Lemma 10.111.7. Let $R \to S$ be a homomorphism of Noetherian rings. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$. Assume the going down property holds for $R \to S$ (for example if $R \to S$ is flat, see Lemma 10.38.19). Then

**Proof.**
By Lemma 10.111.6 we have an inequality $\dim (S_{\mathfrak q}) \leq \dim (R_{\mathfrak p}) + \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q})$. To get equality, choose a chain of primes $\mathfrak pS \subset \mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset \mathfrak q_ d = \mathfrak q$ with $d = \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q})$. On the other hand, choose a chain of primes $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ e = \mathfrak p$ with $e = \dim (R_{\mathfrak p})$. By the going down theorem we may choose $\mathfrak q_{-1} \subset \mathfrak q_0$ lying over $\mathfrak p_{e-1}$. And then we may choose $\mathfrak q_{-2} \subset \mathfrak q_{e-1}$ lying over $\mathfrak p_{e-2}$. Inductively we keep going until we get a chain $\mathfrak q_{-e} \subset \ldots \subset \mathfrak q_ d$ of length $e + d$.
$\square$

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