Homogeneous localization and field of fractions Let $R$ be a graded ring which is also an integral domain. Let $P$ be a prime and homogeneous ideal of $R$. Let $R_{(P)}$ be the homogeneous localization of $R$ in $P$. In other words, let $S:=\{\text{homogeneous elements of $R$ that are not in $P$} \}$. Than $S^{-1}R$ is also a graded ring: if $\frac{a}{s}$ is an element of $S^{-1}R$ with $a\in R$ homogeneous, then put deg$\frac{a}{s}:=$ deg$a-$deg$s$. Then, put $R_{(P)}$ be the subring of elements of zero-degree in $S^{-1}R$. So, a typical element of $R_{(P)}$ is of the form $\frac{a}{s}$ where $a$ and $s$ are homogeneous elements of $R$ of the same degree and $s\notin P$. Moreover, $R_{(P)}$ is also an integral domain.
In particular $(0)$ is a prime and homogeneous ideal of $R$, so with the same above construction, consider $R_{((0))}$. So, a typical element of $R_{((0))}$ is of the form $\frac{a}{s}$ where $a$ and $s$ are homogeneous elements of $R$ of the same degree and $s\neq 0$.  Moreover $R_{((0))}$ is also a field.
The map $$\frac{a}{s}\in R_{(P)} \mapsto \frac{a}{s}\in R_{((0))}$$ is injective. So i can extend it to the map $$f\colon \frac{a}{s} \left( \frac{b}{t} \right)^{-1}\in \text{Quot}R_{(P)}\mapsto \frac{a}{s} \left( \frac{b}{t} \right)^{-1}\in R_{((0))}$$ which is also injective, where $\text{Quot}R_{(P)}$ is the field of fractions of $R_{(P)}$.
My question is

is true that f is an isomorphism? Or, in general, is it true that $\text{Quot}R_{(P)}$ and $R_{((0))}$ are isomoprhic?

I want to show that f is surjective.
Let $\frac{c}{d} \in R_{((0))}$. If $d\notin P$, then i have $\frac{c}{d}=f(\frac{c}{d})$. If $c\notin P$, then i have $\frac{c}{d}=f((\frac{d}{c})^{-1})$.
But what if $c$ and $d$ are both in $P$?
My idea  was searching for an homogeneous element $s$ of $R$ of degree$=$deg$c=$deg$d$, such that $s\notin P$. In this way i have $$\frac{c}{d}=f\left( \frac{c}{s}(\frac{d}{s})^{-1}   \right)$$.
But can how can i find such an element $s\;$ ?
Can it be that $P$ contains all the homogeneous elements of a certain degree and so i can't find a such element $s\;$?
So i want to show that it is not possible. My idea is: let $A=\{n\in\mathbb{N_0}\mid R_n \subseteq P    \}$ with $R_n$ the group of the elements of $R$ of degree $n$. I want to sho that $A=\emptyset$. For absurd suppose that $A\ne \emptyset$. Let $m$ be the minimum of $A$. Then $m\ge 1$ because $1\notin P$. Let assume $m>1$. Let $h,k \in \mathbb{N}$ such that $h,k<m$ and $h+k=m$. Then i can find $x\in R_h$ and $y\in R_k$ that are both not in $P$. But $xy\in R_{h+k}=R_m\subseteq P$ bacause $m\in A$. But $P$ is prime so $x\in P $ or $y\in P$ wich is absurd.
Now, can i prove thah such $m$ must be $>1$? Or my idea may not work?
 A: Your idea is wrong. It's possible to have $A \neq \varnothing$. It's possible to have $m=1$. In fact, it's quite easy.
Example 1: Let $R = k[x]$ with standard grading and let $P = \langle x \rangle$. Then $A = \{n : n \geq 1 \}$ and $m=1$.
Example 2: Let $R = k[x_1,x_2]$ and let $P = \langle x_1 \rangle$. Here, $P$ is prime. But we give $R$ the grading with $\deg(x_1)=1$, $\deg(x_2)=2$. So $R_1 \subset P$. That means $A = \{1\} \neq \varnothing$ and $m=1$.
Example 3: Let $R = k[x_1,\dotsc,x_n]$, each $\deg(x_i)=i$, and $P = \langle x_1,\dotsc,x_k \rangle$.
You cannot prove $A = \varnothing$ or $m>1$ because those statements are not true. Instead of your idea, you will have to do something different.

