The Stacks project

15.41 Regular ring maps

Let $k$ be a field. Recall that a Noetherian $k$-algebra $A$ is said to be geometrically regular over $k$ if and only if $A \otimes _ k k'$ is regular for all finite purely inseparable extensions $k'$ of $k$, see Algebra, Definition 10.166.2. Moreover, if this is the case then $A \otimes _ k k'$ is regular for every finitely generated field extension $k'/k$, see Algebra, Lemma 10.166.1. We use this notion in the following definition.

Definition 15.41.1. A ring map $R \to \Lambda $ is regular if it is flat and for every prime $\mathfrak p \subset R$ the fibre ring

\[ \Lambda \otimes _ R \kappa (\mathfrak p) = \Lambda _\mathfrak p/\mathfrak p\Lambda _\mathfrak p \]

is Noetherian and geometrically regular over $\kappa (\mathfrak p)$.

If $R \to \Lambda $ is a ring map with $\Lambda $ Noetherian, then the fibre rings are always Noetherian.

Lemma 15.41.2 (Regular is a local property). Let $R \to \Lambda $ be a ring map with $\Lambda $ Noetherian. The following are equivalent

  1. $R \to \Lambda $ is regular,

  2. $R_\mathfrak p \to \Lambda _\mathfrak q$ is regular for all $\mathfrak q \subset \Lambda $ lying over $\mathfrak p \subset R$, and

  3. $R_\mathfrak m \to \Lambda _{\mathfrak m'}$ is regular for all maximal ideals $\mathfrak m' \subset \Lambda $ lying over $\mathfrak m$ in $R$.

Proof. This is true because a Noetherian ring is regular if and only if all the local rings are regular local rings, see Algebra, Definition 10.110.7 and a ring map is flat if and only if all the induced maps of local rings are flat, see Algebra, Lemma 10.39.18. $\square$

Proof. Flatness is preserved under any base change, see Algebra, Lemma 10.39.7. Consider a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p \subset R$. The residue field extension $\kappa (\mathfrak p')/\kappa (\mathfrak p)$ is finitely generated as $R'$ is of finite type over $R$. Hence the fibre ring

\[ (\Lambda \otimes _ R R') \otimes _{R'} \kappa (\mathfrak p') = \Lambda \otimes _ R \kappa (\mathfrak p) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \]

is Noetherian by Algebra, Lemma 10.31.8 and the assumption on the fibre rings of $R \to \Lambda $. Geometric regularity of the fibres is preserved by Algebra, Lemma 10.166.1. $\square$

Proof. Let $\mathfrak p \subset A$ be a prime. Let $\kappa (\mathfrak p) \subset k$ be a finite purely inseparable extension. We have to show that $C \otimes _ A k$ is regular. By Lemma 15.41.3 we may assume that $A = k$ and we reduce to proving that $C$ is regular. The assumption is that $B$ is regular and that $B \to C$ is flat with regular fibres. Then $C$ is regular by Algebra, Lemma 10.112.8. Some details omitted. $\square$

Lemma 15.41.5. Let $R$ be a ring. Let $(A_ i, \varphi _{ii'})$ be a directed system of smooth $R$-algebras. Set $\Lambda = \mathop{\mathrm{colim}}\nolimits A_ i$. If the fibre rings $\Lambda \otimes _ R \kappa (\mathfrak p)$ are Noetherian for all $\mathfrak p \subset R$, then $R \to \Lambda $ is regular.

Proof. Note that $\Lambda $ is flat over $R$ by Algebra, Lemmas 10.39.3 and 10.137.10. Let $\kappa (\mathfrak p) \subset k$ be a finite purely inseparable extension. Note that

\[ \Lambda \otimes _ R \kappa (\mathfrak p) \otimes _{\kappa (\mathfrak p)} k = \Lambda \otimes _ R k = \mathop{\mathrm{colim}}\nolimits A_ i \otimes _ R k \]

is a colimit of smooth $k$-algebras, see Algebra, Lemma 10.137.4. Since each local ring of a smooth $k$-algebra is regular by Algebra, Lemma 10.140.3 we conclude that all local rings of $\Lambda \otimes _ R k$ are regular by Algebra, Lemma 10.106.8. This proves the lemma. $\square$

Let's see when a field extension defines a regular ring map.

Lemma 15.41.6. Let $K/k$ be a field extension. Then $k \to K$ is a regular ring map if and only if $K$ is a separable field extension of $k$.

Proof. If $k \to K$ is regular, then $K$ is geometrically reduced over $k$, hence $K$ is separable over $k$ by Algebra, Proposition 10.158.9. Conversely, if $K/k$ is separable, then $K$ is a colimit of smooth $k$-algebras, see Algebra, Lemma 10.158.11 hence is regular by Lemma 15.41.5. $\square$

Lemma 15.41.7. Let $A \to B \to C$ be ring maps. If $A \to C$ is regular and $B \to C$ is flat and surjective on spectra, then $A \to B$ is regular.

Proof. By Algebra, Lemma 10.39.10 we see that $A \to B$ is flat. Let $\mathfrak p \subset A$ be a prime. The ring map $B \otimes _ A \kappa (\mathfrak p) \to C \otimes _ A \kappa (\mathfrak p)$ is flat and surjective on spectra. Hence $B \otimes _ A \kappa (\mathfrak p)$ is geometrically regular by Algebra, Lemma 10.166.3. $\square$

Comments (0)

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 07BY. Beware of the difference between the letter 'O' and the digit '0'.