Understanding problem 2.6 in Hartshornes algebraic geometry book . Can anybody help me in understanding the hint given in the problem $2.6$ in Chapter $1$ of Hartshorne's Algebraic Geometry  book   ?
I cannot see why $A(Y_i) $  can be identified with degree zero elements  of $S(Y)_{x_i}$. Actually I cannot find it out how does  the degree zero  elements in  $S(Y)_{x_i}$ look like.
Please Help me. Thanks.
 A: We know that $Y_{i}=Y\cap A_{i}$ cover $Y$. By definition of the dimension we have $\dim(Y)=\sup \dim (Y_{i})$. Now in local coordinates we have
$$
A[Y_{i}]=S[Y]_{(x_{i})}|^{0}
$$
as Andrew pointed out. To see this let us write $S(Y)=\oplus_{d\ge 0} S_{d}/I(Y)$. For any function $f \in S(Y)$, its affine piece on $A_{i}$ is given by
$$
f(\frac{x_{1}}{x_{i}},\cdots, \frac{x_{n}}{x_{i}})
$$
and since each term $\frac{x_{1}}{x_{i}}$ has degree $0$, $f$ itself must be of degree $0$ as well. Thus every element in $A[Y_{i}]\subset S[Y]_{(x_{i})}|^{0}$, and the reverse inclusion is not difficult. Now we have
$$
S[Y]_{(x_{i})}=A[Y_{i}][x_{i},x_{i}^{-1}]\rightarrow \dim(S[Y]_{(x_{i})})=\dim(A[Y_{i}])+1
$$
And since localization preserves chains of prime ideals, we have $$\dim(S[Y])\ge \dim(S[Y]_{(x_{i})})=\dim(A[Y_{i}])+1$$
Thus $$\dim(S[Y])\ge \sup \dim(Y_{i})+1=\dim(Y)+1$$
And we know $\dim(S[Y])=\dim(S[Y]_{(x_{i})})$(Why?). So we conclude $\dim(S[Y])=\dim(Y)+1$.Mind that the above argument trivially collapses when $Y\cap A_{i}=\emptyset$ or $x_{i}\in I(Y)$, etc. So some extra care is needed. 
