# stacks/stacks-project

Characterization of syntomic ring maps

 @@ -2550,7 +2550,103 @@ \section{Freeness of the conormal module} More on Algebra, Section \ref{more-algebra-section-ideals}). \end{proof} \begin{lemma} \label{lemma-base-change-NL} Consider a cocartesian diagram of rings $$\xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u] & A' \ar[u] }$$ 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. \end{lemma} \begin{proof} 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. \medskip\noindent 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. \end{proof} \begin{lemma} \label{lemma-flat-fp-NL-lci} Let $A \to B$ be a perfect (More on Algebra, Definition \ref{more-algebra-definition-pseudo-coherent-perfect}) ring homomorphism of Noetherian rings. Then the following are equivalent \begin{enumerate} \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 \ref{more-algebra-definition-local-complete-intersection}). \end{enumerate} \end{lemma} \begin{proof} The equivalence of (3) and (4) is More on Algebra, Lemma \ref{more-algebra-lemma-syntomic-lci}. \medskip\noindent If $A \to B$ is syntomic, then we can find a cocartesian diagram $$\xymatrix{ B_0 \ar[r] & B \\ A_0 \ar[r] \ar[u] & A \ar[u] }$$ 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 \ref{more-algebra-lemma-pull-perfect}). This proves that (3) implies (2). \medskip\noindent 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. \end{proof}