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## Tag: 07NQ

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Proposition 14.33.4. Let $A \to B$ be a local homomorphism of Noetherian local rings. Let $k$ be the residue field of $A$ and $\overline{B} = B \otimes_A k$ the special fibre. The following are equivalent
1. $A \to B$ is flat and $\overline{B}$ is geometrically regular over $k$,
2. $A \to B$ is flat and $k \to \overline{B}$ is formally smooth in the $\mathfrak m_{\overline{B}}$-adic topology, and
3. $A \to B$ is formally smooth in the $\mathfrak m_B$-adic topology.

Proof. The equivalence of (1) and (2) follows from Theorem 14.33.1.

Assume (3). By Lemma 14.33.3 we see that $A \to B$ is flat. By Lemma 14.31.8 we see that $k \to \overline{B}$ is formally smooth in the $\mathfrak m_{\overline{B}}$-adic topology. Thus (2) holds.

Assume (2). Lemma 14.31.4 tells us formal smoothness is preserved under completion. The same is true for flatness by Algebra, Lemma 9.93.4. Hence we may replace $A$ and $B$ by their respective completions and assume that $A$ and $B$ are Noetherian complete local rings. In this case choose a diagram $$\xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] }$$ as in Lemma 14.32.3. We will use all of the properties of this diagram without further mention. Fix a regular system of parameters $t_1, \ldots, t_d$ of $R$ with $t_1 = p$ in case the characteristic of $k$ is $p > 0$. Set $\overline{S} = S \otimes_R k$. Consider the short exact sequence $$0 \to J \to S \to B \to 0$$ Since $B$ is flat over $A$ we see that $J \otimes_R k$ is the kernel of $\overline{S} \to \overline{B}$. As $\overline{B}$ and $\overline{S}$ are regular we see that $J \otimes_R k$ is generated by elements $\overline{x}_1, \ldots, \overline{x}_r$ which form part of a regular system of parameters of $\overline{S}$, see Algebra, Lemma 9.101.4. Lift these elements to $x_1, \ldots, x_r \in J$. Then $t_1, \ldots, t_d, x_1, \ldots, x_r$ is part of a regular system of parameters for $S$. Hence $S/(x_1, \ldots, x_r)$ is a power series ring over a field (if the characteristic of $k$ is zero) or a power series ring over a Cohen ring (if the characteristic of $k$ is $p > 0$), see Lemma 14.32.2. Moreover, it is still the case that $R \to S/(x_1, \ldots, x_r)$ maps $t_1, \ldots, t_d$ to a part of a regular system of parameters of $S/(x_1, \ldots, x_r)$. In other words, we may replace $S$ by $S/(x_1, \ldots, x_r)$ and assume we have a diagram $$\xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] }$$ as in Lemma 14.32.3 with moreover $\overline{S} = \overline{B}$. In this case the map $$S \otimes_R A \longrightarrow B$$ is an isomorphism as it is surjective and an isomorphism on special fibres, see Algebra, Lemma 9.94.1. Thus by Lemma 14.31.8 it suffices to show that $R \to S$ is formally smooth in the $\mathfrak m_S$-adic topology. Of course, since $\overline{S} = \overline{B}$, we have that $\overline{S}$ is formally smooth over $k = R/\mathfrak m_R$.

Choose elements $y_1, \ldots, y_m \in S$ such that $t_1, \ldots, t_d, y_1, \ldots, y_m$ is a regular system of parameters for $S$. If the characteristic of $k$ is zero, choose a coefficient field $K \subset S$ and if the characteristic of $k$ is $p > 0$ choose a Cohen ring $\Lambda \subset S$ with residue field $K$. At this point the map $K[[t_1, \ldots, t_d, y_1, \ldots, y_m]] \to S$ (characteristic zero case) or $\Lambda[[t_2, \ldots, t_d, y_1, \ldots, y_m]] \to S$ (characteristic $p > 0$ case) is an isomorphism, see Lemma 14.32.2. From now on we think of $S$ as the above power series ring.

