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Improve exposition finitely presented modules

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aisejohan committed Aug 4, 2013
1 parent 16be7e5 commit cadddd7f06877a86c749c0697b8e2e9d9a32c904
Showing with 84 additions and 55 deletions.
  1. +70 −51 algebra.tex
  2. +1 −1 descent.tex
  3. +2 −2 more-algebra.tex
  4. +10 −0 obsolete.tex
  5. +1 −1 tags/tags
@@ -372,41 +372,6 @@ \section{Finite modules and finitely presented modules}
$\alpha = \beta \circ \gamma$.
\end{proof}

\begin{lemma}
\label{lemma-finite-presentation-module-independent}
Let $M$ be an $R$-module of finite presentation.
For any surjection $\alpha : R^{\oplus n} \to M$ the
kernel of $\alpha$ is a finitely generated $R$-module.
\end{lemma}

\begin{proof}
Choose a presentation
$$
R^{\oplus l} \to R^{\oplus m} \to M \to 0
$$
Let $K = \text{Ker}(\alpha)$. By Lemma \ref{lemma-lift-map}
there exists a map $R^{\oplus m} \to R^{\oplus n}$ such that
the solid diagram
$$
\xymatrix{
& R^{\oplus l} \ar[r] \ar@{..>}[d] & R^{\oplus m} \ar[r] \ar[d] &
M \ar[r] \ar[d]^{id} & 0 \\
0 \ar[r] & K \ar[r] & R^{\oplus n} \ar[r]^\alpha & M \ar[r] & 0
}
$$
commutes. This produces the dotted arrow. By the snake lemma
(Lemma \ref{lemma-snake}) we see that we get an isomorphism
$$
\text{Coker}(R^{\oplus l} \to K)
\cong
\text{Coker}(R^{\oplus m} \to R^{\oplus n})
$$
In particular we conclude that $\text{Coker}(R^{\oplus l} \to K)$
is a finite $R$-module. Hence there are finitely many elements of $K$
which together with the images of the basis vectors of $R^{\oplus l}$
generate $K$, i.e., $K$ is finitely generated.
\end{proof}

\begin{lemma}
\label{lemma-extension}
Let $R$ be a ring.
@@ -429,12 +394,68 @@ \section{Finite modules and finitely presented modules}
\end{lemma}

\begin{proof}
We prove part (5). Assume $M_3$ is finitely presented and $M_2$ finite.
Let $\alpha : R^{\oplus n} \to M_2$ be a surjection. Then we can find
$k_1, \ldots, k_m \in R^{\oplus n}$ which generate the kernel
of the composition $R^{\oplus n} \to M_2 \to M_3$. Then
$\alpha(k_1), \ldots, \alpha(k_m)$ generate $M_1$ as a submodule of $M_2$.
The proofs of the other parts are omitted.
Proof of (1). If $x_1, \ldots, x_n$ are generators of $M_1$ and
$y_1, \ldots, y_m \in M_2$ are elements whose images in $M_3$ are
generators of $M_3$, then $x_1, \ldots, x_n, y_1, \ldots, y_m$
generate $M_2$.

\medskip\noindent
Part (3) is immediate from the definition.

\medskip\noindent
Proof of (5). Assume $M_3$ is finitely presented and $M_2$ finite.
Choose a presentation
$$
R^{\oplus m} \to R^{\oplus n} \to M_3 \to 0
$$
By Lemma \ref{lemma-lift-map} there exists a map
$R^{\oplus n} \to M_2$ such that
the solid diagram
$$
\xymatrix{
& R^{\oplus m} \ar[r] \ar@{..>}[d] & R^{\oplus n} \ar[r] \ar[d] &
M_3 \ar[r] \ar[d]^{id} & 0 \\
0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0
}
$$
commutes. This produces the dotted arrow. By the snake lemma
(Lemma \ref{lemma-snake}) we see that we get an isomorphism
$$
\text{Coker}(R^{\oplus m} \to M_1)
\cong
\text{Coker}(R^{\oplus n} \to M_2)
$$
In particular we conclude that $\text{Coker}(R^{\oplus m} \to M_1)$
is a finite $R$-module. Since $\text{Im}(R^{\oplus m} \to M_1)$
is finite by (3), we see that $M_1$ is finite by part (1).

\medskip\noindent
Proof of (4). Assume $M_2$ is finitely presented and $M_1$ is finite.
Choose a presentation $R^{\oplus m} \to R^{\oplus n} \to M_2 \to 0$.
Choose a surjection $R^{\oplus k} \to M_1$. By Lemma \ref{lemma-lift-map}
there exists a factorization $R^{\oplus k} \to R^{\oplus n} \to M_1$
of this map. Then $R^{\oplus k + m} \to R^{\oplus n} \to M_3 \to 0$
is a presentation.

