The Stacks project

Proof. If the characteristic of $k$ is zero, then the equivalence of (1) and (2) follows from Lemmas 15.38.2 and 15.38.5.

If the characteristic of $k$ is $p > 0$, then it follows from Proposition 15.35.1 that (1), (3), (4), and (5) are equivalent. Assume (2) holds. By Lemma 15.37.8 we see that $k' \to A' = A \otimes _ k k'$ is formally smooth for the $\mathfrak m' = \mathfrak mA'$-adic topology. Hence if $k \subset k'$ is finite purely inseparable, then $A'$ is a regular local ring by Lemma 15.38.2. Thus we see that (1) holds.

Finally, we will prove that (5) implies (2). Choose a solid diagram

as in Definition 15.37.1. As $J^2 = 0$ we see that $J$ has a canonical $B/J$ module structure and via $\bar\psi $ an $A$-module structure. As $\bar\psi $ is continuous for the $\mathfrak m$-adic topology we see that $\mathfrak m^ nJ = 0$ for some $n$. Hence we can filter $J$ by $B/J$-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 $\mathfrak m$. Considering the sequence of ring maps $B \to B/J_1 \to B/J_2 \to \ldots \to B/J$ we see that it suffices to prove the existence of the dotted arrow when $J$ is annihilated by $\mathfrak m$, i.e., when $J$ is a $K$-vector space.

Assume given a diagram as above such that $J$ is annihilated by $\mathfrak m$. By Lemma 15.38.5 we see that $\mathbf{F}_ p \to A$ is formally smooth in the $\mathfrak m$-adic topology. Hence we can find a ring map $\psi : A \to B$ such that $\pi \circ \psi = \bar\psi $. Then $\psi \circ i, \varphi : k \to B$ are two maps whose compositions with $\pi $ are equal. Hence $D = \psi \circ i - \varphi : k \to J$ is a derivation. By Algebra, Lemma 10.131.3 we can write $D = \xi \circ \text{d}$ for some $k$-linear map $\xi : \Omega _{k/\mathbf{F}_ p} \to J$. Using the $K$-vector space structure on $J$ we extend $\xi $ to a $K$-linear map $\xi ' : \Omega _{k/\mathbf{F}_ p} \otimes _ k K \to J$. Using (5) we can find a $K$-linear map $\xi '' : \Omega _{A/\mathbf{F}_ p} \otimes _ A K$ whose restriction to $\Omega _{k/\mathbf{F}_ p} \otimes _ k K$ is $\xi '$. Write

Finally, set $\psi ' = \psi - D' : A \to B$. 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. $\square$

Proof. We may assume that $A$ and $B$ a Noetherian complete local rings by Lemma 15.37.4 and Algebra, Lemma 10.97.6 (this also uses Algebra, Lemma 10.39.9 and 10.97.3 to see that flatness of the map on completions implies flatness of $A \to B$). Choose a commutative diagram

as in Lemma 15.39.3 with $R \to S$ flat. Let $I \subset R$ be the kernel of $R \to A$. Because $B$ is formally smooth over $A$ we see that the $A$-algebra map

has a section, see Lemma 15.37.5. Hence $B$ is a direct summand of the flat $A$-module $S/IS$ (by base change of flatness, see Algebra, Lemma 10.39.7), whence flat. $\square$

Proof. Observe that $\mathfrak m_ B/\mathfrak m_ B^2 = H_1(\mathop{N\! L}\nolimits _{K/B})$ by Algebra, Lemma 10.134.6. By Algebra, Lemma 10.134.4 it remains to show injectivity of $H_1(\mathop{N\! L}\nolimits _{K/A}) \to \mathfrak m_ B/\mathfrak m_ B^2$. With $k$ the residue field of $A$, the Jacobi-Zariski sequence for $A \to k \to K$ gives $\Omega _{K/A} = \Omega _{K/k}$ and an exact sequence

Set $\overline{B} = B \otimes _ A k$. Since $\overline{B}$ is regular the ideal $\mathfrak m_{\overline{B}}$ is generated by a regular sequence. Applying Lemmas 15.30.9 and 15.30.7 to $\mathfrak m_ A B \subset \mathfrak m_ B$ we find $\mathfrak m_ A B / (\mathfrak m_ AB \cap \mathfrak m_ B^2) = \mathfrak m_ A B / \mathfrak m_ A \mathfrak m_ B$ which is equal to $\mathfrak m_ A/\mathfrak m_ A^2 \otimes _ k K$ as $A \to B$ is flat by Lemma 15.40.3. Thus we obtain a short exact sequence

Functoriality of the Jacobi-Zariski sequences shows that we obtain a commutative diagram

The left vertical arrow is injective by Theorem 15.40.1 as $k \to \overline{B}$ is formally smooth in the $\mathfrak m_{\overline{B}}$-adic topology by Lemma 15.37.8. This finishes the proof by the snake lemma. $\square$

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

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

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

as in Lemma 15.39.3 with moreover $\overline{S} = \overline{B}$. In this case the map

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

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

where $\xi '$ is $k$-linear. Using the $K$-vector space structure on $J$ we extend $\xi '$ to a $K$-linear map

Using that $\overline{S}/k$ is formally smooth we see that

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

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. $\square$

Proof. Choose a diagram

as in Lemma 15.39.3. Let $t_1, \ldots , t_ d$ be a regular system of parameters for $R$ with $t_1 = p$ in case the characteristic of $k$ is $p > 0$. As $B$ and $\overline{S} = S \otimes _ R k$ are regular we see that $\mathop{\mathrm{Ker}}(\overline{S} \to B)$ 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 $x_1, \ldots , x_ r \in S$. 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

as in Lemma 15.39.3 with moreover $\overline{S} = B$. In this case $R \to S$ is formally smooth in the $\mathfrak m_ S$-adic topology by Proposition 15.40.5. Hence the base change $C = S \otimes _ R A$ is formally smooth over $A$ in the $\mathfrak m_ C$-adic topology by Lemma 15.37.8. $\square$