Proposition 15.40.5 (07NQ)—The Stacks project

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

$A \to B$ 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. 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. $\square$

Comments (4)

Comment #6484 by Dan Dore on August 06, 2021 at 18:09

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 Johan on September 13, 2021 at 13:17

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

Comment #10917 by yj on October 18, 2025 at 08:58

It seems to me that the proof of the part "(2) $\Rightarrow$ (3)" is a bit complicated or less conceptual. To prove the part "(2) $\Rightarrow$ (3)", we only need to check that for any $k\in\mathbb{N}$ , the base change $A/\mathfrak{m}_A^k\to B/\mathfrak{m}_A^kB$ is formally smooth for the $\mathfrak{m}_B/\mathfrak{m}_A^kB$ -adic topology. Since $A/\mathfrak{m}_A^k\to B/\mathfrak{m}_A^kB$ is flat (by (2)), to prove this, it suffices to prove the following statement, which is similar to Commutative Algebra, Tag 031L (Lemma 10.138.12, as of 2025-10-18):

Lemma A.　Let $R\to A$ be a flat ring homomorphism, $I\subset R$ an ideal such that $I^2=0$ , and $\mathfrak{m}\subset A$ an ideal. Then $R/I\to A/IA$ is formally smooth for the $(\mathfrak{m}+IA)/IA$ -adic topology if and only if $R\to A$ is formally smooth for the $\mathfrak{m}$ -adic topology.

To prove this, I'm going to check two claims. The first of them is the following statement, similar to Commutative Algebra, Tag 031I (Lemma 10.138.7, as of 2025-10-18):

Claim B (see also [EGA IV1, Théorème 0.22.6.1]).　Let $R\to A$ be a ring homomorphism, $P$ a projective $R$ -module, $\pi:R[P]\to A$ a surjective homomorphism of $R$ -algebras from the symmetric $R$ -algebra $R[P]$ obtained by $P$ , and $\mathfrak{m}\subset A$ an ideal. Write $I:=\ker(\pi)$ . Then, $R\to A$ is formally smooth for the $\mathfrak{m}$ -adic topology if and only if for any $k\in \mathbb{N}$ , the natural sequence $0 \to I/I^2\otimes_A A/\mathfrak{m}^k \to \Omega_{R[P]/R}\otimes_{R[P]}A/\mathfrak{m}^k\to \Omega_{A/R}\otimes_AA/\mathfrak{m}^k\to 0$ is exact and split.

Proof of Claim B. 　Write $d:I/I^2\to \Omega_{R[P]/R}\otimes_{R[P]}A$ for the natural morphism.

Necessity:　Take $k\in\mathbb{N}$ and write $\theta:I \to I/I^2\otimes_AA/\mathfrak{m}^k$ for the natural morphism of $A$ -modules. Define $B_{\theta}:=\left( R[P]\oplus (I/I^2\otimes_AA/\mathfrak{m}^k)\right) / \left((f, -\theta(f)), f\in I\right),$ where the multiplication on $B_{\theta}$ is inherited from the symmetric $R[P]$ -algebra $\mathrm{Sym}\_{R[P]}(I/I^2\otimes_AA/\mathfrak{m}^k)$ . Then we obtain a surjective homomorphism of $R$ -algebras $B_{\theta}\to A/\mathfrak{m}^k$ with a square-nilpotent kernel $I/I^2\otimes_A A/\mathfrak{m}^k\subset B_{\theta}$ : $0 \to I/I^2\otimes_A A/\mathfrak{m}^k \to B_{\theta}\to A/\mathfrak{m}^k \to 0.$ Since $R\to A$ is formally smooth for the $\mathfrak{m}$ -adic topology, we obtain a lift $h:A\to B_{\theta}$ of the natural surjection $A\to A/\mathfrak{m}^k$ along $B_{\theta}\to A/\mathfrak{m}^k$ . Then the difference of the natural morphism $R[P]\to B_{\theta}$ and the composite $h\circ \pi: R[P]\to A\to B_{\theta}$ defines an $R[P]$ -derivation $R[P]\to I/I^2\otimes_AA/\mathfrak{m}^k$ such that the corresponding morphism $r:\Omega_{R[P]/R}\otimes\_{R[P]}A/\mathfrak{m}^k\to I/I^2\otimes_AA/\mathfrak{m}^k$ is a retraction of $d\otimes_A\mathrm{id}_{A/\mathfrak{m}^k}$ . This completes the proof of necessity portion of Claim B.

Sufficiency:　Let $B$ be an $R$ -algebra (we regard $B$ as a topological ring by the discrete topology), $J\subset B$ a square-nilpotent ideal, and $f:A\to B/J$ a homomorphism of topological rings. Then, there exists $k\in\mathbb{N}$ such that $f(\mathfrak{m}^k) = 0$ . Write $q:B\to B/J$ for the natural surjection. Since $R\to R[P]$ is formally smooth for the discrete topology, there exists $g:R[P]\to B$ such that $q\circ g = f\circ \pi$ . Since $J^{2}=0$ and $f(\mathfrak{m}^{k})=0$ , $g|\_{I}:I\to J$ factors uniquely through the natural surjection $I\to I/I^2\otimes_{A}A/\mathfrak{m}^k$ . Hence, we obtain a morphism of $A$ -modules $\bar{g}:I/I^2\otimes_A A/\mathfrak{m}^k\to J$ . Since $d\otimes_{A}\mathrm{id}\_{A/\mathfrak{m}^k}$ has a retraction, there exists a morphism $\bar{D}:\Omega_{R[P]/R}\otimes_{R[P]}A/\mathfrak{m}^k\to J$ such that $\bar{D}\circ (d\otimes_A\mathrm{id}\_{A/\mathfrak{m}^k}) = \bar{g}$ . Write $D:P\to J$ for the $R$ -derivation corresponding with the composite of $\bar{D}$ and the natural surjection $\Omega_{R[P]/R}\to \Omega_{R[P]/R}\otimes_{R[P]}A/\mathfrak{m}^k$ . Then $g-D:R[P]\to B$ satisfies that $(g-D)(I)=0$ , so $g-D$ defines a lift $h:A\to B$ of $f:A\to B/J$ . This completes the proof of Claim B.　　　 $\square$

