Martin Orr's Blog

Maths > Algebraic geometry > Functor of points

Affine k-schemes

Posted by Martin Orr on Friday, 25 September 2009 at 15:45

In my last post on functors of points, I showed that functor of points of an affine k-variety is simply the functor \mathop{\mathrm{Hom}}(B, -) for a suitable k-algebra B. Only a restricted class of k-algebras could arise as B however. So in this post I generalise this to allow B to be any k-algebra, and thereby define affine k-schemes.

Note: throughout this post, all Hom-functors are on the category of k-algebras, for a fixed field k.

Categorical definition

Recall that there is a dual equivalence between the category of affine k-varieties and the category of finitely-generated integral-domain k-algebras, sending a k-algebra X to its coordinate ring B. We saw in the last post that the functor of points of X is then \mathop{\mathrm{Hom}}(B, -).

On the algebraic side, the restriction to "finitely-generated integral-domain" seems unnecessary. \mathop{\mathrm{Hom}}(B, -) is a functor k-\textbf{Alg} \to \textbf{Set} for any k-algebra B.

So we shall expand our universe of algebraic-geometric objects to include an object called \mathop{\mathrm{Spec}_k} B for every k-algebra B, whose defining property is that its functor of points is \mathop{\mathrm{Hom}}(B, -). Such objects are called "affine k-schemes."

You might say that I haven't really defined anything, but the ethos behind this series of blog posts is that the functor of points is enough to define any algebraic-geometric object. After all, the usual way to define a geometric object is to start with a set of "points" making up the object. And here I have told you not just one set of points of \mathop{\mathrm{Spec}_k} B, but lots of them: I have told you the set of "points defined over A" for every k-algebra A.

(Note: If you know some commutative algebra or have studied schemes, then the name \mathop{\mathrm{Spec}_k} B might remind you of something. The \mathop{\mathrm{Spec}_k} B I have defined here is closely related to the \mathop{\mathrm{Spec}} B defined in terms of prime ideals, but it is best to set that definition aside for the moment. The relationship between the A-points I am considering here and the prime ideals of B is somewhat complicated; the fact that I am doing everything relative to k adds a further twist.)

Concrete interpretation

I have defined the set of A-points of \mathop{\mathrm{Spec}_k} B to be the set of k-algebra homomorphisms B \to A. That is still rather abstract, but we can give a more concrete interpretation.

Recall the idea that we started with in my first post, that the functor of points of an affine variety tells you solutions to some fixed polynomial equations over a varying ring. The same thing still holds for affine k-schemes: the proof of lemma 1 from last time doesn't depend on B being finitely generated or an integral domain. (Although we now have to look for solutions in A^\Lambda, where \Lambda is an indexing set for the generators of B, and might no longer be finite.)

As an example, I shall compare X = \mathop{\mathrm{Spec}_\mathbb{C}} \mathbb{C}[x]/(x) (which is a variety) with Y = \mathop{\mathrm{Spec}_\mathbb{C}} \mathbb{C}[x]/(x^2) (which is not). X is of course just a point; you might expect Y to be some sort of "double point".

Now both of these have the same set of \mathbb{C}-points: either because x = 0 and x^2 = 0 have the same solutions in \mathbb{C}, or because \mathbb{C}[x]/(x) and \mathbb{C}[x]/(x^2) have the same \mathbb{C}-algebra homomorphisms to \mathbb{C} -- any homomorphism \mathbb{C}[x]/(x^2) \to \mathbb{C} must map x to 0.

However we can look at points of X and Y defined over any \mathbb{C}-algebra. For example, if we take A = \mathbb{C}[\varepsilon]/(\varepsilon^2), then we find that X has just one A-point, while Y has many. In terms of solving equations, these many points arise because x^2 = 0 has not only x = 0 but also x = a\varepsilon as solutions, for any a \in \mathbb{C}. Alternatively, in terms of homomorphisms, you get a \mathbb{C}-homomorphism \mathbb{C}[x]/(x^2) \to \mathbb{C}[\varepsilon]/(\varepsilon^2) by setting x \mapsto a\varepsilon for any a \in \mathbb{C} (including a = 0).

Tags alg-geom, maths, points-func

Trackbacks

  1. Morphisms and functors of points From Martin's Blog

    This post will discuss the fact that -points of an affine -scheme (and more general objects) are the same as morphisms . James already brought this up in his comment last time. As well as proving this in the affine -scheme case, I shall attempt to...

Comments

  1. James Cranch said on Friday, 25 September 2009 at 18:13 :

    Good stuff.

    Except there's a bug at the end. Y has a lot more than two A-points. It's got a whole family parametrised by C: x^2 = 0 admits ae as a solution, for any a in C.

    Geometrically speaking, your "double point" Y manifests itself as being something like a point with a baby one-dimensional tangent space strapped onto it. You're really considering scheme maps from Y into Y when you take A-points.

    Such a scheme map consists of a map on "honest points" (which is the unique map of singleton sets) plus a map on the tangent spaces. This is a map C --> C, and such maps are basically the same thing as complex numbers.

    I'm sure there's some geometric intuition I'm neglecting to tell you, but that's a start.

  2. Martin Orr said on Saturday, 26 September 2009 at 16:31 :

    That's a rather embarrassing mistake! Thanks for pointing it out. I've fixed it now.

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