The Stacks project

Proof. Choose a presentation $A = R[x_1, \ldots , x_ n]/I$ and write $I = (f_1, \ldots , f_ m)$. Define the $A$-module $K$ by the short exact sequence

where the $j$th basis vector $e_ j$ in the middle is mapped to the class of $f_ j$ on the right. Set

The retraction is just the projection onto the degree $0$ part of $C$. We have a surjection $R[x_1, \ldots , x_ n, y_1, \ldots , y_ m] \to C$ which maps $y_ j$ to the class of $f_ j$ in $I/I^2$. The kernel $J$ of this map is generated by the elements $f_1, \ldots , f_ m$ and by elements $\sum h_ j y_ j$ with $h_ j \in R[x_1, \ldots , x_ n]$ such that $\sum h_ j e_ j$ defines an element of $K$. By Algebra, Lemma 10.134.4 applied to $R \to A \to C$ and the presentations above and More on Algebra, Lemma 15.9.12 there is a short exact sequence

of $C$-modules. Let $h \in R[x_1, \ldots , x_ n]$ be an element with image $a \in A$. We will use as presentations for the localized rings

where $I' = (hx_0 - 1, I)$ and $J' = (hx_0 - 1, J)$. Hence $I'/(I')^2 = A_ a \oplus (I/I^2)_ a$ as $A_ a$-modules and $J'/(J')^2 = C_ a \oplus (J/J^2)_ a$ as $C_ a$-modules. Thus we obtain

as the sequence of Algebra, Lemma 10.134.4 corresponding to $R \to A_ a \to C_ a$ and the presentations above.

Next, assume that $a \in A$ is such that $A_ a$ is a local complete intersection over $R$. Then $(I/I^2)_ a$ is finite projective over $A_ a$, see More on Algebra, Lemma 15.32.3. Hence we see $K_ a \oplus (I/I^2)_ a \cong A_ a^{\oplus m}$ is free. In particular $K_ a$ is finite projective too. By More on Algebra, Lemma 15.33.6 the sequence (16.3.1.2) is exact on the left. Hence

This proves (1). Finally, suppose that in addition $A_ a$ is smooth over $R$. Then the same presentation shows that $\Omega _{C_ a/R}$ is the cokernel of the map

The summand $C_ a$ of $J'/(J')^2$ in the decomposition above corresponds to $hx_0 - 1$ and hence maps isomorphically to the summand $C_ a\text{d}x_0$. The summand $I/I^2 \otimes _ A C_ a$ of $J'/(J')^2$ maps injectively to $\bigoplus _{i = 1, \ldots , n} C_ a\text{d}x_ i$ with quotient $\Omega _{A_ a/R} \otimes _{A_ a} C_ a$. The summand $K \otimes _ A C_ a$ maps injectively to $\bigoplus _{j \geq 1} C_ a\text{d}y_ j$ with quotient isomorphic to $I/I^2 \otimes _ A C_ a$. Thus the cokernel of the last displayed map is the module $I/I^2 \otimes _ A C_ a \oplus \Omega _{A_ a/R} \otimes _{A_ a} C_ a$. Since $(I/I^2)_ a \oplus \Omega _{A_ a/R}$ is free (from the definition of smooth ring maps) we see that (2) holds. $\square$

Proof. Assume $R_0 \to A_0$ syntomic, in particular a local complete intersection (More on Algebra, Lemma 15.33.5). Choose a presentation $A_0 = R_0[x_1, \ldots , x_ n]/J_0$. Set $C_0 = \text{Sym}^*_{A_0}(J_0/J_0^2)$. Note that $J_0/J_0^2$ is a finite projective $A_0$-module (Algebra, Lemma 10.136.16). By Lemma 16.3.1 the ring map $A_0 \to C_0$ is smooth and we can find a presentation $C_0 = R_0[y_1, \ldots , y_ m]/K_0$ with $K_0/K_0^2$ free over $C_0$. By Algebra, Lemma 10.136.6 we can assume $C_0 = R_0[y_1, \ldots , y_ m]/(\overline{f}_1, \ldots , \overline{f}_ c)$ where $\overline{f}_1, \ldots , \overline{f}_ c$ maps to a basis of $K_0/K_0^2$ over $C_0$. Choose $f_1, \ldots , f_ c \in R[y_1, \ldots , y_ c]$ lifting $\overline{f}_1, \ldots , \overline{f}_ c$ and set

