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Tag 00S1

Chapter 10: Commutative Algebra > Section 10.132: The naive cotangent complex

Lemma 10.132.2. Suppose given a diagram (10.132.1.1). Let $\alpha : P \to S$ and $\alpha' : P' \to S'$ be presentations.

  1. There exists a morphism of presentations from $\alpha$ to $\alpha'$.
  2. Any two morphisms of presentations induce homotopic morphisms of complexes $\mathop{N\!L}\nolimits(\alpha) \to \mathop{N\!L}\nolimits(\alpha')$.
  3. The construction is compatible with compositions of morphisms of presentations (see proof for exact statement).
  4. If $R \to R'$ and $S \to S'$ are isomorphisms, then for any map $\varphi$ of presentations from $\alpha$ to $\alpha'$ the induced map $\mathop{N\!L}\nolimits(\alpha) \to \mathop{N\!L}\nolimits(\alpha')$ is a homotopy equivalence and a quasi-isomorphism.

In particular, comparing $\alpha$ to the canonical presentation (10.132.0.1) we conclude there is a quasi-isomorphism $\mathop{N\!L}\nolimits(\alpha) \to \mathop{N\!L}\nolimits_{S/R}$ well defined up to homotopy and compatible with all functorialities (up to homotopy).

Proof. Since $P$ is a polynomial algebra over $R$ we can write $P = R[x_a, a \in A]$ for some set $A$. As $\alpha'$ is surjective, we can choose for every $a \in A$ an element $f_a \in P'$ such that $\alpha'(f_a) = \phi(\alpha(x_a))$. Let $\varphi : P = R[x_a, a \in A] \to P'$ be the unique $R$-algebra map such that $\varphi(x_a) = f_a$. This gives the morphism in (1).

Let $\varphi$ and $\varphi'$ morphisms of presentations from $\alpha$ to $\alpha'$. Let $I = \mathop{\rm Ker}(\alpha)$ and $I' = \mathop{\rm Ker}(\alpha')$. We have to construct the diagonal map $h$ in the diagram $$ \xymatrix{ I/I^2 \ar[r]^-{\text{d}} \ar@<1ex>[d]^{\varphi'_1} \ar@<-1ex>[d]_{\varphi_1} & \Omega_{P/R} \otimes_P S \ar@<1ex>[d]^{\varphi'_0} \ar@<-1ex>[d]_{\varphi_0} \ar[ld]_h \\ I'/(I')^2 \ar[r]^-{\text{d}} & \Omega_{P'/R'} \otimes_{P'} S' } $$ where the vertical maps are induced by $\varphi$, $\varphi'$ such that $$ \varphi_1 - \varphi'_1 = h \circ \text{d} \quad\text{and}\quad \varphi_0 - \varphi'_0 = \text{d} \circ h $$ Consider the map $\varphi - \varphi' : P \to P'$. Since both $\varphi$ and $\varphi$ are compatible with $\alpha$ and $\alpha'$ we obtain $\varphi - \varphi' : P \to I'$. This implies that $\varphi, \varphi' : P \to P'$ induce the same $P$-module structure on $I'/(I')^2$, since $\varphi(p)i' - \varphi'(p)i' = (\varphi - \varphi')(p)i' \in (I')^2$. Also $\varphi - \varphi'$ is $R$-linear and $$ (\varphi - \varphi')(fg) = \varphi(f)(\varphi - \varphi')(g) + (\varphi - \varphi')(f)\varphi'(g) $$ Hence the induced map $D : P \to I'/(I')^2$ is a $R$-derivation. Thus we obtain a canonical map $h : \Omega_{P/R} \otimes_P S \to I'/(I')^2$ such that $D = h \circ \text{d}$. A calculation (omitted) shows that $h$ is the desired homotopy.

Suppose that we have a commutative diagram $$ \xymatrix{ S \ar[r]_{\phi} & S' \ar[r]_{\phi'} & S'' \\ R \ar[r] \ar[u] & R' \ar[u] \ar[r] & R'' \ar[u] } $$ and that

  1. $\alpha : P \to S$,
  2. $\alpha' : P' \to S'$, and
  3. $\alpha'' : P'' \to S''$

are presentations. Suppose that

  1. $\varphi : P \to P$ is a morphism of presentations from $\alpha$ to $\alpha'$ and
  2. $\varphi' : P' \to P''$ is a morphism of presentations from $\alpha'$ to $\alpha''$.

Then it is immediate that $\varphi' \circ \varphi : P \to P''$ is a morphism of presentations from $\alpha$ to $\alpha''$ and that the induced map $\mathop{N\!L}\nolimits(\alpha) \to \mathop{N\!L}\nolimits(\alpha'')$ of naive cotangent complexes is the composition of the maps $\mathop{N\!L}\nolimits(\alpha) \to \mathop{N\!L}\nolimits(\alpha')$ and $\mathop{N\!L}\nolimits(\alpha) \to \mathop{N\!L}\nolimits(\alpha')$ induced by $\varphi$ and $\varphi'$.

