Inicio tema 4 (módulos sobre DIPs)

1 files changed, 4894 insertions, 0 deletions

diff --git a/ac/n4.lyx b/ac/n4.lyx
new file mode 100644
index 0000000..2479482
--- /dev/null
+++ b/ac/n4.lyx
@@ -0,0 +1,4894 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input{../defs}
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style french
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+En adelante, salvo que se indique lo contrario, 
+\begin_inset Formula $A$
+\end_inset
+
+ es un DIP y 
+\begin_inset Formula ${\cal P}\subseteq A$
+\end_inset
+
+ es un conjunto irredundante de representantes salvo asociados de los elementos
+ irreducibles o primos de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Por ejemplo, si 
+\begin_inset Formula $A=\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+ podría ser el conjunto de primos positivos, y cuando 
+\begin_inset Formula $A=K[X]$
+\end_inset
+
+ con 
+\begin_inset Formula $K$
+\end_inset
+
+ cuerpo, 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+ podría ser el conjunto de polinomios mónicos irreducibles.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, si 
+\begin_inset Formula $G\leq_{A}F$
+\end_inset
+
+ con 
+\begin_inset Formula $F$
+\end_inset
+
+ libre, entonces 
+\begin_inset Formula $G$
+\end_inset
+
+ es libre y 
+\begin_inset Formula $\text{rg}G\leq\text{rg}F$
+\end_inset
+
+.
+ 
+\series bold
+Demostración
+\series default
+ cuando 
+\begin_inset Formula $F$
+\end_inset
+
+ tiene rango finito
+\series bold
+:
+\series default
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Demostración general en 
+\emph on
+\lang english
+Algebra
+\emph default
+\lang spanish
+ de Thomas W.
+ Hungerfor, IV.6.1, que usa propiedades de los números ordinales.
+\end_layout
+
+\end_inset
+
+ Sean 
+\begin_inset Formula $B=\{x_{1},\dots,x_{n}\}$
+\end_inset
+
+ una base de 
+\begin_inset Formula $F$
+\end_inset
+
+ y, para 
+\begin_inset Formula $j\in\{0,\dots,n\}$
+\end_inset
+
+, 
+\begin_inset Formula $F_{j}\coloneqq\bigoplus_{i=1}^{j}Ax_{i}$
+\end_inset
+
+, tenemos una cadena estrictamente ascendente 
+\begin_inset Formula $0=F_{0}\subsetneq\dots\subsetneq F_{n}=F$
+\end_inset
+
+ donde 
+\begin_inset Formula $\frac{F_{j}}{F_{j-1}}\cong Ax_{j}$
+\end_inset
+
+ para 
+\begin_inset Formula $j\in\{1,\dots,n\}$
+\end_inset
+
+, pero el homomorfismo canónico 
+\begin_inset Formula $A\to Ax_{j}$
+\end_inset
+
+, 
+\begin_inset Formula $a\mapsto ax_{j}$
+\end_inset
+
+, es un isomorfismo (por restricción de 
+\begin_inset Formula $\phi:A^{n}\to F$
+\end_inset
+
+), luego todo submódulo de 
+\begin_inset Formula $\frac{F_{j}}{F_{j-1}}$
+\end_inset
+
+ es isomorfo a un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+, que será principal, y es pues nulo o un 
+\begin_inset Formula $A$
+\end_inset
+
+-módulo libre de rango 1.
+ Intersecando los términos de esta cadena con 
+\begin_inset Formula $G$
+\end_inset
+
+, 
+\begin_inset Formula $0=G\cap F_{0}\subseteq G\cap F_{1}\subseteq\dots\subseteq G\cap F_{n}=G$
+\end_inset
+
+, y el homomorfismo 
+\begin_inset Formula $f_{j}:G\cap F_{j}\hookrightarrow F_{j}\overset{\pi}{\to}\frac{F_{j}}{F_{j-1}}$
+\end_inset
+
+ tiene núcleo 
+\begin_inset Formula $G\cap F_{j-1}$
+\end_inset
+
+, por lo que hay un monomorfismo 
+\begin_inset Formula $\frac{G\cap F_{j}}{G\cap F_{j-1}}\rightarrowtail\frac{F_{j}}{F_{j-1}}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\frac{G\cap F_{j}}{G\cap F_{j-1}}=0$
+\end_inset
+
+ o es un 
+\begin_inset Formula $A$
+\end_inset
+
+-módulo libre de rango 1.
