The Stacks project

Lemma 49.10.1. Let $f : Y \to X$ be a morphism of schemes. The following are equivalent

  1. $f$ is locally quasi-finite and syntomic,

  2. $f$ is locally quasi-finite, flat, and a local complete intersection morphism,

  3. $f$ is locally quasi-finite, flat, locally of finite presentation, and the fibres of $f$ are local complete intersections,

  4. $f$ is locally quasi-finite and for every $y \in Y$ there are affine opens $y \in V = \mathop{\mathrm{Spec}}(B) \subset Y$, $U = \mathop{\mathrm{Spec}}(A) \subset X$ with $f(V) \subset U$ an integer $n$ and $h, f_1, \ldots , f_ n \in A[x_1, \ldots , x_ n]$ such that $B = A[x_1, \ldots , x_ n, 1/h]/(f_1, \ldots , f_ n)$,

  5. for every $y \in Y$ there are affine opens $y \in V = \mathop{\mathrm{Spec}}(B) \subset Y$, $U = \mathop{\mathrm{Spec}}(A) \subset X$ with $f(V) \subset U$ such that $A \to B$ is a relative global complete intersection of the form $B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$,

  6. $f$ is locally quasi-finite, flat, locally of finite presentation, and $\mathop{N\! L}\nolimits _{X/Y}$ has tor-amplitude in $[-1, 0]$, and

  7. $f$ is flat, locally of finite presentation, $\mathop{N\! L}\nolimits _{X/Y}$ is perfect of rank $0$ with tor-amplitude in $[-1, 0]$,

Proof. The equivalence of (1) and (2) is More on Morphisms, Lemma 37.59.8. The equivalence of (1) and (3) is Morphisms, Lemma 29.30.11.

If $A \to B$ is as in (4), then $B = A[x, x_1, \ldots , x_ n]/(xh - 1, f_1, \ldots , f_ n]$ is a relative global complete intersection by see Algebra, Definition 10.136.5. Thus (4) implies (5). It is clear that (5) implies (4).

Condition (5) implies (1): by Algebra, Lemma 10.136.14 a relative global complete intersection is syntomic and the definition of a relative global complete intersection guarantees that a relative global complete intersection on $n$ variables with $n$ equations is quasi-finite, see Algebra, Definition 10.136.5 and Lemma 10.122.2.

Either Algebra, Lemma 10.136.15 or Morphisms, Lemma 29.30.10 shows that (1) implies (5).

More on Morphisms, Lemma 37.59.17 shows that (6) is equivalent to (1). If the equivalent conditions (1) – (6) hold, then we see that affine locally $Y \to X$ is given by a relative global complete intersection $B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$ with the same number of variables as the number of equations. Using this presentation we see that

\[ \mathop{N\! L}\nolimits _{B/A} =\left( (f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2 \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} B \text{d} x_ i\right) \]

By Algebra, Lemma 10.136.13 the module $(f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2$ is free with generators the congruence classes of the elements $f_1, \ldots , f_ n$. Thus $\mathop{N\! L}\nolimits _{B/A}$ has rank $0$ and so does $\mathop{N\! L}\nolimits _{Y/X}$. In this way we see that (1) – (6) imply (7).

Finally, assume (7). By More on Morphisms, Lemma 37.59.17 we see that $f$ is syntomic. Thus on suitable affine opens $f$ is given by a relative global complete intersection $A \to B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$, see Morphisms, Lemma 29.30.10. Exactly as above we see that $\mathop{N\! L}\nolimits _{B/A}$ is a perfect complex of rank $n - m$. Thus $n = m$ and we see that (5) holds. This finishes the proof. $\square$


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