$\dim(A/a) \le \dim A - 1$ for domains $A$

Question

Let $A$ be an integral domain, and let $0 \ne a \in A$. Then I want to show that $\dim(A/a) \le \dim A - 1$.

The obvious idea is to consider any proper chain of prime ideals $\bar{\mathfrak p}_1 \subset \dots \subset \bar{\mathfrak p}_n$ in $A/a$. Contraction yields a proper chain of prime ideals $\mathfrak p_1 \subset \dots \subset \mathfrak p_n$ in $A$ above the ideal $(a)$. My problem is to justify that we can extend the contracted chain by some prime ideal $\mathfrak p_0 \subset \mathfrak p_1$; in particular if $(a) \subseteq \mathfrak p_0$ for all prime ideals $\mathfrak p_0 \subset \mathfrak p_1$.

Answer 1

In the end, I forgot that for a chain in the sense of the Krull dimension, the first prime ideal is allowed to be $(0)$. As $A$ is an integral domain, $\mathfrak p_0 = (0)$ is indeed prime. Furthermore, $\mathfrak p_1 \ne (0)$ since $a \ne 0$, so $(0) \ne (a) \subseteq \mathfrak p_1$.

Answer 2

I will use the notation from your post. Suppose we had indeed that for all prime ideals $\mathfrak{p}\subset A$ with $\mathfrak{p} \subset \mathfrak{p}_1$ it were the case that $(a)\subset \mathfrak{p}$. This means that $a/1 \in \mathfrak{q}$ for all $\mathfrak{q} \subset A_{\mathfrak{p}_1}$. But this means that $a/1$ is nilpotent in $A_{\mathfrak{p}_1}$. This ring however is reduced, so $a/1 = 0$ in $A_{\mathfrak{p}_1}$. As $A$ is an integral domain all its localizations embed into its quotient field. Thus, $a/1=0$ in $A_{\mathfrak{p}_1}$ implies that $a=0$, contradicting the hypothesis.