Homogeneous coordinate rings of product of two projective varieties In Ex 3.15, chapter I of Hartshorne's "Algebraic Geometry", we have shown that $ A(X \times Y) \cong A(X) \otimes A(Y)$, when X and Y are affine varieties. Is the same statement true for projective varieties, i.e. if X and Y are projective varieties then is $ S(X \times Y) \cong S(X) \otimes S(Y)$.     Thank you.

Let $X \subset \mathbb{P}^n, Y \subset \mathbb{P}^m$ be projective varieties with homogeneous coordinate rings $S(X)$ and $S(Y)$ respectively. If we take the equations that define $X$, we see that they are homogeneous polynomials in $n+1$ variables. Naturally, they define an affine variety of $\mathbb{A}^{n+1}$, which is called the affine cone of $X$ and is denoted by $C(X)$. Now, the product $C(X) \times C(Y)$ is an affine variety of $\mathbb{A}^{n+1} \times \mathbb{A}^{m+1} \cong \mathbb{A}^{n+m+2}$ and is given by homogeneous equations. The affine coordinate ring of $C(X) \times C(Y)$ is precisely $S(X) \otimes S(Y)$. Since the equations that define $C(X) \times C(Y)$ are homogeneous, $C(X) \times C(Y)$ has the structure of a projective variety and so we can view it as a projective variety of $\mathbb{P}^{n+m+1}$ with homogeneous coordinate ring $S(X) \otimes S(Y)$. This shows that $S(X) \otimes S(Y)$ is the homogeneous coordinate ring of the projective variety that corresponds to the product of the affine cones of $X$ and $Y$.
To get a sense of what $S(X \times Y)$ is, one needs first to realize that $X \times Y$ is a projective variety of $\mathbb{P}^{(n+1)(m+1)-1}$ by means of the Segre embedding. Let us do an example using the excellent hint by Mariano. Let us take $X=Y= \mathbb{P}^1$. Then $\mathbb{P}^1 \times \mathbb{P}^1$ is a quadratic hypersurface of $\mathbb{P}^3$ given by the equation $z_{00}z_{11}-z_{10}z_{01}=0$ and so its homogeneous coordinate ring is $S(\mathbb{P}^1 \times \mathbb{P}^1) = \frac{k[z_{00},z_{01},z_{10},z_{11}]}{(z_{00}z_{11}-z_{10}z_{01})}$. On the other hand, $S(X)=k[x_0,x_1], S(Y)=k[y_0,y_1]$ and so 
$S(X) \otimes S(Y) = k[x_0,x_1,y_0,y_1]$. This agrees with the fact that $C(X)\cong C(Y)=\mathbb{A}^2$.
