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Add section on base change and NL

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aisejohan committed Nov 27, 2019
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  1. +1 −1 algebra.tex
  2. +13 −51 dpa.tex
  3. +164 −0 more-algebra.tex
  4. +2 −2 more-morphisms.tex
  5. +1 −1 tags/tags
Then $\NL(\alpha) \otimes_R R' = \NL(\alpha) \otimes_S S' = \NL(\alpha')$.
In particular, the canonical map
$$
\NL_{S/R} \otimes_R R' \longrightarrow \NL_{S \otimes_R R'/R'}
\NL_{S/R} \otimes_S S' \longrightarrow \NL_{S \otimes_R R'/R'}
$$
is a homotopy equivalence if $R \to R'$ is flat.
\end{lemma}
64 dpa.tex
@@ -2514,6 +2514,14 @@ \section{Freeness of the conormal module}
we conclude that $I$ is generated by $x_1, \ldots, x_r$.
\end{proof}

\noindent
For any local complete intersection homomorphism $A \to B$
of rings, the naive cotangent complex $\NL_{B/A}$ is perfect
of tor-amplitude in $[-1, 0]$, see
More on Algebra, Lemma \ref{more-algebra-lemma-lci-NL}.
Using the above, we can show that this sometimes
characterizes local complete intersection homomorphisms.

\begin{lemma}
\label{lemma-perfect-NL-lci}
Let $A \to B$ be a perfect (More on Algebra, Definition
@@ -2550,57 +2558,10 @@ \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
Let $A \to B$ be a flat ring map of finite presentation.
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 $D(B)$
@@ -2629,15 +2590,16 @@ \section{Freeness of the conormal module}
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}
By More on Algebra, Lemma \ref{more-algebra-lemma-base-change-NL-flat}
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}
Assume (1). By More on Algebra, Lemma
\ref{more-algebra-lemma-base-change-NL-flat}
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



\section{The naive cotangent complex}
\label{section-netherlander}

\noindent
In this section we continue the discussion started in
Algebra, Section \ref{algebra-section-netherlander}.
We begin with a discussion of base change.
The first lemma shows that taking the naive tensor product
of the naive cotangent complex with a ring extension
isn't quite as naive as one might think.

\begin{lemma}
\label{lemma-tensor-NL}
Let $R \to S$ and $S \to S'$ be ring maps. The canonical map
$\NL_{S/R} \otimes_S^\mathbf{L} S' \to \NL_{S/R} \otimes_S S'$
induces an isomorphism
$\tau_{\geq -1}(\NL_{S/R} \otimes_S^\mathbf{L} S') \to \NL_{S/R} \otimes_S S'$
in $D(S')$. Similarly, given a presentation $\alpha$ of $S$ over $R$
the canonical map
$\NL(\alpha) \otimes_S^\mathbf{L} S' \to \NL(\alpha) \otimes_S S'$
induces an isomorphism $\tau_{\geq -1}(\NL(\alpha) \otimes_S^\mathbf{L} S') \to
\NL(\alpha) \otimes_S S'$ in $D(S')$.
\end{lemma}

\begin{proof}
Let $K^\bullet$ be a complex of $S$-modules with $K^n = 0$
for $n \not \in [-1, 0]$ and $K^0$ flat, for example
$K^\bullet = \NL_{S/R}$ or $K^\bullet = \NL(\alpha)$.
Then we have a distinguished triangle
$$
K^0 \to K^\bullet \to K^{-1}[1] \to K^0[1]
$$
in $D(S)$. This determines a map of distinguished triangles
$$
\xymatrix{
K^0 \otimes_S^\mathbf{L} S' \ar[d] \ar[r] &
K^\bullet \otimes_S^\mathbf{L} S' \ar[r] \ar[d] &
K^{-1} \otimes_S^\mathbf{L} S'[1] \ar[r] \ar[d] &
K^0 \otimes_S^\mathbf{L} S'[1] \ar[d] \\
K^0 \otimes_S S' \ar[r] &
K^\bullet \otimes_S S' \ar[r] &
K^{-1} \otimes_S S'[1] \ar[r] &
K^0 \otimes_S S'[1]
}
$$
The left and right vertical arrows are isomorphisms as $K^0$ is flat.
Since $K^{-1} \otimes_S^\mathbf{L} S' \to K^{-1} \otimes_S S'$
is an isomorphism on cohomology in degree $0$ we conclude.
\end{proof}

