Background
Let $R$ be an integral domain and let $S$ be this set of pairs:
$$S=\{(a,b)\mid a,b\in R, b\neq 0\}$$
Define a relation $\sim$ on the set $S$ by
$$(a,b)\sim (c,d)\text{ means }ad-bc \text{ in }R$$.
Theorem 10.30 Let $R$ be an integral domain. Then there exists a field $F$ whose elements are of the form $a/b$ with $a,b\in R$ and $b\neq 0_R$, subject to the equality condition
$$\frac{a}{b}= \frac{c}{d}\text{ in }F\text{ if and only if }ad=bc\text{ in }R$$
Addition and multiplication in $R$ are given by
$$\frac{a}{b} +\frac{c}{d} =\frac{ad+bc}{bd}, \frac{a}{b}\cdot\frac{c}{d} =\frac{ac}{bd},$$
The set of elements in $F$ of the form $a/1_R(a\in R)$ is an integral domain isomorphic to $R$.
....The ring $R$ will be identified with its isomorphic copy in $F$. Then we can say that $R$ is the subset of $F$ consisting of elements of the form $a/1_R$. The field $F$ is called the field of quotients of $R$.
Theorem 6.1 For an ingeral domain $R$, a field $Q(R)$ can be constructed such that:
$1)$ A field $Q(R)$ contains a subring $P$ which is isomorphic to the ring $R$.
$2)$ Each non-zero element of the field $Q(R)$ can be written in the form $ab^{-1}$ where $a,b\in P, b\neq 0$.
Definition 6.2
The field $Q(R)$ constructed in theorem 6.1 is called the field of fractions of a ring $R$.
Definition for multiplicative closed set. Let $R$ be a ring. A subset $T$ of $R$ is a multiplicative system if $1 \in T$, and if $a, b \in T$ implies that $ab \in T$ — that is, T is multiplicatively closed and contains $1.$
If $T$ is a multiplicative system of a ring $R$, then an equivalence relation can be defined on an appropriate set $S$ of "fractions" using elements of $R$,
$$S=\{a/b\mid a,b\in R, b\in T\}$$
by
$$\frac{a}{b}\cong \frac{c}{d}\text{ if and only if }t(ad-bc)=0$$.
For some $t\in T$.
Definition 6.7 Let $R$ be a ring, and let $T$ be a multiplicative system of $R$. The above ring $R_T$ of equivalence classes of fractions from $R$ whose denominators are elements of $T$, with addition and multiplication defined as above, is called the localization of $R$ with respect to $T$, and which we read as "$R$ localized at $T$."
Theorem 10.6.2. Any integral domain can be embedded in a field. (In the proof, $R$ is used to denoted integral domain, $F$ is used for field.)
Definition 10.6.2. For any integral domain $R$, the field $F$ constructed above is called the field of quotients of $R$.
Exercise 15. Let $R$ be a commutative ring with unity. A subset $S$ of $R$ is said to be multiplicative if $1\in S, 0\not\in S$ and $ab\in S$. Define a binary relation $\theta$ on $R\times S$ by
$$(a,s)\theta (b,t) \text{ iff there exists }u\in S\text{ such that }u(at-bc)=0.$$
Prove that $\theta$ is an equivalence relation on $R\times S$. Let the equivalence containing $(a,s)$ be denoted by $a/s$ and let
$$S^{-1}R=\{\frac{a}{s},a\in R, s\in S\}$$
Define addition and multiplication on $S^{-1}R$ by
$$\frac{a}{s} +\frac{c}{t} =\frac{at+bs}{st}, \text{ and } \frac{a}{s}\cdot\frac{c}{t} =\frac{ac}{st}$$
Then prove that $(S^{-1}R,+,\cdot)$ is a commutative ring with unity. This ring is called the ring of fractions of $R$ by $S$.
Question
For the question to follow, i wrote down definitions and theorems from various sources as references in the above backgroubd section. Suppose I have a single variable polynomial ring $F[x]$, a subset $S$ of $F[x]$ and I have a prime ideal $P=(f(x))$, where $f(x)$ is some specific irreducible polynomial with non-zero constant term over $F$.
If I want to localize $F[x]$ at the prime ideal $P$, the correct way to describe the mulitplicative closed set/system $S$ and the ring of fractions $F[x]_P$ in set builder notation, or since $f(x)$ is some specific polynomial, the best that can be done is the following:
Multiplicative closed set: $$S=\{q(x)\mid q(x)\in F[x], q(x)\not\in P=(f(x))\}$$
and the ring of fractions:
$$F[x]_{P=(f(x))}=S^{-1}F[x]=(F[x]\setminus P)^{-1}F[x]= \{[\frac{p(x)}{q(x)}]\mid p(x),q(x)\in F[x], q(x)\in S\},\text{ and where } S=F[x]\setminus P$$.
Is the way I wrote out $S$ and $F[x]_{P=(f(x))}$ correct?
Thank you in advance
