Definition 10.137.1. A ring map $R \to S$ is *smooth* if it is of finite presentation and the naive cotangent complex $\mathop{N\! L}\nolimits _{S/R}$ is quasi-isomorphic to a finite projective $S$-module placed in degree $0$.

## 10.137 Smooth ring maps

Let us motivate the definition of a smooth ring map by an example. Suppose $R$ is a ring and $S = R[x, y]/(f)$ for some nonzero $f \in R[x, y]$. In this case there is an exact sequence

where the first arrow maps $1$ to $\frac{\partial f}{\partial x} \text{d}x + \frac{\partial f}{\partial y} \text{d}y$ see Section 10.134. We conclude that $\Omega _{S/R}$ is locally free of rank $1$ if the partial derivatives of $f$ generate the unit ideal in $S$. In this case $S$ is smooth of relative dimension $1$ over $R$. But it can happen that $\Omega _{S/R}$ is locally free of rank $2$ namely if both partial derivatives of $f$ are zero. For example if for a prime $p$ we have $p = 0$ in $R$ and $f = x^ p + y^ p$ then this happens. Here $R \to S$ is a relative global complete intersection of relative dimension $1$ which is not smooth. Hence, in order to check that a ring map is smooth it is not sufficient to check whether the module of differentials is free. The correct condition is the following.

In particular, if $R \to S$ is smooth then the module $\Omega _{S/R}$ is a finite projective $S$-module. Moreover, by Lemma 10.137.2 the naive cotangent complex of any presentation has the same structure. Thus, for a surjection $\alpha : R[x_1, \ldots , x_ n] \to S$ with kernel $I$ the map

is a split injection. In other words $\bigoplus _{i = 1}^ n S \text{d}x_ i \cong I/I^2 \oplus \Omega _{S/R}$ as $S$-modules. This implies that $I/I^2$ is a finite projective $S$-module too!

Lemma 10.137.2. Let $R \to S$ be a ring map of finite presentation. If for some presentation $\alpha $ of $S$ over $R$ the naive cotangent complex $\mathop{N\! L}\nolimits (\alpha )$ is quasi-isomorphic to a finite projective $S$-module placed in degree $0$, then this holds for any presentation.

**Proof.**
Immediate from Lemma 10.134.2.
$\square$

Lemma 10.137.3. Let $R \to S$ be a smooth ring map. Any localization $S_ g$ is smooth over $R$. If $f \in R$ maps to an invertible element of $S$, then $R_ f \to S$ is smooth.

**Proof.**
By Lemma 10.134.13 the naive cotangent complex for $S_ g$ over $R$ is the base change of the naive cotangent complex of $S$ over $R$. The assumption is that the naive cotangent complex of $S/R$ is $\Omega _{S/R}$ and that this is a finite projective $S$-module. Hence so is its base change. Thus $S_ g$ is smooth over $R$.

The second assertion follows in the same way from Lemma 10.134.11. $\square$

Lemma 10.137.4. Let $R \to S$ be a smooth ring map. Let $R \to R'$ be any ring map. Then the base change $R' \to S' = R' \otimes _ R S$ is smooth.

**Proof.**
Let $\alpha : R[x_1, \ldots , x_ n] \to S$ be a presentation with kernel $I$. Let $\alpha ' : R'[x_1, \ldots , x_ n] \to R' \otimes _ R S$ be the induced presentation. Let $I' = \mathop{\mathrm{Ker}}(\alpha ')$. Since $0 \to I \to R[x_1, \ldots , x_ n] \to S \to 0$ is exact, the sequence $R' \otimes _ R I \to R'[x_1, \ldots , x_ n] \to R' \otimes _ R S \to 0$ is exact. Thus $R' \otimes _ R I \to I'$ is surjective. By Definition 10.137.1 there is a short exact sequence

and the $S$-module $\Omega _{S/R}$ is finite projective. In particular $I/I^2$ is a direct summand of $\Omega _{R[x_1, \ldots , x_ n]/R} \otimes _{R[x_1, \ldots , x_ n]} S$. Consider the commutative diagram

Since the right vertical map is an isomorphism we see that the left vertical map is injective and surjective by what was said above. Thus we conclude that $\mathop{N\! L}\nolimits (\alpha ')$ is quasi-isomorphic to $\Omega _{S'/R'} \cong S' \otimes _ S \Omega _{S/R}$. And this is finite projective since it is the base change of a finite projective module. $\square$

Lemma 10.137.5. Let $k$ be a field. Let $S$ be a smooth $k$-algebra. Then $S$ is a local complete intersection.

