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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 (16.2.3.1) and (16.2.3.2) 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 (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

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 (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

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 (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

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 (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

\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 (16.2.3.1) and (16.2.3.2) 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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