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. $\square$

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00OG. Beware of the difference between the letter 'O' and the digit '0'.