In the simple case of complexes with 2 terms a quasi-isomorphism is just a map that induces an isomorphism on both the cokernel and the kernel of the maps between the terms. Note that homotopic maps of 2 term complexes (as explained above) define the same maps on kernel and cokernel. Hence if $\varphi$ is a map from a presentation $\alpha$ of $S$ over $R$ to itself, then the induced map $\mathop{N\!L}\nolimits(\alpha) \to \mathop{N\!L}\nolimits(\alpha)$ is a quasi-isomorphism being homotopic to the identity by part (2). To prove (4) in full generality, consider a morphism $\varphi'$ from $\alpha'$ to $\alpha$ which exists by (1). The compositions $\mathop{N\!L}\nolimits(\alpha) \to \mathop{N\!L}\nolimits(\alpha') \to \mathop{N\!L}\nolimits(\alpha)$ and $\mathop{N\!L}\nolimits(\alpha') \to \mathop{N\!L}\nolimits(\alpha) \to \mathop{N\!L}\nolimits(\alpha')$ are homotopic to the identity maps by (3), hence these maps are homotopy equivalences by definition. It follows formally that both maps $\mathop{N\!L}\nolimits(\alpha) \to \mathop{N\!L}\nolimits(\alpha')$ and $\mathop{N\!L}\nolimits(\alpha') \to \mathop{N\!L}\nolimits(\alpha)$ are quasi-isomorphisms. Some details omitted. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 33375–33394 (see updates for more information).

    \begin{lemma}
    \label{lemma-NL-homotopy}
    Suppose given a diagram (\ref{equation-functoriality-NL}).
    Let $\alpha : P \to S$ and $\alpha' : P' \to S'$ be presentations.
    \begin{enumerate}
    \item There exists a morphism of presentations from $\alpha$ to $\alpha'$.
    \item Any two morphisms of presentations induce homotopic
    morphisms of complexes $\NL(\alpha) \to \NL(\alpha')$.
    \item The construction is compatible with compositions of morphisms
    of presentations (see proof for exact statement).
    \item If $R \to R'$ and $S \to S'$ are isomorphisms, then
    for any map $\varphi$ of presentations from $\alpha$ to $\alpha'$
    the induced map $\NL(\alpha) \to \NL(\alpha')$ is a homotopy equivalence
    and a quasi-isomorphism.
    \end{enumerate}
    In particular, comparing $\alpha$ to the canonical presentation
    (\ref{equation-canonical-presentation}) we conclude there is a
    quasi-isomorphism $\NL(\alpha) \to \NL_{S/R}$ well defined
    up to homotopy and compatible with all functorialities (up to homotopy).
    \end{lemma}
    
    \begin{proof}
    Since $P$ is a polynomial algebra over $R$ we can write
    $P = R[x_a, a \in A]$ for some set $A$.
    As $\alpha'$ is surjective, we can choose
    for every $a \in A$ an element $f_a \in P'$
    such that $\alpha'(f_a) = \phi(\alpha(x_a))$. Let
    $\varphi : P = R[x_a, a \in A] \to P'$ be the
    unique $R$-algebra map such that $\varphi(x_a) = f_a$.
    This gives the morphism in (1).
    
    \medskip\noindent
    Let $\varphi$ and $\varphi'$ morphisms of presentations from $\alpha$
    to $\alpha'$. Let $I = \Ker(\alpha)$ and $I' = \Ker(\alpha')$.
    We have to construct the diagonal map $h$ in the diagram
    $$
    \xymatrix{
    I/I^2 \ar[r]^-{\text{d}}
    \ar@<1ex>[d]^{\varphi'_1} \ar@<-1ex>[d]_{\varphi_1}
    &
    \Omega_{P/R} \otimes_P S
    \ar@<1ex>[d]^{\varphi'_0} \ar@<-1ex>[d]_{\varphi_0}
    \ar[ld]_h
    \\
    I'/(I')^2 \ar[r]^-{\text{d}}
    &
    \Omega_{P'/R'} \otimes_{P'} S'
    }
    $$
    where the vertical maps are induced by $\varphi$, $\varphi'$ such that
    $$
    \varphi_1 - \varphi'_1 = h \circ \text{d}
    \quad\text{and}\quad
    \varphi_0 - \varphi'_0 = \text{d} \circ h
    $$
    Consider the map $\varphi - \varphi' : P \to P'$. Since both $\varphi$
    and $\varphi$ are compatible with $\alpha$ and $\alpha'$ we obtain
    $\varphi - \varphi' : P \to I'$. This implies that
    $\varphi, \varphi' : P \to P'$ induce the same $P$-module structure
    on $I'/(I')^2$, since
    $\varphi(p)i' - \varphi'(p)i' = (\varphi - \varphi')(p)i' \in (I')^2$.
    Also $\varphi - \varphi'$ is $R$-linear and
    $$
    (\varphi - \varphi')(fg) =
    \varphi(f)(\varphi - \varphi')(g) + (\varphi - \varphi')(f)\varphi'(g)
    $$
    Hence the induced map $D : P \to I'/(I')^2$ is a $R$-derivation.
    Thus we obtain a canonical map $h : \Omega_{P/R} \otimes_P S \to I'/(I')^2$
    such that $D = h \circ \text{d}$. 
    A calculation (omitted) shows that $h$ is the desired homotopy.
    
