MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

Let $f(x, y), g(x, y) \in \mathbb{C}[x, y]$. Consider the image $S$ of the map

$(f, g): \mathbb{C}^2 \rightarrow \mathbb{C}^2, (z_1, z_2) \mapsto (f(z_1, z_2), g(z_1, z_2)).$

If for an integer $d \in \mathbb{N}$ the set $S$ contains a set $A_1 \times A_2, A_1, A_2 \subset \mathbb{C}$ being such that the minimum of the cardinalities of $|A_1|, |A_2|$ is $d$, one can say that "$S$ contains a product set of width $d$".

If $S$ contains product sets of arbitrarily large width, then one can observe that it follows from the Combinatorial Nullstellensatz that the polynomials $f(x, y)$ and $g(x, y)$ are algebraically independent over $\mathbb{C}$.

I wonder whether it is known if the converse is true (that is, if $f(x, y)$ and $g(x, y)$ are algebraically independent over $\mathbb{C}$, then $S$ contains product sets of arbitrarily large width)?

share|cite|improve this question
You defined $S$ as a certain pair of sets, but I guess you intended a set of pairs, $\{(f(x,y),g(x,y)):(x,y)\in\mathbb C^2\}$. – Andreas Blass Jul 13 '11 at 18:32
The assumption of algebraic independence tells you that $S$ doesn't lie on a curve. Being a constructible set, won't $S$ then have dimension 2 and contain a whole neighborhood of some point (and therefore product sets of any desired width up to the cardinal of the continuum)? – Andreas Blass Jul 14 '11 at 0:03
Thank you; I have corrected the definition; I do not see why $S$ is constructible. – Albertas Jul 15 '11 at 17:47
$S$ is the image of the variety $\{(x,y,z,w):z=f(x,y), w=g(x,y)\}$ under projection to the last two components. Projections of constructible sets are constructible (Tarski-Chevalley theorem). – Andreas Blass Jul 15 '11 at 21:26
Thank you for the answer. Using that projections of constructible sets are constructible, I could prove that the converse is true. – Albertas Sep 12 '11 at 15:09

Your Answer


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

Browse other questions tagged or ask your own question.