I'm currently working on exercise 2.6 in chapter 1 of Algebraic Geometry by Hartshorne. I'm pretty confident with my answer apart from the first bit which I feel I could be "fudging". I'm looking for people to tell me where I have gone wrong… Here is the set up for those who don't have the book:
If $S=k[y_0, \dots ,y_n]$ is a graded ring, let $S_{(f)}$ denote the subring of $S_{f}$ consisting of elements of degree $0$. Let $Y$ be a projective variety embedded in $\mathbb{P}^{n}$ with homogeneous coordinate ring $S(Y)$, let $H$ be the hyperplane $y_0=0$, let $U=\mathbb{P}^{n}-H$, and let $X=Y\cap U$. The claim is that the affine coordinate ring $A(X)$ is isomorphic to $S(Y)_{(y_0)}$.
My answer with help from the book: define an isomorphism $\theta$ from $k[x_1, \dots ,x_n]$ to $S_{(y_0)}$ by mapping $f(x_1, \dots ,x_n)$ to $f(y_1/y_0, \dots ,y_n/y_0)$. The image of $I(X)$ under this map is equal to $I(Y)_{(y_0)}$. Indeed, if $f(x_1, \dots ,x_n)$ vanishes on $X$, then $y_0^{d}f(y_1/y_0, \dots ,y_n/y_0)$ vanishes on $Y$, and hence, $f(y_1/y_0, \dots ,y_n/y_0)=y_0^{d}f(y_1/y_0, \dots ,y_n/y_0)/y_0^{d}$ is an element of $I(Y)_{(y_0)}$. To show surjectivity, it is enough to there is an element of $I(X)$ which maps to each generator of $I(Y)_{(y_0)}$. If $F(y_0, \dots ,y_n)/y_0^{d}=y_0^{d}F(1,y_1/y_0 \dots ,y_n/y_0)/y_0^{d}$ is such a generator, then $F(x_1 \dots x_n)$ is an element of $I(X)$ which maps to it. Thus, $\theta$ induces an isomorphism from $A(X)$ to $S_{(y_0)}/I(Y)_{(y_0)}$, and the latter is clearly isomorphic to $S(Y)_{(y_0)}$.
Also, is there an easier way to see/think about this? For example, some books I have read have stated that $A(X)$ can be identified with $k[\xi_1/\xi_0 \dots \xi_n/\xi_0]$ (where the $\xi_i$ are the coordinate functions of $Y$) without further justification.
Thanks for any help
Best Answer
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.