The rest of the proof is analogous to the argument in the proof of Theorem 14.33.1. Choose a solid diagram $$\xymatrix{ S \ar[r]_{\bar\psi} \ar@{-->}[rd] & N/J \\ R \ar[u]^i \ar[r]^\varphi & N \ar[u]_\pi }$$ as in Definition 14.31.1. As $J^2 = 0$ we see that $J$ has a canonical $N/J$ module structure and via $\bar\psi$ a $S$-module structure. As $\bar\psi$ is continuous for the $\mathfrak m_S$-adic topology we see that $\mathfrak m_S^nJ = 0$ for some $n$. Hence we can filter $J$ by $N/J$-submodules $0 \subset J_1 \subset J_2 \subset \ldots \subset J_n = J$ such that each quotient $J_{t + 1}/J_t$ is annihilated by $\mathfrak m_S$. Considering the sequence of ring maps $N \to N/J_1 \to N/J_2 \to \ldots \to N/J$ we see that it suffices to prove the existence of the dotted arrow when $J$ is annihilated by $\mathfrak m_S$, i.e., when $J$ is a $K$-vector space.

Assume given a diagram as above such that $J$ is annihilated by $\mathfrak m_S$. As $\mathbf{Q} \to S$ (characteristic zero case) or $\mathbf{Z} \to S$ (characteristic $p > 0$ case) is formally smooth in the $\mathfrak m_S$-adic topology (see Lemma 14.32.1), we can find a ring map $\psi : S \to N$ such that $\pi \circ \psi = \bar \psi$. Since $S$ is a power series ring in $t_1, \ldots, t_d$ (characteristic zero) or $t_2, \ldots, t_d$ (characteristic $p > 0$) over a subring, it follows from the universal property of power series rings that we can change our choice of $\psi$ so that $\psi(t_i)$ equals $\varphi(t_i)$ (automatic for $t_1 = p$ in the characteristic $p$ case). Then $\psi \circ i$ and $\varphi : R \to N$ are two maps whose compositions with $\pi$ are equal and which agree on $t_1, \ldots, t_d$. Hence $D = \psi \circ i - \varphi : R \to J$ is a derivation which annihilates $t_1, \ldots, t_d$. By Algebra, Lemma 9.125.3 we can write $D = \xi \circ \text{d}$ for some $R$-linear map $\xi : \Omega_{R/\mathbf{Z}} \to J$ which annihilates $\text{d}t_1, \ldots, \text{d}t_d$ (by construction) and $\mathfrak m_R \Omega_{R/\mathbf{Z}}$ (as $J$ is annihilated by $\mathfrak m_R$). Hence $\xi$ factors as a composition $$\Omega_{R/\mathbf{Z}} \to \Omega_{k/\mathbf{Z}} \xrightarrow{\xi'} J$$ where $\xi'$ is $k$-linear. Using the $K$-vector space structure on $J$ we extend $\xi'$ to a $K$-linear map $$\xi'' : \Omega_{k/\mathbf{Z}} \otimes_k K \longrightarrow J.$$ Using that $\overline{S}/k$ is formally smooth we see that $$\Omega_{k/\mathbf{Z}} \otimes_k K \to \Omega_{\overline{S}/\mathbf{Z}} \otimes_S K$$ is injective by Theorem 14.33.1 (this is true also in the characteristic zero case as it is even true that $\Omega_{k/\mathbf{Z}} \to \Omega_{K/\mathbf{Z}}$ is injective in characteristic zero, see Algebra, Proposition 9.144.8). Hence we can find a $K$-linear map $\xi''' : \Omega_{\overline{S}/\mathbf{Z}} \otimes_S K \to J$ whose restriction to $\Omega_{k/\mathbf{Z}} \otimes_k K$ is $\xi''$. Write $$D' : S \xrightarrow{\text{d}} \Omega_{S/\mathbf{Z}} \to \Omega_{\overline{S}/\mathbf{Z}} \to \Omega_{\overline{S}/\mathbf{Z}} \otimes_S K \xrightarrow{\xi'''} J.$$ Finally, set $\psi' = \psi - D' : S \to N$. The reader verifies that $\psi'$ is a ring map such that $\pi \circ \psi' = \bar \psi$ and such that $\psi' \circ i = \varphi$ as desired. $\square$

\begin{proposition}
\label{proposition-fs-flat-fibre-fs}
Let $A \to B$ be a local homomorphism of Noetherian local rings.
Let $k$ be the residue field of $A$ and $\overline{B} = B \otimes_A k$
the special fibre. The following are equivalent
\begin{enumerate}
\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}
The equivalence of (1) and (2) follows from Theorem \ref{theorem-regular-fs}.