\medskip\noindent
Proof of (2). Assume that $M_1$ and $M_3$ are finitely presented.
The argument in the proof of part (1) produces a commutative diagram
$$
\xymatrix{
0 \ar[r] & R^{\oplus n} \ar[d] \ar[r] & R^{\oplus n + m} \ar[d] \ar[r] &
R^{\oplus m} \ar[d] \ar[r] & 0 \\
0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0
}
$$
with surjective vertical arrows. By the snake lemma we obtain a short
exact sequence
$$
0 \to \text{Ker}(R^{\oplus n} \to M_1) \to
\text{Ker}(R^{\oplus n + m} \to M_2) \to
\text{Ker}(R^{\oplus m} \to M_3) \to 0
$$
By part (5) we see that the outer two modules are finite. Hence the
middle one is finite too. By (4) we see that $M_2$ is of finite presentation.
\end{proof}

\begin{lemma}
@@ -4057,11 +4078,11 @@ \section{More glueing results}
$$
0 \rightarrow K \rightarrow R^n \rightarrow M \rightarrow 0
$$
Since finite presentation does not depend on the chosen presentation
(Lemma \ref{lemma-finite-presentation-module-independent})
and localization is an exact functor $K_{f_i}$ is finitely
generated for all $1 \leq i \leq n$. By 2. this implies that $K$ is a
finitely generated $R$-module and therefore $M$ is finitely presented.
Since localization is an exact functor and $M_{f_i}$ is finitely
presented we see that $K_{f_i}$ is finitely generated for all
$1 \leq i \leq n$ by Lemma \ref{lemma-extension}.
By (2) this implies that $K$ is a finite $R$-module and therefore
$M$ is finitely presented.

\item By Proposition \ref{proposition-localize-twice}
the assumption implies that the induced morphism
these elements generate $M_g$ for some $g \in R$, $g \not \in \mathfrak p$.
The corresponding surjection $\varphi : R_g^{\oplus r} \to M_g^{\oplus r}$
has the following two properties: (a) $\text{Ker}(\varphi)$ is a finite
$R_g$-module (see
Lemma \ref{lemma-finite-presentation-module-independent})
$R_g$-module (see Lemma \ref{lemma-extension})
and (b) $\text{Ker}(\varphi) \otimes \kappa(\mathfrak p) = 0$
by flatness of $M_g$ over $R_g$ (see
Lemma \ref{lemma-flat-tor-zero}).
Lemma \ref{lemma-finite-projective}
we see that (4) is equivalent to the assertion that $I$ is pure
and $R/I$ finitely presented. Moreover, $R/I$ is finitely presented
if and only if $I$ is finitely generated, see
Lemma \ref{lemma-finite-presentation-module-independent}.
if and only if $I$ is finitely generated, see Lemma \ref{lemma-extension}.
Hence (4) is equivalent to (1).
\end{proof}

$S^{-1}M$ as an $S^{-1}R$-module. Let
$K = \text{Ker}(R^{\oplus n} \to M)$ where the map is given by
the rule $(a_1, \ldots, a_n) \mapsto \sum a_i x_i$. By
Lemma \ref{lemma-finite-presentation-module-independent}
Lemma \ref{lemma-extension}
we see that $S^{-1}K$ is a finite $S^{-1}R$-module.
By (1) we can find a finite type submodule $K' \subset K$
with $S^{-1}K' = S^{-1}K$. Take
@@ -1366,7 +1366,7 @@ \subsection{Descent for properties of modules}
To see (b) assume $M \otimes_R S$ is finitely presented. Then $M$ is finitely
generated by (a). Choose a surjection $R^{\oplus n} \to M$ with kernel $K$.
Then $K \otimes_R S \to S^{\oplus r} \to M \otimes_R S \to 0$ is exact.
By Algebra, Lemma \ref{algebra-lemma-finite-presentation-module-independent}
By Algebra, Lemma \ref{algebra-lemma-extension}
the kernel of $S^{\oplus r} \to M \otimes_R S$
is a finite $S$-module. Thus we can find finitely many elements
$k_1, \ldots, k_t \in K$ such that the images of $k_i \otimes 1$ in
@@ -687,7 +687,7 @@ \section{Fibre products of rings}
there is also a
canonical map $K_D \times_{K_C} K_{C'} \to K'$ inverse to the displayed
arrow. Hence the displayed map is an isomorphism. By
Algebra, Lemma \ref{algebra-lemma-finite-presentation-module-independent}
Algebra, Lemma \ref{algebra-lemma-extension}
the modules $K_D$ and $K_{C'}$ are finite. We conclude from
Lemma \ref{lemma-relative-finite-module-over-fibre-product}
that $K'$ is a finite $D'$-module provided that $K_D \to K_C$ and
@@ -11595,7 +11595,7 @@ \section{Relatively finitely presented modules}
$A_{f_i}$ we see by
Lemma \ref{lemma-relatively-finitely-presented}
and
Algebra, Lemma \ref{algebra-lemma-finite-presentation-module-independent}
Algebra, Lemma \ref{algebra-lemma-extension}
that the localization $K_{y_i}$ is a finitely generated
$P_{y_i}$-module. Choose elements
$k_{i, j} \in K$, $i = 1, \ldots, r$, $j = 1, \ldots, s_i$ such
@@ -28,6 +28,16 @@ \section{Introduction}
\section{Obsolete algebra lemmas}
\label{section-algebra}

\begin{lemma}
\label{lemma-finite-presentation-module-independent}
Let $M$ be an $R$-module of finite presentation.
For any surjection $\alpha : R^{\oplus n} \to M$ the
kernel of $\alpha$ is a finite $R$-module.
\end{lemma}

\begin{proof}
This is a special case of Algebra, Lemma \ref{algebra-lemma-extension}.
\end{proof}

\noindent
The following technical lemma says that you can lift any sequence
@@ -6685,7 +6685,7 @@
055W,more-morphisms-lemma-fibre-geometrically-connected-reduced
055X,conventions-section-notation
055Y,algebra-item-annihilator
055Z,algebra-lemma-finite-presentation-module-independent
055Z,obsolete-lemma-finite-presentation-module-independent
0560,algebra-lemma-finite-over-subring
0561,algebra-lemma-finitely-presented-over-subring
0562,algebra-section-finite

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