Corollary 1

For every prime ideal p of A, the set of prime ideals of B lying above p is finite.

Suppose first that $p = (0)$; then the only prime ideal $q$ of $B$ such that $q \cap A = (0)$ is $(0)$; otherwise, writing $S = A - \{0\}, S^{-1}q$ would be a non-zero prime ideal of $S^{-1}B$ (Chapter 11, § 2, no. 5, Proposition 11) and $S^{-1}B$ is just the field of fractions of $B$, for it is a subring of $L$ containing $K$ (\emph{Algebra}, Chapter V, § 3, no. 2, Proposition 3); whence an absurd conclusion. If now $p \neq (0)$, it follows from Proposition 5 that $B/pB$ is a finite-dimensional vector space over the field $A/p$, hence an \emph{Artinian} ring and therefore has only a finite number of prime ideals (Chapter IV, § 2, no. 5, Proposition 9), which proves that there is only a finite number of prime ideals of $B$ containing $p$.