Proposition 15.40.5. 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 15.40.1.

Assume (3). By Lemma 15.40.3 we see that $A \to B$ is flat. By Lemma 15.37.8 we see that $k \to \overline{B}$ is formally smooth in the $\mathfrak m_{\overline{B}}$-adic topology. Thus (2) holds.

Assume (2). Lemma 15.37.4 tells us formal smoothness is preserved under completion. The same is true for flatness by Algebra, Lemma 10.97.3. 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 15.39.3. 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

$0 \to J \to S \to B \to 0$

As $\overline{B}$ and $\overline{S}$ are regular, the kernel of $\overline{S} \to \overline{B}$ 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 10.106.4. 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 15.39.2. 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 15.39.3 with moreover $\overline{S} = \overline{B}$. In this case the map

$S \otimes _ R A \longrightarrow B$

is an isomorphism as it is surjective, an isomorphism on special fibres, and source and target are flat over $A$ (for example use Algebra, Lemma 10.99.1 or use that tensoring the short exact sequence $0 \to I \to S \otimes _ R A \to B \to 0$ over $A$ with $k$ we find $I \otimes _ A k = 0$ hence $I = 0$ by Nakayama). Thus by Lemma 15.37.8 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 15.39.2. 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 15.40.1. 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 15.37.1. 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 15.39.1), 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 10.131.3 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 15.40.1 (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 10.158.9). 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. $\square$

Comment #6484 by Dan Dore on

I think the first part of the proof that (2) => (3) has some issues: the sequence $0 \rightarrow J \rightarrow S \rightarrow B \rightarrow 0$ isn't a sequence of $A$-modules, just a sequence of $R$-modules, and $B$ isn't $R$-flat in general. But the claim that $J \otimes_R k$ is the kernel of $\overline{S} \rightarrow \overline{B}$ isn't used anywhere: you can always lift $\overline{x}_1, \ldots, \overline{x}_r \in \mathrm{ker}(\overline{S} \rightarrow \overline{B})$ to elements of $J$, no flatness needed.

So you can reduce as written to the case that $\overline{S} \xrightarrow{\sim} \overline{B}$. Then flatness is needed only for the appeal to \ref[00ME]{https://stacks.math.columbia.edu/tag/00ME}. Alternatively, since $S \otimes_R A \rightarrow B$ is surjective by design, you can apply flatness to the sequence $0 \rightarrow I \rightarrow S \otimes_R A \rightarrow B$ of $A$-modules to conclude that $I = 0$.

Comment #6556 by on

Dear Dan, thanks very much for pointing this out. I fixed it exactly as you suggested here.

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