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 -variety is simply the functor
for a suitable
-algebra
. Only a restricted class of
-algebras could arise as
however. So in this post I generalise this to allow
to be any
-algebra, and thereby define affine
-schemes.
Note: throughout this post, all Hom-functors are on the category of -algebras, for a fixed field
.
Categorical definition
Recall that there is a dual equivalence between the category of affine -varieties and the category of finitely-generated integral-domain
-algebras, sending a
-algebra
to its coordinate ring
. We saw in the last post that the functor of points of
is then
.
On the algebraic side, the restriction to "finitely-generated integral-domain" seems unnecessary. is a functor
for any
-algebra
.
So we shall expand our universe of algebraic-geometric objects to include an object called for every
-algebra
, whose defining property is that its functor of points is
. Such objects are called "affine
-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 , but lots of them: I have told you the set of "points defined over
" for every
-algebra
.
(Note: If you know some commutative algebra or have studied schemes, then the name might remind you of something. The
I have defined here is closely related to the
defined in terms of prime ideals, but it is best to set that definition aside for the moment. The relationship between the
-points I am considering here and the prime ideals of
is somewhat complicated; the fact that I am doing everything relative to
adds a further twist.)
Concrete interpretation
I have defined the set of -points of
to be the set of
-algebra homomorphisms
. 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 -schemes: the proof of lemma 1 from last time doesn't depend on
being finitely generated or an integral domain. (Although we now have to look for solutions in
, where
is an indexing set for the generators of
, and might no longer be finite.)
As an example, I shall compare (which is a variety) with
(which is not).
is of course just a point; you might expect
to be some sort of "double point".
Now both of these have the same set of -points: either because
and
have the same solutions in
, or because
and
have the same
-algebra homomorphisms to
-- any homomorphism
must map
to
.
However we can look at points of and
defined over any
-algebra. For example, if we take
, then we find that
has just one
-point, while
has many. In terms of solving equations, these many points arise because
has not only
but also
as solutions, for any
. Alternatively, in terms of homomorphisms, you get a
-homomorphism
by setting
for any
(including
).
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.
That's a rather embarrassing mistake! Thanks for pointing it out. I've fixed it now.