Let $X$ be an irreducible smooth projective variety of dimension $d$. Do there exist irreducible smooth projective curves $C_1, C_2,\ldots, C_d$, an open subset $U\subset C_1\times C_2\times\ldots\times C_d$ and a dominant morphism $f:U\to X$.

8$\begingroup$ This is not true. For example, this does not hold for sufficiently general hypersurfaces of large degree (and dimension $> 1$) by results of C. Schoen "Varieties dominated by product varieties." Internat. J. Math. 7 (1996), no. 4, 541–571. $\endgroup$ – ulrich Jun 4 '12 at 14:30

$\begingroup$ Tony Scholl's comment at mathoverflow.net/questions/33665/… looks relevant. $\endgroup$ – David E Speyer Jun 4 '12 at 18:51

1$\begingroup$ @Ulrich: please add that as an answer. $\endgroup$ – Steven Gubkin Jun 4 '12 at 19:55

$\begingroup$ @David: Thanks for pointing out that comment. $\endgroup$ – Rex Jun 5 '12 at 18:51
Ulrich observes in the comments that C. Schoen provides counterexamples in his paper "Varieties dominated by product varieties". Beyond Schoen's work, this problem has some interesting history, which I learned from this note of Oort.
Grothendieck, in attempting to prove the Weil conjectures, had hoped that every variety was rationally dominated by a product of curves. He asked Serre if this was true; Serre showed that a sufficiently general surface contained in an explicit Abelian variety of dimension $5$ is a counterexample. (See p. 145 of the GrothendieckSerre Correspondence, which is a really amazing book.) The counterexample is quite beautiful and very simple, and of a rather different nature than Schoen's.
Essentially Serre observes that of $S\subset A$ is a smooth surface passing through the origin and satisfying the following property:
$(*)$ If $C, C'$ are curves contained in $S$, then $C+C'$ is not contained in $S$
then $S$ cannot be rationally dominated by a product of curves. This is because the rational map must extend to a morphism (as it is a map into an Abelian variety) given by adding two maps $C\to A, C'\to A$. So it suffices to find a surface $S$ satisfying $(*)$. Serre does this by writing down an explicit analytic germ at the origin satisfying $(*)$ (not too hard) and then approximating this germ by an honest surface (which one may take to be a complete intersection, for example).
Oort notes that Schoen seems not to have been aware of Serre's counterexample.

$\begingroup$ The link to the Oort note is broken. The title is Did earlier thoughts inspire Grothendieck?, and it is now available at this link. $\endgroup$ – R. van Dobben de Bruyn Feb 9 '17 at 17:21