**Proof.**
By Lemmas 10.137.4 and 10.135.11 it suffices to prove this when $k$ is algebraically closed. Choose a presentation $\alpha : k[x_1, \ldots , x_ n] \to S$ with kernel $I$. Let $\mathfrak m$ be a maximal ideal of $S$, and let $\mathfrak m' \supset I$ be the corresponding maximal ideal of $k[x_1, \ldots , x_ n]$. We will show that condition (5) of Lemma 10.135.4 holds (with $\mathfrak m$ instead of $\mathfrak q$). We may write $\mathfrak m' = (x_1 - a_1, \ldots , x_ n - a_ n)$ for some $a_ i \in k$, because $k$ is algebraically closed, see Theorem 10.34.1. By our assumption that $k \to S$ is smooth the $S$-module map $\text{d} : I/I^2 \to \bigoplus _{i = 1}^ n S \text{d}x_ i$ is a split injection. Hence the corresponding map $I/\mathfrak m' I \to \bigoplus \kappa (\mathfrak m') \text{d}x_ i$ is injective. Say $\dim _{\kappa (\mathfrak m')}(I/\mathfrak m' I) = c$ and pick $f_1, \ldots , f_ c \in I$ which map to a $\kappa (\mathfrak m')$-basis of $I/\mathfrak m' I$. By Nakayama's Lemma 10.20.1 we see that $f_1, \ldots , f_ c$ generate $I_{\mathfrak m'}$ over $k[x_1, \ldots , x_ n]_{\mathfrak m'}$. Consider the commutative diagram

(proof commutativity omitted). The middle vertical map is the one defining the naive cotangent complex of $\alpha $. Note that the right lower horizontal arrow induces an isomorphism $\bigoplus \kappa (\mathfrak m') \text{d}x_ i \to \mathfrak m'/(\mathfrak m')^2$. Hence our generators $f_1, \ldots , f_ c$ of $I_{\mathfrak m'}$ map to a collection of elements in $k[x_1, \ldots , x_ n]_{\mathfrak m'}$ whose classes in $\mathfrak m'/(\mathfrak m')^2$ are linearly independent over $\kappa (\mathfrak m')$. Therefore they form a regular sequence in the ring $k[x_1, \ldots , x_ n]_{\mathfrak m'}$ by Lemma 10.106.3. This verifies condition (5) of Lemma 10.135.4 hence $S_ g$ is a global complete intersection over $k$ for some $g \in S$, $g \not\in \mathfrak m$. As this works for any maximal ideal of $S$ we conclude that $S$ is a local complete intersection over $k$. $\square$

Definition 10.137.6. Let $R$ be a ring. Given integers $n \geq c \geq 0$ and $f_1, \ldots , f_ c \in R[x_1, \ldots , x_ n]$ we say

is a *standard smooth algebra over $R$* if the polynomial

maps to an invertible element in $S$.

Lemma 10.137.7. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c) = R[x_1, \ldots , x_ n]/I$ be a standard smooth algebra. Then

the ring map $R \to S$ is smooth,

the $S$-module $\Omega _{S/R}$ is free on $\text{d}x_{c + 1}, \ldots , \text{d}x_ n$,

the $S$-module $I/I^2$ is free on the classes of $f_1, \ldots , f_ c$,

for any $g \in S$ the ring map $R \to S_ g$ is standard smooth,

for any ring map $R \to R'$ the base change $R' \to R'\otimes _ R S$ is standard smooth,

if $f \in R$ maps to an invertible element in $S$, then $R_ f \to S$ is standard smooth, and

the ring $S$ is a relative global complete intersection over $R$.

