Proposition 2

Let $\rho : A \to B$ be a homomorphism of nonzero rings possessing the property (PM). One has the inequality

(1)

where S is the set of maximal ideals of A.

We know that $\dim(A) = \sup_{m \in S} \dim(A_m)$ (§ 1, No. 3, prop. 8). It is therefore enough to establish the inequality

(2)

for every maximal ideal m of A. In other words, it is a matter of proving the inequality

(3)

if $p_0 \subset ... \subset p_n$ is a chain of prime ideals of A contained in m and $\overline{q}_0 \subset ... \subset \overline{q}_r$ is a chain of prime ideals of B/mB. For $0 \leq i \leq r$, there exists a prime ideal $q_{n+i}$ of B lying above m such that $\overline{q}_i = q_{n+i}/mB$, and $q_n \subset ... \subset q_{n+r}$ is a chain of prime ideals of B. Put $p'_i = p_i$ for $0 \leq i \leq n-1$ and $p'_n = m$, so that $p'_0 \subset ... \subset p'_n$ is a chain of prime ideals of A and q_n lies above $p'_n$. If $q_i$ is a prime ideal of B lying above $p'_i$ ($1 \leq i \leq n$), prop. 1, c) proves that there exists a prime ideal $q_{i-1}$ of B lying above $p'_{i-1}$ and contained in $q_i$. By descending induction, we thus construct a chain $q_0 \subset ... \subset q_n$ of prime ideals of B such that $q_i$ lies above $p_i$ for $0 \leq i \leq n$. As $q_0 \subset ... \subset q_{n+r}$ is a chain of prime ideals of B, we have proved the inequality (3).