Can the zero set of irreducible polynomial, with complex coefficients, in more than one variable, ever be compact in Euclidean topology?

Question

Let $f \in \mathbb C[x_1,...,x_n]$ be an irreducible polynomial, where $n>1$.

Let $Z(f):=\{(a_1,...,a_n)\in \mathbb C^n: f(a_1,...,a_n)=0\}$.

Is it possible that $Z(f)$ is compact in the Euclidean topology of $\mathbb C^n$ ?

Answer 1

Depending on whether you already know some basic algebraic geometry, here’s one way to see this:

Let $X = \operatorname{Spec} A$ and $A = ℂ[X_1, …, X_n]/(f)$. By Noether normalization, going-up and Krull’s principal ideal theorem, there is a surjective and closed map of varieties $X → \mathbb A^{n-1}_ℂ$. As $ℂ$ is algebraically closed, this induces a surjective map of classically algebraic sets $Z(f) → \mathbb A^{n-1} (ℂ)$. Being polynomial in nature, this map is continuous both with the Zariski and the Euclidean topology.

Thus, if $Z(f)$ was compact in the Euclidean topology, so was $ℂ^{n-1}$, which is not for $n > 1$.

Answer 2

We don't need to assume irreducibility (although the general case follows from this one by decomposing $f$ in irreducible parts). Of course the result hold for the zero polynomial, but not for any other constant polynomial, so we assume $f$ is not constant.

Up to permuting variables, $x_1$ has at least one non-zero exponent in standard form. Then $f(x_1,\ldots,x_n)=\sum_{i=0}^\infty x_1^ip_i(x_2,\ldots,x_n)$ for certain polynomials $p_i$.

Let us show that for all $k$, there exist $|a_2|,\ldots,|a_n|>k$ such that $f(x,a_2,\ldots,a_n)$ is not constant. Suppose otherwise, i.e. for all $|a_2|,\ldots,|a_n|>k$, the polynomial $$f(x,a_2,\ldots,a_n)=\sum_i x^ip_i(a_2,\ldots,a_n)$$ is constant, which means that $p_i(a_2,\ldots,a_n)=0$ for $i\geq 1$.

It follows that for $i\geq 1$, the polynomial $p_i$ is zero outside of a large ball of $\mathbb{C}^{n-1}$, which implies $p_i=0$ for $i\geq 1$ (as is probably known beforehand; otherwise you can prove it by induction with a similar argument).

Therefore $f(x_1,\ldots,x_n)=p_0(x_2,\ldots,x_n)$, contradicting the fact that $f$ has a nonzero power of $x_1$ in standard form.

Then for $k>0$, choose $a_2,\ldots,a_n>k$ such that $f(x,a_2,\ldots,a_n)$ is not constant. Let $a_1$ be any root of this polynomial. The tuple $(a_1,\ldots,a_n)$ is a root of $f$ of euclidean norm $>k$. Therefore $f^{-1}(0)$ is unbounded and in particular non-compact.