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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