Proposition 16

Let $B$ be a ring and $A$ a subring of $B$. For every minimal prime ideal $p$ of $A$ there exists a minimal prime ideal $q$ of $B$ such that $q \cap A = p$.

We write $S = A - p$; then the ring $A_+ = S^{-1}A$ is identified with a subring of $S^{-1}B$ (no. 1, Proposition 2) and on the other hand $A_+$ has only a single prime ideal $p'$ since $p$ is minimal (no. 5, Proposition 11). As $S^{-1}B$ is not reduced to 0 (since it contains $A_+$), it has at least one prime ideal $r'$ and therefore $r' \cap A_+ = p'$; if $j = i_B^S$ and we write $r = j'(t')$, then

hence $r \cap A \subset p$ and, as $p$ is minimal, $t \cap A = p$; moreover, $r$ is prime in $B$; if $q$ is a minimal prime ideal of $B$ contained in $r$ (Lemma 1), then a fortiori $q \cap A = p$ since $p$ is minimal.