Counterexample for going up theorem

Question

I am searching for an example which shows that integral extensions are necessary for going up theorem.

Basically I want rings $A\subset B$ (not integral extension) such that lying over holds, but going up does not hold.

Answer 1

Let $O$ be a discrete valuation ring and let $t$ be its prime element. The Lying-Over-Theorem holds in the polynomial extension $O\subset O[x]$, but not the Going-Up-Theorem: the ideal generated by the polynomial $tx-1$ is prime, lies over $0$, but is maximal since $O[1/t]$ equals the fraction field of $O$.

Note also that the Going-Down-Theorem holds, since the extension is faithfully flat.

Answer 2

Integrality is NOT a necessary condition for GU (going up). A simple counterexample is $k\subset k[X]$, where $k$ is a field.

Indeed, GU is satisfied by default, because $0$ is the only prime ideal of $k$ (and LO and GD hold for the same reason). But clearly, $k[X]$ is not integral over $k$.