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Say a bit more about det for flat fp modules

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aisejohan committed Sep 7, 2019
1 parent 7c3552b commit d4117efdf077e291d44a831d68a1023cdab0d4d1
Showing with 62 additions and 15 deletions.
  1. +62 −15 modules.tex
@@ -3978,7 +3978,7 @@ \section{Rank and determinant}
Thus the rank defines a homomorphism
$$
K_0(\textit{Vec}(X)) \longrightarrow \text{Map}_{cont}(X, \mathbf{Z}),\quad
[\mathcal{E}] \longmapsto r_\mathcal{E}
[\mathcal{E}] \longmapsto \text{rank}_\mathcal{E}
$$

\medskip\noindent
@@ -4043,26 +4043,73 @@ \section{Rank and determinant}
\end{proof}

\noindent
{\bf Determinants, reprise.} Let $(X, \mathcal{O}_X)$ be a ringed space.
Instead of looking at finite locally free $\mathcal{O}_X$-modules we could
look at those $\mathcal{O}_X$-modules $\mathcal{F}$ which are locally on $X$
Let $(X, \mathcal{O}_X)$ be a ringed space. Instead of looking at
finite locally free $\mathcal{O}_X$-modules we can look at those
$\mathcal{O}_X$-modules $\mathcal{F}$ which are locally on $X$
a direct summand of a finite free $\mathcal{O}_X$-module. This is the same
thing as asking $\mathcal{F}$ to be a flat $\mathcal{O}_X$-module of
finite presentation, see Lemma \ref{lemma-flat-locally-finite-presentation}.
For such a module in general the rank function is undefined; for
If all the stalks $\mathcal{O}_{X, x}$ are local, then such a module
$\mathcal{F}$ is finite locally free, see
Lemma \ref{lemma-direct-summand-of-locally-free-is-locally-free}.
In general however this will not be the case; for
example $X$ could be a point and $\Gamma(X, \mathcal{O}_X)$ could
be the product $A \times B$ of two nonzero rings and $\mathcal{F}$
could correspond to $A \times 0$. On the other hand, for
$\mathcal{F}$ flat and of finite presentation we can still
define $\det(\mathcal{F})$ and this will be an invertible
$\mathcal{O}_X$-module in the sense of
could correspond to $A \times 0$.
Thus for such a module the rank function is undefined.
However, it turns out we can still define $\det(\mathcal{F})$
and this will be an invertible $\mathcal{O}_X$-module in the sense of
Definition \ref{definition-invertible} (not necessarily locally
free of rank $1$).
This is done using the construction of
More on Algebra, Section \ref{more-algebra-section-determinants}
on the values of $\mathcal{F}$ on sufficiently small opens of $X$.
If we ever need this we will precisely state and prove the relevant
lemmas here.
free of rank $1$). Our construction will agree with the one above
in the case that $\mathcal{F}$ is finite locally free.
We urge the reader to skip the rest of this section.

\begin{lemma}
\label{lemma-determinant-as-socle}
Let $(X, \mathcal{O}_X)$ be a ringed space. Let $\mathcal{F}$
be a flat and finitely presented $\mathcal{O}_X$-module.
Denote
$$
\det(\mathcal{F}) \subset
\wedge^*_{\mathcal{O}_X}(\mathcal{F})
$$
the annihilator of $\mathcal{F} \subset \wedge^*_{\mathcal{O}_X}(\mathcal{F})$.
Then $\det(\mathcal{F})$ is an invertible $\mathcal{O}_X$-module.
\end{lemma}

\begin{proof}
To prove this we may work locally on $X$. Hence we may assume $\mathcal{F}$
is a direct summand of a finite free module, see
Lemma \ref{lemma-flat-locally-finite-presentation}. Say
$\mathcal{F} \oplus \mathcal{G} = \mathcal{O}_X^{\oplus n}$.
Set $R = \mathcal{O}_X(X)$. Then we see
$\mathcal{F}(X) \oplus \mathcal{G}(X) = R^{\oplus n}$
and correspondingly
$\mathcal{F}(U) \oplus \mathcal{G}(U) = \mathcal{O}_X(U)^{\oplus n}$
for all opens $U \subset X$.
We conclude that $\mathcal{F} = \mathcal{F}_M$ as in
Lemma \ref{lemma-construct-quasi-coherent-sheaves}
with $M = \mathcal{F}(X)$ a finite projective $R$-module.
In other words, we have $\mathcal{F}(U) = M \otimes_R \mathcal{O}_X(U)$.
This implies that
$\det(M) \otimes_R \mathcal{O}_X(U) = \det(\mathcal{F}(U))$
for all open $U \subset X$ with $\det$ as in
More on Algebra, Section \ref{more-algebra-section-determinants}. By
More on Algebra, Remark \ref{more-algebra-remark-determinant-as-socle}
we see that
$$
\det(M) \otimes_R \mathcal{O}_X(U) =
\det(\mathcal{F}(U)) \subset
\wedge^*_{\mathcal{O}_X(U)}(\mathcal{F}(U))
$$
is the annihilator of $\mathcal{F}(U)$. We conclude that
$\det(\mathcal{F})$ as defined in the statement of the lemma
is equal to $\mathcal{F}_{\det(M)}$. Some details omitted; one has
to be careful as annihilators cannot be defined as the sheafification
of taking annihilators on sections over opens. Thus $\det(\mathcal{F})$
is the pullback of an invertible module and we conclude.
\end{proof}




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