**Proof.**
Consider the naive cotangent complex of the given presentation

Let us compose this map with the projection onto the first $c$ direct summands of the direct sum. According to the definition of a standard smooth algebra the classes $f_ i \bmod (f_1, \ldots , f_ c)^2$ map to a basis of $\bigoplus _{i = 1}^ c S\text{d}x_ i$. We conclude that $(f_1, \ldots , f_ c)/(f_1, \ldots , f_ c)^2$ is free of rank $c$ with a basis given by the elements $f_ i \bmod (f_1, \ldots , f_ c)^2$, and that the homology in degree $0$, i.e., $\Omega _{S/R}$, of the naive cotangent complex is a free $S$-module with basis the images of $\text{d}x_{c + j}$, $j = 1, \ldots , n - c$. In particular, this proves $R \to S$ is smooth.

The proofs of (4) and (6) are omitted. But see the example below and the proof of Lemma 10.136.9.

Let $\varphi : R \to R'$ be any ring map. Denote $S' = R'[x_1, \ldots , x_ n]/(f_1^\varphi , \ldots , f_ c^\varphi )$ where $f^\varphi $ is the polynomial obtained from $f \in R[x_1, \ldots , x_ n]$ by applying $\varphi $ to all the coefficients. Then $S' \cong R' \otimes _ R S$. Moreover, the determinant of Definition 10.137.6 for $S'/R'$ is equal to $g^\varphi $. Its image in $S'$ is therefore the image of $g$ via $R[x_1, \ldots , x_ n] \to S \to S'$ and hence invertible. This proves (5).

To prove (7) it suffices to show that $S \otimes _ R \kappa (\mathfrak p)$ has dimension $n - c$ for every prime $\mathfrak p \subset R$. By (5) it suffices to prove that any standard smooth algebra $k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ over a field $k$ has dimension $n - c$. We already know that $k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a local complete intersection by Lemma 10.137.5. Hence, since $I/I^2$ is free of rank $c$ we see that $k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ has dimension $n - c$, by Lemma 10.135.4 for example. $\square$

Example 10.137.8. Let $R$ be a ring. Let $f_1, \ldots , f_ c \in R[x_1, \ldots , x_ n]$. Let

Set $S = R[x_1, \ldots , x_{n + 1}]/(f_1, \ldots , f_ c, x_{n + 1}h - 1)$. This is an example of a standard smooth algebra, except that the presentation is wrong and the variables should be in the following order: $x_1, \ldots , x_ c, x_{n + 1}, x_{c + 1}, \ldots , x_ n$.

Lemma 10.137.9. A composition of standard smooth ring maps is standard smooth.

**Proof.**
Suppose that $R \to S$ and $S \to S'$ are standard smooth. We choose presentations $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ and $S' = S[y_1, \ldots , y_ m]/(g_1, \ldots , g_ d)$. Choose elements $g_ j' \in R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]$ mapping to the $g_ j$. In this way we see $S' = R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]/ (f_1, \ldots , f_ c, g'_1, \ldots , g'_ d)$. To show that $S'$ is standard smooth it suffices to verify that the determinant

is invertible in $S'$. This is clear since it is the product of the two determinants which were assumed to be invertible by hypothesis. $\square$

Lemma 10.137.10. Let $R \to S$ be a smooth ring map. There exists an open covering of $\mathop{\mathrm{Spec}}(S)$ by standard opens $D(g)$ such that each $S_ g$ is standard smooth over $R$. In particular $R \to S$ is syntomic.

**Proof.**
Choose a presentation $\alpha : R[x_1, \ldots , x_ n] \to S$ with kernel $I = (f_1, \ldots , f_ m)$. For every subset $E \subset \{ 1, \ldots , m\} $ consider the open subset $U_ E$ where the classes $f_ e, e\in E$ freely generate the finite projective $S$-module $I/I^2$, see Lemma 10.79.4. We may cover $\mathop{\mathrm{Spec}}(S)$ by standard opens $D(g)$ each completely contained in one of the opens $U_ E$. For such a $g$ we look at the presentation

mapping $x_{n + 1}$ to $1/g$. Setting $J = \mathop{\mathrm{Ker}}(\beta )$ we use Lemma 10.134.12 to see that $J/J^2 \cong (I/I^2)_ g \oplus S_ g$ is free. We may and do replace $S$ by $S_ g$. Then using Lemma 10.136.6 we may assume we have a presentation $\alpha : R[x_1, \ldots , x_ n] \to S$ with kernel $I = (f_1, \ldots , f_ c)$ such that $I/I^2$ is free on the classes of $f_1, \ldots , f_ c$.

