Definition 10.136.1. A ring map $R \to S$ is called *syntomic*, or we say $S$ is a *flat local complete intersection over $R$* if it is flat, of finite presentation, and if all of its fibre rings $S \otimes _ R \kappa (\mathfrak p)$ are local complete intersections, see Definition 10.135.1.

## 10.136 Syntomic morphisms

Syntomic ring maps are flat finitely presented ring maps all of whose fibers are local complete intersections. We discuss general local complete intersection ring maps in More on Algebra, Section 15.33.

Clearly, an algebra over a field is syntomic over the field if and only if it is a local complete intersection. Here is a pleasing feature of this definition.

Lemma 10.136.2. Let $R \to S$ be a ring map. Let $R \to R'$ be a faithfully flat ring map. Set $S' = R'\otimes _ R S$. Then $R \to S$ is syntomic if and only if $R' \to S'$ is syntomic.

**Proof.**
By Lemma 10.126.2 and Lemma 10.39.8 this holds for the property of being flat and for the property of being of finite presentation. The map $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is surjective, see Lemma 10.39.16. Thus it suffices to show given primes $\mathfrak p' \subset R'$ lying over $\mathfrak p \subset R$ that $S \otimes _ R \kappa (\mathfrak p)$ is a local complete intersection if and only if $S' \otimes _{R'} \kappa (\mathfrak p')$ is a local complete intersection. Note that $S' \otimes _{R'} \kappa (\mathfrak p') = S \otimes _ R \kappa (\mathfrak p) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p')$. Thus Lemma 10.135.11 applies.
$\square$

Lemma 10.136.3. Any base change of a syntomic map is syntomic.

**Proof.**
This is true for being flat, for being of finite presentation, and for having local complete intersections as fibres by Lemmas 10.39.7, 10.6.2 and 10.135.11.
$\square$

Lemma 10.136.4. Let $R \to S$ be a ring map. Suppose we have $g_1, \ldots g_ m \in S$ which generate the unit ideal such that each $R \to S_{g_ i}$ is syntomic. Then $R \to S$ is syntomic.

**Proof.**
This is true for being flat and for being of finite presentation by Lemmas 10.39.18 and 10.23.3. The property of having fibre rings which are local complete intersections is local on $S$ by its very definition, see Definition 10.135.1.
$\square$

Definition 10.136.5. Let $R \to S$ be a ring map. We say that $R \to S$ is a *relative global complete intersection* if there exists a presentation $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ and every nonempty fibre of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ has dimension $n - c$. We will say “let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ be a relative global complete intersection” to indicate this situation.

The following lemma is occasionally useful to find global presentations.

Lemma 10.136.6. Let $S$ be a finitely presented $R$-algebra which has a presentation $S = R[x_1, \ldots , x_ n]/I$ such that $I/I^2$ is free over $S$. Then $S$ has a presentation $S = R[y_1, \ldots , y_ m]/(f_1, \ldots , f_ c)$ such that $(f_1, \ldots , f_ c)/(f_1, \ldots , f_ c)^2$ is free with basis given by the classes of $f_1, \ldots , f_ c$.

**Proof.**
Note that $I$ is a finitely generated ideal by Lemma 10.6.3. Let $f_1, \ldots , f_ c \in I$ be elements which map to a basis of $I/I^2$. By Nakayama's lemma (Lemma 10.20.1) there exists a $g \in 1 + I$ such that

and $I_ g \cong (f_1, \ldots , f_ c)_ g$. Hence we see that

as desired. It follows that $f_1, \ldots , f_ c,gx_{n + 1} - 1$ form a basis for $(f_1, \ldots , f_ c, gx_{n + 1} - 1)/(f_1, \ldots , f_ c, gx_{n + 1} - 1)^2$ for example by applying Lemma 10.134.12. $\square$

Example 10.136.7. Let $n , m \geq 1$ be integers. Consider the ring map

In other words, this is the unique ring map of polynomial rings as indicated such that the polynomial factorization

holds. Note that $S$ is generated by $n + m$ elements over $R$ (namely, $b_ i, c_ j$) and that there are $n + m$ equations (namely $a_ k = a_ k(b_ i, c_ j)$). In order to show that $S$ is a relative global complete intersection over $R$ it suffices to prove that all fibres have dimension $0$.

