# stacks/stacks-project

Add section on base change and NL

aisejohan committed Nov 27, 2019
1 parent 8f6f00f commit f18336aa820bdac6970146be1c42009c18fe0d71
Showing with 181 additions and 55 deletions.
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 $$K^0 \to K^\bullet \to K^{-1} \to K^0$$ in $D(S)$. This determines a map of distinguished triangles $$\xymatrix{ K^0 \otimes_S^\mathbf{L} S' \ar[d] \ar[r] & K^\bullet \otimes_S^\mathbf{L} S' \ar[r] \ar[d] & K^{-1} \otimes_S^\mathbf{L} S' \ar[r] \ar[d] & K^0 \otimes_S^\mathbf{L} S' \ar[d] \\ K^0 \otimes_S S' \ar[r] & K^\bullet \otimes_S S' \ar[r] & K^{-1} \otimes_S S' \ar[r] & K^0 \otimes_S S' }$$ 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