Next, I'm going to check the following claim:

Claim C (see [EGA IV1, Théorème 0.22.6.1] or [MatCA, Section (29.B), Lemma]).　Let $A$ be a ring, $I\subset A$ a nilpotent ideal, $M$ a projective $A$ -module, $L$ an $A$ -module, $s:L\to M$ a morphism of $A$ -modules, and $\bar{r}:M/IM\to L/IL$ a morphism of $A/I$ -modules. Write $\bar{s}:L/IL\to M/IM$ for the morphism of $A/I$ -modules determined by $s$ . Assume that $\bar{r}\circ \bar{s} = \mathrm{id}\_{L/IL}$ . Then, there exists a morphism of $A$ -modules $r:M\to L$ such that the morphism of $A/I$ -modules $M/IM\to L/IL$ deteremind by $r$ is equal to $\bar{r}$ .

Proof of Claim C.　Write $p:M\to M/IM$ and $q:L\to L/IL$ for the natural morphisms. To prove Claim C, by induction, we may assume $I^2=0$ . Since $M$ is projective, there exists $r':M\to L$ such that $q\circ r' = \bar{r}\circ p$ . Write $w:=r'\circ s$ . Then $w(L)+IL=L$ . Hence, by Nakayama, $w$ is surjective. Moreover, since $w$ induces the identity morphism on $L/IL$ and $I^2=0$ , it holds that $w|\_{IL}=\mathrm{id}\_{IL}$ . Hence $w$ is an isomorphism. Now $w^{-1}\circ r'$ is the desired morphism.　　　 $\square$

Now I'm going to prove Lemma A:

Proof of Lemma A.　Note that $\mathfrak{m}$ and $\mathfrak{m}+IA$ define the same topology on $A$ . Hence we may assume that $IA\subset \mathfrak{m}$ . Take a surjective homomorphism $\pi:P\to A$ of $R$ -algebras from a polynomial $R$ -algebra $P$ and write $J:=\ker(\pi)$ . Write $\overline{R}:=R/I$ , $\overline{P}:=P/IP$ , $\overline{A}:=A/IA$ , $\overline{\mathfrak{m}}:=\mathfrak{m}/IA$ , $\overline{J}:=\ker(\overline{P}\to \overline{A})$ , $d:J/J^2\to \Omega\_{P/R}\otimes\_PA$ for the natural morphism of $A$ -modules, and $\overline{d}:\overline{J}/\overline{J}^2\to \Omega\_{\overline{P}/\overline{R}}\otimes_{\overline{P}}\overline{A}$ for the natural morphism of $\overline{A}$ -modules. By the first step of the proof of Commutative Algebra, Tag 031I, the flatness of $R\to A$ implies that $\overline{d}$ is equal to $d\otimes\_A\mathrm{id}\_{\overline{A}}$ via the natural isomorphism (in particular, $J/J^2\otimes\_A\overline{A} \cong \overline{J}/\overline{J}^2$ ). Thus the followings are equivalent:

(1) 　 $R\to A$ is formally smooth for the $\mathfrak{m}$ -adic topology. (2) 　For any $k\in\mathbb{N}$ , $d\otimes\_AA/\mathfrak{m}^k$ has a retraction. (3) 　 $d\otimes\_AA/\mathfrak{m}$ has a retraction. (4) 　 $\overline{d}\otimes\_{\overline{A}}\overline{A}/\overline{\mathfrak{m}}$ has a retraction. (5) 　For any $k\in\mathbb{N}$ , $\overline{d}\otimes\_A\overline{A}/\overline{\mathfrak{m}}^k$ has a retraction. (6) 　 $\overline{R}\to \overline{A}$ is formally smooth for the $\overline{\mathfrak{m}}$ -adic topology.

Here, (1)⇔(2) and (5)⇔(6) follow from Claim B, (2)⇔(3) and (4)⇔(5) follow from Claim C (since the $\Omega$ of a polynomial ring is free), and finally, (3)⇔(4) follows from $d\otimes\_A\mathrm{id}\_{\overline{A}} \cong \overline{d}$ . This completes the proof of Lemma A. 　　　 $\square$

Final note: Using Claim B and following the same method as in the proof of Commutative Algebra, Tag 06CM (Lemma 10.138.16, as of 2025-10-18), one sees that formal smoothness for a given (not necessarily discrete) topology descends along faithfully flat morphisms.

Comment #10939 by yj on October 25, 2025 at 07:09

I apologize... I have misunderstood EGA 0.22.6.1 and made an error in the proof of the necessity part of Claim B above. I now have doubts that the necessity part is hold. That said, I believe there is a better way to prove the implication (2) ⇒ (3): namely, by giving a proof based on a theory parallel to Section 00TH, in particular, by proving a topological-ring version of Lemma 031L. I will comment again once I have a better method worked out. I should have been more careful when commenting; I am sorry for the confusion and any trouble this may have caused.

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