The statement you are asking for is false. It is not true that $f$ is an isomorphism and it is not true in general that $\operatorname{Quot}(R_{(P)})$ and $R_{((0))}$ are isomorphic.
First I'll prove a positive result under an additional hypothesis. At the end I'll give the negative result.

Let $R$ be a graded ring which is also an integral domain. More specifically, suppose that $R$ is $\mathbb{Z}_{\geq 0}$-graded: that is, $R = \bigoplus_{n=0}^{\infty} R_n$, so there is a graded piece for each nonnegative integer. (As usual each $R_n$ is an abelian group and $R_m R_n \subseteq R_{m+n}$.) Let $P \subset R$ be a homogeneous prime ideal.
Let's start with a simplifying assumption that every degree $R_n$ is nonzero.
Claim 4: Suppose that $R_n \neq 0$ for each $n$ and suppose that there exists some $n \geq 1$ such that $R_n \not\subset P$. Then $\operatorname{Quot}(R_{(P)}) \cong R_{((0))}$.
It is equivalent to assume that $R_1 \neq 0$, since then $R_1^n \subseteq R_n$, and $R_1^n \neq 0$ by the assumption that $R$ is an integral domain.
Proof: Let $s$ be a homogeneous element of degree $n \geq 1$ such that $s \notin P$. Let $c,d \neq 0$ be homogeneous elements of the same degree, say $m \geq 0$. Write $m = qn-r$ where $q \geq 0$, $r \geq 0$. The point is that $s^q$ has degree $qn$, so $\deg(s^q) \geq \deg(c),\deg(d)$. Let $t$ be any nonzero element in $R$ of degree $r$. Then in $R_{((0))}$ we have
$$
  \frac{c}{d} = \frac{ct}{s^q} \left(\frac{dt}{s^q}\right)^{-1} .
$$
This proves that $\operatorname{Quot}(R_{(P)}) \to R_{((0))}$ is surjective, so these fields are isomorphic. $\Box$
Now discard the assumption that $R_n \neq 0$ for each $n$. For example, say $R=k[t^2,t^3] \subset k[t]$; or the (frivolously named) "Chicken McNugget" example $R = k[t^6,t^9,t^{20}] \subset k[t]$, which has elements of degrees $6,9,12,15,18,20,21,24,\dotsc$.
Claim 5: Suppose there is some $n \geq 1$ such that $R_n \not\subset P$. Then the same result holds.
Proof: The idea is the same. Given $c,d \neq 0$ with $\deg(c)=\deg(d)=m \geq 0$ and $s \notin P$, $\deg(s) = n \geq 1$, we are looking for some element $t \neq 0$ such that $\deg(t) = r = qn-m$. The problem is that $R_{qn-m}$ might be zero for some values of $q$. Let $g = \gcd(m,n)$. Since $R$ has nonzero elements of degrees $m$ and $n$, then $R$ has nonzero elements of degree $am+bn$ for all $a,b\geq 0$, so $R$ has nonzero elements of degree $fg$ for all sufficiently large $f$. Now $qn-m$ is a multiple of $g$; just choose a sufficiently large value of $q$ so that $R$ contains nonzero elements $t$ of degree $r=qn-m$. The rest of the proof is the same. $\Box$
None of these claims end up having anything to do with whether $R$ is finitely generated. That was a total red herring. Oops, sorry about that.

I did assume in both claims that the value of $n$ giving $R_n \not\subset P$ has to be $n > 0$. If $R_n \subset P$ for all $n \geq 1$—so that the only $R_n \not\subset P$ is $n=0$—then $\operatorname{Quot}(R_{(P)}) \cong \operatorname{Quot}(R_0)$, not $R_{((0))}$.
Example 6: Let $R = \mathbb{Z}[x]$ and $P = \langle 2,x \rangle$. Then the only homogeneous elements not in $P$ are odd integers (in degree $0$), so $R_{(P)} = \mathbb{Z}_{(2)}$ and $\operatorname{Quot}(R_{(P)}) = \mathbb{Q} = \operatorname{Quot}(R_0)$. Here $R_0$ means the $0$th graded piece of $R$. It is different from $R_{((0))} = \mathbb{Q}(x)$.