+ Tras eliminar términos repetidos de la cadena, existen 
+\begin_inset Formula $k_{i}\in\{1,\dots,n\}$
+\end_inset
+
+, 
+\begin_inset Formula $k_{1}<\dots<k_{t}$
+\end_inset
+
+, con 
+\begin_inset Formula $0=G\cap F_{k_{0}}\subsetneq G\cap F_{k_{1}}\subsetneq\dots\subsetneq G\cap F_{k_{t}}=G$
+\end_inset
+
+ y donde cada 
+\begin_inset Formula $\frac{G\cap F_{k_{i}}}{G\cap F_{k_{i-1}}}$
+\end_inset
+
+ es un 
+\begin_inset Formula $A$
+\end_inset
+
+-módulo libre de rango 1.
+ Si 
+\begin_inset Formula $t=0$
+\end_inset
+
+ hemos terminado.
+ En otro caso, sean 
+\begin_inset Formula $H_{i}\coloneqq G\cap F_{k_{i}}$
+\end_inset
+
+, 
+\begin_inset Formula $y_{1}$
+\end_inset
+
+ un generador de 
+\begin_inset Formula $H_{1}$
+\end_inset
+
+ y, para 
+\begin_inset Formula $i\in\{2,\dots,n\}$
+\end_inset
+
+, 
+\begin_inset Formula $y_{i}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $y_{i}+H_{i-1}$
+\end_inset
+
+ sea un generador de 
+\begin_inset Formula $\frac{H_{i}}{H_{i-1}}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $(y_{1},\dots,y_{i})$
+\end_inset
+
+ es base de 
+\begin_inset Formula $H_{i}$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $i=1$
+\end_inset
+
+ esto es claro.
+ Para 
+\begin_inset Formula $i>1$
+\end_inset
+
+, por inducción, para 
+\begin_inset Formula $m\in H_{i}$
+\end_inset
+
+ existe 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ tal que, en 
+\begin_inset Formula $\frac{H_{i}}{H_{i-1}}$
+\end_inset
+
+, 
+\begin_inset Formula $\overline{a}\overline{y_{i}}=\overline{m}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $ay_{i}-m\in H_{i-1}$
+\end_inset
+
+, que por inducción se puede expresar unívocamente como combinación lineal
+ de 
+\begin_inset Formula $(y_{1},\dots,y_{i-1})$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $\{y_{1},\dots,y_{i}\}$
+\end_inset
+
+ es generador de 
+\begin_inset Formula $H_{j}$
+\end_inset
+
+, y es linealmente independiente porque, al ser 
+\begin_inset Formula $\frac{H_{i}}{H_{i-1}}=(\overline{y_{i}})$
+\end_inset
+
+ de rango 1, 
+\begin_inset Formula $\phi:A\to(\overline{y_{i}})$
+\end_inset
+
+ dado por 
+\begin_inset Formula $\phi(a)\coloneqq a\overline{y_{i}}$
+\end_inset
+
+ es un isomorfismo y, si 
+\begin_inset Formula $n\in H_{i-1}$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ cumplen 
+\begin_inset Formula $n+ay_{i}=0$
+\end_inset
+
+, 
+\begin_inset Formula $ay_{i}=-n\in H_{i-1}$
+\end_inset
+
+ y 
+\begin_inset Formula $\phi(a)=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $a=0$
+\end_inset
+
+ y 
+\begin_inset Formula $n=0$
+\end_inset
+
+.
+ Por tanto, para 
+\begin_inset Formula $i=t$
+\end_inset
+
+, 
+\begin_inset Formula $H_{t}=G$
+\end_inset
+
+ tiene base 
+\begin_inset Formula $(y_{1},\dots,y_{t})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Todo epimorfismo de 
+\begin_inset Formula $A$
+\end_inset
+
+-módulos 
+\begin_inset Formula $p:M\twoheadrightarrow F$
+\end_inset
+
+ con 
+\begin_inset Formula $F$
+\end_inset
+
+ libre tiene inverso por la derecha, un 
+\begin_inset Formula $f:F\to M$
+\end_inset
+
+ con 
+\begin_inset Formula $p\circ f=1_{F}$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $M=\text{Im}f\oplus\ker p$
+\end_inset
+
+ y 
+\begin_inset Formula $M\cong F\times\ker p$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Dada una base 
+\begin_inset Formula $X=\{x_{i}\}_{i\in I}$
+\end_inset
+
+ de 
+\begin_inset Formula $F$
+\end_inset
+
+, y para 
+\begin_inset Formula $i\in I$
+\end_inset
+
+, 
+\begin_inset Formula $m_{i}\in M$
+\end_inset
+
+ con 
+\begin_inset Formula $p(m_{i})=x_{i}$
+\end_inset
+
+, existe un único 
+\begin_inset Formula $f:F\to M$
+\end_inset
+
+ con 
+\begin_inset Formula $f(x_{i})=m_{i}$
+\end_inset
+
+ para cada 
+\begin_inset Formula $i$
+\end_inset
+
+, luego el homomorfismo 
+\begin_inset Formula $p\circ f$
+\end_inset
+
+ es la identidad sobre los elementos de 
+\begin_inset Formula $X$
+\end_inset
+
+ y por tanto sobre todos los de 
+\begin_inset Formula $F$
+\end_inset
+
+.