To prove this, let $R \to k$ be a ring map into a field $k$. Say $a_ i$ maps to $\alpha _ i \in k$. Consider the fibre ring $S_ k = k \otimes _ R S$. Let $k \to K$ be a field extension. A $k$-algebra map of $S_ k \to K$ is the same thing as finding $\beta _1, \ldots , \beta _ n, \gamma _1, \ldots , \gamma _ m \in K$ such that

Hence we see there are at most finitely many choices of such $n + m$-tuples in $K$. This proves that all fibres have finitely many closed points (use Hilbert's Nullstellensatz to see they all correspond to solutions in $\overline{k}$ for example) and hence that $R \to S$ is a relative global complete intersection.

Another way to argue this is to show $\mathbf{Z}[a_1, \ldots , a_{n + m}] \to \mathbf{Z}[b_1, \ldots , b_ n, c_1, \ldots , c_ m]$ is actually also a *finite* ring map. Namely, by Lemma 10.38.5 each of $b_ i, c_ j$ is integral over $R$, and hence $R \to S$ is finite by Lemma 10.36.4.

Example 10.136.8. Consider the ring map

In other words this is the unique ring map of polynomial rings as indicated such that

holds in $\mathbf{Z}[\alpha _ i, x]$. Another way to say this is that $a_ i$ maps to the $i$th elementary symmetric function in $\alpha _1, \ldots , \alpha _ n$. Note that $S$ is generated by $n$ elements over $R$ subject to $n$ equations. Hence to show that $S$ is a relative global complete intersection over $R$ we have to show that the fibre rings $S \otimes _ R \kappa (\mathfrak p)$ have dimension $0$. This follows as in Example 10.136.7 because the ring map $\mathbf{Z}[a_1, \ldots , a_ n] \to \mathbf{Z}[\alpha _1, \ldots , \alpha _ n]$ is actually *finite* since each $\alpha _ i \in S$ satisfies the monic equation $x^ n - a_1 x^{n - 1} + \ldots + (-1)^ n a_ n$ over $R$.

Lemma 10.136.9. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ be a relative global complete intersection (Definition 10.136.5)

For any $R \to R'$ the base change $R' \otimes _ R S = R'[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection.

For any $g \in S$ which is the image of $h \in R[x_1, \ldots , x_ n]$ the ring $S_ g = R[x_1, \ldots , x_ n, x_{n + 1}]/(f_1, \ldots , f_ c, hx_{n + 1} - 1)$ is a relative global complete intersection.

If $R \to S$ factors as $R \to R_ f \to S$ for some $f \in R$. Then the ring $S = R_ f[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection over $R_ f$.

**Proof.**
By Lemma 10.116.5 the fibres of a base change have the same dimension as the fibres of the original map. Moreover $R' \otimes _ R R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c) = R'[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$. Thus (1) follows. The proof of (2) is that the localization at one element can be described as $S_ g \cong S[x_{n + 1}]/(gx_{n + 1} - 1)$. Assertion (3) follows from (1) since under the assumptions of (3) we have $R_ f \otimes _ R S \cong S$.
$\square$

Lemma 10.136.10. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$. We will find $h \in R[x_1, \ldots , x_ n]$ which maps to $g \in S$ such that

is a relative global complete intersection with a presentation as in Definition 10.136.5 in each of the following cases:

Let $I \subset R$ be an ideal. If the fibres of $\mathop{\mathrm{Spec}}(S/IS) \to \mathop{\mathrm{Spec}}(R/I)$ have dimension $n - c$, then we can find $(h, g)$ as above such that $g$ maps to $1 \in S/IS$.

