Going up lifts every prime chain in $A$ to one in $B$, so $\dim B\geq\dim A$. Conversely, contracting a strict chain of primes of $B$ gives a chain in $A$, and the <incomparability theorem for integral extensions> ensures that no strict inclusion contracts to equality. Thus $\dim A\geq\dim B$, and
$$
\boxed{\dim A=\dim B.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 5 / b / iii / Solution

If $A\subseteq B$ is integral and $\mathfrak q_1\subseteq\mathfrak q_2$ are primes of $B$ with equal contractions to $A$, then $\mathfrak q_1=\mathfrak q_2$. Consequently strict prime chains remain strict under contraction.


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 Incoming links: Incomparability theorem for integral extensions