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)?

`$\{(f(x,y),g(x,y)):(x,y)\in\mathbb C^2\}$`

. – Andreas Blass Jul 13 '11 at 18:32`$\{(x,y,z,w):z=f(x,y), w=g(x,y)\}$`

under projection to the last two components. Projections of constructible sets are constructible (Tarski-Chevalley theorem). – Andreas Blass Jul 15 '11 at 21:26