Equations in algebraically closed fields.

Let K be an algebraically closed field and f∈K[X,Y] non-constant. Let us prove that f has infinitely many zeros in K x K. Let us define the following sets:

A={a ∈K : f(a,y) is constant in K[y]}
B={b∈K : f(x,b) is constant in K[x] }

if one of them is finite we are done. In fact if for example A is finite ∀ x0 in K \ A f(x0,y) is not constant in K[y] and therefore since K is algebraically closed there will exist a y0 such that f(x0,y0)=0. But since K \ A is infinite (because K is algebraically closed and therefore infinite) there exist an infinity of such pairs.

Let us then assume for the sake of argument that A and B are both infinite. Let a∈A, a’∈A and b∈ B and say that f(a,y)=c_a, f(a’,y)=c_a’ and f(x,b)=c_b with c_a,c_a’,c_b∈K. Then we have that:

c_a’=f(a’,b)=c_b=f(a,b)=c_a

For every a in A we then set c_a=c. We show that f is the constant polynomial c. Let x0,y0 be in generic K and we prove that f(x0,y0)=c. We consider the polynomial:

g(x)=f(x,y0)-c

For every a in A we have g(a)=f(a,y0)-c=c_a-c=c-c=0 therefore g has infinite zeros in K it follows that it is the null polynomial, that is for every x∈K we have g(x)=0 in particular g(x0)=0 and that is

f(x0,y0)-c=0

and then

f(x0,y0)=c

but then f is constant, absurd. It follows that one of A and B is finite from which the thesis as announced. □