Let $\mathfrak p \subset R$ be a prime. If $\dim (S \otimes _ R \kappa (\mathfrak p)) = n - c$, then we can find $(h, g)$ as above such that $g$ maps to a unit of $S \otimes _ R \kappa (\mathfrak p)$.

Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. If $\dim _{\mathfrak q}(S/R) = n - c$, then we can find $(h, g)$ as above such that $g \not\in \mathfrak q$.

**Proof.**
Ad (1). By Lemma 10.125.6 there exists an open subset $W \subset \mathop{\mathrm{Spec}}(S)$ containing $V(IS)$ such that all fibres of $W \to \mathop{\mathrm{Spec}}(R)$ have dimension $\leq n - c$. Say $W = \mathop{\mathrm{Spec}}(S) \setminus V(J)$. Then $V(J) \cap V(IS) = \emptyset $ hence we can find a $g \in J$ which maps to $1 \in S/IS$. Let $h \in R[x_1, \ldots , x_ n]$ be any preimage of $g$.

Ad (2). By Lemma 10.125.6 there exists an open subset $W \subset \mathop{\mathrm{Spec}}(S)$ containing $\mathop{\mathrm{Spec}}(S \otimes _ R \kappa (\mathfrak p))$ such that all fibres of $W \to \mathop{\mathrm{Spec}}(R)$ have dimension $\leq n - c$. Say $W = \mathop{\mathrm{Spec}}(S) \setminus V(J)$. Then $V(J \cdot S \otimes _ R \kappa (\mathfrak p)) = \emptyset $. Hence we can find a $g \in J$ which maps to a unit in $S \otimes _ R \kappa (\mathfrak p)$ (details omitted). Let $h \in R[x_1, \ldots , x_ n]$ be any preimage of $g$.

Ad (3). By Lemma 10.125.6 there exists a $g \in S$, $g \not\in \mathfrak q$ such that all nonempty fibres of $R \to S_ g$ have dimension $\leq n - c$. Let $h \in R[x_1, \ldots , x_ n]$ be any element that maps to $g$. $\square$

The following lemma says we can do absolute Noetherian approximation for relative global complete intersections.

Lemma 10.136.11. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ be a relative global complete intersection (Definition 10.136.5). There exist a finite type $\mathbf{Z}$-subalgebra $R_0 \subset R$ such that $f_ i \in R_0[x_1, \ldots , x_ n]$ and such that

is a relative global complete intersection.

**Proof.**
Let $R_0 \subset R$ be the $\mathbf{Z}$-algebra of $R$ generated by all the coefficients of the polynomials $f_1, \ldots , f_ c$. Let $S_0 = R_0[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$. Clearly, $S = R \otimes _{R_0} S_0$. Pick a prime $\mathfrak q \subset S$ and denote $\mathfrak p \subset R$, $\mathfrak q_0 \subset S_0$, and $\mathfrak p_0 \subset R_0$ the primes it lies over. Because $\dim (S \otimes _ R \kappa (\mathfrak p) ) = n - c$ we also have $\dim (S_0 \otimes _{R_0} \kappa (\mathfrak p_0)) = n - c$, see Lemma 10.116.5. By Lemma 10.125.6 there exists a $g \in S_0$, $g \not\in \mathfrak q_0$ such that all nonempty fibres of $R_0 \to (S_0)_ g$ have dimension $\leq n - c$. As $\mathfrak q$ was arbitrary and $\mathop{\mathrm{Spec}}(S)$ quasi-compact, we can find finitely many $g_1, \ldots , g_ m \in S_0$ such that (a) for $j = 1, \ldots , m$ the nonempty fibres of $R_0 \to (S_0)_{g_ j}$ have dimension $\leq n - c$ and (b) the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(S_0)$ is contained in $D(g_1) \cup \ldots \cup D(g_ m)$. In other words, the images of $g_1, \ldots , g_ m$ in $S = R \otimes _{R_0} S_0$ generate the unit ideal. After increasing $R_0$ we may assume that $g_1, \ldots , g_ m$ generate the unit ideal in $S_0$. By (a) the nonempty fibres of $R_0 \to S_0$ all have dimension $\leq n - c$ and we conclude.
$\square$

