Ok, I know I haven’t posted in a bit, but it’s the end of my first semester of my first year, so hopefully I can be forgiven. Today, we’re going to start a (potentially doomed from the start) project: algebraic geometry without prerequisites, or at least, with minimal prerequisites. The goal is to get as much done as possible, with detours into many of the more interesting theorems, and today we’re going to start with affine varieties.

Rather than do things completely abstractly, let us begin by deciding that we’re going to work over the complex numbers. We denote by the collection of polynomials with complex coefficients. What can we say about them algebraically? Well, for one thing, they factor completely into linear terms. That’s just the fundamental theorem of algebra.

But what about geometrically? They’re functions on the complex plane, so we can ask about solutions, that is, when is a given polynomial zero? The geometric version of the fundamental theorem would be the fact that we get a finite collection of points and that the number of points is bounded by the degree of the polynomial. Sadly, this case doesn’t do much else. However, if we take two collections of finitely many points and take their union, we get finitely many points. If we take any number of these collections and take their intersection, we get finitely many points. Now, technically, there are two other possibilities. We can get every single point by looking at or we can get no points by looking at . This defines what is called a *topology* on the complex plane. It’s not terribly interesting here, but just watch what happens next.

Let’s look at now. That is, complex coefficient polynomials in two variables. These don’t necessarily factor into linear terms, but they do factor (uniquely) into what are called irreducible polynomials. So now we’ll define an *algebraic set* to be the common zeroes of any finite collection of polynomials. Now, algebraic sets satisfy the same conditions that the finite sets did before. Namely that finite unions, arbitrary intersections, the empty set and the whole complex plane are all algebraic sets, so we get another topology! We call this the *Zariski topology*.

We call an algebraic set *irreducible* if it cannot be written as the union of two algebraic sets contained in it, in which case, we will call it a *variety*. So, for instance, if we look at the collection of two points, it is reducible, because we can write it as a union of two individual points, which are algebraic sets. However, a single point, or a line, is irreducible. Now we need a bit of algebra.

Let be an algebraic set. We define to be the *ideal* of the algebraic set. In general, an ideal is a collection of polynomials such that if we take any two of them, their sum is in the collection, and if we take any polynomial in the ideal and any polynomial at all, their product is in the ideal. It turns out that there is a very special type of ideal that corresponds to an algebraic set: a *radical ideal*. That is, an ideal such that if then . That sounds a bit funny though, so let’s think about it for a moment. The real content here is that if a function, when raised to some power, is zero, then the function itself had to be zero. So it shouldn’t be surprising if these are what come up when looking at polynomials on algebraic sets.

In general, we define an algebraic set to be the zeroes of finitely many polynomials on , the -dimensional complex space.

We’ll need two facts in general, both of which have fairly involved proofs which (if I do them) deserve to be posts in and of themselves. The first is the Hilbert Basis Theorem, which says that any ideal of polynomials is finitely generated, that is, there exist in the ideal such that every element in the ideal can be written for some polynomials . The second is Hilber’s Nullstellensatz, which says that if you take an ideal, and then look at the common zeroes of every polynomial in it, and then look at the ideal of polynomials vanishing on this algebraic set, and the ideal has a close connection to the original: it is the radical of this ideal. That is, it is the ideal define by if and only if for some .

These two facts which seem fairly straightforward, are in fact rather difficult to prove (here is a proof that Tao wrote up), and are essential to algebraic geometry. The first tells us that we only need to worry about finitely many polynomials, and the second tells us that we only need to worry about reduced polynomials (that is, polynomials with no repeated factor).

That’s quite enough algebra for the moment, though. Let’s see what sort of geometry we can do. Let’s denote the zero set of a collection of polynomials by , for variety. Then let’s stick to the plane a bit more, and look at . This is just , by definition, and so what does this look like? It’s a parabola intersecting a horizontal line, and so we get two points, in fact, they’ll be points with and with , so we get . In fact, if you hit the parabola with a line, it almost always intersects in two points. There are only two exceptions. The first is the tangential line . But this is simple to solve, we should just count that point twice, just like we do in the fundamental theorem of algebra. But what about the vertical line ? It only intersects the parabola once, and there is no way that we should count it twice.

This turns out to be a fundamental problem with working in . Not all of the points of intersection always exist. For another example, two lines intersect in a point…except when they are parallel. This may not seem like a huge problem at first, but math is supposed to be elegant, and having theorems that say “two things intersect in a certain number of points, except when they don’t” or the like is terrible. Next time, we’ll address this problem a bit, and make the first baby steps towards proving Bezout’s Theorem, a generalization of the theorem of algebra and, quite frankly, one of my all time favorite theorems.

December 18, 2007 at 4:29 am

Don’t let the fact that it’s doomed from the start stop you. People told me my own (similar, but wider-ranging) project was doomed from the start and I’ve been blogging almost a year now on it with usually 300-500 hits a day. I’m looking forward to good things here, especially if I can mine them for references when I eventually get around to algebraic geometry :D

December 18, 2007 at 11:06 am

I apologise for nitpicking, but you wrote the definition of a primary ideal instead of the definition of a radical ideal.

A radical ideal is an ideal such that so, using the definition of the radical (which, by the way, contains a typo) you get that an ideal is radical if

December 18, 2007 at 4:56 pm

Yes, I apologize. I had been reading up on primary ideals earlier and wrote one thing while typing another. Thanks for pointing it out, I’ve made the correction.

December 18, 2007 at 6:07 pm

[...] Yesterday, there was talk of affine varieties, and I pointed out the fact that intersections don’t work out nicely in this case. Today [...]

December 21, 2007 at 2:46 am

[...] the first post in this series we defined the Zariski topology in terms of algebraic sets. Now we need to sit down [...]

January 29, 2008 at 3:45 pm

[...] Armstrong’s exposition as a basic reference. I talked a bit about the topology relevant in algebraic geometry (which is rather different than that in most point set topology settings, note that we make use of [...]

March 24, 2009 at 7:06 am

[...] describe the basic phenomenon more generally first, though. Take a variety , it can be covered by affine varieties, which are the same things as nice rings. However, there’s a way to make the nice rings [...]

August 8, 2010 at 1:44 pm

Small correction: You say a radical ideal is an ideal *I* such that if $$f^n \in I$$ then $$f \in \sqrt{I}$$, but I don’t think the radical sign is necessary- because every ideal satisfies the condition you wrote.

August 8, 2010 at 1:44 pm

Whoops, I don’t know how to do Tex on wordpress… excuse the illegibility.

August 8, 2010 at 2:09 pm

Thanks! Corrected that one, can’t believe I missed it.

April 21, 2013 at 5:23 pm

Shouldn’t an algebraic set be the common zeroes of an arbitrary (not necessarily finite) collection of sets? Otherwise I don’t see how they’re closed under arbitrary intersection.

April 21, 2013 at 7:10 pm

I perhaps should, but thanks to the Hilbert Basis Theorem, which says that every ideal in is finitely generated, we have that any infinite set of polynomials generates an ideal of all things that vanish on their common zeros (up to taking radicals), and then that ideal is actually finitely generated, so only finitely many polynomials suffice.

May 16, 2013 at 4:18 pm

Aah! Thank you for spelling that out.