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
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
$R \to \Lambda $ is regular,
$R_\mathfrak p \to \Lambda _\mathfrak q$ is regular for all $\mathfrak q \subset \Lambda $ lying over $\mathfrak p \subset R$, and
$R_\mathfrak m \to \Lambda _{\mathfrak m'}$ 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. $\square$
Lemma 15.41.3 (Regular maps and base change). Let $R \to \Lambda $ be a regular ring map. For any finite type ring map $R \to R'$ the base change $R' \to \Lambda \otimes _ R R'$ is regular too.
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
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. $\square$
Lemma 15.41.4 (Composition of regular maps). Let $A \to B$ and $B \to C$ be regular ring maps. If the fibre rings of $A \to C$ are Noetherian, then $A \to C$ is regular.
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
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. $\square$
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