Lemma 10.136.12. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ be a relative global complete intersection (Definition 10.136.5). For every prime $\mathfrak q$ of $S$, let $\mathfrak q'$ denote the corresponding prime of $R[x_1, \ldots , x_ n]$. Then

$f_1, \ldots , f_ c$ is a regular sequence in the local ring $R[x_1, \ldots , x_ n]_{\mathfrak q'}$,

each of the rings $R[x_1, \ldots , x_ n]_{\mathfrak q'}/(f_1, \ldots , f_ i)$ is flat over $R$, and

the $S$-module $(f_1, \ldots , f_ c)/(f_1, \ldots , f_ c)^2$ is free with basis given by the elements $f_ i \bmod (f_1, \ldots , f_ c)^2$.

**Proof.**
By Lemma 10.69.2 part (3) follows from part (1).

Assume $R$ is Noetherian. Let $\mathfrak p = R \cap \mathfrak q'$. By Lemma 10.135.4 for example we see that $f_1, \ldots , f_ c$ form a regular sequence in the local ring $R[x_1, \ldots , x_ n]_{\mathfrak q'} \otimes _ R \kappa (\mathfrak p)$. Moreover, the local ring $R[x_1, \ldots , x_ n]_{\mathfrak q'}$ is flat over $R_{\mathfrak p}$. Since $R$, and hence $R[x_1, \ldots , x_ n]_{\mathfrak q'}$ is Noetherian we see from Lemma 10.99.3 that (1) and (2) hold.

Let $R$ be general. Write $R = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } R_\lambda $ as the filtered colimit of finite type $\mathbf{Z}$-subalgebras (compare with Section 10.127). We may assume that $f_1, \ldots , f_ c \in R_\lambda [x_1, \ldots , x_ n]$ for all $\lambda $. Let $R_0 \subset R$ be as in Lemma 10.136.11. Then we may assume $R_0 \subset R_\lambda $ for all $\lambda $. It follows that $S_\lambda = R_\lambda [x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection (as base change of $S_0$ via $R_0 \to R_\lambda $, see Lemma 10.136.9). Denote $\mathfrak p_\lambda $, $\mathfrak q_\lambda $, $\mathfrak q'_\lambda $ the prime of $R_\lambda $, $S_\lambda $, $R_\lambda [x_1, \ldots , x_ n]$ induced by $\mathfrak p$, $\mathfrak q$, $\mathfrak q'$. With this notation, we have (1) and (2) for each $\lambda $. Since

we deduce flatness in (2) over $R$ from Lemma 10.39.6. Since we have

and since filtered colimits are exact (Lemma 10.8.8) we conclude that we have (1). $\square$

Lemma 10.136.13. A relative global complete intersection is syntomic, i.e., flat.

**Proof.**
Let $R \to S$ be a relative global complete intersection. The fibres are global complete intersections, and $S$ is of finite presentation over $R$. Thus the only thing to prove is that $R \to S$ is flat. This is true by (2) of Lemma 10.136.12.
$\square$

Lemma 10.136.14. Suppose that $A$ is a ring, and $P(x) = x^ n + b_1 x^{n-1} + \ldots + b_ n \in A[x]$ is a monic polynomial over $A$. Then there exists a syntomic, finite locally free, faithfully flat ring extension $A \subset A'$ such that $P(x) = \prod _{i = 1, \ldots , n} (x - \beta _ i)$ for certain $\beta _ i \in A'$.

