What is the general geometric interpretation of modules in algebraic geometry?

Question

Algebraic geometry is quite new for me, so this question may be too naive. therefore, I will also be happy to get answers explaining why this is a bad question.

I understand that the basic philosophy begins with considering an abstract commutative ring as a function space of a certain "geometric" object (the spectrum of the ring). I also understand that at least certain types of modules correspond to well known geometric constructions. For example, projective modules should be thought of as vector bundles over the spectrum (and that there are formal statements such as the Serre-Swan theorem which make this correspondence precise in certain categories). My question is, what is the general geometric counterpart of modules?

This is not a formal mathematical question, and I am not looking for the formal scheme-theoretic concept (of sheaves of certain type and so on), but for the geometric picture that I should keep in mind when working with modules.

I will appreciate any kind of insight or even just a particularly Enlightening example.

Answer 1

Roughly a module can be thought of as a vector bundle on the spectrum, where the dimension of fibers may vary. Let me give some examples and facts:

• A free modules corresponds to trival vector bundles, or more generally projective modules correspond to vector bundles as you already pointed out.

• Let $R$ be the coordinate ring of a variety and $I$ a radical. Then the $R$ module $R/I$ corresponds to attaching a one dimensional vector space on each point of $Z(I)$ and the zero vector space everywhere else. For Example $R=k[x,y]$ and $I=(x,y)$ gives the skyscraper sheaf at the origin. $I=(x)$ gives the trivial one dimensional bundle on the y-axis etc. If your Ideal is not a radical, the situation is slightly more complicated. $R/I$ can be thought of as the trival bundle on an infinitesimal neighborhood of Z(I).

• Another nice example is a geometric explanation why the tensorproduct $\mathbb Z/p \otimes_{\mathbb Z} \mathbb Z/q$ for say $p,q$ without common divisor vanishes. Our space $spec(\mathbb Z)$ consists just of a point for each prime (and a generic point). Now with the above intuition in mind, our two modules are geometrically just one dimensional vectorspaces attached to infinitesimal neighborhoods of the prime divisors of $p,q$. Since $p$ and $q$ have no common divisor there is no point where both $\mathbb Z/p$ and $\mathbb Z/q$ have nonzero fiber. As with vectorbundles, tensor produduct of modules can be thought of geometrically as fiberwise tensor product ($i^*$ commutes with $\otimes$). But of course the fiberwise tensor product vanishes because there are no points where both modules have nonzero fibers, so $\mathbb Z/p \otimes \mathbb Z/q=0$ .

• Finally any finitely generated module (more generally a coherent sheaf on a noetherian scheme) is built up of vector bundles on subspaces in the following way: There exists a stratification of spec(R) such that the module pulled back to the strata is a vector bundle. This follows from Hartshorne Ex II.5.8.

Answer 2

As you say, projective modules correspond in a highly moral way to vector bundles. Bundles pull back but do not push forward, in topologists' terms. This might be a good point at which to start. Sheaves, on the other hand, push forward easily. You don't get something for nothing (well, I've known mathematicians argue that you can elegantly do just that ...) and so there is a "picture deficit": you can think of the pullback of a fibre bundle easily in terms of the fibres you already had being attached to inverse image points. You don't have quite the same easy access to the pushforward of the sheaf of sections of a bundle.

Following up this line of thought, algebraic geometry as practised since Serre's FAC is full of pushforwards of sheaves that start off locally free. They don't end up locally free - really can't because you can't do that successfully just with vector bundles. This is more like a Grothendieck justification of what the general quasicoherent sheaves are doing in the theory.

Historically it seems that the concept of module evolved because "ideal theory", as commutative algebra was once called, needed examples that were less rigid than ideals of a ring.

Answer 3

These somewhat naive remarks are motivated by the desire to think about the question myself.

Sections of bundles are natural examples of (locally) free modules and the standard examples are the tangent and cotangent bundles of smooth varieties. The need for more general modules is the fact that kernels and cokernels of maps of (locally) free modules need not be so. In addition to ideal sheaves of subvarieties, as noted above the kahler differentials are a standard example of not necessarily locally free sheaves.

These two remarks are connected by the standard exact sequence of a closed subscheme, which exhibits the sheaf of kahler differentials of a local complete intersection, as the cokernel of a map of locally free sheaves. I,.e, the kahler differentials on a closed subvariety of a smooth variety, are the quotient of the kahler differentials on the ambient variety by the conormal sheaf of the subvariety.

Thus, the standard examples of modules are ideals of subvarieties and kahler differentials, and at least for lci subvarieties these arise as either kernels or cokernels of maps of bundles. So geometrically one could say one wants to consider differentials and subschemes, and algebraically one wants to be able to take kernels and cokernels.

Answer 4

Here is a (hopefully) enlightening example. Let $X$ be an irreducible variety over a perfect field, and consider its sheaf of Kähler differentials $\Omega_X$. If $X$ is smooth, then $\Omega_X$ is locally free, and corresponds to the cotangent bundle. If $X$ is not smooth, then the notion of "cotangent bundle" does not really make sense, but the sheaf $\Omega_X$ serves as a sort of substitute:
-The open subset of $X$ on which $\Omega_X$ is locally free consists precisely of the set of points at which $X$ is smooth.
-If $x \in X$ is a point with residue field $\Bbbk(x)$, then any reasonable definition of the cotangent space at $x$ will turn out to be equivalent to $\Omega_{X,x} \otimes \Bbbk(x)$. The singular points of $X$ are precisely those points at which the dimension of the cotangent space is "too large."

Answer 5

Have a look at Serre's definition of a sheaf [FAC], namely via étale spaces. This gives a geometric picture of a sheaf. Over every point of our topological space $X$, there sits the fiber of the sheaf. Of course we make some compatibility conditions on these fibers, namely that they vary continuously. Topologists call this a bundle on $X$. Now if $X$ has some extra structure, it is reasonable to study the bundles with some appropriate extra structure. Namely, if $X$ is a ringed space, then the fibers should be modules. The corresponding sheafes are called module sheaves. If $X$ is a scheme, then we restrict to quasi-coherent sheaves in order to involve the local affine charts. In every case, you get a special type of a bundle over $X$.

In the same way as the structure of a ring $A$ may be studied by means of the modules over $A$, the structure of a scheme $X$ may be studied by means of quasi-coherent sheaves on $X$. Actually the Reconstruction Theorem by Rosenberg justifies this. Even in the affine case this helps to enlighten some concepts of module theory. An example is the support of a module.