Saturation of a multiplicatively closed subset Exercise 3.7 of Atiyah-MacDonald asks the reader: if $A$ is a commutative ring and $\mathfrak{a} \triangleleft A$ an ideal, find the saturation of $1 + \mathfrak{a}$.
Previously we have shown that the saturation of a multiplicatively closed subset $S$ is the complement of the union of prime ideals not meeting $S$.
Are they just looking for the fact that the saturation is the complement of the union of all $\mathfrak{p}$ prime such that $\mathfrak{p} + \mathfrak{a} \neq R$? Or is there more to say? 
 A: You can say something more precise: $\overline S=A-\bigcup_{\mathfrak m\in\operatorname{Max}(A);\ \mathfrak a\subseteq\mathfrak m}\mathfrak m$.
A: What I give as a description of the saturation of $S$ is:
$$\overline{S}=\{s\in A |\bar{s} \text{ is a unit in } A/\mathfrak{a} \}.$$
One can easily see that  $\overline{S}$ is saturated and contains $S$.
To prove that $\overline{S}$ is indeed the saturation of $S$, from 3.7(ii) of the book it suffices to prove that for any prime ideal $\mathfrak{p}\cap S=\varnothing$, we have $\mathfrak{p}\cap \overline{S}=\varnothing$. (If this is proved, then $A\setminus \overline{S}\supseteq\bigcup_{\mathfrak{p}\cap S=\varnothing}\mathfrak{p}$, which in turn shows that $\overline{S}$ is contained in the saturation of $S$, and thus coincide with the saturation of $S$. )
We assume the contrary that $s\in \mathfrak{p}\cap \overline{S}$. By the definition of $\overline{S}$, there exists $t\in A$ and $a\in \mathfrak{a}$ such that $s\cdot t=1+a.$ So $1+a\in\mathfrak{p}\cap S$, a contradiction.
