# Homogeneous Components of the Homogeneous Coordinate Ring of a Product of Projective Varieties

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.

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 $$X,Y$$.
• 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? May 10 '20 at 5:13
• 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). May 10 '20 at 5:26
• 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 $x_iy_j$? May 10 '20 at 5:37