Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why is the zero set in $\mathbb{C}$ of a polynomial $f(x,y)$ in two complex variables always non-discrete (no zero of $f$ is isolated)?

share|cite|improve this question
Are you asking for a proof? – Rasmus Mar 7 '11 at 16:57
Don't you mean the zero set in $\mathbb{C}\times\mathbb{C}$? – joriki Mar 7 '11 at 17:01
consider $f(x,a)$ for a fixed value of $y$. this has $\text{deg}_xf$ zeros. varying $a$ continuously gives curves of zeros (the zeros of $f(x,a)$ depend continuously on $a$). – yoyo Mar 7 '11 at 17:06

Write $f(x,y)$ as a polynomial in $y$ with coefficients in $\mathbb C[x]$, so that $f(x,y)=\sum_{i=0}^nf_i(x)y^i$. If $x\in\mathbb C$ is such that $f_n(x)\neq0$, then there exist $n$ values of $y$ such that $f(x,y)=0$. It follows that if $X\subseteq \mathbb C\times\mathbb C$ is the zero locus of $f$, then the first projection $\pi_x:X\to\mathbb C$ has an image with finite complement. This implies, in particular, that $X$ is uncountable, and no uncountable subset of $\mathbb C\times\mathbb C$ is discrete.

share|cite|improve this answer
The uncountability of $X$ implies that $X$ has at least one nonisolated point, but a little more is needed to show $X$ has no isolated points. – Zarrax Mar 8 '11 at 0:30
@Zarrax: the question in fact "explains" incorrectly non-discrete as meaning no isolated points. I chose to ignore the parenthetical remark. – Mariano Suárez-Alvarez Mar 8 '11 at 0:33

This actually holds for any analytic function of two complex variables on an open set. Suppose $f(x_0,y_0) = 0$. Then viewed as a function of $y$, $f(x_0,y)$ must have an isolated zero of some order $k$ at $y = y_0$. By complex analysis (the residue theorem will work for example), for some small $r$ one has $${1 \over 2\pi i} \int_{|z - y_0| = r} {f'(x_0, z) \over f(x_0,z)}\,dz = k$$ But the integrand above is a continuous function of $x$ near $x = x_0$ and takes integer values. Hence for $x$ close enough to $x_0$ one also has $${1 \over 2\pi i} \int_{|z - y_0| = r} {f'(x, z) \over f(x,z)}\,dz = k$$ This in turn implies that whenever $x$ is close enough to $x_0$, $f(x,y)$ (viewed as a function of $y$) has $k$ zeroes inside $|z - y_0| = r$. Since this holds for arbitrarily small $r$, we conclude the zero of $f(x,y)$ at $(x_0,y_0)$ is not isolated.

This is actually a watered-down version of the proof of the famous Weierstrass Preparation Theorem (which incidentally also implies this fact in pretty short order).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.