Using the presentation $\alpha $ obtained at the end of the previous paragraph, we more or less repeat this argument with the basis elements $\text{d}x_1, \ldots , \text{d}x_ n$ of $\Omega _{R[x_1, \ldots , x_ n]/R}$. Namely, for any subset $E \subset \{ 1, \ldots , n\} $ of cardinality $c$ we may consider the open subset $U_ E$ of $\mathop{\mathrm{Spec}}(S)$ where the differential of $\mathop{N\! L}\nolimits (\alpha )$ composed with the projection

is an isomorphism. Again we may find a covering of $\mathop{\mathrm{Spec}}(S)$ by (finitely many) standard opens $D(g)$ such that each $D(g)$ is completely contained in one of the opens $U_ E$. By renumbering, we may assume $E = \{ 1, \ldots , c\} $. For a $g$ with $D(g) \subset U_ E$ we look at the presentation

mapping $x_{n + 1}$ to $1/g$. Setting $J = \mathop{\mathrm{Ker}}(\beta )$ we conclude from Lemma 10.134.12 that $J = (f_1, \ldots , f_ c, fx_{n + 1} - 1)$ where $\alpha (f) = g$ and that the composition

is an isomorphism. Reordering the coordinates as $x_1, \ldots , x_ c, x_{n + 1}, x_{c + 1}, \ldots , x_ n$ we conclude that $S_ g$ is standard smooth over $R$ as desired.

This finishes the proof as standard smooth algebras are syntomic (Lemmas 10.137.7 and 10.136.13) and being syntomic over $R$ is local on $S$ (Lemma 10.136.4). $\square$

Definition 10.137.11. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$. We say $R \to S$ is *smooth at $\mathfrak q$* if there exists a $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is smooth.

For ring maps of finite presentation we can characterize this as follows.

Lemma 10.137.12. Let $R \to S$ be of finite presentation. Let $\mathfrak q$ be a prime of $S$. The following are equivalent

$R \to S$ is smooth at $\mathfrak q$,

$H_1(L_{S/R})_\mathfrak q = 0$ and $\Omega _{S/R, \mathfrak q}$ is a finite free $S_\mathfrak q$-module,

$H_1(L_{S/R})_\mathfrak q = 0$ and $\Omega _{S/R, \mathfrak q}$ is a projective $S_\mathfrak q$-module, and

$H_1(L_{S/R})_\mathfrak q = 0$ and $\Omega _{S/R, \mathfrak q}$ is a flat $S_\mathfrak q$-module.

**Proof.**
We will use without further mention that formation of the naive cotangent complex commutes with localization, see Section 10.134, especially Lemma 10.134.13. Note that $\Omega _{S/R}$ is a finitely presented $S$-module, see Lemma 10.131.15. Hence (2), (3), and (4) are equivalent by Lemma 10.78.2. It is clear that (1) implies the equivalent conditions (2), (3), and (4). Assume (2) holds. Writing $S_\mathfrak q$ as the colimit of principal localizations we see from Lemma 10.127.6 that we can find a $g \in S$, $g \not\in \mathfrak q$ such that $(\Omega _{S/R})_ g$ is finite free. Choose a presentation $\alpha : R[x_1, \ldots , x_ n] \to S$ with kernel $I$. We may work with $\mathop{N\! L}\nolimits (\alpha )$ instead of $\mathop{N\! L}\nolimits _{S/R}$, see Lemma 10.134.2. The surjection

has a right inverse after inverting $g$ because $(\Omega _{S/R})_ g$ is projective. Hence the image of $\text{d} : (I/I^2)_ g \to \Omega _{R[x_1, \ldots , x_ n]/R} \otimes _ R S_ g$ is a direct summand and this map has a right inverse too. We conclude that $H_1(L_{S/R})_ g$ is a quotient of $(I/I^2)_ g$. In particular $H_1(L_{S/R})_ g$ is a finite $S_ g$-module. Thus the vanishing of $H_1(L_{S/R})_{\mathfrak q}$ implies the vanishing of $H_1(L_{S/R})_{gg'}$ for some $g' \in S$, $g' \not\in \mathfrak q$. Then $R \to S_{gg'}$ is smooth by definition. $\square$

Lemma 10.137.13. Let $R \to S$ be a ring map. Then $R \to S$ is smooth if and only if $R \to S$ is smooth at every prime $\mathfrak q$ of $S$.

