And we’re back, hope everyone enjoyed their holidays and their new year, and we’ll start out 2008 by significantly generalizing our study. So far, we’ve been working over the complex numbers out of convenience and familiarity. However, we don’t HAVE to restrict ourselves this way. In the future, we won’t make this assumption. We will use the complex numbers for intuition, however.

To generalize, first we need to define an object called a field. We have defined rings, and so we’ll start with that. A ring (we just say ring when we mean commutative ring with identity) k is a field if the collection of nonzero elements along with multiplication forms a commutative group. That is, for each nonzero element a there exists b such that ab=ba=1. This actually severely restricts things.

Some examples of fields include \mathbb{F}_2, the field of two elements. That is, we take the integers and quotient out the ideal of even integers. More generally, we get \mathbb{F}_p for each prime number is a field. These along with \mathbb{Q} are called prime fields, because any other field must contain one of them. To see this, we note that every ring with identity has a natural homomorphism \mathbb{Z}\to R. Now, either everything is sent to different things, that is, we have an injection, or else some integer is sent to zero (and so multiples of it will) which induces an isomorphism \mathbb{Z}/n\mathbb{Z} with a subring of R. If n has any nontrivial factors then we’d have n=n_1n_2, and so we’d have n_1n_2=0 in R. There’s no way to multiply n_1 by anything to get one, and so we can’t have a field.

Thus, any field contains \mathbb{F}_p or \mathbb{Q}. But we know these aren’t the only fields. For one thing, \mathbb{R},\mathbb{C} are both fields. However, what about other finite ones? Well, if we take \mathbb{F}_p[x], that is, polynomials in one variable over a finite field, and then quotient out by a prime ideal (say, one generated by an irreducible polynomial), we get another field, and it has p^2 elements.

In fact, the only finite fields are those with a number of elements equal to p^n for some prime. To prove this, let k be a field, and let \mathbb{F}_p be the prime field it contains. Then k is a vector space over \mathbb{F}_p, with vector addition just addition addition in k and scalar multiplication just multiplication by elements of the subfield. It must be a finite dimensional vector space, and so it has a finite basis. The elements are the just n-tuples of elements of \mathbb{F}_p, and so there will be p^n of them. We define the characteristic of a field to be the number of elements of the prime subfield, if finite, or zero if the prime subfield is \mathbb{Q}.

So now let’s talk about algebraic geometry. We can define affine and projective space over any field in exactly the way we did before. So now we can define morphisms in just the same way, and so we get categories \mathcal{V}_k of varieties over a field. However, the algebra and the geometry don’t connect quite as well. For instance, let’s work over \mathbb{R} for a moment. The polynomial x^2+y^2+1 then defines the empty subvariety of \mathbb{R}^2. This should bother us, because it violates the Nullstellensatz! This tells us that the Nullstellensatz doesn’t hold over every field.

The type of fields that it does hold over are called algebraically closed fields. A field k is algebraically closed if every polynomial in k[x] has a root in k. In light of this definition, we can rephrase the Fundamental Theorem of Algebra as saying that \mathbb{C} is algebraically closed. Because the Nullstellensatz (and some other fundamental things) work best over algebraically closed fields, we tend to restrict ourselves to this situation.

A theorem that we won’t prove is that every field can be embedded into an algebraically closed field, and the smallest such field is called the algebraic closure of k. So though I won’t bother saying it, we’ll be working over an arbitrary algebraically closed field for awhile. Sometimes we’ll need restrictions on the characteristic (later, when we do Elliptic Curves, we’ll say characteristic not two or three) or we might need an uncountable field, or else specifically the complex numbers. But generally, we’ll just assume algebraically closed.

That’s all for now, we do a bit more algebra next time, and then right back into the geometry and towards a proof of Bezout’s Theorem, which is our immediate goal.