Lemma 10.112.1. Suppose $R \to S$ is a ring map satisfying either going up, see Definition 10.41.1, or going down see Definition 10.41.1. Assume in addition that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective. Then $\dim (R) \leq \dim (S)$.

**Proof.**
Assume going up. Take any chain $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ e$ of prime ideals in $R$. By surjectivity we may choose a prime $\mathfrak q_0$ mapping to $\mathfrak p_0$. By going up we may extend this to a chain of length $e$ of primes $\mathfrak q_ i$ lying over $\mathfrak p_ i$. Thus $\dim (S) \geq \dim (R)$. The case of going down is exactly the same. See also Topology, Lemma 5.19.9 for a purely topological version.
$\square$

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