Definition 10.138.1. Let $R \to S$ be a ring map. We say $S$ is *formally smooth over $R$* if for every commutative solid diagram

where $I \subset A$ is an ideal of square zero, a dotted arrow exists which makes the diagram commute.

In this section we define formally smooth ring maps. It will turn out that a ring map of finite presentation is formally smooth if and only if it is smooth, see Proposition 10.138.13.

Definition 10.138.1. Let $R \to S$ be a ring map. We say $S$ is *formally smooth over $R$* if for every commutative solid diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[r] \ar[u] & A \ar[u] } \]

where $I \subset A$ is an ideal of square zero, a dotted arrow exists which makes the diagram commute.

Lemma 10.138.2. Let $R \to S$ be a formally smooth ring map. Let $R \to R'$ be any ring map. Then the base change $S' = R' \otimes _ R S$ is formally smooth over $R'$.

**Proof.**
Let a solid diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rrd] & R' \otimes _ R S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[u] \ar[r] & R' \ar[r] \ar[u] & A \ar[u] } \]

as in Definition 10.138.1 be given. By assumption the longer dotted arrow exists. By the universal property of tensor product we obtain the shorter dotted arrow. $\square$

Lemma 10.138.3. A composition of formally smooth ring maps is formally smooth.

**Proof.**
Omitted. (Hint: This is completely formal, and follows from considering a suitable diagram.)
$\square$

Lemma 10.138.4. A polynomial ring over $R$ is formally smooth over $R$.

**Proof.**
Suppose we have a diagram as in Definition 10.138.1 with $S = R[x_ j; j \in J]$. Then there exists a dotted arrow simply by choosing lifts $a_ j \in A$ of the elements in $A/I$ to which the elements $x_ j$ map to under the top horizontal arrow.
$\square$

Lemma 10.138.5. Let $R \to S$ be a ring map. Let $P \to S$ be a surjective $R$-algebra map from a polynomial ring $P$ onto $S$. Denote $J \subset P$ the kernel. Then $R \to S$ is formally smooth if and only if there exists an $R$-algebra map $\sigma : S \to P/J^2$ which is a right inverse to the surjection $P/J^2 \to S$.

**Proof.**
Assume $R \to S$ is formally smooth. Consider the commutative diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & P/J \\ R \ar[r] \ar[u] & P/J^2\ar[u] } \]

By assumption the dotted arrow exists. This proves that $\sigma $ exists.

Conversely, suppose we have a $\sigma $ as in the lemma. Let a solid diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[r] \ar[u] & A \ar[u] } \]

as in Definition 10.138.1 be given. Because $P$ is formally smooth by Lemma 10.138.4, there exists an $R$-algebra homomorphism $\psi : P \to A$ which lifts the map $P \to S \to A/I$. Clearly $\psi (J) \subset I$ and since $I^2 = 0$ we conclude that $\psi (J^2) = 0$. Hence $\psi $ factors as $\overline{\psi } : P/J^2 \to A$. The desired dotted arrow is the composition $\overline{\psi } \circ \sigma : S \to A$. $\square$

Remark 10.138.6. Lemma 10.138.5 holds more generally whenever $P$ is formally smooth over $R$.

Lemma 10.138.7. Let $R \to S$ be a ring map. Let $P \to S$ be a surjective $R$-algebra map from a polynomial ring $P$ onto $S$. Denote $J \subset P$ the kernel. Then $R \to S$ is formally smooth if and only if the sequence

\[ 0 \to J/J^2 \to \Omega _{P/R} \otimes _ P S \to \Omega _{S/R} \to 0 \]

of Lemma 10.131.9 is a split exact sequence.

**Proof.**
Assume $S$ is formally smooth over $R$. By Lemma 10.138.5 this means there exists an $R$-algebra map $S \to P/J^2$ which is a right inverse to the canonical map $P/J^2 \to S$. By Lemma 10.131.11 we have $\Omega _{P/R} \otimes _ P S = \Omega _{(P/J^2)/R} \otimes _{P/J^2} S$. By Lemma 10.131.10 the sequence is split.

