= Solution
An extension $A\subseteq B$ is an <integral extension> when every $b\in B$ is an <integral element> over $A$, meaning that
$$
b^n+a_{n-1}b^{n-1}+\cdots+a_0=0\qquad(a_i\in A)
$$
for some monic polynomial. The <Krull dimension> of a <ring> is the supremum of the lengths $r$ of strict chains $\mathfrak p_0\subsetneq\cdots\subsetneq\mathfrak p_r$ 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 $\mathfrak p\in\operatorname{Spec}A$ 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 $\mathfrak pA_{\mathfrak p}$. 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 $\mathfrak p\subseteq\mathfrak p'$ 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 $\mathfrak p'/\mathfrak p$. Pulling the resulting prime back gives $\mathfrak q'\supseteq\mathfrak q$ with $\mathfrak q'\cap A=\mathfrak p'$.
For the <incomparability theorem for integral extensions>, suppose $\mathfrak q\subseteq\mathfrak q'$ 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 $\mathfrak q'/\mathfrak q$ 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 $\mathfrak q'/\mathfrak q=0$, proving $\mathfrak q' =\mathfrak q$.
A strict prime chain in $B$ therefore contracts to a strict chain in $A$, giving $\dim B\leq\dim A$. 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
$$
\boxed{\dim B=\dim A.}
$$
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 $(0)\subsetneq(2)$. The preceding argument yields
$$
\boxed{\dim\mathcal O=\dim\mathbb Z=1.}
$$
This <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.
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