Corollary 3

Let $A, B$ be two rings, $\rho$ a homomorphism from $A$ to $B$ and $p$ a prime ideal of $A$. For there to exist a prime ideal $p'$ of $B$ such that $\overline{\rho}^{-1}(p') = p$, it is necessary and sufficient that $\overline{\rho}^{-1}(B\rho(p)) = p$.

If there exists an ideal $p'$ of $B$ such that $\bar{\rho}^{-1}(p') = p$, then $\rho(p) \subset p'$, whence $B_{\rho}(p) \subset p'$ and $\bar{\rho}^{-1}(B_{\rho}(p)) \subset \bar{\rho}^{-1}(p') \subset p$; as the converse inclusion is obvious, $\bar{\sigma}^{-1}(B_{\rho}(p)) = p$. Conversely, suppose that $\bar{\rho}^{-1}(B_{\rho}(p)) = p$ and consider the multiplicative subset $S = \rho(A - p)$ of $B$; the hypothesis shows that

by Corollary 2 there exists a prime ideal $p'$ of $B$ containing $B_{\rho}(p)$ and not meeting $S$; then $\bar{\rho}^{-1}(p')$ contains $p$ and cannot contain any element of $A - p$ and hence is equal to $p$.