Corollary

Let A be an integral domain and $\mathfrak{p}_1, \ldots, \mathfrak{p}_n$ prime ideals of A, no two of which are comparable with respect to inclusion. If $A = \bigcap_1^n A_{\mathfrak{p}_i}$ in the field of fractions of A, the maximal ideals of A are $\mathfrak{p}_1, \ldots, \mathfrak{p}_n$.

Setting $S = \bigcap_{i=1}^n (\mathbf{A} - p_i)$, $S^{-1}\mathbf{A} = \mathbf{A}$ by Proposition 17 (c); hence the elements of $S$ are invertible in $\mathbf{A}$ and $S^{-1}p_i = p_i$ for all $i$. Our assertion then follows by virtue of Proposition 17 (a).