Projective varieties with isomorphic coordinate rings Let $X$, $Y$ be projective varieties with isomorphic homogeneous coordinate rings (as graded rings). What can we say about $X$ and $Y$? Are they necessarily isomorphic? 
 A: I think this is true. Here is my attempt at a proof: Suppose $f:S(X)\rightarrow S(Y)$ is a coordinate ring isomorphism between projective varieties $X$ and $Y$. Suppose the coordinates of $X$ are $x_0,\dots, x_n$, and suppose $X$ is defined by the ideal $I$. Then $I_i:=I+\langle x_i - 1\rangle$ defines an affine variety $X_i$. It should be the case that
$$k[X_i]=k[x_0,\dots, x_n]/I_i = S(X)/\langle x_i - 1\rangle$$
and if so, we can take $f(x_i-1)\in S(Y)$ and embed this into $k[y_0,\dots, y_m]$. I'll use $f(x_i-1)$ to refer to this polynomial. If $Y$ is defined by an ideal $J$, then we get an affine subvariety $Y_i:=V(J,f(x_i-1))$, with coordinate ring given by
$$ S(Y)/\langle f(x_i-1)\rangle = k[y_0,\dots, y_m]/\langle J,f(x_i-1)\rangle=k[Y_i]$$
Since $S(Y)\cong S(X)$, this means $k[X_i]\cong k[Y_i]$. Since affine varieties with isomorphic coordinate rings are isomorphic, we have $X_i\cong Y_i$.
We can do this for all $i$, and we get an open cover of $X$ such that every set in this cover is isomorphic to a set in $Y$. If they agree on their intersection, we should be done, because then we can glue them into a regular map.
We should have that $f$ can be restricted to $f_i:k[X_i]\rightarrow k[Y_i]$, and then it will form the required isomorphism. Then $f_i^*:Y_i\rightarrow X_i$ will be an isomorphism from $Y_i$ to $X_i$. Since $f_i^*$ and $f_j^*$ are just restrictions of the same map, they should agree (in fact this seems to say that $f^*:Y\rightarrow X$ is an isomorphism).
A: Question: "Let X, Y be projective varieties with isomorphic homogeneous coordinate rings (as graded rings). What can we say about X and Y? Are they necessarily isomorphic?"
Answer: Yes: If $X:=Proj(T), Y:=Proj(S)$ with $T:=\oplus_{d \geq 0}T_d, S:=\oplus_{d \geq 0}S_d$ and $T \cong S$ as graded rings, it follows $Proj(S) \cong Proj(T)$ holds.
