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Let $X\subseteq \mathbb{P}^r,Y \subseteq \mathbb{P}^s$ be two projectve varieties,what is the coordinate ring of $X\times Y$(segre embedding)?Is it true that
$$S(X\times Y)=S(X)\otimes_k S(Y)?$$

I also want to know what is the dimension of $ X\times Y$?

Edit:As Zhen Lin has indicated ,the equality $S(X\times Y)=S(X)\otimes_k S(Y)$ is not true.But what is the relationship between them,can we express $S(X\times Y)$ in terms of $S(X),S(Y)$?

Wei Xia
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  • Did you calculate any examples? Say, $X = Y = \mathbb{P}^1$? – Zhen Lin Jul 26 '14 at 17:40
  • I know that $S(\mathbb{P}^1)=k[x,y]$,but what is $S(\mathbb{P}^1\times \mathbb{P}^1)$? – Wei Xia Jul 27 '14 at 09:32
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    It doesn't matter. $S (\mathbb{P}^1) \otimes_k S (\mathbb{P}^1)$ is then $k [x, y, z, w]$ with the usual grading, and so the associated projective variety is $\mathbb{P}^3$, which doesn't even have the correct dimension. – Zhen Lin Jul 27 '14 at 10:09
  • Note that it doesn't make sense to talk about the coordinate ring, as it is not unique for a given projective variety. As you mentionned Segre embedding, I guess you want to decribe $S(X\times Y)$ as a quotient of the canonical coordinte ring of $\mathbb P^{rs+r+s-1}$ ? – Cantlog Jul 30 '14 at 16:11
  • @Cantlog:Yes.I also want to know the dimension of $X\times Y$. – Wei Xia Aug 02 '14 at 13:49
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    @WeiXia: as a general fact, the dimension of the product of two algebraic varieties is always the sum of the dimensions. – Cantlog Aug 10 '14 at 11:08

1 Answers1

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It turns out that the coordinate ring for $X\times Y$ is

$$S(X\times Y) = \bigoplus_{i} S(X)_i\otimes S(Y)_i \subset S(X)\otimes S(Y)$$

where $S(X)_i$ is the degree $i$ component of $S(X)$.

Exercise 13.14 in Eisenbud's Commutative Algebra text goes through this fact.

Sergio Da Silva
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