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