**Proof.**
The direct implication is trivial. Suppose that $R \to S$ is smooth at every prime $\mathfrak q$ of $S$. Since $\mathop{\mathrm{Spec}}(S)$ is quasi-compact, see Lemma 10.17.8, there exists a finite covering $\mathop{\mathrm{Spec}}(S) = \bigcup D(g_ i)$ such that each $S_{g_ i}$ is smooth. By Lemma 10.23.3 this implies that $S$ is of finite presentation over $R$. According to Lemma 10.134.13 we see that $\mathop{N\! L}\nolimits _{S/R} \otimes _ S S_{g_ i}$ is quasi-isomorphic to a finite projective $S_{g_ i}$-module. By Lemma 10.78.2 this implies that $\mathop{N\! L}\nolimits _{S/R}$ is quasi-isomorphic to a finite projective $S$-module.
$\square$

Lemma 10.137.14. A composition of smooth ring maps is smooth.

**Proof.**
You can prove this in many different ways. One way is to use the snake lemma (Lemma 10.4.1), the Jacobi-Zariski sequence (Lemma 10.134.4), combined with the characterization of projective modules as being direct summands of free modules (Lemma 10.77.2). Another proof can be obtained by combining Lemmas 10.137.10, 10.137.9 and 10.137.13.
$\square$

Lemma 10.137.15. Let $R$ be a ring. Let $S = S' \times S''$ be a product of $R$-algebras. Then $S$ is smooth over $R$ if and only if both $S'$ and $S''$ are smooth over $R$.

**Proof.**
Omitted. Hints: By Lemma 10.137.13 we can check smoothness one prime at a time. Since $\mathop{\mathrm{Spec}}(S)$ is the disjoint union of $\mathop{\mathrm{Spec}}(S')$ and $\mathop{\mathrm{Spec}}(S'')$ by Lemma 10.21.2 we find that smoothness of $R \to S$ at $\mathfrak q$ corresponds to either smoothness of $R \to S'$ at the corresponding prime or smoothness of $R \to S''$ at the corresponding prime.
$\square$

Lemma 10.137.16. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ be a relative global complete intersection. Let $\mathfrak q \subset S$ be a prime. Then $R \to S$ is smooth at $\mathfrak q$ if and only if there exists a subset $I \subset \{ 1, \ldots , n\} $ of cardinality $c$ such that the polynomial

does not map to an element of $\mathfrak q$.

**Proof.**
By Lemma 10.136.12 we see that the naive cotangent complex associated to the given presentation of $S$ is the complex

The maximal minors of the matrix giving the map are exactly the polynomials $g_ I$.

Assume $g_ I$ maps to $g \in S$, with $g \not\in \mathfrak q$. Then the algebra $S_ g$ is smooth over $R$. Namely, its naive cotangent complex is quasi-isomorphic to the complex above localized at $g$, see Lemma 10.134.13. And by construction it is quasi-isomorphic to a free rank $n - c$ module in degree $0$.

Conversely, suppose that all $g_ I$ end up in $\mathfrak q$. In this case the complex above tensored with $\kappa (\mathfrak q)$ does not have maximal rank, and hence there is no localization by an element $g \in S$, $g \not\in \mathfrak q$ where this map becomes a split injection. By Lemma 10.134.13 again there is no such localization which is smooth over $R$. $\square$

Lemma 10.137.17. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. Assume

there exists a $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is of finite presentation,

the local ring homomorphism $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat,

the fibre $S \otimes _ R \kappa (\mathfrak p)$ is smooth over $\kappa (\mathfrak p)$ at the prime corresponding to $\mathfrak q$.

Then $R \to S$ is smooth at $\mathfrak q$.

