Corollary 2

Let $A$ be a local noetherian ring. a) Let $x \in m_A$. Then $\dim(A/xA)$ is equal to $\dim(A)$ or to $\dim(A) - 1$. In order that one have $\dim(A/xA) = \dim(A) - 1$, it is necessary and sufficient that $x$ belong to none of the minimal prime ideals $p$ of $A$ such that $\dim(A/p) = \dim(A)$, and it suffices that $x$ be not a zero divisor in $A$. b) Let $a$ be an ideal of $A$ distinct from $A$ such that $\dim(A/a) < \dim(A)$. There exists $x \in a$ such that $\dim(A/xA) = \dim(A) - 1$. c) If $\dim(A) \geqslant 1$, there exists $x \in m_A$ such that $\dim(A/xA) = \dim(A) - 1$. By Prop. 2, $\dim(A/xA)$ is equal to $\dim(A)$ or to $\dim(A) - 1$. In order that one have $\dim(A/xA) = \dim(A)$, it is necessary and sufficient that there exist a chain $p_0 \subset ... \subset p_n$ of prime ideals of $A$ such that $x \in p_0$ and $n = \dim(A)$, that is to say, that there exist a prime ideal $p_0$ of $A$ containing $x$ such that $\dim(A/p_0) = \dim(A)$. But such a prime ideal $p_0$ is necessarily minimal, and every element of $p_0$ is therefore a zero divisor in $A$ (IV, § 1, No. 1, Cor. 3 of Prop. 2 and No. 4, Th. 2). This proves a).

Let $\Phi$ be the set of minimal prime ideals of $A$, and $\Phi'$ the set of $p \in \Phi$ such that $\dim(A/p) = \dim(A)$. We know (II, § 4, No. 3, Corollary 3 of Proposition 14) that $\Phi$ is finite, hence $\Phi'$ is finite. Let $a$ be an ideal of $A$ such that $\dim(A/a) < \dim(A)$. For every $p \in \Phi'$, one has $\dim(A/a) < \dim(A/p)$, hence $a \not\subset p$. By Proposition 2 of II, § 1, No. 1, there exists therefore an element $x$ of $a$ which does not belong to any of the $p \in \Phi'$, and one then has $\dim(A/xA) = \dim(A) - 1$ by a). This proves b).

The assertion c) is the particular case $a = m_A$ of b).