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We will work over an infinite field $\Bbbk$. Let $U\subseteq\Bbbk^m\times\Bbbk^n=\Bbbk^{m+n}$ be a Zariski dense subset and for all $x\in\Bbbk^m$, consider $$U_x := \{\, y \in \Bbbk^n \mid (x,y)\in U \,\}.$$ I would like to show that there exists some $x\in\Bbbk^m$ with the property that $U_x$ is Zariski dense in $\Bbbk^n$. I have tried to attack this from different angles, but neither using $I(U)=0$ and working with ideals, nor trying to come up with a topological proof has been successful. I would be truly grateful for either a proof or a counterexample (which I cannot come up with, either).

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I claim that this is false. Proof:

Let $\mathbb{k} = \mathbb{Q}$. Let $$ U = \{ (\frac{p}{q}, \frac{p'}{q'}) | \gcd(p,q)=\gcd(p',q')=1, p',q' \leq q \} \subseteq \mathbb{Q}^2 $$

We first show that $U$ is Zariski dense. Let $C = \{ (x, y) | f(x,y) = 0\} \neq \mathbb{Q}$ be some proper Zariski closed set. We need to find some $(x,y) \in U \setminus C$. Note first of all that there must be some $(x_0, y_0) \in \mathbb{Q} \setminus C$. Now, take $y_0$ and view $f(x, y_0)$ as a polynomial in $x$. Since $f(x_0, y_0) \neq 0$, $f(x, y_0)$ cannot be the zero polynomial. Now write $y_0$ as $\frac{p'}{q'}$ where $\gcd(p',q') = 1$ and choose $n$ large enough such that $p',q' \leq 2^n$. Since $f(x, y_0) \not \equiv 0$, there must be some odd $p$ such that $f(\frac{p}{2^n}, y_0) \neq 0$. But now we have $(\frac{p}{2^n}, y_0) \in U \setminus C$ as required.

However, note that for any $x = \frac{p}{q}$, $$ U_x = \{ y \in \mathbb{Q} | (x, y) \in U \} = \{ \frac{p'}{q'} | p',q' \leq q \} $$ In particular, $U_x$ is finite, and hence can't possibly be dense in $\mathbb{Q}$.

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Nice counterexample. Thanks! – user38830 Feb 3 '13 at 12:25
up vote 1 down vote accepted

The answer is positive if $U$ is open. Otherwise consider $U$ equal to the graph of the function $\sin x$ in $\mathbb R^2$. It is Zariski-dense in $\mathbb R^2$ because it is not contained in an algebraic curve (otherwise it would meet $x$-axis only at finitely many points). But the fibers $U_x$ are all reduced to one point.

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