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Proposition 14.35.2. Let $A \to B$ be a local homomorphism of Noetherian complete local rings. The following are equivalent
- $A \to B$ is regular,
- $A \to B$ is flat and $\overline{B}$ is geometrically regular over $k$,
- $A \to B$ is flat and $k \to \overline{B}$ is formally smooth in the $\mathfrak m_{\overline{B}}$-adic topology, and
- $A \to B$ is formally smooth in the $\mathfrak m_B$-adic topology.
Proof. We have seen the equivalence of (2), (3), and (4) in Proposition 14.27.4. 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 14.25.8 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 9.147.10. 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 14.27.6. (Completion in the tensor product is not needed as $A_0 \to A$ is finite, see Algebra, Lemma 9.93.2.) Hence it suffices to show that $C \otimes_{A_0} f.f.(A_0)$ is a geometrically regular algebra over $f.f.(A_0)$.
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 9.107.8. Hence $B \otimes_A f.f.(A)$ is a regular ring too and we win if the characteristic of $f.f.(A)$ is zero.
Thus we are left with the case where $A = k[[x_1, \ldots, x_n]]$ and $k$ is a field of characteristic $p > 0$. Set $K = f.f.(A)$. Let $L \supset K$ 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 14.34.4 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 14.25.9 we can use Lemma 14.35.1 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 14.34.3 and we win. $\square$
\begin{proposition}
\label{proposition-fs-regular}
Let $A \to B$ be a local homomorphism of Noetherian complete local rings.
The following are equivalent
\begin{enumerate}
\item $A \to B$ is regular,
\item $A \to B$ is flat and $\overline{B}$ is geometrically regular
over $k$,
\item $A \to B$ is flat and $k \to \overline{B}$ is formally smooth
in the $\mathfrak m_{\overline{B}}$-adic topology, and
\item $A \to B$ is formally smooth in the $\mathfrak m_B$-adic
topology.
\end{enumerate}
\end{proposition}
\begin{proof}
We have seen the equivalence of (2), (3), and (4) in
Proposition \ref{proposition-fs-flat-fibre-fs}.
It is clear that (1) implies (2).
Thus we assume the equivalent conditions (2), (3), and (4) hold
and we prove (1).
\medskip\noindent
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 \ref{lemma-base-change-fs}
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 $\mathfrak p = (0)$.
\medskip\noindent
Choose $A_0 \subset A$ as in Algebra, Lemma
\ref{algebra-lemma-complete-local-Noetherian-domain-finite-over-regular}.
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 \ref{remark-what-does-it-mean}.
(Completion in the tensor product is not needed as $A_0 \to A$ is
finite, see Algebra, Lemma \ref{algebra-lemma-completion-tensor}.)
Hence it suffices to show that $C \otimes_{A_0} f.f.(A_0)$
is a geometrically regular algebra over $f.f.(A_0)$.
\medskip\noindent
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 \ref{algebra-lemma-flat-over-regular-with-regular-fibre}.
Hence $B \otimes_A f.f.(A)$ is a regular ring too and we win
if the characteristic of $f.f.(A)$ is zero.
\medskip\noindent
Thus we are left with the case where $A = k[[x_1, \ldots, x_n]]$
and $k$ is a field of characteristic $p > 0$. Set $K = f.f.(A)$.
Let $L \supset K$ 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 \ref{lemma-find-D} 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 \ref{lemma-descent-fs}
we can use
Lemma \ref{lemma-lift-derivation-through-fs}
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 \ref{lemma-degree-p-extension-regular} and we win.
\end{proof}
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