Lemma 23.11.4. Let $A \to B$ be a perfect (More on Algebra, Definition 15.82.1) ring homomorphism of Noetherian rings. Then the following are equivalent

1. $\mathop{N\! L}\nolimits _{B/A}$ has tor-amplitude in $[-1, 0]$,

2. $\mathop{N\! L}\nolimits _{B/A}$ is a perfect object of $D(B)$ with tor-amplitude in $[-1, 0]$, and

3. $A \to B$ is a local complete intersection (More on Algebra, Definition 15.33.2).

Proof. Write $B = A[x_1, \ldots , x_ n]/I$. Then $\mathop{N\! L}\nolimits _{B/A}$ is represented by the complex

$I/I^2 \longrightarrow \bigoplus B \text{d}x_ i$

of $B$-modules with $I/I^2$ placed in degree $-1$. Since the term in degree $0$ is finite free, this complex has tor-amplitude in $[-1, 0]$ if and only if $I/I^2$ is a flat $B$-module, see More on Algebra, Lemma 15.66.2. Since $I/I^2$ is a finite $B$-module and $B$ is Noetherian, this is true if and only if $I/I^2$ is a finite locally free $B$-module (Algebra, Lemma 10.78.2). Thus the equivalence of (1) and (2) is clear. Moreover, the equivalence of (1) and (3) also follows if we apply Proposition 23.11.3 (and the observation that a regular ideal is a Koszul regular ideal as well as a quasi-regular ideal, see More on Algebra, Section 15.32). $\square$

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