Lemma 1

Let $A$ be a noetherian ring, $p_0 \subset ... \subset p_n$ a chain of prime ideals of $A$ of length $n \geqslant 1$, and $x$ an element of $p_n$. There exists a chain $p'_0 \subset ... \subset p'_n$ with $p'_0 = p_0$, $p'_n = p_n$ and $x \in p'_1$.

We reason by induction on $n$, the case $n = 1$ being trivial. Suppose therefore that one has $n \geqslant 2$ and that $x$ does not belong to $p_{n-1}$. Let $p'_{n-1}$ be a minimal element of the set of prime ideals of $A$ contained in $p'_n = p_n$ and containing $p_{n-2} + Ax$ (II, § 2, No. 6, lemma 2). By prop. 1, the ideal $p'_{n-1}/p_{n-2}$ of the ring $A/p_{n-2}$ is of height 1, and since $p_{n-2} \subset p_{n-1} \subset p_n$ is a chain of length 2, the same is true of $p_{n-2} \subset p'_{n-1} \subset p'_n$. One has $x \in p'_{n-1}$. The induction hypothesis applied to the chain $p_0 \subset p_1 \subset ... \subset p_{n-2} \subset p'_{n-1}$ shows that there exists a chain $p'_0 \subset p'_1 \subset ... \subset p'_{n-2} \subset p'_{n-1}$ with $x \in p'_1$ and $p'_0 = p_0$. The chain

satisfies the required conditions.