## 10.112 Homomorphisms and dimension

This section contains a collection of easy results relating dimensions of rings when there are maps between them.

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. $\square$

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 $\dim (R) = \dim (S)$.

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

1. $R$ is regular,

2. $S/\mathfrak m_ RS$ is regular, and

3. $R \to S$ 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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).