Going-Up Theorem

In an integral extension, chains of prime ideals lift upward.

Going-Up Theorem

Statement

Let $A \subseteq B$ be an integral extension (equivalently, every $b\in B$ is integral over $A$ ).

Suppose $\mathfrak p \subseteq \mathfrak q$ are prime ideals of $A$ , and let $\mathfrak P$ be a prime ideal of $B$ with

\mathfrak P \cap A = \mathfrak p.

Then there exists a prime ideal $\mathfrak Q \subseteq B$ such that

\mathfrak P \subseteq \mathfrak Q \quad\text{and}\quad \mathfrak Q \cap A = \mathfrak q.

Equivalently: for the map on spectra

Spec (B)\to Spec (A),

prime chains in $A$ can be lifted to prime chains in $B$ once you start with a prime lying over the first one.

Gaussian integers.
Take $A=\mathbb Z \subseteq B=\mathbb Z[i]$ , an integral extension.
Consider the chain $(0)\subset (5)$ in $\mathbb Z$ . Start with $\mathfrak P=(0)\subset \mathbb Z[i]$ lying over $(0)$ .
Going-up produces a prime $\mathfrak Q\supset (0)$ lying over $(5)$ ; concretely, $(5)$ factors in $\mathbb Z[i]$ and one can take $\mathfrak Q=(2+i)$ (or $(2-i)$ ).
A simple normalization map.
Let $A=k[t^2]\subseteq B=k[t]$ . This is integral since $t$ satisfies $T^2-t^2=0$ over $A$ .
The chain $(0)\subset (t^2)$ in $A$ lifts: starting with $\mathfrak P=(0)$ in $B$ lying over $(0)$ , going-up gives $\mathfrak Q=(t)$ in $B$ lying over $(t^2)$ , and indeed $(t)\cap k[t^2]=(t^2)$ .
General picture in terms of spectra.
If $A\subseteq B$ is integral, then the induced map Spec (B) $\to$ Spec (A) is surjective (lying-over ) and satisfies going-up: specializations in $A$ lift to specializations in $B$ .