By construction $C_0 = C/IC$. By Algebra, Lemma 10.136.10 we can after replacing $C$ by $C_ g$ assume that $C$ is a relative global complete intersection over $R$. We conclude that there exists a finite projective $A_0$-module $P_0$ such that $C_0 = \text{Sym}^*_{A_0}(P_0)$ is isomorphic to $C/IC$ for some syntomic $R$-algebra $C$.

Choose an integer $n$ and a direct sum decomposition $A_0^{\oplus n} = P_0 \oplus Q_0$. By More on Algebra, Lemma 15.9.11 we can find an étale ring map $C \to C'$ which induces an isomorphism $C/IC \to C'/IC'$ and a finite projective $C'$-module $Q$ such that $Q/IQ$ is isomorphic to $Q_0 \otimes _{A_0} C/IC$. Then $D = \text{Sym}_{C'}^*(Q)$ is a smooth $C'$-algebra (see More on Algebra, Lemma 15.9.13). Picture

Observe that our choice of $Q$ gives

Choose $f_1, \ldots , f_ n \in D$ which map to $x_1, \ldots , x_ n$ in $D/ID = A_0[x_1, \ldots , x_ n]$. Set $A = D/(f_1, \ldots , f_ n)$. Note that $A_0 = A/IA$. We claim that $R \to A$ is syntomic in a neighbourhood of $V(IA)$. If the claim is true, then we can find a $f \in A$ mapping to $1 \in A_0$ such that $A_ f$ is syntomic over $R$ and the proof of (1) is finished.

Proof of the claim. Observe that $R \to D$ is syntomic as a composition of the syntomic ring map $R \to C$, the étale ring map $C \to C'$ and the smooth ring map $C' \to D$ (Algebra, Lemmas 10.136.17 and 10.137.10). The question is local on $\mathop{\mathrm{Spec}}(D)$, hence we may assume that $D$ is a relative global complete intersection (Algebra, Lemma 10.136.15). Say $D = R[y_1, \ldots , y_ m]/(g_1, \ldots , g_ s)$. Let $f'_1, \ldots , f'_ n \in R[y_1, \ldots , y_ m]$ be lifts of $f_1, \ldots , f_ n$. Then we can apply Algebra, Lemma 10.136.10 to get the claim.

Proof of (2). Since a smooth ring map is syntomic, we can find a syntomic ring map $R \to A$ such that $A_0 = A/IA$. By assumption the fibres of $R \to A$ are smooth over primes in $V(I)$ hence $R \to A$ is smooth in an open neighbourhood of $V(IA)$ (Algebra, Lemma 10.137.17). Thus we can replace $A$ by a localization to obtain the result we want. $\square$

Proof. Apply Lemma 16.3.1 to get $A \to C$. By Algebra, Lemma 10.136.6 we can write $C = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ with $f_ i$ mapping to a basis of $J/J^2$. The ring map $R \to C$ is syntomic (hence flat) as it is a composition of a syntomic and a smooth ring map. The dimension of the fibres is $n - c$ by Algebra, Lemma 10.135.4 (the fibres are local complete intersections, so the lemma applies). $\square$

Proof. Apply Lemma 16.3.3 to get a smooth $R$-algebra map $A \to C$ with a retraction such that $C = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection over $R$. As $C$ is smooth over $R$ we have a short exact sequence

