In Algebraic Geometry by Hartshorne in the proof of theorem 3.4 in Chapter 1 he gives an isomorphism of $k[y_1,...,y_n]$ with $k[x_0,...,x_n]_{(x_i)}$ by sending $f(y_1,...,y_n)$ to $f(x_0/x_i,...,x_n/x_i)$ leaving out $x_i/x_i$.
Then he says that this isomorphism sends $I(Y_i)$ to $I(Y)S_{(x_i)}$. I don't understand this last bit.
Can some one help?
Thanks
First of all, the notation Hartshorne uses is a bit confusing: by $I(Y)S(Y)_{(x_i)}$ Hartshorne means the ideal $(I(Y) \cdot S(Y)_{x_i}) \cap S(Y)_{(x_i)}$ (see here). If $g \in S(Y)$ is homogeneous and $e$ its degree, the fraction $g/x_i^e \in S(Y)_{(x_i)}$ is contained in this ideal if and only if $x_i^rg \in I(Y)$ for some $r>0$.
As in the proof of Proposition 2.2, let $\beta: k[y_1,\ldots,y_n] \rightarrow k[x_0,\ldots,x_n]$ be the "homogenisation map". It sends a polynomial $f \in k[y_1,\ldots,y_n]$ of degree, say, $d$, to $\beta(f)=x_i^{d}f(x_0/x_i,\ldots,x_n/x_i)$, omitting $x_i/x_i$. Thus, the isomorphism maps $f$ to $\beta(f)/x_i^d$.
We have to show that $f \in I(Y_i)$ if and only if $\beta(f)/x_i^d \in I(Y)S(Y)_{(x_i)}$. In view of the first paragraph, latter condition is equivalent to $x_i^r\beta(f) \in I(Y)$ for some $r>0$. This means that $Y \subseteq Z(x_i^r\beta(f))=Z(\beta(f))\cup Z(x_i)$, i.e. $Y \cap U_i = Y \setminus Z(x_i) \subseteq Z(\beta(f))$. As in Proposition 2.2, this occurs if and only if $Y_i \subseteq Z(f)$, i.e., $f \in I(Y_i)$.