\medskip\noindent
Assume (3). By Lemma \ref{lemma-formally-smooth-flat} we see that
$A \to B$ is flat. By Lemma \ref{lemma-base-change-fs} we see that
$k \to \overline{B}$ is formally smooth in the
$\mathfrak m_{\overline{B}}$-adic topology. Thus (2) holds.

\medskip\noindent
Assume (2). Lemma \ref{lemma-formally-smooth-completion}
tells us formal smoothness is preserved under completion. The same is true
for flatness by Algebra, Lemma \ref{algebra-lemma-completion-faithfully-flat}.
Hence we may replace $A$ and $B$ by their respective completions and
assume that $A$ and $B$ are Noetherian complete local rings.
In this case choose a diagram
$$\xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] }$$
as in Lemma \ref{lemma-embed-map-Noetherian-complete-local-rings}.
We will use all of the properties of this diagram without further mention.
Fix a regular system of parameters $t_1, \ldots, t_d$ of $R$
with $t_1 = p$ in case the characteristic of $k$ is $p > 0$.
Set $\overline{S} = S \otimes_R k$. Consider the short exact sequence
$$0 \to J \to S \to B \to 0$$
Since $B$ is flat over $A$ we see that $J \otimes_R k$ is the kernel
of $\overline{S} \to \overline{B}$. As $\overline{B}$ and $\overline{S}$
are regular we see that $J \otimes_R k$ is generated by elements
$\overline{x}_1, \ldots, \overline{x}_r$ which form part of a regular
system of parameters of $\overline{S}$, see
Algebra, Lemma \ref{algebra-lemma-regular-quotient-regular}.
Lift these elements to $x_1, \ldots, x_r \in J$. Then
$t_1, \ldots, t_d, x_1, \ldots, x_r$ is part of a regular system of
parameters for $S$. Hence $S/(x_1, \ldots, x_r)$ is a power
series ring over a field (if the characteristic of $k$ is zero)
or a power series ring over a Cohen ring (if the characteristic of
$k$ is $p > 0$), see
Lemma \ref{lemma-quotient-power-series-ring-over-Cohen}.
Moreover, it is still the case that $R \to S/(x_1, \ldots, x_r)$
maps $t_1, \ldots, t_d$ to a part of a regular system of parameters
of $S/(x_1, \ldots, x_r)$. In other words, we may replace $S$ by
$S/(x_1, \ldots, x_r)$ and assume we have a diagram
$$\xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] }$$
as in Lemma \ref{lemma-embed-map-Noetherian-complete-local-rings}
with moreover $\overline{S} = \overline{B}$. In this case the map
$$S \otimes_R A \longrightarrow B$$
is an isomorphism as it is surjective and an isomorphism on special
fibres, see Algebra, Lemma \ref{algebra-lemma-mod-injective}. Thus by
Lemma \ref{lemma-base-change-fs}
it suffices to show that $R \to S$ is formally
smooth in the $\mathfrak m_S$-adic topology.
Of course, since $\overline{S} = \overline{B}$, we have
that $\overline{S}$ is formally smooth over $k = R/\mathfrak m_R$.

\medskip\noindent
Choose elements $y_1, \ldots, y_m \in S$ such that
$t_1, \ldots, t_d, y_1, \ldots, y_m$ is a regular system of parameters
for $S$. If the characteristic of $k$ is zero, choose a coefficient
field $K \subset S$ and if the characteristic of $k$ is $p > 0$
choose a Cohen ring $\Lambda \subset S$ with residue field $K$.
At this point the map $K[[t_1, \ldots, t_d, y_1, \ldots, y_m]] \to S$
(characteristic zero case) or
$\Lambda[[t_2, \ldots, t_d, y_1, \ldots, y_m]] \to S$
(characteristic $p > 0$ case) is an isomorphism, see
Lemma \ref{lemma-quotient-power-series-ring-over-Cohen}.
From now on we think of $S$ as the above power series ring.