Since $\Omega _{C/R}$ is a projective $C$-module this sequence is split. Choose a left inverse $t$ to the first map. Say $t(\text{d}x_ i) = \sum c_{ij} f_ j$ so that $\sum _ i \frac{\partial f_ j}{\partial x_ i} c_{i\ell } = \delta _{j\ell }$ (Kronecker delta). Let

The $R$-algebra map $C \to B'$ has a retraction given by mapping $y_ j$ to zero. We claim that the map

is étale at every point in the image of $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(B')$. In $\Omega _{B'/R[z_1, \ldots , z_ n]}$ we have

Since $0 = \text{d}z_ i = \text{d}x_ i$ modulo $\sum B'\text{d}y_ j + (y_1, \ldots , y_ c)\Omega _{B'/R[z_1, \ldots , z_ n]}$ we conclude that

As $\Omega _{B'/R[z_1, \ldots , z_ n]}$ is a finite $B'$-module by Nakayama's lemma there exists a $g \in 1 + (y_1, \ldots , y_ c)$ that $(\Omega _{B'/R[z_1, \ldots , z_ n]})_ g = 0$. This proves that $R[z_1, \ldots , z_ n] \to B'_ g$ is unramified, see Algebra, Definition 10.151.1. For any ring map $R \to k$ where $k$ is a field we obtain an unramified ring map $k[z_1, \ldots , z_ n] \to (B'_ g) \otimes _ R k$ between smooth $k$-algebras of dimension $n$. It follows that $k[z_1, \ldots , z_ n] \to (B'_ g) \otimes _ R k$ is flat by Algebra, Lemmas 10.128.1 and 10.140.2. By the critère de platitude par fibre (Algebra, Lemma 10.128.8) we conclude that $R[z_1, \ldots , z_ n] \to B'_ g$ is flat. Finally, Algebra, Lemma 10.143.7 implies that $R[z_1, \ldots , z_ n] \to B'_ g$ is étale. Set $B = B'_ g$. Note that $C \to B$ is smooth and has a retraction, so also $A \to B$ is smooth and has a retraction. Moreover, $R[z_1, \ldots , z_ n] \to B$ is étale. By Algebra, Lemma 10.143.2 we can write

with $\det (\partial g_ j/\partial w_ i)$ invertible in $B$. This proves the lemma. $\square$

Proof. Let $A \to \Lambda $ be an $R$-algebra map with $A$ of finite presentation over $R$. According to Algebra, Lemma 10.127.4 we have to factor this map through a standard smooth algebra, and we know we can factor it as $A \to B \to \Lambda $ with $B$ smooth over $R$. Choose an $R$-algebra map $B \to C$ with a retraction $C \to B$ such that $C$ is standard smooth over $R$, see Lemma 16.3.4. Then the desired factorization is $A \to B \to C \to B \to \Lambda $. $\square$

Proof. Choose a presentation $A = R[y_1, \ldots , y_ m]/(g_1, \ldots , g_ d)$ such that the image of $\det (\partial g_ j/\partial y_ i)_{i, j = 1, \ldots , d}$ is invertible in $A$. Choose an enumerations $E = \{ a_1, \ldots , a_ n\} $ and choose $h_ i \in R[y_1, \ldots , y_ m]$ whose image in $A$ is $a_ i$. Consider the presentation

and set $c = n + d$. $\square$

Proof. Part (a) is clear from the definitions and Algebra, Lemma 10.137.7. Part (b) is clear from Definition 16.2.3.

Proof of (c). Choose a presentation $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ such that (16.2.3.1) and (16.2.3.2) hold. Choose $h \in R[x_1, \ldots , x_ n]$ mapping to $a$. Then

Write $J = (x_0h - 1, f_1, \ldots , f_ m)$. By (16.2.3.2) we see that the $A_ a$-module $J/J^2$ is generated by $x_0h - 1, f_1, \ldots , f_ c$ over $A_ a$. Hence, as in the proof of Algebra, Lemma 10.136.6, we can choose a $g \in 1 + J$ such that

At this point (16.2.3.1) implies that $R \to A_ a$ is standard smooth (use the coordinates $x_0, x_1, \ldots , x_ c, x_{n + 1}$ to take derivatives).

