Let $n>1$, $K$ be an uncountable algebraically closed field and $f(X_1,…,X_n)\in K[X_1,…,X_n] $ be a non-constant polynomial. Then it is known that $Z(f):=\{\bar a \in K^n : f(\bar a)=0\}$ is infinite. My question is, can $Z(f)$ be countable ?
If $K=\mathbb C$, it is known that $Z(f)$ contains a connected (in the Euclidean topology of $\mathbb C^n$) subset (corresponding to an irreducible factor of $f$), hence is always uncountable.
Best Answer
There are tons and tons of ways to prove this. Here's probably the lowest-tech way.
Since $f$ is nonconstant, some variable appears in $f$; suppose without loss of generality that it is $X_1$. Consider $f$ as a polynomial in $X_1,\dots,X_{n-1}$ with coefficients in $K[X_n]$. Since $X_1$ appears in $f$, this polynomial with coefficients in $K[X_n]$ is nonconstant. For all but finitely many values of $a\in K$, then, $f(X_1,\dots,X_{n-1},a)$ will still be nonconstant (since $a$ just needs to not be a simultaneous root of all the coefficients besides the constant term). Since such a nonconstant polynomial has a root, this gives uncountably many roots of $f$.
(Or, if you don't know that a nonconstant polynomial has a root, you can continue this argument by induction on $n$ to prove it.)