nicely: the map from $I\otimes I \to I^2$ takes $f\otimes g$ to $fg$, and the kernel is generated by the element $u$, which is killed by $I$, so it is isomorphic to $R/I \cong \mathbb C$.

To summarize: tensor product, despite being a fundamental operation, is actually quite bad, and Tor helps you to understand it.

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I can sympathize with this question because I am about to teach a first (graduate level) course in commutative algebra. No homological algebra of any sort is a prerequisite: I'll be happy if all of my students are comfortable with exact sequences.

On the other hand, just a little bit of Tor is extremely helpful when studying commutative algebra (and homological methods become increasingly important later on in the subject). Just now I am re/writing my notes on flat modules, including the "Equational Criterion for Flatness". The best source I have found so far is Matsumura's Commutative Ring Theory which does use a little Tor in the proof. After having asked a question about this on MO and mulling over the answers, I think that avoiding Tor entirely is not a good approach. So where am I going to tell my students to go to learn about Tor?

If you know me, you're probably expecting me to say that I've written up my own notes on Tor. But I didn't want to do that -- basic homological algebra has a kind of "just do it" aspect to it, such that my own personal take on the subject is only going to be different from someone else's if I happen to screw it up (which is a non-negligible possibility if I try to write up something quick and short). Since Matsumura uses Tor, I looked to see what he did.

In fact he has written an essentially optimal $9$-page precis of homological algebra, Tor and Ext as Appendix B of his text: pages 274-282. (There is also Appendix A on tensor products, direct and inverse limits, which is fantastic as well and should be read first.) I do not aspire to do any better than his treatment, so I will make it available to my students when the course begins.

If anyone wishes to have a copy of this passage, please let me know.

(Perhaps I should add that on the first page of my copy of Matsumura's book it is written "Pete L. Clark, 2/1/98". So I've had this one for a while, and my general engagement with the subject of commutative algebra has also been there, off and on but most recently on, for almost exactly $13$ years. Nevertheless I still find portions of his book to be intimidating: he covers ground so quickly and with little enough explanation that I often have to take a deep breath and gird myself to read something out of his book. I mention this because I do not feel this way at all about Appendices A and B: they read like a novel.)

Finally, I think it is a little weird that Atiyah and Macdonald introduce Tor in their book so as only to use it in a few exercises. The most significant application that I can recall is the characterization of rings over which every module is flat (i.e., absolutely flat rings), which is interesting but hardly seems essential. Let me reiterate that on a first reading you can absolutely just ignore all the Tor, and it will go fine.

Added: As of today there is now a treatment of Tor (and Ext) in my commutative algebra notes: see $\S 3.10$. This was intended to fill out a whirlwind lecture I gave on derived functors in my course a couple of weeks ago, and I tried to give enough detail (especially, on definitions) so that the student can see what is really going on, without getting bogged down in any of the nontrivial diagram chases that are so unpleasant to both write up and (I've often found, at least) to read. I had my copy of Weibel's Homological Algebra at my side throughout, and a lot of the more technical proofs are farmed out to there.

As a side remark, that lecture would probably have been better if I had taken at least part of it verbatim from Akhil Mathew's short answer to this question. In general I like an expository challenge, but this homological stuff kind of tires me out...

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Very few of the classical books on commutative algebra have explanations! It is an ingredient that our commutative friends have only recently began using. –  Mariano Suárez-Alvarez Jan 4 '11 at 22:54
    
@Mariano: I think I agree with you, but the lack of explanation combined with the fast pace (and also perhaps the somewhat random order of the topics) makes Matsumura's book especially hard. For instance, Bourbaki's text has close to the minimum possible amount of explanation, but it is thoughtfully organized and doesn't make any sharp turns, so it's not so hard to see where it's going. –  Pete L. Clark Jan 4 '11 at 22:58
    
By the way, of course right after I said something slightly dismissive about absolutely flat rings, I started editing my notes on commutative algebra and encountered them in a fundamental way: that $R/\operatorname{nil}(R)$ is absolutely flat is a useful equivalent condition for $\operatorname{Spec}(R)$ to be Hausdorff. –  Pete L. Clark Jan 4 '11 at 23:00
    
In the course I took last semester, the teacher did not use Tor (in fact, he talked rather little about flatness), but he made free use of Ext when concluding the course with the homological theory (when, in fact, the people in the course -- almost all undergrads -- often didn't know what exactness meant at the start). The approach he took was to give the definition via a projective resolution, state the properties of Ext in one lecture, and prove a few of them as people were curious. After that, he just went ahead and referred to them without saying anything general about derived functors. –  Akhil Mathew Jan 4 '11 at 23:43
    
@Pete: heh. Absolutely flat rings, but non-commutative (thus von Neumann regular, rather) were a good part of my afternoon here :P –  Mariano Suárez-Alvarez Jan 5 '11 at 1:24
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