The spectrum of a commutative ring, pt. 1

Introduction

Spec_Z

In this post, I will introduce the notion of the spectrum of a commutative ring which is a fundamental component in the definition of the objects known as schemes in algebraic geometry, twisted creations of none other than the great Alexandre Grothendieck himself (who incidentally now lives in isolation from human society). Hence, this post is supposed to be about algebraic geometry, but there are so many different perspectives one can take towards spectra, so I’ll be trying to make sense of all of them.

An overwhelmingly pervasive theme in mathematics is that one can understand the geometry of some space X by looking closely at algebraic data, namely certain kinds of functions defined on X. A good example of this is the Banach-Stone theorem, which says a compact Hausdorff space can be recovered from the algebra of continuous functions defined on it. In fact, it turns out that compact Hausdorff spaces “are the same thing as” commutative, unital C*-algebras (that is, there is a so-called equivalence of categories). This is amazing since it means we can transport constructions, ideas, and intuitions from one setting to the other. The Serre-Swan theorem provides a similar analogy between vector bundles and projective modules which I heard about from this document.

To name another example: sheaf theory, whose associated “cohomology theory” is so indispensable to modern algebraic geometry, was also born of this insight, or rather, the more refined insight that one should look at functions defined on every little “piece” of X (that is, on the open sets U \subseteq X), not just the ones defined on all of X, because there may be very few of those. Although sheaves on X are often defined simply as certain contravariant functors, it is common to come across textbooks that define them as special topological spaces “over X“: étalé spaces. Lurking behind this multi-faceted concept, once again, is a rather nice equivalence of categories.

The classical case

In classical algebraic geometry, one looks at so-called affine n-space \mathbb{A}^n(k) over some algebraically closed field k, and notices that to every affine algebraic set X \subseteq \mathbb{A}^n, one can associate an algebraic object: its so-called coordinate ring \Gamma(X) = k[x_1,\ldots,x_n]/I(X), in other words, the “ring of polynomial functions defined on X“. As one would expect, we have \Gamma(\mathbb{A}^n) = k[x_1,\ldots,x_n].

Notice that \Gamma(X) is not just some random commutative ring off the street: since we have an inclusion k \hookrightarrow \Gamma(X) (namely, the constant polynomials!), \Gamma(X) is in fact a k-algebra, indeed a finitely generated one, simply because k[x_1,\ldots,x_n] is polynomially generated by the indeterminates x_1, \ldots, x_n (“duh”). Further, because I(X) is always a radical ideal of k[x_1,\ldots,x_n], this translates into the quotient ring \Gamma(X) lacking nilpotent elements. Hence \Gamma(X) is a finitely generated, reduced k-algebra!

That’s not even the end of the story: any “homomorphism of algebraic sets” (polynomial map, to be precise) f : X \to Y induces a homomorphism of k-algebras \Gamma(Y) \to \Gamma(X) between their coordinate rings, going in the opposite direction: just take \phi \mapsto \phi \circ f. So we have a (contravariant) functor. One of the crown jewels of classical algebraic geometry was the discovery that in fact, this functor constitutes an equivalence of categories. Hence, studying affine algebraic sets is the same as studying finitely generated reduced k-algebras.

Points and maximal ideals

In the sense above, it seems like we now know how to convert geometry into algebra. The point of the “spectrum” idea is that we want to “go the other way”: from algebra back to geometry, the catch being that we now want to do it for any commutative ring, not just these special k-algebras. Thus, hoping to always obtain something as simple as an affine variety is no longer reasonable. However, it will be a geometric object: to be precise, it will be a topological space, further equipped with a “sheaf of rings” (but let’s not get into that for now).

Hilbert’s celebrated Nullstellensatz tells us that since k is algebraically closed, every maximal ideal of k[x_1,\ldots,x_n] is of the form (x_1-a_1,\ldots,x_n-a_n) for some point (a_1,\ldots,a_n) \in \mathbb{A}^n. More generally, the maximal ideals of \Gamma(X) correspond exactly to the points of X. So in this case, we don’t have any lame stuff like the maximal ideal (x^2 + 1) \subseteq \mathbb{R}[x] which “doesn’t correspond to a point”. One final time for emphasis: points = maximal ideals.

Armed with this observation, one might guess we can recover X, as a geometric object, merely as the “set of all maximal ideals of the ring \Gamma(X)“. Thus, given a commutative ring R, your first idea might be to take the set of all the maximal ideals of R and define a topology on it somehow. Well, that’s a reasonable guess, but unfortunately, it has shitty functorial properties. [Thanks to Erik Crevier for this motivation].

To elaborate, suppose we defined \text{Spec } R to be the set of all maximal ideals of R. If this is the “right” thing, then to any ring homomorphism f : R \to R' we should be able to obtain a map between \text{Spec } R and \text{Spec } R' in some natural way. The first thing that comes to mind is to pull back a maximal ideal of R' to R along f, but this fails miserably, simply because the preimage of a maximal ideal of R' need not be maximal in R. Therefore, we don’t actually get a map \text{Spec } R' \to \text{Spec } R. It takes a trained eye to perceive the graveness of this signal, but at least now, what we do next won’t seem completely bizarre. Hopefully.

Prime ideals

Since pullbacks of prime ideals along ring homomorphisms are still prime, and recalling that the ideal I(X) of a variety (that is, irreducible algebraic set) X was prime, perhaps the answer is really to look at prime ideals, not maximal ideals. This turns out to be the case — we define (as a set) the spectrum of R, denoted \text{Spec } R, to just consist of the prime ideals of R. Note immediately that if R=\Gamma(X) for some variety X then \text{Spec } R has elements corresponding to each point of X as well as an “extra” point for each subvariety of X.

Thus the spectrum of \mathbb{Z} is just (2), (3), (5), (7), and so on… plus this extra point (0), which people call the “generic point” for reasons I don’t want to get into. It’s pictured at the top of this post.

Now let’s look at an (unrealistically nice) example, where R is a familiar polynomial ring.

specCx

The field \mathbb{C} of complex numbers is algebraically closed. Furthermore, the polynomial ring k[x] in one variable is a PID when k is a field, so one can show that all nonzero prime ideals are maximal. As a result, the Krull dimension of k[x] (length of the longest chain of prime ideals in that ring) is always 1. It is a fact that the algebraic (that is, Krull) dimension coincides with the intuitive, geometric notion of dimension. The prime ideals of \mathbb{C}[x] are pictured above: it’s just a “complex line” (the “line” terminology being justified by our previous remark) worth of maximal ideals, sitting above the (non-maximal) prime (0).

For a more intricate example, we can try looking at the ring \mathbb{Z}[x] of integer-coefficient polynomials. Here is a reproduction of a sketch of \text{Spec } \mathbb{Z}[x] that David Mumford, one of the “big names” in algebraic geometry, drew in his famous “Red Book” of varieties and schemes (which ironically is now published by Springer — with a yellow cover).

mumford-21

He called it the arithmetic surface; the Krull dimension of \mathbb{Z}[x] is 2 (you probably know this, since this ring is a classic example of a non-PID). If you’re interested, it is explained in great depth here, and even related to a bunch of crazy \mathbb{F}_1-mathematics here. The longer you stare at it, the more you realize how much information it’s encoding.

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About mlbaker

just another guy trying to make the diagrams commute.
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One Response to The spectrum of a commutative ring, pt. 1

  1. Ehsaan says:

    Alexandre

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