Proof of (d). Choose a presentation $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ such that (16.2.3.3) and (16.2.3.4) hold. Write $I = (f_1, \ldots , f_ m)$. We already know that $A_ a$ is smooth over $R$, see Lemma 16.2.5. By Lemma 16.2.4 we see that $(I/I^2)_ a$ is free on $f_1, \ldots , f_ c$ and maps isomorphically to a direct summand of $\bigoplus A_ a \text{d}x_ i$. Since $\Omega _{A_ a/R} = (\Omega _{A/R})_ a$ is the cokernel of the map $(I/I^2)_ a \to \bigoplus A_ a \text{d}x_ i$ we conclude that it is stably free.

Proof of (e). Choose a presentation $A = R[x_1, \ldots , x_ n]/I$ with $I$ finitely generated. By assumption we have a short exact sequence

which is split exact. Hence we see that $(I/I^2)_ a \oplus \Omega _{A_ a/R}$ is a free $A_ a$-module. Since $\Omega _{A_ a/R}$ is stably free we see that $(I/I^2)_ a$ is stably free as well. Thus replacing the presentation chosen above by $A = R[x_1, \ldots , x_ n, x_{n + 1}, \ldots , x_{n + r}]/J$ with $J = (I, x_{n + 1}, \ldots , x_{n + r})$ for some $r$ we get that $(J/J^2)_ a$ is (finite) free. Choose $f_1, \ldots , f_ c \in J$ which map to a basis of $(J/J^2)_ a$. Extend this to a list of generators $f_1, \ldots , f_ m \in J$. Consider the presentation $A = R[x_1, \ldots , x_{n + r}]/(f_1, \ldots , f_ m)$. Then (16.2.3.4) holds for $a^ e$ for all sufficiently large $e$ by construction. Moreover, since $(J/J^2)_ a \to \bigoplus \nolimits _{i = 1, \ldots , n + r} A_ a\text{d}x_ i$ is a split injection we can find an $A_ a$-linear left inverse. Writing this left inverse in terms of the basis $f_1, \ldots , f_ c$ and clearing denominators we find a linear map $\psi _0 : A^{\oplus n + r} \to A^{\oplus c}$ such that

is multiplication by $a^{e_0}$ for some $e_0 \geq 1$. By Lemma 16.2.4 we see (16.2.3.3) holds for all $a^{ce_0}$ and hence for $a^ e$ for all $e$ with $e \geq ce_0$.

Proof of (f). Choose a presentation $A_ a = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ such that $\det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c}$ is invertible in $A_ a$. We may assume that for some $m < n$ the classes of the elements $x_1, \ldots , x_ m$ correspond $a_ i/1$ where $a_1, \ldots , a_ m \in A$ are generators of $A$ over $R$, see Lemma 16.3.6. After replacing $x_ i$ by $a^ Nx_ i$ for $m < i \leq n$ we may assume the class of $x_ i$ is $a_ i/1 \in A_ a$ for some $a_ i \in A$. Consider the ring map

This is a surjective ring map. By replacing $f_ j$ by $a^ Nf_ j$ we may assume that $f_ j \in R[x_1, \ldots , x_ n]$ and that $\Psi (f_ j) = 0$ (since after all $f_ j(a_1/1, \ldots , a_ n/1) = 0$ in $A_ a$). Let $J = \mathop{\mathrm{Ker}}(\Psi )$. Then $A = R[x_1, \ldots , x_ n]/J$ is a presentation and $f_1, \ldots , f_ c \in J$ are elements such that $(J/J^2)_ a$ is freely generated by $f_1, \ldots , f_ c$ and such that $\det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c}$ maps to an invertible element of $A_ a$. It follows that (16.2.3.1) and (16.2.3.2) hold for $a^ e$ and all large enough $e$ as desired. $\square$