Ideal in ring of fractions I got some problems with the following:
Let $R$ be a ring with $1\in R$. Let $S\subset R$ a subset which is closed under multiplication and contains $1$. On the set $R \times S$ we define an equivalence relation:
$(a,s)$~$(b,t) \iff$ there exists $u\in S$ such that $(at-sb)u=0$.
We denote the equivalence class of $(a,s)$ as $\frac{a}{s}$ and the set of all equivalence classes with $S^{-1}R$.
Finally we define the following operations: $\frac{a}{s}+\frac{b}{t}=\frac{at+bs}{st}$ and $\frac{a}{s}\cdot \frac{b}{t}=\frac{ab}{st}$ $\rightarrow$ with these operations $S^{-1}R$ becomes an commutative ring with $1$.
Now the actual task I want to solve:

Show that $S^{-1}I=\{\frac{a}{s}:a\in I,s\in S\}$ is an ideal in $S^{-1}R$.

The problem for me is to find the inverse element to given element (in terms of additive group).
I guess $e=\frac{0}{1}$, because $\frac{a}{s}+ \frac{0}{1}=\frac{a\cdot 1+0\cdot s}{s\cdot 1}=\frac{a}{s}$
So, I have to find an element to a given $\frac{a}{s}$, such that $\frac{at+bs}{st}=\frac{0}{1}$.
But I can't assume that there exists a $t\in S^{-1}$ such that $st=1$, so what should I do?
I am thankful for any kind of help!
 A: To show the first part: 
Theorem: If $I$ is an ideal in $R$, then $S^{-1}I$ is an ideal in $S^{-1}R$.
Proof: We first show that $S^{-1}I$ is an additive group:


*

*$\displaystyle \frac01 \in S^{-1}I$ since $0 \in I$ because $I$ is an ideal.

*Let $\displaystyle \frac{i}{s}, \frac{j}{t} \in S^{-1}I$, then $\displaystyle \frac{i}{s} + \frac{j}{t} = \frac{it + js}{st}$. Notice that $st \in S$ since $S$ is a multiplicatively closed set. Since $i, j \in I$, then $it, js \in I$ since $I$ absorbs and $it + js \in I$ since $I$ is an additive group. Conclude that $\displaystyle \frac{it + js}{st} \in S^{-1}I$.

*Let $\displaystyle \frac{i}{s} \in S^{-1}I$, where $i \in I$, then $-i \in I$ since $I$ is an additive group, so $\displaystyle \frac{-i}{s} \in S^{-1}I$.


Now we show that $S^{-1}I$ absorbs elements of $S^{-1}R$. Let $\displaystyle \frac{r}{t} \in S^{-1}{R}$ be arbitrary and let $\displaystyle \frac{i}{s} \in S^{-1}I$. Observe that $\displaystyle \frac{r}{t}\frac{i}{s} = \frac{ri}{st}$ where $st \in S$ since $S$ is a multiplicatively closed set and $ri \in I$ since $I$ is an ideal and absorbs elements from $R$.

Now, for the second question you asked in the comments:
Theorem: $S^{-1}I$ is a proper ideal if and only if $S \cap I = \emptyset$.
Proof: Suppose $S^{-1}I$ is a proper ideal, then there exists $\displaystyle \frac{r}{s} \in S^{-1}R$ such that $\displaystyle \frac{r}{s} \notin S^{-1}I$. If we suppose, by way of contradiction, that $S \cap I \neq \emptyset$, then we can choose $t \in S \cap I$. Observe that $\displaystyle \frac{r}{s} = \frac{rt}{st} \in S^{-1}I$, a contradiction. Conclude that $S \cap I = \emptyset$.
Conversely, suppose that $S \cap I = \emptyset$. If we suppose, by way of contradiction, that $S^{-1}I = S^{-1}R$, then $\displaystyle \frac11 = \frac{i}{s}$ where $i \in I$ and $s \in S$, which means that $(s - i)t = 0$ for some $t \in S$ but this means $st = it$ where $st \in S$ and $it \in I$. Hence $S \cap I \neq \emptyset$, a contradiction. Conclude that $S^{-1}I \subsetneq S^{-1}R$.