**Proof.**
By Lemmas 10.136.15 and 10.137.5 we see that there exists a $g \in S$ such that $S_ g$ is a relative global complete intersection. Replacing $S$ by $S_ g$ we may assume $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection. For any subset $I \subset \{ 1, \ldots , n\} $ of cardinality $c$ consider the polynomial $g_ I = \det (\partial f_ j/\partial x_ i)_{j = 1, \ldots , c, i \in I}$ of Lemma 10.137.16. Note that the image $\overline{g}_ I$ of $g_ I$ in the polynomial ring $\kappa (\mathfrak p)[x_1, \ldots , x_ n]$ is the determinant of the partial derivatives of the images $\overline{f}_ j$ of the $f_ j$ in the ring $\kappa (\mathfrak p)[x_1, \ldots , x_ n]$. Thus the lemma follows by applying Lemma 10.137.16 both to $R \to S$ and to $\kappa (\mathfrak p) \to S \otimes _ R \kappa (\mathfrak p)$.
$\square$

Note that the sets $U, V$ in the following lemma are open by definition.

Lemma 10.137.18. Let $R \to S$ be a ring map of finite presentation. Let $R \to R'$ be a flat ring map. Denote $S' = R' \otimes _ R S$ the base change. Let $U \subset \mathop{\mathrm{Spec}}(S)$ be the set of primes at which $R \to S$ is smooth. Let $V \subset \mathop{\mathrm{Spec}}(S')$ the set of primes at which $R' \to S'$ is smooth. Then $V$ is the inverse image of $U$ under the map $f : \mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S)$.

**Proof.**
By Lemma 10.134.8 we see that $\mathop{N\! L}\nolimits _{S/R} \otimes _ S S'$ is homotopy equivalent to $\mathop{N\! L}\nolimits _{S'/R'}$. This already implies that $f^{-1}(U) \subset V$.

Let $\mathfrak q' \subset S'$ be a prime lying over $\mathfrak q \subset S$. Assume $\mathfrak q' \in V$. We have to show that $\mathfrak q \in U$. Since $S \to S'$ is flat, we see that $S_{\mathfrak q} \to S'_{\mathfrak q'}$ is faithfully flat (Lemma 10.39.17). Thus the vanishing of $H_1(L_{S'/R'})_{\mathfrak q'}$ implies the vanishing of $H_1(L_{S/R})_{\mathfrak q}$. By Lemma 10.78.6 applied to the $S_{\mathfrak q}$-module $(\Omega _{S/R})_{\mathfrak q}$ and the map $S_{\mathfrak q} \to S'_{\mathfrak q'}$ we see that $(\Omega _{S/R})_{\mathfrak q}$ is projective. Hence $R \to S$ is smooth at $\mathfrak q$ by Lemma 10.137.12. $\square$

Lemma 10.137.19. Let $K/k$ be a field extension. Let $S$ be a finite type algebra over $k$. Let $\mathfrak q_ K$ be a prime of $S_ K = K \otimes _ k S$ and let $\mathfrak q$ be the corresponding prime of $S$. Then $S$ is smooth over $k$ at $\mathfrak q$ if and only if $S_ K$ is smooth at $\mathfrak q_ K$ over $K$.

**Proof.**
This is a special case of Lemma 10.137.18.
$\square$

Lemma 10.137.20. Let $R$ be a ring and let $I \subset R$ be an ideal. Let $R/I \to \overline{S}$ be a smooth ring map. Then there exists elements $\overline{g}_ i \in \overline{S}$ which generate the unit ideal of $\overline{S}$ such that each $\overline{S}_{g_ i} \cong S_ i/IS_ i$ for some (standard) smooth ring $S_ i$ over $R$.

**Proof.**
By Lemma 10.137.10 we find a collection of elements $\overline{g}_ i \in \overline{S}$ which generate the unit ideal of $\overline{S}$ such that each $\overline{S}_{g_ i}$ is standard smooth over $R/I$. Hence we may assume that $\overline{S}$ is standard smooth over $R/I$. Write $\overline{S} = (R/I)[x_1, \ldots , x_ n]/(\overline{f}_1, \ldots , \overline{f}_ c)$ as in Definition 10.137.6. Choose $f_1, \ldots , f_ c \in R[x_1, \ldots , x_ n]$ lifting $\overline{f}_1, \ldots , \overline{f}_ c$. Set $S = R[x_1, \ldots , x_ n, x_{n + 1}]/(f_1, \ldots , f_ c, x_{n + 1}\Delta - 1)$ where $\Delta = \det (\frac{\partial f_ j}{\partial x_ i})_{i, j = 1, \ldots , c}$ as in Example 10.137.8. This proves the lemma.
$\square$

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