\medskip\noindent
The rest of the proof is analogous to the argument
in the proof of Theorem \ref{theorem-regular-fs}. Choose a solid diagram
$$\xymatrix{ S \ar[r]_{\bar\psi} \ar@{-->}[rd] & N/J \\ R \ar[u]^i \ar[r]^\varphi & N \ar[u]_\pi }$$
as in Definition \ref{definition-formally-smooth}. As $J^2 = 0$ we see
that $J$ has a canonical $N/J$ module structure and via $\bar\psi$ a
$S$-module structure. As $\bar\psi$ is continuous for the
$\mathfrak m_S$-adic topology we see that $\mathfrak m_S^nJ = 0$ for some
$n$. Hence we can filter $J$ by $N/J$-submodules
$0 \subset J_1 \subset J_2 \subset \ldots \subset J_n = J$
such that each quotient $J_{t + 1}/J_t$ is annihilated by $\mathfrak m_S$.
Considering the sequence of ring maps
$N \to N/J_1 \to N/J_2 \to \ldots \to N/J$
we see that it suffices to prove the existence of the dotted arrow when
$J$ is annihilated by $\mathfrak m_S$, i.e., when $J$ is a
$K$-vector space.

\medskip\noindent
Assume given a diagram as above such that $J$ is annihilated by
$\mathfrak m_S$. As $\mathbf{Q} \to S$ (characteristic zero case)
or $\mathbf{Z} \to S$ (characteristic $p > 0$ case)
is formally smooth in the $\mathfrak m_S$-adic topology (see
Lemma \ref{lemma-power-series-ring-over-Cohen-fs}), we can find
a ring map $\psi : S \to N$ such that $\pi \circ \psi = \bar \psi$.
Since $S$ is a power series ring in $t_1, \ldots, t_d$ (characteristic zero)
or $t_2, \ldots, t_d$ (characteristic $p > 0$) over
a subring, it follows from the universal property of power series rings
that we can change our choice of $\psi$ so that $\psi(t_i)$ equals
$\varphi(t_i)$ (automatic for $t_1 = p$ in the characteristic $p$ case).
Then $\psi \circ i$ and $\varphi : R \to N$ are two maps whose
compositions with $\pi$ are equal and which agree on $t_1, \ldots, t_d$.
Hence $D = \psi \circ i - \varphi : R \to J$ is a derivation which
annihilates $t_1, \ldots, t_d$.
By Algebra, Lemma \ref{algebra-lemma-universal-omega} we can write
$D = \xi \circ \text{d}$ for some $R$-linear map
$\xi : \Omega_{R/\mathbf{Z}} \to J$ which annihilates
$\text{d}t_1, \ldots, \text{d}t_d$ (by construction) and
$\mathfrak m_R \Omega_{R/\mathbf{Z}}$ (as $J$ is annihilated by
$\mathfrak m_R$). Hence $\xi$ factors as a composition
$$\Omega_{R/\mathbf{Z}} \to \Omega_{k/\mathbf{Z}} \xrightarrow{\xi'} J$$
where $\xi'$ is $k$-linear. Using the $K$-vector space structure on $J$ we
extend $\xi'$ to a $K$-linear map
$$\xi'' : \Omega_{k/\mathbf{Z}} \otimes_k K \longrightarrow J.$$
Using that $\overline{S}/k$ is formally smooth we see that
$$\Omega_{k/\mathbf{Z}} \otimes_k K \to \Omega_{\overline{S}/\mathbf{Z}} \otimes_S K$$
is injective by Theorem \ref{theorem-regular-fs} (this is true also
in the characteristic zero case as it is even true that
$\Omega_{k/\mathbf{Z}} \to \Omega_{K/\mathbf{Z}}$ is injective
in characteristic zero, see Algebra,
Proposition \ref{algebra-proposition-characterize-separable-field-extensions}).
Hence we can find a $K$-linear map
$\xi''' : \Omega_{\overline{S}/\mathbf{Z}} \otimes_S K \to J$ whose
restriction to $\Omega_{k/\mathbf{Z}} \otimes_k K$ is $\xi''$. Write
$$D' : S \xrightarrow{\text{d}} \Omega_{S/\mathbf{Z}} \to \Omega_{\overline{S}/\mathbf{Z}} \to \Omega_{\overline{S}/\mathbf{Z}} \otimes_S K \xrightarrow{\xi'''} J.$$
Finally, set $\psi' = \psi - D' : S \to N$. The reader verifies that $\psi'$
is a ring map such that $\pi \circ \psi' = \bar \psi$ and such that
$\psi' \circ i = \varphi$ as desired.
\end{proof}

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