I know that there are isomorphic projective varieties which have nonisomorphic coordinate rings, but I'm a little mystified as to "why" this is the case. Why doesn't a usual functoriality proof go through to prove this, and is there any insight into this other than that the definitions just work out this way?
Algebraic Geometry – Why the Coordinate Ring of a Projective Variety is Not Determined by the Isomorphism Class
algebraic-geometry
Related Solutions
I must admit i looked at your question only quickly, but i do believe there's an error: to show surjectivity, you must show that every element in $S_{(y_o)}$ has a preimage, not every element in $I(Y)_{(y_0)}$.
From your question i get that you're working your way through Hartshorne. In that case i think with your current knowledge your solution is good. However have a look at the following reasoning:
Oh by the way, on this page you can find solutions to many of the exercises in Hartshorne, http://www.math.northwestern.edu/~jcutrone/research.html, they helped me a lot.
The point you need to consider, is that $X$ is not immediately affine, at first it is just quasi projective. We need the well known map $$ \phi: U \longmapsto \mathbb{A}^n $$ which maps $[a_0:\ldots:a_n]$ to $(a_1/a_0,\ldots,a_n/a_o)$. This gives an isomorphism (check Hartshorne) of varieties, and gives $X$ the structure of an affine variety. It is only then that we can calculate its affine ideal and hence its coordinate ring $A(X)$.
Now here's comes a quick argument for your result that does require theory from some later chapters of Hartshorne..
By the map above, the quasi projective variety $Y\cap U$ is isomorphic to its image, lets say $Z$ (this is NOT the same thing as your $X$, although they are of course isomorphic). Then the rings of global regular functions are isomorphic $\mathcal{O}_Z(Z) \cong \mathcal{O}_X(X)$. Since $Z$ is affine, $\mathcal{O}_Z(Z) \cong A(Z)$, and by a theorem that Hartshorne unfortunately only states for schemes, (II 2.5), we have $\mathcal{O}_X(X) \cong S(Y)_{(y_0)}$, hence your result. The result is quite immediate, i believe that that is the reason some authors just state it as "known".
Let me know if you still have questions.
Thanks to @Heinrich's comments, I realized that we cannot define the regular map $\alpha$ as I stated in my proof. I think the following proof works.
An isomorphism of the $\mathbb{C}$-algebras would correspond to an isomorphism between the affine cones over the two projective varieties. Let $V,W$ be projective varieties in $\mathbb{P}^n$. Let $\phi:V\rightarrow W$ be the linear change of coordinates map and $\psi:W\rightarrow V$ be its inverse map. Suppose $V=V(f_1,…,f_r)$. Then $W=V(f_1\circ \psi,…,f_r\circ \psi)$. As affine varieties they are apparently isomorphic.
(Notice that this proof does not work for general isomorphic projective varieties.)
Best Answer
The homogeneous coordinate ring of a projective variety embedded in $\mathbb{P}^n$ is not the "ring of functions" on the variety but actually the direct sum of sections of symmetric powers of some line bundle. The point is that this construction actually depends on the line bundle, and the same variety can have many line bundles which give rise to non-isomorphic rings.
I'm not sure what "a usual functoriality proof" is, but you should find either that you end up proving something about the category of projective varieties together with an embedding or that you can't define the homogeneous coordinate ring at all without a choice of embedding.
I'm not totally comfortable with this answer since I can't claim to have much experience with line bundles, so here's a more straightforward version. First, the reason this issue doesn't come up for affine varieties is that once you have a definition of a morphism of affine varieties, the ring of functions $k[V]$ on an affine variety $V$ really is just the set of morphisms $V \to \mathbb{A}^1(k)$, and this definition is clearly functorial.
The same cannot be said for the homogeneous coordinate ring of a projective variety $V \subset \mathbb{P}^n$. First of all, its elements are not functions on the variety. Quotients of two elements of the same degree do define
morphismsrational maps $V \dashrightarrow \mathbb{P}^1$, but you're considering much more than just such quotients. When $V$ is irreducible you might be tempted to replace the ring of functions with the field $k(V)$ of rational functions, but $k(V)$ does not completely capture $V$ (for example it ignores the removal of finitely many points).After you think about this problem for awhile you should realize that the problem here is that the homogeneous coordinate ring really is defined in terms of the embedding, and so there's no reason to expect it to be independent of the choice of embedding. So next you might look for a definition of projective variety that is independent of the choice of embedding, and then sooner or later you have to start studying schemes.