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An extension $A\subseteq B$ is an integral extension when every $b\in B$ is an integral element over $A$, meaning thatfor some monic polynomial. The Krull dimension of a ring is the supremum of the lengths $r$ of strict chains ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_r$ of prime ideals; the number counted is the number of inclusions.

We prove the prime-chain facts needed for integral extensions preserve Krull dimension. Taking quotients preserves an integral equation. Localizing both rings by the same multiplicative subset of $A$ also preserves integrality: for $b/s$, divide the equation for $b$ by $s^n$, so its new coefficients are $a_i/s^{n-i}$. Localizing just $B$ at all elements outside one of its primes would not be the same operation.

Two elementary field observations are useful. If a field $C$ is integral over a subdomain $R$, then $R$ is a field, by the inverse argument given in Question 2. Conversely, if an integral domain $C$ is integral over a field $R$, every nonzero $c\in C$ has a monic equation with nonzero constant term: cancel a factor of $c$ from an equation if necessary. Dividing by $c$ and by that constant term expresses $c^{-1}$ as a polynomial in $c$. Thus $C$ is a field. In particular, the contraction of a maximal ideal under an integral extension is maximal.

For the Lying-over theorem, fix $\mathfrak{p}\in \mathrm{Spec}A$ and localize by $S=A\setminus \mathfrak{p}$. The nonzero domain $S^{-1}B$ is integral over the local ring $A_{\mathfrak{p}}$. Choose a maximal ideal $\mathfrak{n}$ of $S^{-1}B$. Its contraction is maximal in $A_{\mathfrak{p}}$, so it is the unique maximal ideal $\mathfrak{p}{A}_{\mathfrak{p}}$. The prime ideal correspondence for localization now pulls $\mathfrak{n}$ back to a prime ideal $\mathfrak{q}$ of $B$ with $\mathfrak{q}\cap A=\mathfrak{p}$.

For the going-up theorem, suppose $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$ and $\mathfrak{q}\cap A=\mathfrak{p}$. The injection $A/\mathfrak{p}\hookrightarrow B/\mathfrak{q}$ is still integral. Apply the just-proved Lying-over theorem to ${\mathfrak{p}}^{\prime }/\mathfrak{p}$. Pulling the resulting prime back gives ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ with ${\mathfrak{q}}^{\prime }\cap A={\mathfrak{p}}^{\prime }$.

For the incomparability theorem for integral extensions, suppose $\mathfrak{q}\subseteq {\mathfrak{q}}^{\prime }$ have the same contraction $\mathfrak{p}$. Write $R=A/\mathfrak{p}$ and $C=B/\mathfrak{q}$. The latter is a domain integral over $R$. Localizing both by $R\setminus \{0\}$ makes $S^{-1}C$ integral over the fraction field of $R$, so $S^{-1}C$ is a field. The prime ${\mathfrak{q}}^{\prime }/\mathfrak{q}$ is disjoint from $S$ and therefore localizes to a proper prime, which must be zero in this field. Since $C$ is a domain and all denominators are nonzero, an element of $C$ whose localization is zero was already zero. Hence ${\mathfrak{q}}^{\prime }/\mathfrak{q}=0$, proving ${\mathfrak{q}}^{\prime }=\mathfrak{q}$.

A strict prime chain in $B$ therefore contracts to a strict chain in $A$, giving $\dim B\le \dim A$. Conversely, start over the bottom of any finite prime chain in $A$ using Lying-over theorem, and successively use going-up theorem to lift the entire chain. Distinct contractions ensure that the lifted inclusions are strict, giving the reverse inequality. Taking suprema, including when arbitrarily long chains exist, proves

For an algebraic extension $K/\mathbb{Q}$, its ring $\mathcal{O}$ of algebraic integers is an integral extension of $\mathbb{Z}$. The primes of $\mathbb{Z}$ are $(0)$ and the maximal ideals $(p)$, so the longest strict chains have length one; for example $(0)⊊(2)$. The preceding argument yieldsThis Krull dimension of rings of algebraic integers argument also covers infinite algebraic extensions. It uses integrality, not an assumption that $K$ is a number field or that $\mathcal{O}$ is Noetherian.