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So here's my problem: I have no intuition for how a "generic" module over a commutative ring should behave. (I think I should never have been told "modules are like vector spaces.") The only examples I'm really comfortable with are

  • vector spaces,
  • finitely generated modules over a PID, and
  • modules over a group algebra.

But when I try to apply these examples to understanding something like Nakayama's lemma I don't have any intuition to bring to the table. So, what other examples of modules should I keep in mind so that

  • I'm not fooled by my intuition about vector spaces, and
  • I can concretely understand what something like Nakayama's lemma means, at least in an important special case?
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Revisited the question to add jsmath to my answer, wanted to add this: "(locally flat) modules are like vector spaces spread over the base" (that is, bundles over Spec R as said before). –  Ilya Nikokoshev Nov 6 '09 at 17:28

9 Answers 9

up vote 10 down vote accepted

Yes, there is a big class of modules that have an intuition different from the abstract algebra, namely the ones that come from an algebraic geometry. If $R$ is a (say, Noetherian) commutative ring, then you consider a scheme $\mathrm{Spec}\, R$ and (finitely generated) modules over $R$ correspond to coherent sheaves on $\mathrm{Spec}\, R$.

For example, a skyscraper sheaf on point $p$ corresponds to the module $R/m_p$, where $ m_p$ is a maximal ideal that defines point. More interestingly, a locally trivial bundle over $\mathrm{Spec}\, R$ corresponds to a projective module (this is exactly the statement locally free = projective).

The objects that seem unnatural on algebraic side, like support of a module, in fact have a clear geometric meaning — support of a sheaf is where the sheaf sits, that's it. And this is the geometric setting that explains Nakayama's lemma: the lemma says you can consider things locally for a coherent sheaf (see wikipedia).

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Here's a bonus: I was thinking about good books that would emphasize this connection between commutative algebra and schemes and the name 'Eisenbud' came to my mind. –  Ilya Nikokoshev Oct 22 '09 at 17:54
Some additional comment is the following. The type of vector bundles you are considering above are algebraic bundles. To give a sense that $\mathcal{C}^{\infty}$-bundles may be very distinct objects from those that you are considering above, here is a result due to Quillen (1978) Algebraic vector bundles over $\mathcal{C}^{\infty}$ are trivial. this result is trivial on the topological category. –  Csar Lozano Huerta Nov 13 '09 at 6:12

I think a good philosophy for finding interesting modules is to find interesting rings.

The polynomial ring k[x1,x2,x3,...] is a great source of counterexamples to theorems that require a Noetherian hypothesis, because for something "pathological", it's pretty easy to think about. You can think of what the graded version of Nakayama means about this ring, or you can localize at the maximal ideal to get a local ring, and think about the usual Nakayama lemma.

Also, since you're comfortable with group rings (or group "algebras"), you should keep in mind that semigroup rings are just as easy to make. If S is any semigroup (written multiplicatively), and R is a ring, you can make a new ring R[S] by "adjoining the elements of S" in just the way you'd expect. If S is typically written additively, I usually think of the elements of S as exponents of a formal variable, so (x^s)(x^t)=x^(s+t).

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Really? Finitely-generated abelian groups didn't disabuse you of the notion that "modules are like vector spaces?" That's usually my first line of defense against bad intuitions about modules -- I think, does this hold for this next-easiest-to-think-about case?

After that, I agree with Joel that ideals are the best things to think about. Plus, ideals of the polynomial ring have the obvious geometric interpretation, so that's nice.

One important special corollary of Nakayama's lemma, in particular, is the going up theorem. There are nice geometric ways of thinking about going up, which I'll let someone else explain since I only just learned this stuff and am liable to screw it up.

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I guess I should be more specific. The property all of the examples I gave have in common is that I understand their structure theorems. –  Qiaochu Yuan Oct 22 '09 at 7:12
Oh, right -- if you haven't done at least a good chunk of the exercises in Atiyah-Macdonald chapter 2, or the equivalent in Eisenbud or Matsumura, do so. –  Harrison Brown Oct 22 '09 at 7:14

Another thing you can do is look at the frontispiece of Miles Reid's commutative algebra book (click the frontispiece link in the contents; Google books won't link directly to the right page). It's subtitled "Let A be a ring and M an A module". I think the picture shows pretty much everything, how the module looks over the generic point, it's support and so on. If you look at the picture it's really easy to remember Nakayama's lemma, just as ilya said.

