Integral extensions preserve Krull dimension

If $A\subseteq B$ is an integral extension, then $\dim A=\dim B$. The incomparability theorem for integral extensions contracts every strict prime chain in $B$ to a strict chain in $A$, giving $\dim B\le \dim A$. Conversely, the Lying-over theorem starts over the bottom of any prime chain in $A$, and repeated use of the going-up theorem lifts the whole chain to $B$.

For any algebraic field extension $K/\mathbb{Q}$, including an infinite one, its ring ${\mathcal{O}}_K$ of elements integral over $\mathbb{Z}$ is an integral extension of $\mathbb{Z}$. The Lying-over theorem, going-up theorem and incomparability theorem for integral extensions give $\dim {\mathcal{O}}_K=\dim \mathbb{Z}=1$. The argument does not require that ${\mathcal{O}}_K$ be Noetherian or finitely generated over $\mathbb{Z}$.

If $f\in k[Y][X]$ is monic of positive degree in $X$, monic division makes $k[Y][X]/(f)$ free of rank ${\mathrm{deg}}_{X}f$ over $k[Y]$. The inclusion of $k[Y]$ is injective and the quotient is an integral extension, so its Krull dimension is one. This argument applies even when the polynomial factors or the field has positive characteristic.