## Curves

Definition A curve over a field k is the set of zeros in k × k of a single polynomial equation in two variables, f(x, y) = 0.

### Examples

Let k=R and consider the following curves. (Here we should insert the graphs of the following curves.) These examples raise several important questions about the naive definition. For example
1. Are (i) and (ii) the same or different curves?
2. Should (v) and (vi) even be called curves at all?
3. Why are (iii), (ix) and (xi) disconnected?
4. What's special about (vii), (viii), and (x)?
The questions are of two kinds. The first two questions are metaphysical; they ask if we have the correct definition. The final two questions are analytic; they begin to study the intrinsic geometry of curves.

To elaborate on question (A), consider the families of curves (Pictures again need to be inserted.)

1. y = ax.
2. y2 = ax.
Intersect each family of curves with the line x = 1. When a 0, curves in the first family intersect the line in exactly one point, but curves in the second family intersect in two points. As the parameter a |→ 0, the limiting curve of the first family is curve (i), and it still intersects in one point. However, the limiting curve in the second family is curve (ii), and the two points of intersection coalesce in the limit. Should the limit curve intersect the line in one point, or in two points "properly counted"?

Regarding question (B), there are two schools of thought (which will be described shortly). In either case, the following theorem shows that the underlying difficulty is that R is not algebraically closed.

Theorem Let k be an algebraically closed field and let f k[x,y] be a nonconstant polynomial. Then there are infinitely many solutions to f(x,y)=0.

Proof: Write f(x,y) = a0(y) + a1(y)x + + an(y)xn , with an(y) 0, as a polynomial of degree n in x with coefficients from k[y]. If n = 0, then we simply have f(x,y) = a0(y), a nonconstant polynomial in one variable. Since k is algebraically closed, this polynomial has a root y0. Then all of the infinitely many pairs (x, y0) as x ranges over the (infinite) algebraically closed field k provide solutions to f(x,y) = 0.

So, suppose n > 0. Then an(y) = 0 has only finitely many solutions. There are infinitely many points y0 where an(y0) 0; at each one of them, we get a polynomial f(x,y0) of degree n in the single variable x. This polynomial has a solution x0, yielding infinitely many points (x0,y0) where f(x,y) = 0.

Remark The proof also shows that all but finitely many vertical lines hit the curve in exactly n points, properly counted.

The restrictive school of thought says that you can avoid the difficulties posed by examples (v) and (vi) by considering only algebraically closed fields. This seems to be rather narrow-minded, since it foregoes any chance of applying geometric techniques to some interesting problems that arise in number theory. A more open-minded approach is to "fix" the naive definition to make it useful over arbitrary fields. Nevertheless, we will acquiesce (at least temporarily), and work over algebraically closed fields for the remainder of this chapter.

Definition As a set, affine n-space over a field k is defined to be the Cartesian product

Ank = k × k × × k

of k with itself n times.

Definition Let k be algebraically closed. An affine algebraic set over k is the set of common zeros in Ank of some set of polynomials S k[x1, , xn]. Given such a set S, its algebraic set of zeros will be denoted Z(S).

Remark If I is the ideal generated by S, then it is easy to see that Z(I)=Z(S).

Definition The Zariski topology on An is defined by taking the closed sets to be the algebraic sets. The Zariski toplogy on an algebraic set is defined as the topology induced from its embedding in An.

Remark The algebraic sets really do form the closed sets of a topology, but it does not fully reflect the structure of affine space. For instance, the topological space A2 is not the same thing as the space A1 × A1 with the product topology.