Is $A_{S}$ a ring of fractions or a localized ring? Let $$A_{S}:=\Biggl\{\frac{a}{b}\in K\mid a\in A,b=1\vee b=st,\;\forall s,t\in S\Biggr\},$$
where $K=Frac(A)$, $A\subset K$, $A$-integral domain and $0\not\in S\subset A$. 
Is $A_{S}$ a ring of fractions or is it a localized ring at $S$ or neither?
I think it is a ring of fractions but would like someone to check if I am correct.
I have learnt that a ring of fractions is a construction of this type 
$$S^{-1}A:=\Biggl\{\frac{a}{s}\in K\mid a\in A,s\in S\Biggr\}$$ where we insist on $S$ to be a multiplicative set, i.e., $1\in S$ and $\forall s,t\in S\Rightarrow st\in S$.
Does it mean in the above notation that $b=1\in S$? Because then $S$ is a multiplicative set and $A_{S}=S^{-1}A$ right? 
While a localized ring is a structure of a type 
$$A_{P}:=\Biggl\{\frac{a}{b}\in K\mid a,b\in A \wedge b\not\in P\Biggr\}.$$
But here we localize a ring at a prime ideal $P$.
In my book I have found a statement that if $P$ is a prime ideal then $S=A\setminus P$ is a multiplicative set. This somehow indicates that localizing at $P$ is a way of creating a ring of fractions. Is it true?
Any ideas gratefully appreciated.
 A: There is some confusion in notation. Some texts use $A_S$ where others use $S^{-1}A$ for the same thing, when $S$ is a multiplicative subset of $A$.
I too was a bit confused when my teacher said that when $P$ is a prime ideal, then we would write $A_P$ as an abbreviation of $A_{A\setminus P}$. Since then, I have always used the clearer notation $S^{-1}A$.
Localization at a prime ideal is a special case of making a ring of fractions. In the case $A$ is a domain, everything is simpler. If $S$ is a multiplicative subset of $A$ (that is, $st\in S$ whenever $s\in S$ and $t\in S$), one can define
$$
S^{-1}A=\left\{\frac{a}{b}:a\in A,\; b\in S\cup\{1\}\right\}
$$
and show that this is a subring of the quotient field of $A$, precisely the smallest subring where every element of $S$ is invertible. One has to assume $0\notin S$, of course, otherwise such a subring would not exist. Note also that it's always not restrictive to assume that $1\in S$, because
$$
S_1^{-1}A=S^{-1}A
$$
if $S_1=S\cup\{1\}$; $S_1$ is obviously a multiplicative subset if $S$ is.
When $P$ is a prime ideal, $S=A\setminus P$ is a multiplicative subset not containing $0$, so one defines
$$
A_P=(A\setminus P)^{-1}A
$$
because the emphasis is on $P$ rather than its complement. Just a question of notation.

Note that the operations can be carried over also when $A$ is not a domain. However, fraction must be defined more carefully. We consider the set $A\times S$ and the equivalence relation on it defined by $(a,s)\sim(b,t)$ if there exists $u\in S$ with $u(at-bs)=0$. We assume that $S$ is a multiplicative subset.
One can show that denoting by $\dfrac{a}{s}$ the equivalence class of $(a,s)$ and defining
$$
\frac{a}{s}+\frac{b}{t}=\frac{at+bs}{st},\qquad
\frac{a}{s}\frac{b}{t}=\frac{ab}{st}
$$
one obtains a ring structure on the quotient set, denoted by $S^{-1}A$.
A: That is sort of where the intuition comes from, it's more of a way of creating a field of fractions off of a certain prime ideal. However we can recover the concept of the "field of fractions" because $(0)$ is prime, and thus its compliment closed under multiplication. Thus when we localize at $(0)$ we get the set of all $p/q$, $p \in R$, $q \in R^*$ as desired. Hope this is enlightening! It's a good question to ask.
To answer your first question, I believe that what you're doing is essentially localizing at the compliment of the multiplicative closure of S.