Assume the exact sequence of the lemma is split exact. Choose a splitting $\sigma : \Omega _{S/R} \to \Omega _{P/R} \otimes _ P S$. For each $\lambda \in S$ choose $x_\lambda \in P$ which maps to $\lambda $. Next, for each $\lambda \in S$ choose $f_\lambda \in J$ such that

\[ \text{d}f_\lambda = \text{d}x_\lambda - \sigma (\text{d}\lambda ) \]

in the middle term of the exact sequence. We claim that $s : \lambda \mapsto x_\lambda - f_\lambda \mod J^2$ is an $R$-algebra homomorphism $s : S \to P/J^2$. To prove this we will repeatedly use that if $h \in J$ and $\text{d}h = 0$ in $\Omega _{P/R} \otimes _ R S$, then $h \in J^2$. Let $\lambda , \mu \in S$. Then $\sigma (\text{d}\lambda + \text{d}\mu - \text{d}(\lambda + \mu )) = 0$. This implies

\[ \text{d}(x_\lambda + x_\mu - x_{\lambda + \mu } - f_\lambda - f_\mu + f_{\lambda + \mu }) = 0 \]

which means that $x_\lambda + x_\mu - x_{\lambda + \mu } - f_\lambda - f_\mu + f_{\lambda + \mu } \in J^2$, which in turn means that $s(\lambda ) + s(\mu ) = s(\lambda + \mu )$. Similarly, we have $\sigma (\lambda \text{d}\mu + \mu \text{d}\lambda - \text{d}\lambda \mu ) = 0$ which implies that

\[ \mu (\text{d}x_\lambda - \text{d}f_\lambda ) + \lambda (\text{d}x_\mu - \text{d}f_\mu ) - \text{d}x_{\lambda \mu } + \text{d}f_{\lambda \mu } = 0 \]

in the middle term of the exact sequence. Moreover we have

\[ \text{d}(x_\lambda x_\mu ) = x_\lambda \text{d}x_\mu + x_\mu \text{d}x_\lambda = \lambda \text{d}x_\mu + \mu \text{d} x_\lambda \]

in the middle term again. Combined these equations mean that $x_\lambda x_\mu - x_{\lambda \mu } - \mu f_\lambda - \lambda f_\mu + f_{\lambda \mu } \in J^2$, hence $(x_\lambda - f_\lambda )(x_\mu - f_\mu ) - (x_{\lambda \mu } - f_{\lambda \mu }) \in J^2$ as $f_\lambda f_\mu \in J^2$, which means that $s(\lambda )s(\mu ) = s(\lambda \mu )$. If $\lambda \in R$, then $\text{d}\lambda = 0$ and we see that $\text{d}f_\lambda = \text{d}x_\lambda $, hence $\lambda - x_\lambda + f_\lambda \in J^2$ and hence $s(\lambda ) = \lambda $ as desired. At this point we can apply Lemma 10.138.5 to conclude that $S/R$ is formally smooth. $\square$

Proposition 10.138.8. Let $R \to S$ be a ring map. Consider a formally smooth $R$-algebra $P$ and a surjection $P \to S$ with kernel $J$. The following are equivalent

$S$ is formally smooth over $R$,

for some $P \to S$ as above there exists a section to $P/J^2 \to S$,

for all $P \to S$ as above there exists a section to $P/J^2 \to S$,

for some $P \to S$ as above the sequence $0 \to J/J^2 \to \Omega _{P/R} \otimes S \to \Omega _{S/R} \to 0$ is split exact,

for all $P \to S$ as above the sequence $0 \to J/J^2 \to \Omega _{P/R} \otimes S \to \Omega _{S/R} \to 0$ is split exact, and

the naive cotangent complex $\mathop{N\! L}\nolimits _{S/R}$ is quasi-isomorphic to a projective $S$-module placed in degree $0$.

**Proof.**
It is clear that (1) implies (3) implies (2), see first part of the proof of Lemma 10.138.5. It is also true that (3) implies (5) implies (4) and that (2) implies (4), see first part of the proof of Lemma 10.138.7. Finally, Lemma 10.138.7 applied to the canonical surjection $R[S] \to S$ (10.134.0.1) shows that (1) implies (6).