**Proof.**
Take $A' = A \otimes _ R S$, where $R$ and $S$ are as in Example 10.136.8, where $R \to A$ maps $a_ i$ to $b_ i$, and let $\beta _ i = -1 \otimes \alpha _ i$. Observe that $R \to S$ is syntomic (Lemma 10.136.13), $R \to S$ is finite by construction, and $R$ is Noetherian (so any finite $R$-module is finitely presented). Hence $S$ is finite locally free as an $R$-module by Lemma 10.78.2. We omit the verification that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective, which shows that $S$ is faithfully flat over $R$ (Lemma 10.39.16). These properties are inherited by the base change $A \to A'$; some details omitted.
$\square$

Lemma 10.136.15. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. The following are equivalent:

There exists an element $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is syntomic.

There exists an element $g \in S$, $g \not\in \mathfrak q$ such that $S_ g$ is a relative global complete intersection over $R$.

There exists an element $g \in S$, $g \not\in \mathfrak q$, such that $R \to S_ g$ is of finite presentation, the local ring map $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat, and the local ring $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is a complete intersection ring over $\kappa (\mathfrak p)$ (see Definition 10.135.5).

**Proof.**
The implication (1) $\Rightarrow $ (3) is Lemma 10.135.8. The implication (2) $\Rightarrow $ (1) is Lemma 10.136.13. It remains to show that (3) implies (2).

Assume (3). After replacing $S$ by $S_ g$ for some $g \in S$, $g\not\in \mathfrak q$ we may assume $S$ is finitely presented over $R$. Choose a presentation $S = R[x_1, \ldots , x_ n]/I$. Let $\mathfrak q' \subset R[x_1, \ldots , x_ n]$ be the prime corresponding to $\mathfrak q$. Write $\kappa (\mathfrak p) = k$. Note that $S \otimes _ R k = k[x_1, \ldots , x_ n]/\overline{I}$ where $\overline{I} \subset k[x_1, \ldots , x_ n]$ is the ideal generated by the image of $I$. Let $\overline{\mathfrak q}' \subset k[x_1, \ldots , x_ n]$ be the prime ideal generated by the image of $\mathfrak q'$. By Lemma 10.135.8 the equivalent conditions of Lemma 10.135.4 hold for $\overline{I}$ and $\overline{\mathfrak q}'$. Say the dimension of $\overline{I}_{\overline{\mathfrak q}'}/ \overline{\mathfrak q}'\overline{I}_{\overline{\mathfrak q}'}$ over $\kappa (\overline{\mathfrak q}')$ is $c$. Pick $f_1, \ldots , f_ c \in I$ mapping to a basis of this vector space. The images $\overline{f}_ j \in \overline{I}$ generate $\overline{I}_{\overline{\mathfrak q}'}$ (by Lemma 10.135.4). Set $S' = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$. Let $J$ be the kernel of the surjection $S' \to S$. Since $S$ is of finite presentation $J$ is a finitely generated ideal (Lemma 10.6.2). Consider the short exact sequence

As $S_\mathfrak q$ is flat over $R$ we see that $J_{\mathfrak q'} \otimes _ R k \to S'_{\mathfrak q'} \otimes _ R k$ is injective (Lemma 10.39.12). However, by construction $S'_{\mathfrak q'} \otimes _ R k$ maps isomorphically to $S_\mathfrak q \otimes _ R k$. Hence we conclude that $J_{\mathfrak q'} \otimes _ R k = J_{\mathfrak q'}/\mathfrak pJ_{\mathfrak q'} = 0$. By Nakayama's lemma (Lemma 10.20.1) we conclude that there exists a $g \in R[x_1, \ldots , x_ n]$, $g \not\in \mathfrak q'$ such that $J_ g = 0$. In other words $S'_ g \cong S_ g$. After further localizing we see that $S'$ (and hence $S$) becomes a relative global complete intersection by Lemma 10.136.10 as desired. $\square$

Lemma 10.136.16. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]/I$ for some finitely generated ideal $I$. If $g \in S$ is such that $S_ g$ is syntomic over $R$, then $(I/I^2)_ g$ is a finite projective $S_ g$-module.

