Localization of graded rings: Prove $k[x_0,...,x_n]_{(x_i)}$ is isomorphic to $k[\frac{x_0}{x_i},...,\frac{x_n}{x_i}]$ I was reading Vakil's FOAG in section 4.5.7, there is an example which shows the projective space $\Bbb{P}^n_k$ is the special case of $\text{Proj} (S_\cdot)$.
Let $ k[x_0,...,x_n]$ be the polynomial ring over $k$, and the localization at $f = x_i$ will be $k[x_0,...,x_n,1/x_i]$,then $k[x_0,...,x_n,1/x_i]$ is naturally a graded ring,with degree of $f/x_i^n$  being $\deg f - n$.
Denote the degree zero part of it be $k[x_0,...,x_n,1/x_i]_0$
Try to prove $k[x_{0/i},...,x_{n/i}]/(x_{i/i}- 1) \cong k[x_0,...,x_n,1/x_i]_0$

My attempt easy to define the map $$k[x_{0/i},...,x_{n/i}] \to k[x_0,...,x_n,1/x_i]_0\\ x_{j/i}\mapsto x_j/x_i$$
it can be checked it is a surjective ring homomorphism, the hard part is to show the kernel of the map is $(x_{i/i}-1)$.
How can I show this?
 A: Question: My attempt easy to define the map $$k[x_{0/i},...,x_{n/i}] \to k[x_0,...,x_n,1/x_i]_0\\ x_{j/i}\mapsto x_j/x_i$$
easy to see it is a surjective ring homomorphism, the hard part is to show the kernel of the map is $(x_{i/i}-1)$.
How can I show this?
Answer: You ring consists of equivalence classes of quotients
$$f(x_j)/x_n^d$$
where $f:=f_0+f_1+\cdots +f_d$ and where $f_i(x_0,..,x_n)$ is homogeneous of degree $i$.
You may write
$$f_d/x_n^d= \frac{\sum_I \alpha_I x_0^{j_0} \cdots x_n^{j_n}}{x_n^d}.$$
Since $\sum j_k =d$ it follows $j_n=d-j_0-j_1-\cdots -j_{n-1}$ and you get
$$f_d/x_n^d= \frac{\sum_I \alpha_I  x_0^{j_0} \cdots x_n^{d-j_0-j_1-\cdots -j_{n-1}}}{x_n^d} =\sum_I \alpha_I (\frac{x_0}{x_n})^{j_0}\cdots (\frac{x_0}{x_{n-1}})^{j_{n-1}}.$$
from this it follows there is an isomorphism
$$T^{-1}S_{(0)}\cong k[\frac{x_0}{x_n},\cdots , \frac{x_{n-1}}{x_n}].$$
All calculations are done inside the quotient field $k(x_i)$ the claim follows from this: As an example: If $n=2$ and if you consider the surjection
$$\phi:k[y_0,y_1] \rightarrow k[x_0/x_2,x_1/x_2],$$
with
$$\phi(f(y_0,y_1))= \sum_{(i,j)} a_{(i,j)}(x_0/x_1)^i(x_1/x_2)^j$$
where $f$ is homogeneous of degree $d$, you get
$$\phi(f)= \frac{ \sum_{(i,j)} a_{(i,j)}x_0^ix_1^j}{x_2^d}=0$$
and this is zero in the quotientfield $k(x_j)$ iff the coefficients $a_{(i,j)}=0$
for all $(i,j)$. Hence the original $f=0$ is zero. Hence the map $\phi$ is injective.
Response: "thank you @hm2020 , in the last line why it's zero in the quotient field k(xj) do you mean k(x2)[x0,x1]?"
Answer: Since $A:=k[x_i]$ is an integral domain it follows any localization $S^{-1}A \subseteq K(A) \cong k(x_i)$ for any $S$, is a sub ring of the quotient field. This holds for any integral domain.
Response: "and why iff the coefficient $a(i,j)=0$ "
Answer: If
$$\phi(f)= \frac{ \sum_{(i,j)} a_{(i,j)}x_0^ix_1^j}{x_2^d}=0$$
it follows since $k[x_i]$ is a domain that
$$\sum_{(i,j)} a_{(i,j)}x_0^ix_1^j=0$$
but this is iff $a_{(i,j)}=0$ for all $(i,j)$ since $f(x_0,x_1)=\sum_{(i,j)} a_{(i,j)}x_0^ix_1^j$ is homogeneous of degree $d$ in the variables $x_0,x_1$.
