Lemma 49.11.1. Let $f : Y \to X$ be a morphism of schemes. The following are equivalent

1. $f$ is finite and syntomic,

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

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

4. $f$ is finite and for every $x \in X$ there is an affine open $x \in U = \mathop{\mathrm{Spec}}(A) \subset X$ an integer $n$ and $f_1, \ldots , f_ n \in A[x_1, \ldots , x_ n]$ such that $f^{-1}(U)$ is isomorphic to the spectrum of $A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$,

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

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

Proof. The equivalence of (1), (2), (3), (5), and (6) and the implication (4) $\Rightarrow$ (1) follow immediately from Lemma 49.10.1. Assume the equivalent conditions (1), (2), (3), (5), (6) hold. Choose a point $x \in X$ and an affine open $U = \mathop{\mathrm{Spec}}(A)$ of $x$ in $X$ and say $x$ corresponds to the prime ideal $\mathfrak p \subset A$. Write $f^{-1}(U) = \mathop{\mathrm{Spec}}(B)$. Write $B = A[x_1, \ldots , x_ n]/I$. Since $\mathop{N\! L}\nolimits _{B/A}$ is perfect of tor-amplitude in $[-1, 0]$ by (6) we see that $I/I^2$ is a finite locally free $B$-module of rank $n$. Since $B_\mathfrak p$ is semi-local we see that $(I/I^2)_\mathfrak p$ is free of rank $n$, see Algebra, Lemma 10.78.7. Thus after replacing $A$ by a principal localization at an element not in $\mathfrak p$ we may assume $I/I^2$ is a free $B$-module of rank $n$. Thus by Algebra, Lemma 10.136.6 we can find a presentation of $B$ over $A$ with the same number of variables as equations. In other words, we may assume $B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$. This proves (4). $\square$

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