Algebraic maps from products of affine varieties I have a question which might be fairly elementary, but I could not find an answer in the literature yet. Any pointers are very welcome :)
Let $X$, $Y$ and $Z$ be affine algebraic varieties. I have a map 
$f:X\times Y \rightarrow Z$ and I know that for fixed $x_0\in X$, the map 
$Y\rightarrow Z, y \mapsto f(x_0,y)$ is a morphism. Also, for fixed
$y_0\in Y$, the map $X\rightarrow Z, x \mapsto f(x,y_0)$ is a morphism. I regard 
$X\times Y$ as the product variety of $X$ and $Y$. 
Is $f$ itself a morphism then? Can I prove this by stating a few known results from algebraic geometry? (Pointers to literature would be very helpful). Or do I have to prove it "by hand"? Or is the statement even false?
Thank you!
 A: This question is answered by [Richard S. Palais, Some Analogues of Hartogs' Theorem in an Algebraic Setting. American Journal of Mathematics, Vol. 100, No. 2 (Apr., 1978), pp. 387-405]
For examnple, he proves that a necesary and sufficient condition on a field $K$ for every separately polynomial map $K\times K\to K$ to be polynomial is that it be finite or uncountable. In particular, there do exist fields which do not have this property.
More generally, he shows that if one of $X$ or $Y$ is algebraically of the second category —that is, it cannot be written as a countable union of proper subvarieties— and $f:X\times Y\to Z$ is a separately polynomial map, then $f$ is in fact a polynomial map. Moreover, an uncountable algebraically closed field has the property that all affine algebraic varieties over it are algebraically of the second category, and the same holds for perfect complete non-discrete valued fields.
Again, there are fields for which separately polynomial maps are not polynomial.
The paper is worth a read :)