Assume (4) and let's prove (6). Consider the sequence of Lemma 10.134.4 associated to the ring maps $R \to P \to S$. By the implication (1) $\Rightarrow $ (6) proved above we see that $\mathop{N\! L}\nolimits _{P/R} \otimes _ R S$ is quasi-isomorphic to $\Omega _{P/R} \otimes _ P S$ placed in degree $0$. Hence $H_1(\mathop{N\! L}\nolimits _{P/R} \otimes _ P S) = 0$. Since $P \to S$ is surjective we see that $\mathop{N\! L}\nolimits _{S/P}$ is homotopy equivalent to $J/J^2$ placed in degree $1$ (Lemma 10.134.6). Thus we obtain the exact sequence $0 \to H_1(L_{S/R}) \to J/J^2 \to \Omega _{P/R} \otimes _ P S \to \Omega _{S/R} \to 0$. By assumption we see that $H_1(L_{S/R}) = 0$ and that $\Omega _{S/R}$ is a projective $S$-module. Thus (6) follows.

Finally, let's prove that (6) implies (1). The assumption means that the complex $J/J^2 \to \Omega _{P/R} \otimes S$ where $P = R[S]$ and $P \to S$ is the canonical surjection (10.134.0.1). Hence Lemma 10.138.7 shows that $S$ is formally smooth over $R$. $\square$

Lemma 10.138.9. Let $A \to B \to C$ be ring maps. Assume $B \to C$ is formally smooth. Then the sequence

\[ 0 \to \Omega _{B/A} \otimes _ B C \to \Omega _{C/A} \to \Omega _{C/B} \to 0 \]

of Lemma 10.131.7 is a split short exact sequence.

**Proof.**
Follows from Proposition 10.138.8 and Lemma 10.134.4.
$\square$

Lemma 10.138.10. Let $A \to B \to C$ be ring maps with $A \to C$ formally smooth and $B \to C$ surjective with kernel $J \subset B$. Then the exact sequence

\[ 0 \to J/J^2 \to \Omega _{B/A} \otimes _ B C \to \Omega _{C/A} \to 0 \]

of Lemma 10.131.9 is split exact.

**Proof.**
Follows from Proposition 10.138.8, Lemma 10.134.4, and Lemma 10.131.9.
$\square$

Lemma 10.138.11. Let $A \to B \to C$ be ring maps. Assume $A \to C$ is surjective (so also $B \to C$ is) and $A \to B$ formally smooth. Denote $I = \mathop{\mathrm{Ker}}(A \to C)$ and $J = \mathop{\mathrm{Ker}}(B \to C)$. Then the sequence

\[ 0 \to I/I^2 \to J/J^2 \to \Omega _{B/A} \otimes _ B B/J \to 0 \]

of Lemma 10.134.7 is split exact.

**Proof.**
Since $A \to B$ is formally smooth there exists a ring map $\sigma : B \to A/I^2$ whose composition with $A \to B$ equals the quotient map $A \to A/I^2$. Then $\sigma $ induces a map $J/J^2 \to I/I^2$ which is inverse to the map $I/I^2 \to J/J^2$.
$\square$

Lemma 10.138.12. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Assume

$I^2 = 0$,

$R \to S$ is flat, and

$R/I \to S/IS$ is formally smooth.

Then $R \to S$ is formally smooth.

**Proof.**
Assume (1), (2) and (3). Let $P = R[\{ x_ t\} _{t \in T}] \to S$ be a surjection of $R$-algebras with kernel $J$. Thus $0 \to J \to P \to S \to 0$ is a short exact sequence of flat $R$-modules. This implies that $I \otimes _ R S = IS$, $I \otimes _ R P = IP$ and $I \otimes _ R J = IJ$ as well as $J \cap IP = IJ$. We will use throughout the proof that

\[ \Omega _{(S/IS)/(R/I)} = \Omega _{S/R} \otimes _ S (S/IS) = \Omega _{S/R} \otimes _ R R/I = \Omega _{S/R} / I\Omega _{S/R} \]

and similarly for $P$ (see Lemma 10.131.12). By Lemma 10.138.7 the sequence

10.138.12.1

\begin{equation} \label{algebra-equation-split} 0 \to J/(IJ + J^2) \to \Omega _{P/R} \otimes _ P S/IS \to \Omega _{S/R} \otimes _ S S/IS \to 0 \end{equation}

is split exact. Of course the middle term is $\bigoplus _{t \in T} S/IS \text{d}x_ t$. Choose a splitting $\sigma : \Omega _{P/R} \otimes _ P S/IS \to J/(IJ + J^2)$. For each $t \in T$ choose an element $f_ t \in J$ which maps to $\sigma (\text{d}x_ t)$ in $J/(IJ + J^2)$. This determines a unique $S$-module map

