Lemma 57.17.1. Let $(R, \mathfrak m, \kappa ) \to (A, \mathfrak n, \lambda )$ be a flat local ring homorphism of local rings which is essentially of finite presentation. Let $\overline{f}_1, \ldots , \overline{f}_ r \in \mathfrak n/\mathfrak m A \subset A/\mathfrak m A$ be a regular sequence. Let $K \in D(A)$. Assume

1. $K$ is perfect,

2. $K \otimes _ A^\mathbf {L} A/\mathfrak m A$ is isomorphic in $D(A/\mathfrak m A)$ to the Koszul complex on $\overline{f}_1, \ldots , \overline{f}_ r$.

Then $K$ is isomorphic in $D(A)$ to a Koszul complex on a regular sequence $f_1, \ldots , f_ r \in A$ lifting the given elements $\overline{f}_1, \ldots , \overline{f}_ r$. Moreover, $A/(f_1, \ldots , f_ r)$ is flat over $R$.

Proof. Let us use chain complexes in the proof of this lemma. The Koszul complex $K_\bullet (\overline{f}_1, \ldots , \overline{f}_ r)$ is defined in More on Algebra, Definition 15.28.2. By More on Algebra, Lemma 15.75.4 we can represent $K$ by a complex

$K_\bullet : A \to A^{\oplus r} \to \ldots \to A^{\oplus r} \to A$

whose tensor product with $A/\mathfrak mA$ is equal (!) to $K_\bullet (\overline{f}_1, \ldots , \overline{f}_ r)$. Denote $f_1, \ldots , f_ r \in A$ the components of the arrow $A^{\oplus r} \to A$. These $f_ i$ are lifts of the $\overline{f}_ i$. By Algebra, Lemma 10.128.6 $f_1, \ldots , f_ r$ form a regular sequence in $A$ and $A/(f_1, \ldots , f_ r)$ is flat over $R$. Let $J = (f_1, \ldots , f_ r) \subset A$. Consider the diagram

$\xymatrix{ K_\bullet \ar[rd] \ar@{..>}[rr]_{\varphi _\bullet } & & K_\bullet (f_1, \ldots , f_ r) \ar[ld] \\ & A/J }$

Since $f_1, \ldots , f_ r$ is a regular sequence the south-west arrow is a quasi-isomorphism (see More on Algebra, Lemma 15.30.2). Hence we can find the dotted arrow making the diagram commute for example by Algebra, Lemma 10.71.4. Reducing modulo $\mathfrak m$ we obtain a commutative diagram

$\xymatrix{ K_\bullet (\overline{f}_1, \ldots , \overline{f}_ r) \ar[rd] \ar[rr]_{\overline{\varphi }_\bullet } & & K_\bullet (\overline{f}_1, \ldots , \overline{f}_ r) \ar[ld] \\ & (A/\mathfrak m A)/(\overline{f}_1, \ldots , \overline{f}_ r) }$

by our choice of $K_\bullet$. Thus $\overline{\varphi }$ is an isomorphism in the derived category $D(A/\mathfrak m A)$. It follows that $\overline{\varphi } \otimes _{A/\mathfrak m A}^\mathbf {L} \lambda$ is an isomorphism. Since $\overline{f}_ i \in \mathfrak n / \mathfrak m A$ we see that

$\text{Tor}_ i^{A/\mathfrak m A}( K_\bullet (\overline{f}_1, \ldots , \overline{f}_ r), \lambda ) = K_ i(\overline{f}_1, \ldots , \overline{f}_ r) \otimes _{A/\mathfrak m A} \lambda$

Hence $\varphi _ i \bmod \mathfrak n$ is invertible. Since $A$ is local this means that $\varphi _ i$ is an isomorphism and the proof is complete. $\square$

Comment #5408 by Shogōki on

In the first 4 lines of proof, all the "R"'s should be replaced by "A".

Comment #5409 by Shogōki on

When demonstrating the existence of \phi, the sentence perhaps should read: "Since f1,…,fr is a regular sequence the \emph{south-west} arrow is a quasi-isomorphism" Also, you referred to Lemma 00LS to construct \phi, but that Lemma is for resolutions, and we haven't shown (yet) that K_{\bullet} is a resolution of A/J, we just know A/J is its \pi_0. (This is only a small issue, as K_{\bullet}(F_i) is a resolution of A/J, and K_{\bullet} consists of free things, we can just pick the arrows \phi_{\bullet} one at a time.)

Comment #5639 by on

OK, I fixed the typos and replaced east by west. The issue pointed out by you about lifting the map is fixed because I just upgraded Lemma 10.71.4 due to an issue mentioned by Rankeya. Thanks! See this corresponding commit.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).