The Stacks project

Lemma 10.112.7. 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 10.39.19). Then

\[ \dim (S_{\mathfrak q}) = \dim (R_{\mathfrak p}) + \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}). \]

Proof. By Lemma 10.112.6 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 $e + d$. $\square$


Comments (1)

Comment #10080 by René Bruin on

Am I right that there is a typo in the sentence And then we may choose lying over ?

If I understand it correctly, then it should be , since by the going down property, one defines a chain of prime ideals lying over the chain of prime ideals in the sense that the preimage of under the map equals .


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