The Stacks project

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10.160 Geometrically regular algebras

Let $k$ be a field. Let $A$ be a Noetherian $k$-algebra. Let $k \subset K$ be a finitely generated field extension. Then the ring $K \otimes _ k A$ is Noetherian as well, see Lemma 10.30.8. Thus the following lemma makes sense.

Lemma 10.160.1. Let $k$ be a field. Let $A$ be a $k$-algebra. Assume $A$ is Noetherian. The following properties of $A$ are equivalent:

  1. $k' \otimes _ k A$ is regular for every finitely generated field extension $k \subset k'$, and

  2. $k' \otimes _ k A$ is regular for every finite purely inseparable extension $k \subset k'$.

Here regular ring is as in Definition 10.109.7.

Proof. The lemma makes sense by the remarks preceding the lemma. It is clear that (1) $\Rightarrow $ (2).

Assume (2) and let $k \subset K$ be a finitely generated field extension. By Lemma 10.44.3 we can find a diagram

\[ \xymatrix{ K \ar[r] & K' \\ k \ar[u] \ar[r] & k' \ar[u] } \]

where $k \subset k'$, $K \subset K'$ are finite purely inseparable field extensions such that $k' \subset K'$ is separable. By Lemma 10.152.10 there exists a smooth $k'$-algebra $B$ such that $K'$ is the fraction field of $B$. Now we can argue as follows: Step 1: $k' \otimes _ k A$ is a regular ring because we assumed (2). Step 2: $B \otimes _{k'} k' \otimes _ k A$ is a regular ring as $k' \otimes _ k A \to B \otimes _{k'} k' \otimes _ k A$ is smooth (Lemma 10.135.4) and ascent of regularity along smooth maps (Lemma 10.157.10). Step 3. $K' \otimes _{k'} k' \otimes _ k A = K' \otimes _ k A$ is a regular ring as it is a localization of a regular ring (immediate from the definition). Step 4. Finally $K \otimes _ k A$ is a regular ring by descent of regularity along the faithfully flat ring map $K \otimes _ k A \to K' \otimes _ k A$ (Lemma 10.158.4). This proves the lemma. $\square$

Definition 10.160.2. Let $k$ be a field. Let $R$ be a Noetherian $k$-algebra. The $k$-algebra $R$ is called geometrically regular over $k$ if the equivalent conditions of Lemma 10.160.1 hold.

It is clear from the definition that $K \otimes _ k R$ is a geometrically regular algebra over $K$ for any finitely generated field extension $K$ of $k$. We will see later (More on Algebra, Proposition 15.34.1) that it suffices to check $R \otimes _ k k'$ is regular whenever $k \subset k' \subset k^{1/p}$ (finite).

slogan

Lemma 10.160.3. Let $k$ be a field. Let $A \to B$ be a faithfully flat $k$-algebra map. If $B$ is geometrically regular over $k$, so is $A$.

Proof. Assume $B$ is geometrically regular over $k$. Let $k \subset k'$ be a finite, purely inseparable extension. Then $A \otimes _ k k' \to B \otimes _ k k'$ is faithfully flat as a base change of $A \to B$ (by Lemmas 10.29.3 and 10.38.7) and $B \otimes _ k k'$ is regular by our assumption on $B$ over $k$. Then $A \otimes _ k k'$ is regular by Lemma 10.158.4. $\square$

Lemma 10.160.4. Let $k$ be a field. Let $A \to B$ be a smooth ring map of $k$-algebras. If $A$ is geometrically regular over $k$, then $B$ is geometrically regular over $k$.

Proof. Let $k \subset k'$ be a finitely generated field extension. Then $A \otimes _ k k' \to B \otimes _ k k'$ is a smooth ring map (Lemma 10.135.4) and $A \otimes _ k k'$ is regular. Hence $B \otimes _ k k'$ is regular by Lemma 10.157.10. $\square$

Lemma 10.160.5. Let $k$ be a field. Let $A$ be an algebra over $k$. Let $k = \mathop{\mathrm{colim}}\nolimits k_ i$ be a directed colimit of subfields. If $A$ is geometrically regular over each $k_ i$, then $A$ is geometrically regular over $k$.

Proof. Let $k \subset k'$ be a finite purely inseparable field extension. We can get $k'$ by adjoining finitely many variables to $k$ and imposing finitely many polynomial relations. Hence we see that there exists an $i$ and a finite purely inseparable field extension $k_ i \subset k_ i'$ such that $k_ i = k \otimes _{k_ i} k_ i'$. Thus $A \otimes _ k k' = A \otimes _{k_ i} k_ i'$ and the lemma is clear. $\square$

Lemma 10.160.6. Let $k \subset k'$ be a separable algebraic field extension. Let $A$ be an algebra over $k'$. Then $A$ is geometrically regular over $k$ if and only if it is geometrically regular over $k'$.

Proof. Let $k \subset L$ be a finite purely inseparable field extension. Then $L' = k' \otimes _ k L$ is a field (see material in Fields, Section 9.28) and $A \otimes _ k L = A \otimes _{k'} L'$. Hence if $A$ is geometrically regular over $k'$, then $A$ is geometrically regular over $k$.

Assume $A$ is geometrically regular over $k$. Since $k'$ is the filtered colimit of finite extensions of $k$ we may assume by Lemma 10.160.5 that $k'/k$ is finite separable. Consider the ring maps

\[ k' \to A \otimes _ k k' \to A. \]

Note that $A \otimes _ k k'$ is geometrically regular over $k'$ as a base change of $A$ to $k'$. Note that $A \otimes _ k k' \to A$ is the base change of $k' \otimes _ k k' \to k'$ by the map $k' \to A$. Since $k'/k$ is an ├ętale extension of rings, we see that $k' \otimes _ k k' \to k'$ is ├ętale (Lemma 10.141.3). Hence $A$ is geometrically regular over $k'$ by Lemma 10.160.4. $\square$


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