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).
10.112 Homomorphisms and dimension
This section contains a collection of easy results relating dimensions of rings when there are maps between them.
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
Lemma 10.112.2. Suppose that R \to S is a ring map with the going up property, see Definition 10.41.1. If \mathfrak q \subset S is a maximal ideal. Then the inverse image of \mathfrak q in R is a maximal ideal too.
Proof. Trivial. \square
Lemma 10.112.3. Suppose that R \to S is a ring map such that S is integral over R. Then \dim (R) \geq \dim (S), and every closed point of \mathop{\mathrm{Spec}}(S) maps to a closed point of \mathop{\mathrm{Spec}}(R).
Proof. Immediate from Lemmas 10.36.20 and 10.112.2 and the definitions. \square
Lemma 10.112.4. Suppose R \subset S and S integral over R. Then \dim (R) = \dim (S).
Proof. This is a combination of Lemmas 10.36.22, 10.36.17, 10.112.1, and 10.112.3. \square
Definition 10.112.5. Suppose that R \to S is a ring map. Let \mathfrak q \subset S be a prime lying over the prime \mathfrak p of R. The local ring of the fibre at \mathfrak q is the local ring
S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} = (S/\mathfrak pS)_{\mathfrak q} = (S \otimes _ R \kappa (\mathfrak p))_{\mathfrak q}
Lemma 10.112.6. Let R \to S be a homomorphism of Noetherian rings. Let \mathfrak q \subset S be a prime lying over the prime \mathfrak p. Then
\dim (S_{\mathfrak q}) \leq \dim (R_{\mathfrak p}) + \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}).
Proof. We use the characterization of dimension of Proposition 10.60.9. Let x_1, \ldots , x_ d be elements of \mathfrak p generating an ideal of definition of R_{\mathfrak p} with d = \dim (R_{\mathfrak p}). Let y_1, \ldots , y_ e be elements of \mathfrak q generating an ideal of definition of S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} with e = \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}). It is clear that S_{\mathfrak q}/(x_1, \ldots , x_ d, y_1, \ldots , y_ e) has a nilpotent maximal ideal. Hence x_1, \ldots , x_ d, y_1, \ldots , y_ e generate an ideal of definition of S_{\mathfrak q}. \square
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
Lemma 10.112.8. Let R \to S be a local homomorphism of local Noetherian rings. Assume
R is regular,
S/\mathfrak m_ RS is regular, and
R \to S is flat.
Then S is regular.
Proof. By Lemma 10.112.7 we have \dim (S) = \dim (R) + \dim (S/\mathfrak m_ RS). Pick generators x_1, \ldots , x_ d \in \mathfrak m_ R with d = \dim (R), and pick y_1, \ldots , y_ e \in \mathfrak m_ S which generate the maximal ideal of S/\mathfrak m_ RS with e = \dim (S/\mathfrak m_ RS). Then we see that x_1, \ldots , x_ d, y_1, \ldots , y_ e are elements which generate the maximal ideal of S and e + d = \dim (S). \square
The lemma below will later be used to show that rings of finite type over a field are Cohen-Macaulay if and only if they are quasi-finite flat over a polynomial ring. It is a partial converse to Lemma 10.128.1.
Lemma 10.112.9. Let R \to S be a local homomorphism of Noetherian local rings. Assume R Cohen-Macaulay. If S is finite flat over R, or if S is flat over R and \dim (S) \leq \dim (R), then S is Cohen-Macaulay and \dim (R) = \dim (S).
Proof. Let x_1, \ldots , x_ d \in \mathfrak m_ R be a regular sequence of length d = \dim (R). By Lemma 10.68.5 this maps to a regular sequence in S. Hence S is Cohen-Macaulay if \dim (S) \leq d. This is true if S is finite flat over R by Lemma 10.112.4. And in the second case we assumed it. \square
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