\begin{lemma}
\label{lemma-base-change-NL}
Let $R \to S$ and $R \to R'$ be ring maps.
Let $\alpha : P \to S$ be a presentation of $S$ over $R$.
Then $\alpha' : P \otimes_R R' \to S \otimes_R R'$ is a
presentation of $S' = S \otimes_R R'$ over $R'$.
The canonical map
$$
NL(\alpha) \otimes_S S' \to \NL(\alpha')
$$
is an isomorphism on $H^0$ and surjective on $H^{-1}$. In particular,
the canonical map
$$
\NL_{S/R} \otimes_S S' \to \NL_{S'/R'}
$$
is an isomorphism on $H^0$ and surjective on $H^{-1}$.
\end{lemma}

\begin{proof}
Denote $I = \Ker(P \to S)$. Denote $P' = P \otimes_R R'$ and
$I' = \Ker(P' \to S')$. Suppose $P$ is a polynomial algebra
on $x_j$ for $j \in J$. The map displayed in the lemma becomes
$$
\xymatrix{
\bigoplus_{j \in J} S' \text{d}x_j \ar[r] &
\bigoplus_{j \in J} S' \text{d}x_j \\
I/I^2 \otimes_S S' \ar[r] \ar[u] &
I'/(I')^2 \ar[u]
}
$$
where the left column is $\NL(\alpha) \otimes_S S'$ and the right
column is $\NL(\alpha')$. By right exactness of tensor product we
see that $I \otimes_R R' \to I'$ is surjective. Hence the bottom arrow
is a surjection. This proves the first statement of the lemma.
The statement for $\NL_{S/R} \otimes_S S' \to \NL_{S'/R'}$
follows as these complexes are homotopic to $\NL(\alpha) \otimes_S S'$ and
$\NL(\alpha')$.
\end{proof}

\begin{lemma}
\label{lemma-base-change-NL-flat}
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 an isomorphism of complexes 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
Lemma \ref{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-lci-NL}
Let $A \to B$ be a local complete intersection as in
Definition \ref{definition-local-complete-intersection}.
Then $\NL_{B/A}$ is a perfect object of
$D(B)$ with tor amplitude in $[-1, 0]$.
\end{lemma}

\begin{proof}
Write $B = A[x_1, \ldots, x_n]/I$. Then $\NL_{B/A}$ 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
Lemma \ref{lemma-last-one-flat}. By definition $I$ is a Koszul regular
ideal and hence a quasi-regular ideal, see Section \ref{section-ideals}.
Thus $I/I^2$ is a finite projective $B$-module
(Lemma \ref{lemma-quasi-regular-ideal-finite-projective})
and we conclude both that $\NL_{B/A}$ is perfect and that it has tor amplitude
in $[-1, 0]$.
\end{proof}








@@ -3829,7 +3829,7 @@ \section{The naive cotangent complex}
Consider a cartesian diagram of schemes
$$
\xymatrix{
X' \ar[r] \ar[d]_{g'} & X \ar[d] \\
X' \ar[d] \ar[r]_{g'} & X \ar[d] \\
Y' \ar[r] & Y
}
$$
@@ -3859,7 +3859,7 @@ \section{The naive cotangent complex}

\begin{proof}
Translated into algebra this is
Divided Power Algebra, Lemma \ref{dpa-lemma-base-change-NL}.
More on Algebra, Lemma \ref{more-algebra-lemma-base-change-NL-flat}.
To do the translation use Lemma \ref{lemma-NL-affine}
and Derived Categories of Schemes, Lemmas
\ref{perfect-lemma-affine-compare-bounded} and
0FJR,dpa-lemma-vasconcelos
0FJS,dpa-proposition-regular-ideal
0FJT,dpa-lemma-perfect-NL-lci
0FJU,dpa-lemma-base-change-NL
0FJU,more-algebra-lemma-base-change-NL-flat
0FJV,dpa-lemma-flat-fp-NL-lci
0FJW,perfect-section-det-two-terms
0FJX,perfect-lemma-determinant-two-term-complexes

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