How many roots does a polynomial in 2 variables have?

Question

We know that in general a polynomial in one variable of degree n has n roots. What about in two variables?

It seems instead of discrete points we have whole regions. For example:

$xy = 0$ has as solutions the $x$ axis and the $y$ axis.

How about higher degree/more complicated polynomials in two variables?

Answer 1

The degree of a polynomial's term is the sum of the exponents for all variables in that term. So I'd find which term in the polynomial has a maximum exponent sum for all variables and that's the polynomial's degree, which the Fundamental Theorem of Algebra says has that many roots. (I think - 95% sure but not 100% sure.)

Answer 2

Since you say a single-variable polynomial of degree $n$ has $n$ roots, let us first discuss the complex case. In $\mathbf{C}^d$ with $d > 1$, a single polynomial of degree $n > 0$ always has infinitely many zeros, and the set of zeros (often called a hypersurface) is always $d - 1$ dimensional for an appropriate sense of dimension (for example, near "most" of its points it looks like $\mathbf{C}^{d-1}$). That it has infinitely many solutions is not so hard to see by induction on $d$, essentially all but finitely many values of $x_d$ will keep the polynomial non-constant, and then it must have zeros.

In $\mathbf{R}^d$ the situation is not as uniform. For example in $\mathbf{R}^2$, the polynomial $$((x-a_1)^2 + (y-b_1)^2) \dotsm ((x-a_r)^2 + (y - b_r)^2)$$ has exactly $r$ zeros $(a_i, b_i)$, and you already have examples where the set of zeros is infinite. Using a similar idea, and depending on your definition of dimension, the set of zeros of a (non-constant) polynomial in $\mathbf{R}^d$ can have dimension any integer from $0$ to $d-1$.

The case of finite fields (where of course the set of zeros has to be finite) was summarized in a comment by Carl Schildkraut. For other infinite fields the definition of dimension is more complicated (or at least different than what I wrote above) but depending on whether the field is algebraically closed then the situation is similar to that of $\mathbf{C}$ or $\mathbf{R}$.