Suppose $X \subset \mathbb P^n$ and $Y \subset \mathbb P^m$ are projective varieties, and let $S(X)$ and $S(Y)$ be their homogeneous coordinate rings. Consider the projective variety $X \times Y$ in $\mathbb P^N$ via the Segre embedding. If subscript $d$ denotes the $d^{th}$ homogeneous component of a graded algebra, I am trying to show that

$$S(X \times Y)_d \simeq S(X)_d \otimes S(Y)_d$$

as $k$-algebras for algebraically closed field $k$.

The closest I have been able to find is this answer:

Hilbert polynomial of product of projective varieties

However, I don't see how the map given there

$$S(X)_d \times S(Y)_d \to S(X \times Y)_d$$

actually lands in $S(X \times Y)_d$ since the members of its image seem to have degree $2d$. Assuming I am simply misunderstanding that (and please correct me if I am), I still don't see what the induced map

$$S(X)_d \otimes S(Y)_d \to S(X \times Y)_d$$

explicitly is, nor why it's, in particular, surjective.


1 Answer 1


First, you should remember what the coordinate ring of the Segre embedding looks like: if we're embedding $\Bbb P^m$ and $\Bbb P^n$ with coordinate algebras $k[x_0,\cdots,x_m]$ and $k[y_0,\cdots,y_n]$, respectively, then the coordinate algebra of their product inside $\Bbb P^{nm+n+m}$ is $k[x_iy_j]_{0\leq i \leq m,0\leq j\leq n}$, where we take the degree of all the generating monomials $x_iy_j$ to be one and enforce the obvious relations $x_iy_j\cdot x_ky_l = x_iy_l\cdot x_ky_j$. Now it is clear why $S(X)\times S(Y)\to S(X\times Y)$ sends the degree $(d,d)$ piece to the degree $d$ piece, and it's also clear why it's surjective: we can write a degree $d$ polynomial in the target as a $k$-linear combination of monomials $x_{i_1}\cdots x_{i_d}y_{j_1}\cdots y_{j_d}$, and there's an obvious choice of preimage for each of these basis elements. This argument easily descends to any quotient you'd like, which implies the result for a general choice of $X,Y$.

  • $\begingroup$ Is the preimage just the product of the $x_i$s in the first coordinates and the product of the $y_j$s in the second coordinate? $\endgroup$ May 10, 2020 at 5:13
  • $\begingroup$ Yes, that's the preimage of the basis element. $\endgroup$
    – KReiser
    May 10, 2020 at 5:22
  • $\begingroup$ I guess I'm having trouble visualizing the actual map does to a general element. Suppose our coordinate rings were polynomial rings in one variable, and I wanted the image of $(x^2, y^3)$. How would I get that based on your definition of what the map does to the coordinates themselves? I don't see how to write that as a product of points whose images we actually know by definition (which means I'm not understanding something). $\endgroup$ May 10, 2020 at 5:26
  • $\begingroup$ Your question is unclear (what do you mean "a product of points"?) and the example you've chosen (a polynomial ring in one variable, polynomials of different degrees) is not useful, because the projective variety corresponding to that ring is a point. Perhaps you could rephrase? $\endgroup$
    – KReiser
    May 10, 2020 at 5:30
  • $\begingroup$ My apologies. I'll shorten and rework the question: suppose my individual coordinate rings are polynomial rings in two variables, $x_1$, $x_2$ and $y_1$, $y_2$, resp. What's the image of, say, $(x_1^2 + x_1x_2, y_1y_2 + y_2^2)$ under the identification you've made in your answer, which, as I understand it, is simply $(x_i, y_j)$ \to $x_iy_j$? $\endgroup$ May 10, 2020 at 5:37

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