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Unfortunately, the Google books preview doesn't seem to show the frontispiece. –  Qiaochu Yuan Oct 22 '09 at 13:49
@Qiaochu: yes, it does. The link I added takes you to the contents, from which you can click the "frontispiece" link. –  Anton Geraschenko Oct 22 '09 at 19:57
Can someone explain this picture? –  Dinakar Muthiah Nov 3 '09 at 1:40
@Dinakar: the top part of the picture is a module M. The labels M_P and M_Q denote the localizations at primes. The bottom part of the picture is the ring R, with primes P and Q labeled. The curves drawn in R represent the support of M. Presumably the thicker lines are indicating embedded primes (non-minimal associated primes), and eta is a generic point (non-maximal prime). –  Justin DeVries Nov 6 '09 at 18:24

Following the idea of Andrew of looking at interesting rings to find interesting modules, I suggest you take a look at rings which do not satisfy the IBN (invariant basis number property). Those are rings for which exist left (right) modules $M$ such that $M^n \cong M^m$ for some different positive integers $m,n$; this is, rings for which a notion of "dimension" (rank) of modules cannot be properly defined.

In the late 1950s, W.G. Leavitt ((1),(2),(3)) constructed explicit examples of rings $R$ that do not have the IBN property, that in rigor asks for any two bases (i.e., linearly independent spanning sets) of a free left $R$-module to have the same number of elements, generalizing a well-known property of fields. Noetherian rings and commutative rings are included among the many classes of rings having this property (so that if you really want to find modules quite different in nature to vector spaces, you must move to the non-commutative setting). It can be shown that if $R$ is a unital ring, then $R^1 \cong R^n$ (as left $R$-modules, for example) for some $n>1$ if and only if there is a set of $2n$ elements in $R$ which produce the appropriate isomorphisms as matrix multiplications by an $n$-row vector and an $n$-column vector with entries in $R$, that is, if there exist elements $x_1, ... , x_n \in R$ and $y_1,..., y_n \in R$ such that $x_iy_j = \delta_{ij}1_R$ for all $i,j$, and $\sum_{i=1}^n y_ix_i = 1_R$. If a unital ring $R$ does not have IBN, it is said that $R$ has module type $(m,n)$, where $m\in {\mathbb N}$ is the minimal number such that $R^m \cong R^n$ for some $n>m$ and $n$ is minimal given $m$. In his seminal paper (3), Leavitt proved that for each pair of positive integers $n>m$ and any field $K$ there always exists a $K$-algebra of module type $(m,n)$, namely the quotient of the unital free associative $K$-algebra in the appropriate number of variables satisfying the relations described above. This algebra is denoted by $L_K(m,n)$ and called the Leavitt $K$-algebra of type $(m,n)$.The $K$-algebras of module type $(m,n)$ need not be unique: for example, if $V$ is an infinite dimensional vector space and $R={\rm End}_K V$ is its ring of endomorphisms, then it is easily seen that $R^1 \cong R^2$; but it can be shown that $R$ is not isomorphic to $L_K(1,2)$.

(1) W.G. Leavitt. Modules over rings of words. Proc. Amer. Math. Soc. 7 (1956), 188-193.

(2) W.G. Leavitt. Modules without invariant basis number. Proc. Amer. Math. Soc. 8 (1957), 322-328.

(3) W.G. Leavitt. The module type of a ring. Trans. Amer. Math. Soc. 42 (1962), 113-130.

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Any ring comes equipped with a whole bunch of 'natural' modules. Namely, if J is an ideal of R then J is an R module and so is R/J. To my mind these are really the modules that we can understand well.

Notice that in the classification theorem for f.g. modules over a PID, we say that every module is isomorphic to a direct sum of these ones.

You can certainly use these modules as test cases for understanding theorems about general modules even though, in general, it won't really be true that every module comes from these ones.

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I'll add to Ilya's answer that it is instructive to look at the "provocative" picture in the front piece of Miles Reid's undergraduate commutative algebra. I can't post it here because it is copyrighted material, But you can look it up in Amazon. Search for the phrase: "Frontispiece: Let A be a ring and M be an A-module..."

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You can view the frontispiece on Google books: books.google.com/… –  Anton Geraschenko Oct 22 '09 at 19:59
@Anton: thanks for the link ! –  David Lehavi Oct 22 '09 at 20:21

It's not over a commutative ring, but if you have a space X and a covering space Y of X, then the homology of Y is an R-module where R is the integral group ring of the group of deck transformations. These things are all over the place.

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Not to follow you around leaving comments about this or anything, but one of the best examples of this is the Alexander module of a knot K in S^3: here we let X be the complement S^3-K and Y its universal abelian cover. This is a Z-fold cyclic cover, so the Alexander module is a module over R=Z[Z]=Z[t,t^{-1}]. Its 1st elementary ideal (the ideal generated by the mxm minors of an nxm presentation matrix) is always principal, and its generator is the Alexander polynomial of K. –  Steven Sivek Nov 6 '09 at 14:26
:) That's ok. Sounds like you're on Alexander module patrol. P.S. I *heart* the Fox Jacobian. –  Richard Kent Nov 6 '09 at 14:52

There's also vector spaces as K[X]-modules via a linear transformation T (perhaps I should say a K[T]-module).

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I preferred when you said K[X]-modules. –  Jonas Meyer Nov 11 '09 at 9:57

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