The Stacks project

Lemma 16.3.7. Let $R \to A$ be a ring map of finite presentation. Let $a \in A$. Consider the following conditions on $a$:

  1. $A_ a$ is smooth over $R$,

  2. $A_ a$ is smooth over $R$ and $\Omega _{A_ a/R}$ is stably free,

  3. $A_ a$ is smooth over $R$ and $\Omega _{A_ a/R}$ is free,

  4. $A_ a$ is standard smooth over $R$,

  5. $a$ is strictly standard in $A$ over $R$,

  6. $a$ is elementary standard in $A$ over $R$.

Then we have

  1. (4) $\Rightarrow $ (3) $\Rightarrow $ (2) $\Rightarrow $ (1),

  2. (6) $\Rightarrow $ (5),

  3. (6) $\Rightarrow $ (4),

  4. (5) $\Rightarrow $ (2),

  5. (2) $\Rightarrow $ the elements $a^ e$, $e \geq e_0$ are strictly standard in $A$ over $R$,

  6. (4) $\Rightarrow $ the elements $a^ e$, $e \geq e_0$ are elementary standard in $A$ over $R$.

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 ( and ( hold. Choose $h \in R[x_1, \ldots , x_ n]$ mapping to $a$. Then

\[ A_ a = R[x_0, x_1, \ldots , x_ n]/(x_0h - 1, f_1, \ldots , f_ m). \]

Write $J = (x_0h - 1, f_1, \ldots , f_ m)$. By ( 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

\[ A_ a = R[x_0, \ldots , x_ n, x_{n + 1}]/ (x_0h - 1, f_1, \ldots , f_ m, gx_{n + 1} - 1). \]

At this point ( 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 ( and ( 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

\[ 0 \to (I/I^2)_ a \to \bigoplus \nolimits _{i = 1, \ldots , n} A_ a\text{d}x_ i \to \Omega _{A_ a/R} \to 0 \]

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 ( 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

\[ A^{\oplus c} \xrightarrow {(f_1, \ldots , f_ c)} J/J^2 \xrightarrow {f \mapsto \text{d}f} \bigoplus \nolimits _{i = 1, \ldots , n + r} A \text{d}x_ i \xrightarrow {\psi _0} A^{\oplus c} \]

is multiplication by $a^{e_0}$ for some $e_0 \geq 1$. By Lemma 16.2.4 we see ( 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

\[ \Psi : R[x_1, \ldots , x_ n] \longrightarrow A,\quad x_ i \longmapsto a_ i. \]

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 ( and ( hold for $a^ e$ and all large enough $e$ as desired. $\square$

Comments (4)

Comment #2659 by Anonymous on

Proof of d) Why is there a presentation of with being free?

Proof of e) It should say that is a split injection. Also in the sequence below the sum should be from and should map from .

Comment #2675 by on

OK, I agree that the proof of (d) wasn't optimal. Also, thanks for pointing out the problem with indices in the proof of (e). The fixes are here.

Comment #2676 by Anonymous on

Oh, I forgot, in the proof of c): A few times it says instead of .

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