Maybe this is a dumb question, but better safe than sorry.

In Hartshorne, Exercise I.3.15 and I.3.16 we are asked to examine products of affine and projective varieties. I.3.15 has been smooth sailing. The reader is asked to prove:

a) $X \times Y$ is irreducible

b) $A(X \times Y) = A(X) \otimes_k A(Y)$

c) $X \times Y$ is a categorical product

d) $\dim (X \times Y) = \dim X + \dim Y$

None of these have been particularly difficult.

In the next exercise, we are asked to prove some similar claims about projective varieties, but (d) does not have an analogous statement above. It would seem plausible to me that something like this is true. We have many analogous theorems about the dimension of a projective variety from I.2, so I'd be surprised if there was no such relationship.

Can we find a proof of this claim, or a counter-example?

  • $\begingroup$ It is correct. Did you try to prove it? $\endgroup$ – Mohan Jun 20 '17 at 20:09
  • $\begingroup$ @Mohan I've been playing with it, but Hartshorne is my first exposure to projective anything (which might not be so good, now that I think of it), so I'm not all that confident that an analogous argument to the affine case would be okay. $\endgroup$ – Alfred Yerger Jun 20 '17 at 20:10
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    $\begingroup$ If $Z$ is an irreducible variety and $U\subset X$ a non-empty open subset, then $\dim Z=\dim U$. $\endgroup$ – Mohan Jun 21 '17 at 0:25

Let $X$ and $Y$ be irreducible projective varieties; there exists (irreducible) open subsets $U$ and $V$ of $X$ and $Y$, respectively, such that $\dim X=\dim U$ and $\dim Y=\dim V$.

Because \begin{equation} \overline{U}=X,\,\overline{V}=Y\Rightarrow X\times Y=\overline{U\times V} \end{equation} then $\dim(X\times Y)=\dim(U\times V)$, by hypothesis and previous exercise $\dim(X\times Y)=\dim X+\dim Y$.

  • $\begingroup$ The comment by Mohan convinced me when I asked this question a year ago. I hope that this explanation satisfies the person who put out the bounty. $\endgroup$ – Alfred Yerger Aug 6 '18 at 14:56
  • $\begingroup$ I didn't read the Mohan's comment, otherwise I would have avoid to write "another answer". By the way, I thank you for accepting my answer. $\endgroup$ – Armando j18eos Aug 6 '18 at 15:01
  • $\begingroup$ Well the hope is that you get the bounty (because the problem really is this simple, I was just new to Hartshorne) and that this question gets moved out of the "unanswered" bin. $\endgroup$ – Alfred Yerger Aug 6 '18 at 15:05

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