Prove that a complex valued polynomial over two variables has infinitely many zeroes

Question

This is a homework question that I am struggling with. Given a polynomial over the complex numbers in two variables, show that the polynomial has infinitely many zeroes.

So let's say that the polynomial is a functions of $u$ and $v$. Let's consider the polynomial as a function of $u$, with $v$ as a parameter: $P_v(u)$. Then for each $v$ there is going to be a finite set of values. So now I should be able to proceed by contradiction by finding a suitable value or constraint on $v$, but I'm not sure how to proceed.

Answer 1

Since you included the (algebraic-geometry) tag, let me answer in an algebro-geometric fashion:

The (closed) points of $\mathbb{A}^2_\mathbb{C}$ correspond to maximal ideals in the polynomial ring $\mathbb{C}[u,v]$. A polynomial $P \in \mathbb{C}[u,v]$ vanishes at a point $(u_0, v_0) \in \mathbb{A}^2_\mathbb{C}$ iff $P \in (u - u_0, v - v_0)$. If $P$ is nonconstant, then we can factor $P$ into irreducibles, and replacing $P$ by one of its irreducible factors, we may assume $P$ is irreducible.

Thus, we wish to show that the quotient ring $R = \mathbb{C}[u,v]/(P)$, which is a domain but not a field, has infinitely many maximal ideals. But $R$ is a Jacobson ring by the general Nullstellensatz, so the prime ideal $(0)$ in $R$ is an intersection of maximal ideals. If there were only finitely many maximal ideals, then $(0)$ would be a finite intersection of maximal ideals; but then $(0)$ would be maximal by the following proposition, contradiction.

Proposition: If a prime ideal $p$ is equal to a finite intersection of ideals, $p = I_1 \cap ... \cap I_n$, then $p = I_j$ for some $j$ (use the fact that if $IJ \subseteq p$, then $I \subseteq p$ or $J \subseteq p$).

Note 1: The same reasoning above works over any algebraically closed field, not just $\mathbb{C}$.

Note 2: This also works in higher dimensions: if $P$ is a nonconstant polynomial in $n$ variables over an algebraically closed field $k$, then the zero set $V(P)$ contains infinitely many closed points of $\mathbb{A}^n_k$.

Note 3: This was surprisingly complicated reasoning, requiring use of heavy theorems. I would be interested in seeing more elementary proofs of the statement in Note 2.

Answer 2

Let $f\in \mathbb{C}[x,y]$ and suppose $V(f)$ is finite. Take, $(a,b)\in V(f)$ and write $f$ as $$f(x,y)=\sum_{k=0}^{n}f_k(x)y^{k}.$$

Notice that $f(a,y)=\sum_{k=0}^{n}f_k(a)y^{k}$ is a non-constant polynomial. Otherwise, as $f(a,b)=0$ we would get $f(a,y)=0$ and therefore $V(f)$ would be infinite. As $f(a,y)$ is not constant we assure that there exists $0<k\leq n$ (if the only $k$ such that $f_k(a)\neq 0$ is k=0 then the polynomial would be again constant) such that $f_k(a)\neq 0$ and therefore $f_k(c)\neq 0$ for all $c\in \mathbb{C}$ except for finitely many exceptions.

But this means that the polynomial $f(c,y)$ is not constant for infinitely many $c$, and by the fundamental theorem of algebra, a root $y_c$ must exist! Therefore, for every $c\in \mathbb{C}$, there exists $y_c\in\mathbb{C}$ such that $(c,y_c)\in V(f)$. Therefore, $V(f)$ must be infinite! (in fact, uncountable)

Answer 3

Here is an elementary proof. The idea is straightforward, the details a little tedious.

Suppose $p$ is a non-constant polynomial. By translating if necessary, we may assume $p(0,0) \neq 0$.

Let $Z= p^{-1}\{0\} \subset \mathbb{C}^2$ be the zero set, and note that $(0,0) \notin Z$. Furthermore, since $p$ is non-constant, we can find some other point $(x_0,y_0) \notin Z$ such that $p(x_0,y_0) \neq p(0,0)$.

Hence we can find some $\epsilon>0$ such that if $(x,y) \in B((x_0,y_0), \epsilon)$, then $p(x,y) \neq 0$ and $p(x,y) \neq p(0,0)$.

Choose $\theta \in [0,\pi)$ and let $\lambda_\theta: \mathbb{C} \to \mathbb{C}^2$ be given by $\lambda_\theta(t) = (t \cos \theta, t \sin \theta)$. Note that if $\theta \neq \phi$ with $\theta,\phi \in [0,\pi)$, then $\lambda_\theta(\mathbb{C}) \cap \lambda_\phi(\mathbb{C}) = \{(0,0)\}$.

Define the polynomials $p_\theta:\mathbb{C} \to \mathbb{C}$ by $p_\theta = p \circ \lambda_\theta$, and let $Z_\theta = p_\theta^{-1}\{0\} \subset \mathbb{C}$. Note that $0 \notin Z_\theta$.

Let $\Theta = \{ \theta \in [0,\pi) | \lambda_\theta(\mathbb{C}) \cap B((x_0,y_0), \epsilon) \neq \emptyset \}$, and note that $\Theta$ must contain some interval $(\theta_0,\theta_1)$.

Note that if $\theta \in (\theta_0,\theta_1)$, then $p_\theta$ is a non-constant polynomial (in one variable) and hence $Z_\theta \neq \emptyset$.

Finally, let $W=\{\lambda_\theta(t) | \theta \in (\theta_0,\theta_1), t \in Z_\theta\}$, and note that $W$ is infinite (since $(\theta_0,\theta_1)$ is uncountable) and $W \subset Z$.