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Characterization of syntomic ring maps

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aisejohan committed Sep 11, 2019
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@@ -2550,7 +2550,103 @@ \section{Freeness of the conormal module}
More on Algebra, Section \ref{more-algebra-section-ideals}).

Consider a cocartesian diagram of rings
B \ar[r] & B' \\
A \ar[r] \ar[u] & A' \ar[u]
If $B$ is flat over $A$, then the canonical map
$\NL_{B/A} \otimes_B B' \to \NL_{B'/A'}$ is a quasi-isomorphism.
If in addition $\NL_{B/A}$ has tor-amplitude in $[-1, 0]$
then $\NL_{B/A} \otimes_B^\mathbf{L} B' \to \NL_{B'/A'}$
is a quasi-isomorphism too.

Choose a presentation $\alpha : P \to B$ as in
Algebra, Section \ref{algebra-section-netherlander}.
Let $I = \Ker(\alpha)$. Set $P' = P \otimes_A A'$ and denote
$\alpha' : P' \to B'$ the corresponding presentation of $B'$ over $A'$.
As $B$ is flat over $A$ we see that $I' = \Ker(\alpha')$ is equal
to $I \otimes_A A'$. Hence
I'/(I')^2 = \Coker(I^2 \otimes_A A' \to I \otimes_A A') =
I/I^2 \otimes_A A' = I/I^2 \otimes_B B'
We have $\Omega_{P'/A'} = \Omega_{P/A} \otimes_A A'$
because both sides have the same basis. It follows that
$\Omega_{P'/A'} \otimes_{P'} B' = \Omega_{P/A} \otimes_P B \otimes_B B'$.
This proves that $\NL(\alpha) \otimes_B B' \to \NL(\alpha')$
is a quasi-isomorphism and hence the first statement holds.

We have
\NL(\alpha) = I/I^2 \longrightarrow \Omega_{P/A} \otimes_P B
as a complex of $B$-modules with $I/I^2$ placed in degree $-1$.
Since the term in degree $0$ is 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 \ref{more-algebra-lemma-last-one-flat}.
If this holds, then $\NL(\alpha) \otimes_B^\mathbf{L} B' =
\NL(\alpha) \otimes_B B'$ and we get the second statement.

Let $A \to B$ be a perfect (More on Algebra, Definition
ring homomorphism of Noetherian rings. Then the following are equivalent
\item $\NL_{B/A}$ has tor-amplitude in $[-1, 0]$,
\item $\NL_{B/A}$ is a perfect object of $D(B)$
with tor-amplitude in $[-1, 0]$,
\item $A \to B$ is syntomic
(Algebra, Definition \ref{algebra-definition-lci}), and
\item $A \to B$ is a local complete intersection
(More on Algebra, Definition

The equivalence of (3) and (4) is More on Algebra, Lemma

If $A \to B$ is syntomic, then we can find a cocartesian diagram
B_0 \ar[r] & B \\
A_0 \ar[r] \ar[u] & A \ar[u]
such that $A_0 \to B_0$ is syntomic and $A_0$ is Noetherian, see
Algebra, Lemmas \ref{algebra-lemma-limit-module-finite-presentation} and
\ref{algebra-lemma-colimit-lci}. By Lemma \ref{lemma-perfect-NL-lci}
we see that $\NL_{B_0/A_0}$ is perfect of tor-amplitude in $[-1, 0]$.
By Lemma \ref{lemma-base-change-NL}
we conclude the same thing is true for
$\NL_{B/A} = \NL_{B_0/A_0} \otimes_{B_0}^\mathbf{L} B$ (see
also More on Algebra, Lemmas \ref{more-algebra-lemma-pull-tor-amplitude} and
This proves that (3) implies (2).

Assume (1). By Lemma \ref{lemma-base-change-NL}
for every ring map $A \to k$ where
$k$ is a field, we see that $\NL_{B \otimes_A k/k}$ has
tor-amplitude in $[-1, 0]$ (see
More on Algebra, Lemma \ref{more-algebra-lemma-pull-tor-amplitude}).
Hence by Lemma \ref{lemma-perfect-NL-lci} we see that $k \to B \otimes_A k$ is
a local complete intersection homomorphism. Thus $A \to B$
is syntomic by definition. This proves (1) implies (3)
and finishes the proof.

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