**Proof.**
By Lemma 10.136.15 there exist finitely many elements $g_1, \ldots , g_ m \in S$ which generate the unit ideal in $S_ g$ such that each $S_{gg_ j}$ is a relative global complete intersection over $R$. Since it suffices to prove that $(I/I^2)_{gg_ j}$ is finite projective, see Lemma 10.78.2, we may assume that $S_ g$ is a relative global complete intersection. In this case the result follows from Lemmas 10.134.16 and 10.136.12.
$\square$

Lemma 10.136.17. Let $R \to S$, $S \to S'$ be ring maps.

If $R \to S$ and $S \to S'$ are syntomic, then $R \to S'$ is syntomic.

If $R \to S$ and $S \to S'$ are relative global complete intersections, then $R \to S'$ is a relative global complete intersection.

**Proof.**
Proof of (2). Say $R \to S$ and $S \to S'$ are relative global complete intersections and we have presentations $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ and $S' = S[y_1, \ldots , y_ m]/(h_1, \ldots , h_ d)$ as in Definition 10.136.5. Then

for some lifts $h_ j' \in R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]$ of the $h_ j$. Hence it suffices to bound the dimensions of the fibre rings. Thus we may assume $R = k$ is a field. In this case we see that we have a ring, namely $S$, which is of finite type over $k$ and equidimensional of dimension $n - c$, and a finite type ring map $S \to S'$ all of whose nonempty fibre rings are equidimensional of dimension $m - d$. Then, by Lemma 10.112.6 for example applied to localizations at maximal ideals of $S'$, we see that $\dim (S') \leq n - c + m - d$ as desired.

We will reduce part (1) to part (2). Assume $R \to S$ and $S \to S'$ are syntomic. Let $\mathfrak q' \subset S$ be a prime ideal lying over $\mathfrak q \subset S$. By Lemma 10.136.15 there exists a $g' \in S'$, $g' \not\in \mathfrak q'$ such that $S \to S'_{g'}$ is a relative global complete intersection. Similarly, we find $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is a relative global complete intersection. By Lemma 10.136.9 the ring map $S_ g \to S_{gg'}$ is a relative global complete intersection. By part (2) we see that $R \to S_{gg'}$ is a relative global complete intersection and $gg' \not\in \mathfrak q'$. Since $\mathfrak q'$ was arbitrary combining Lemmas 10.136.15 and 10.136.4 we see that $R \to S'$ is syntomic (this also uses that the spectrum of $S'$ is quasi-compact, see Lemma 10.17.10). $\square$

The following lemma will be improved later, see Smoothing Ring Maps, Proposition 16.3.2.

Lemma 10.136.18. Let $R$ be a ring and let $I \subset R$ be an ideal. Let $R/I \to \overline{S}$ be a syntomic 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 relative global complete intersection $S_ i$ over $R$.

**Proof.**
By Lemma 10.136.15 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 a relative global complete intersection over $R/I$. Hence we may assume that $\overline{S}$ is a relative global complete intersection. Write $\overline{S} = (R/I)[x_1, \ldots , x_ n]/(\overline{f}_1, \ldots , \overline{f}_ c)$ as in Definition 10.136.5. Choose $f_1, \ldots , f_ c \in R[x_1, \ldots , x_ n]$ lifting $\overline{f}_1, \ldots , \overline{f}_ c$. Set $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$. Note that $S/IS \cong \overline{S}$. By Lemma 10.136.10 we can find $g \in S$ mapping to $1$ in $\overline{S}$ such that $S_ g$ is a relative global complete intersection over $R$. Since $\overline{S} \cong S_ g/IS_ g$ this finishes the proof.
$\square$

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