\[ \tilde\sigma : \Omega _{P/R} \otimes _ R S = \bigoplus S\text{d}x_ t \longrightarrow J/J^2 \]

with the property that $\tilde\sigma (\text{d}x_ t) = f_ t$. As $\sigma $ is a section to $\text{d}$ the difference

\[ \Delta = \text{id}_{J/J^2} - \tilde\sigma \circ \text{d} \]

is a self map $J/J^2 \to J/J^2$ whose image is contained in $(IJ + J^2)/J^2$. In particular $\Delta ((IJ + J^2)/J^2) = 0$ because $I^2 = 0$. This means that $\Delta $ factors as

\[ J/J^2 \to J/(IJ + J^2) \xrightarrow {\overline{\Delta }} (IJ + J^2)/J^2 \to J/J^2 \]

where $\overline{\Delta }$ is a $S/IS$-module map. Using again that the sequence (10.138.12.1) is split, we can find a $S/IS$-module map $\overline{\delta } : \Omega _{P/R} \otimes _ P S/IS \to (IJ + J^2)/J^2$ such that $\overline{\delta } \circ d$ is equal to $\overline{\Delta }$. In the same manner as above the map $\overline{\delta }$ determines an $S$-module map $\delta : \Omega _{P/R} \otimes _ P S \to J/J^2$. After replacing $\tilde\sigma $ by $\tilde\sigma + \delta $ a simple computation shows that $\Delta = 0$. In other words $\tilde\sigma $ is a section of $J/J^2 \to \Omega _{P/R} \otimes _ P S$. By Lemma 10.138.7 we conclude that $R \to S$ is formally smooth. $\square$

Proposition 10.138.13. Let $R \to S$ be a ring map. The following are equivalent

$R \to S$ is of finite presentation and formally smooth,

$R \to S$ is smooth.

**Proof.**
Follows from Proposition 10.138.8 and Definition 10.137.1. (Note that $\Omega _{S/R}$ is a finitely presented $S$-module if $R \to S$ is of finite presentation, see Lemma 10.131.15.)
$\square$

Lemma 10.138.14. Let $R \to S$ be a smooth ring map. Then there exists a subring $R_0 \subset R$ of finite type over $\mathbf{Z}$ and a smooth ring map $R_0 \to S_0$ such that $S \cong R \otimes _{R_0} S_0$.

**Proof.**
We are going to use that smooth is equivalent to finite presentation and formally smooth, see Proposition 10.138.13. Write $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and denote $I = (f_1, \ldots , f_ m)$. Choose a right inverse $\sigma : S \to R[x_1, \ldots , x_ n]/I^2$ to the projection to $S$ as in Lemma 10.138.5. Choose $h_ i \in R[x_1, \ldots , x_ n]$ such that $\sigma (x_ i \bmod I) = h_ i \bmod I^2$. The fact that $\sigma $ is an $R$-algebra homomorphism $R[x_1, \ldots , x_ n]/I \to R[x_1, \ldots , x_ n]/I^2$ is equivalent to the condition that

\[ f_ j(h_1, \ldots , h_ n) = \sum \nolimits _{j_1 j_2} a_{j_1 j_2} f_{j_1} f_{j_2} \]

for certain $a_{kl} \in R[x_1, \ldots , x_ n]$. Let $R_0 \subset R$ be the subring generated over $\mathbf{Z}$ by all the coefficients of the polynomials $f_ j, h_ i, a_{kl}$. Set $S_0 = R_0[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$, with $I_0 = (f_1, \ldots , f_ m)$. Let $\sigma _0 : S_0 \to R_0[x_1, \ldots , x_ n]/I_0^2$ defined by the rule $x_ i \mapsto h_ i \bmod I_0^2$; this works since the $a_{lk}$ are defined over $R_0$ and satisfy the same relations. Thus by Lemma 10.138.5 the ring $S_0$ is formally smooth over $R_0$. $\square$

Lemma 10.138.15. Let $A = \mathop{\mathrm{colim}}\nolimits A_ i$ be a filtered colimit of rings. Let $A \to B$ be a smooth ring map. There exists an $i$ and a smooth ring map $A_ i \to B_ i$ such that $B = B_ i \otimes _{A_ i} A$.

