Corollary

Let F be a closed subset of Spec(A). If Y is an irreducible component of the inverse image of F by the mapping “ρ : Spec(B) → Spec(A), then the closure of “ρ(Y) is an irreducible component of F.

Let indeed a be an ideal of A such that F = V(a) and q the prime ideal of B such that Y = V(q). The inverse image by “ρ of F is the subset V(ρ(a)B) of Spec(B) and the closure of “ρ(Y) is the irreducible closed subset V(ρ^{-1}(q)) of Spec(A).

It is a matter of proving that if q is minimal among the prime ideals of B containing $\rho(a)$, then $\rho^{-1}(q)$ is minimal among the prime ideals of A containing a. Otherwise, there would exist a prime ideal $p'$ of A with $a \subset p' \subset \rho^{-1}(q)$ and $p' \neq \rho^{-1}(q)$; according to prop. 1, c), there would exist a prime ideal $q'$ of B such that $q' \subset q$ and $p' = \rho^{-1}(q')$, whence $\rho(a) \subset q' \subset q$ and $q' \neq q$ contrary to the assumption made on q.