    \medskip\noindent
    Suppose that we have a commutative diagram
    $$
    \xymatrix{
    S \ar[r]_{\phi} & S' \ar[r]_{\phi'} & S'' \\
    R \ar[r] \ar[u] & R' \ar[u] \ar[r] & R'' \ar[u]
    }
    $$
    and that
    \begin{enumerate}
    \item $\alpha : P \to S$,
    \item $\alpha' : P' \to S'$, and
    \item $\alpha'' : P'' \to S''$
    \end{enumerate}
    are presentations. Suppose that
    \begin{enumerate}
    \item $\varphi : P \to P$ is a morphism of presentations from
    $\alpha$ to $\alpha'$ and
    \item $\varphi' : P' \to P''$
    is a morphism of presentations from $\alpha'$ to $\alpha''$.
    \end{enumerate}
    Then it is immediate that
    $\varphi' \circ \varphi : P \to P''$
    is a morphism of presentations from $\alpha$ to $\alpha''$ and that
    the induced map $\NL(\alpha) \to \NL(\alpha'')$ of naive cotangent complexes
    is the composition of the maps $\NL(\alpha) \to \NL(\alpha')$ and
    $\NL(\alpha) \to \NL(\alpha')$ induced by $\varphi$ and $\varphi'$.
    
    \medskip\noindent
    In the simple case of complexes with 2 terms a quasi-isomorphism
    is just a map that induces an isomorphism on both the cokernel
    and the kernel of the maps between the terms. Note that homotopic
    maps of 2 term complexes (as explained above) define the same maps on
    kernel and cokernel. Hence if $\varphi$ is a map from a presentation
    $\alpha$ of $S$ over $R$ to itself, then the induced map
    $\NL(\alpha) \to \NL(\alpha)$ is a quasi-isomorphism being homotopic
    to the identity by part (2). To prove (4) in full generality, consider
    a morphism $\varphi'$ from $\alpha'$ to $\alpha$ which exists by (1).
    The compositions $\NL(\alpha) \to \NL(\alpha') \to \NL(\alpha)$ and
    $\NL(\alpha') \to \NL(\alpha) \to \NL(\alpha')$ are homotopic to the identity
    maps by (3), hence these maps are homotopy equivalences by definition.
    It follows formally that both maps
    $\NL(\alpha) \to \NL(\alpha')$ and $\NL(\alpha') \to \NL(\alpha)$ are
    quasi-isomorphisms. Some details omitted.
    \end{proof}

    Comments (5)

    Comment #693 by Keenan Kidwell on June 16, 2014 a 7:30 pm UTC

    In the first part of the lemma, "there exist" should be "there exists."

    Comment #1717 by Yogesh More on December 3, 2015 a 3:56 pm UTC

    Trivial remark but might be worth adding, after the line "we conclude that $\phi -\phi:P \to I'$.":

    This implies that $\phi, \phi':P \to P'$ induce the same $P$-module structure on $I'/(I')^2$, since $\phi(p)i'-\phi'(p)i'=(\phi-\phi')(p)i' \in (I')^2$.

    I suggest adding this (obvious) remark because you are implicitly using it in the equation showing that $(\phi-\phi')(fg)$ satisfies Leibniz rule (which I feel is the key to the result), and also in using the universal property for $\Omega_{P/R}$, i.e. to get the map $\Omega_{P/R} \to I'/(I')^2$, you are considering $I'/(I')^2$ as a $P$-module

    Comment #1757 by Johan (site) on December 15, 2015 a 7:05 pm UTC

    Thanks, fixed here.

    Comment #2789 by Dario Weißmann on August 27, 2017 a 9:10 pm UTC

    Typo in the proof of (2): "Since both $\varphi$ and $\varphi$ are compatible with..." should read "Since both $\varphi$ and $\varphi'$ are compatible with..."

    Comment #2791 by Dario Weißmann on August 30, 2017 a 8:40 am UTC

    In the proof of (3) the last line should read: "...and NL$(\alpha') \to $NL$(\alpha'')$ induced by $\varphi$ and $\varphi'$."

    There are also 7 comments on Section 10.132: Commutative Algebra.

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