**Proof.**
Follows from Lemma 10.138.14 since $R_0 \to A$ will factor through $A_ i$ for some $i$ by Lemma 10.127.3.
$\square$

Lemma 10.138.16. Let $R \to S$ be a ring map. Let $R \to R'$ be a faithfully flat ring map. Set $S' = S \otimes _ R R'$. Then $R \to S$ is formally smooth if and only if $R' \to S'$ is formally smooth.

**Proof.**
If $R \to S$ is formally smooth, then $R' \to S'$ is formally smooth by Lemma 10.138.2. To prove the converse, assume $R' \to S'$ is formally smooth. Note that $N \otimes _ R R' = N \otimes _ S S'$ for any $S$-module $N$. In particular $S \to S'$ is faithfully flat also. Choose a polynomial ring $P = R[\{ x_ i\} _{i \in I}]$ and a surjection of $R$-algebras $P \to S$ with kernel $J$. Note that $P' = P \otimes _ R R'$ is a polynomial algebra over $R'$. Since $R \to R'$ is flat the kernel $J'$ of the surjection $P' \to S'$ is $J \otimes _ R R'$. Hence the split exact sequence (see Lemma 10.138.7)

\[ 0 \to J'/(J')^2 \to \Omega _{P'/R'} \otimes _{P'} S' \to \Omega _{S'/R'} \to 0 \]

is the base change via $S \to S'$ of the corresponding sequence

\[ J/J^2 \to \Omega _{P/R} \otimes _ P S \to \Omega _{S/R} \to 0 \]

see Lemma 10.131.9. As $S \to S'$ is faithfully flat we conclude two things: (1) this sequence (without ${}'$) is exact too, and (2) $\Omega _{S/R}$ is a projective $S$-module. Namely, $\Omega _{S'/R'}$ is projective as a direct sum of the free module $\Omega _{P'/R'} \otimes _{P'} S'$ and $\Omega _{S/R} \otimes _ S {S'} = \Omega _{S'/R'}$ by what we said above. Thus (2) follows by descent of projectivity through faithfully flat ring maps, see Theorem 10.95.6. Hence the sequence $0 \to J/J^2 \to \Omega _{P/R} \otimes _ P S \to \Omega _{S/R} \to 0$ is exact also and we win by applying Lemma 10.138.7 once more. $\square$

It turns out that smooth ring maps satisfy the following strong lifting property.

Lemma 10.138.17. Let $R \to S$ be a smooth ring map. Given a commutative solid diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[r] \ar[u] & A \ar[u] } \]

where $I \subset A$ is a locally nilpotent ideal, a dotted arrow exists which makes the diagram commute.

**Proof.**
By Lemma 10.138.14 we can extend the diagram to a commutative diagram

\[ \xymatrix{ S_0 \ar[r] & S \ar[r] \ar@{-->}[rd] & A/I \\ R_0 \ar[r] \ar[u] & R \ar[r] \ar[u] & A \ar[u] } \]

with $R_0 \to S_0$ smooth, $R_0$ of finite type over $\mathbf{Z}$, and $S = S_0 \otimes _{R_0} R$. Let $x_1, \ldots , x_ n \in S_0$ be generators of $S_0$ over $R_0$. Let $a_1, \ldots , a_ n$ be elements of $A$ which map to the same elements in $A/I$ as the elements $x_1, \ldots , x_ n$. Denote $A_0 \subset A$ the subring generated by the image of $R_0$ and the elements $a_1, \ldots , a_ n$. Set $I_0 = A_0 \cap I$. Then $A_0/I_0 \subset A/I$ and $S_0 \to A/I$ maps into $A_0/I_0$. Thus it suffices to find the dotted arrow in the diagram

\[ \xymatrix{ S_0 \ar[r] \ar@{-->}[rd] & A_0/I_0 \\ R_0 \ar[r] \ar[u] & A_0 \ar[u] } \]

The ring $A_0$ is of finite type over $\mathbf{Z}$ by construction. Hence $A_0$ is Noetherian, whence $I_0$ is nilpotent, see Lemma 10.32.5. Say $I_0^ n = 0$. By Proposition 10.138.13 we can successively lift the $R_0$-algebra map $S_0 \to A_0/I_0$ to $S_0 \to A_0/I_0^2$, $S_0 \to A_0/I_0^3$, $\ldots $, and finally $S_0 \to A_0/I_0^ n = A_0$. $\square$

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