Show that $S^{-1}(I+J)=S^{-1}I+S^{-1}J$

Question

Let $R$ be a ring, $S$ a multiplicative subset of $R$ and $I,J$ ideals of $R$. Then $$S^{-1}(I+J)=S^{-1}I+S^{-1}J\,.$$

Proof. Note $I,J\subset I+J.$ So $S^{-1}(I+J)\supset S^{-1}I+S^{-1}J.$

I need help with the rest of the proof. And I'm not sure that the first part is correct.

Thanks in advance.

Answer 1

Hints:

$\subseteq$: a typical element of $S^{-1}(I + J)$ has the form $s^{-1}(i + j)$ where $s \in S$, $i \in I$ and $j \in J$. Then we have

$$s^{-1}(i + j) = s^{-1}i + s^{-1}j \in S^{-1}I + S^{-1}J$$

$\supseteq$: a typical element of $S^{-1}I + S^{-1}J$, has the form $s^{-1}i + t^{-1}j$, where $s, t\in S$, $i \in I$ and $j \in J$. Then we have $$s^{-1}i + t^{-1}j = (st)^{-1}(ti + sj) \in S^{-1}(I + J)$$

(because $S$ is multiplicatively closed and $I$ and $J$ are ideals).

All this is taking place in the ring of fractions $S^{-1}R$ of $R$ with respect to $S$. This is defined as the set of "fractions" $s^{-1}x$ (or $x/s$) with $s \in S$ and $x \in R$ modulo an equivalence relation. So for a detailed verification you need to check that the algebraic reasoning in the above argument respects that equivalence relation. This isn't too difficult using the fact that the mapping $x \mapsto x/1$ is a ring homomorphism $R \to S^{-1}R$ which sends every $s$ in $S$ to a unit in $S^{-1}R$.