# 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!

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Welcome! This is an interesting question which I'd love to know the answer to. – Bruno Joyal Apr 12 '12 at 22:32

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 :)

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Thank you for your answer - but I don't understand it :( For example, I don't know what "seperately polynomial" means or what finite or uncountable maps are. Also, I don't see how your $K\times K \rightarrow K$ example helps me. I don't have access to the paper you mentioned now, but in a few days I might - I will surely have a look. Is the answer to my question positive or negative - is $f$ algebraic? For now, I understand that it does not seem to have a trivial answer :) – Sh4pe Apr 12 '12 at 23:58
A separately polynomial map $f:K\times K\to K$ is a function such that for each $a\in K$ the maps $f(a,-):K\to K$ and $f(-,a):K\to K$ are polynomials. You misread my first sentence: nowhere do I say "finite or uncountable maps"... those qualifiers apply to the field $K$ of that statement. – Mariano Suárez-Alvarez Apr 13 '12 at 0:01
The example helps you because the affine line over a field is the field. – Mariano Suárez-Alvarez Apr 13 '12 at 0:02
Thank you :) That clarifies some of my confusions, and I'm sure that the rest will be clear when I had a look at the paper. I'm waiting to close this question until then though :) – Sh4pe Apr 13 '12 at 0:04
I've read the paper. It is indeed very helpful and accessible - thank you for the hint. That answers my question completely. – Sh4pe Apr 15 '12 at 15:22