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Update section on rank and determinant in modules

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aisejohan committed Sep 7, 2019
1 parent 0b2b5dd commit 3f8bf28efc6083a9bab5596e287fd23bd004d31e
Showing with 51 additions and 18 deletions.
  1. +51 −18 modules.tex
@@ -2741,8 +2741,8 @@ \section{Flat modules}

\begin{lemma}
\label{lemma-flat-locally-finite-presentation}
Let $(X, \mathcal{O}_X)$ be a ringed space. Let $\mathcal{F}$
be locally of finite presentation and flat. Then $\mathcal{F}$ is
Let $(X, \mathcal{O}_X)$ be a ringed space. Let $\mathcal{F}$ be a
flat $\mathcal{O}_X$-module of finite presentation. Then $\mathcal{F}$ is
locally a direct summand of a finite free $\mathcal{O}_X$-module.
\end{lemma}

@@ -3943,16 +3943,16 @@ \section{Rank and determinant}

\noindent
Let $(X, \mathcal{O}_X)$ be a ringed space. Consider the category
$\mathcal{C}$ of finite locally free $\mathcal{O}_X$-modules.
$\textit{Vect}(X)$ of finite locally free $\mathcal{O}_X$-modules.
This is an exact category
(see Injectives, Remark \ref{injectives-remark-embed-exact-category})
whose admissible epimorphisms are
surjections and whose admissible monomorphisms are kernels of
surjections. Moreover, there is a set of isomorphism classes
of objects of $\mathcal{C}$ (proof omitted). Thus we can form
the Grothendieck $K$-group $K(\mathcal{C})$, which is often denoted
$K_0^{naive}(X)$. Explicitly, in this case $K_0^{naive}(X)$ is
the abelian group generated by $[\mathcal{E}]$ for $\mathcal{E}$
of objects of $\textit{Vect}(X)$ (proof omitted). Thus we can form
the zeroth Grothendieck $K$-group $K_0(\textit{Vect}(X))$.
Explicitly, in this case $K_0(\textit{Vect}(X))$
is the abelian group generated by $[\mathcal{E}]$ for $\mathcal{E}$
a finite locally free $\mathcal{O}_X$-module, subject to the relations
$$
[\mathcal{E}'] = [\mathcal{E}] + [\mathcal{E}'']
@@ -3962,44 +3962,46 @@ \section{Rank and determinant}
of finite locally free $\mathcal{O}_X$-modules.

\medskip\noindent
Ranks. Given a finite locally free $\mathcal{O}_X$-module $\mathcal{E}$,
{\bf Ranks.} Assume all stalks $\mathcal{O}_{X, x}$ are nonzero rings.
Given a finite locally free $\mathcal{O}_X$-module $\mathcal{E}$,
the {\it rank} is a locally constant function
$$
r = r_\mathcal{E} : X \longrightarrow \mathbf{Z}_{\geq 0},\quad
\text{rank}_\mathcal{E} : X \longrightarrow \mathbf{Z}_{\geq 0},\quad
x \longmapsto \text{rank}_{\mathcal{O}_{X, x}} \mathcal{E}_x
$$
This makes sense as $\mathcal{E}_x \cong \mathcal{O}_{X, x}^{\oplus r(x)}$
and this uniquely determines $r(x)$. By definition of locally free
modules the function $r$ is locally constant. If
See Lemma \ref{lemma-rank}. By definition of locally free
modules the function $\text{rank}_\mathcal{E}$ is locally constant. If
$0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{E}'' \to 0$
is a short exact sequence of finite locally free $\mathcal{O}_X$-modules,
then $r_\mathcal{E} = r_{\mathcal{E}'} + r_{\mathcal{E}''}$,
then $\text{rank}_\mathcal{E} = \text{rank}_{\mathcal{E}'} +
\text{rank}_{\mathcal{E}''}$,
Thus the rank defines a homomorphism
$$
K_0^{naive}(X) \longrightarrow \text{Map}_{cont}(X, \mathbf{Z}),\quad
K_0(\textit{Vec}(X)) \longrightarrow \text{Map}_{cont}(X, \mathbf{Z}),\quad
[\mathcal{E}] \longmapsto r_\mathcal{E}
$$

\medskip\noindent
Determinants. Given a finite locally free
{\bf Determinants.} Given a finite locally free
$\mathcal{O}_X$-module $\mathcal{E}$ we obtain a disjoint union
decomposition
$$
X = X_0 \amalg X_1 \amalg X_2 \amalg \ldots
$$
with $X_i$ open and closed, such that $\mathcal{E}$ is finite locally
free of rank $i$ on $X_i$ (this is exactly the same as saying the
$r_\mathcal{E}$ is locally constant). In this case we define
$\text{rank}_\mathcal{E}$ is locally constant). In this case we define
$\det(\mathcal{E})$ as the invertible sheaf on $X$ which is equal to
$\wedge^i(\mathcal{E}|_{X_i})$ on $X_i$ for all $i \geq 0$.
Since the decomposition above is disjoint, there are no glueing
conditions to check. By Lemma \ref{lemma-det-ses} below
this defines a homomorphism
$$
\det : K_0^{naive}(X) \longrightarrow \Pic(X),\quad
\det : K_0(\textit{Vect}(X)) \longrightarrow \Pic(X),\quad
[\mathcal{E}] \longmapsto \det(\mathcal{E})
$$
of abelian groups.
of abelian groups. The elements of $\Pic(X)$ we get in this manner
are locally free of rank $1$ (see below the lemma for a generalization).

\begin{lemma}
\label{lemma-det-ses}
@@ -4040,6 +4042,37 @@ \section{Rank and determinant}
$s''_i$ to a section of $\mathcal{E}$. We omit the details.
\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 $X$
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
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
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.













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