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
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 (0)