**Proof.**
We have seen the equivalence of (2), (3), and (4) in Proposition 15.40.5. It is clear that (1) implies (2). Thus we assume the equivalent conditions (2), (3), and (4) hold and we prove (1).

Let $\mathfrak p$ be a prime of $A$. We will show that $B \otimes _ A \kappa (\mathfrak p)$ is geometrically regular over $\kappa (\mathfrak p)$. By Lemma 15.37.8 we may replace $A$ by $A/\mathfrak p$ and $B$ by $B/\mathfrak pB$. Thus we may assume that $A$ is a domain and that $\mathfrak p = (0)$.

Choose $A_0 \subset A$ as in Algebra, Lemma 10.160.11. We will use all the properties stated in that lemma without further mention. As $A_0 \to A$ induces an isomorphism on residue fields, and as $B/\mathfrak m_ A B$ is geometrically regular over $A/\mathfrak m_ A$ we can find a diagram

\[ \xymatrix{ C \ar[r] & B \\ A_0 \ar[r] \ar[u] & A \ar[u] } \]

with $A_0 \to C$ formally smooth in the $\mathfrak m_ C$-adic topology such that $B = C \otimes _{A_0} A$, see Remark 15.40.7. (Completion in the tensor product is not needed as $A_0 \to A$ is finite, see Algebra, Lemma 10.97.1.) Hence it suffices to show that $C \otimes _{A_0} K_0$ is a geometrically regular algebra over the fraction field $K_0$ of $A_0$.

The upshot of the preceding paragraph is that we may assume that $A = k[[x_1, \ldots , x_ n]]$ where $k$ is a field or $A = \Lambda [[x_1, \ldots , x_ n]]$ where $\Lambda $ is a Cohen ring. In this case $B$ is a regular ring, see Algebra, Lemma 10.112.8. Hence $B \otimes _ A K$ is a regular ring too (where $K$ is the fraction field of $A$) and we win if the characteristic of $K$ is zero.

Thus we are left with the case where $A = k[[x_1, \ldots , x_ n]]$ and $k$ is a field of characteristic $p > 0$. Let $L/K$ be a finite purely inseparable field extension. We will show by induction on $[L : K]$ that $B \otimes _ A L$ is regular. The base case is $L = K$ which we've seen above. Let $K \subset M \subset L$ be a subfield such that $L$ is a degree $p$ extension of $M$ obtained by adjoining a $p$th root of an element $f \in M$. Let $A'$ be a finite $A$-subalgebra of $M$ with fraction field $M$. Clearing denominators, we may and do assume $f \in A'$. Set $A'' = A'[z]/(z^ p -f)$ and note that $A' \subset A''$ is finite and that the fraction field of $A''$ is $L$. By induction we know that $B \otimes _ A M$ ring is regular. We have

\[ B \otimes _ A L = B \otimes _ A M[z]/(z^ p - f) \]

By Lemma 15.48.5 we know there exists a derivation $D : A' \to A'$ such that $D(f) \not= 0$. As $A' \to B \otimes _ A A'$ is formally smooth in the $\mathfrak m$-adic topology by Lemma 15.37.9 we can use Lemma 15.49.1 to extend $D$ to a derivation $D' : B \otimes _ A A' \to B \otimes _ A A'$. Note that $D'(f) = D(f)$ is a unit in $B \otimes _ A M$ as $D(f)$ is not zero in $A' \subset M$. Hence $B \otimes _ A L$ is regular by Lemma 15.48.4 and we win.
$\square$

## Comments (2)

Comment #6294 by Ehsan on

Comment #6407 by Johan on