+ Para la descomposición, 
+\begin_inset Formula $\text{Im}f\cap\ker p=0$
+\end_inset
+
+ porque sus elementos son de la forma 
+\begin_inset Formula $f(x)$
+\end_inset
+
+ con 
+\begin_inset Formula $x\in F$
+\end_inset
+
+ y 
+\begin_inset Formula $p(f(x))=0$
+\end_inset
+
+, pero 
+\begin_inset Formula $p(f(x))=x$
+\end_inset
+
+, e 
+\begin_inset Formula $\text{Im}f+\ker p=M$
+\end_inset
+
+ porque, para 
+\begin_inset Formula $m\in M$
+\end_inset
+
+, 
+\begin_inset Formula $m=m-f(p(m))+f(p(m))$
+\end_inset
+
+ con 
+\begin_inset Formula $f(p(m))\in\text{Im}f$
+\end_inset
+
+ y 
+\begin_inset Formula $m-f(p(m))\in\ker p$
+\end_inset
+
+ ya que 
+\begin_inset Formula $p(m-f(p(m)))=p(m)-p(m)=0$
+\end_inset
+
+.
+ Finalmente, como 
+\begin_inset Formula $f$
+\end_inset
+
+ es inyectiva, su restricción a la imagen es un isomorfismo e 
+\begin_inset Formula $\text{Im}f\cong F$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Grupos abelianos
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{reminder}{GyA}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+orden
+\series default
+ de [un grupo] 
+\begin_inset Formula $G$
+\end_inset
+
+ al cardinal del conjunto.
+ [...]
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un anillo, 
+\begin_inset Formula $(A,+)$
+\end_inset
+
+ es su 
+\series bold
+grupo aditivo
+\series default
+, que es abeliano, y 
+\begin_inset Formula $(A^{*},\cdot)$
+\end_inset
+
+ es su 
+\series bold
+grupo de unidades
+\series default
+, que es abeliano cuando el anillo es conmutativo.
+ [...]
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+orden
+\series default
+ de 
+\begin_inset Formula $a\in G$
+\end_inset
+
+ al orden de 
+\begin_inset Formula $\langle a\rangle$
+\end_inset
+
+, 
+\begin_inset Formula $|a|\coloneqq|\langle a\rangle|$
+\end_inset
+
+, y escribimos 
+\begin_inset Formula $\langle a\rangle_{n}$
+\end_inset
+
+ para referirnos a 
+\begin_inset Formula $\langle a\rangle$
+\end_inset
+
+ indicando que tiene orden 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ El orden de 
+\begin_inset Formula $a$
+\end_inset
+
+ divide al de 
+\begin_inset Formula $G$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f:\mathbb{Z}\to G$
+\end_inset
+
+ el homomorfismo dado por 
+\begin_inset Formula $f(n)\coloneqq a^{n}$
+\end_inset
+
+, 
+\begin_inset Formula $\ker f=n\mathbb{Z}$
+\end_inset
+
+ para algún 
+\begin_inset Formula $n\geq0$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $n=0$
+\end_inset
+
+, 
+\begin_inset Formula $f$
+\end_inset
+
+ es inyectivo y 
+\begin_inset Formula $(\mathbb{Z},+)\cong\langle a\rangle$
+\end_inset
+
+, y en otro caso 
+\begin_inset Formula $\mathbb{Z}_{n}\cong\langle a\rangle$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $n=|a|$
+\end_inset
+
+ y 
+\begin_inset Formula $a^{n}=1\iff|a|\mid n$
+\end_inset
+
+.
+ De aquí, 
+\begin_inset Formula $a^{k}=a^{l}\iff k\equiv l\bmod n$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $|a|$
+\end_inset
+
+ es el menor entero positivo con 
+\begin_inset Formula $a^{n}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $a$
+\end_inset
+
+ tiene orden finito y 
+\begin_inset Formula $n>0$
+\end_inset
+
+, 
+\begin_inset Formula 
+\[
+|a^{n}|=\frac{|a|}{\text{mcd}\{|a|,n\}}.
+\]
+
+\end_inset
+
+Si 
+\begin_inset Formula $G=\langle a\rangle$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $G$
+\end_inset
+
+ tiene orden infinito, 
+\begin_inset Formula $G\cong(\mathbb{Z},+)\cong C_{\infty}$
+\end_inset
+
+ y los subgrupos de 
+\begin_inset Formula $G$
+\end_inset
+
+ son los 
+\begin_inset Formula $\langle a^{n}\rangle$
+\end_inset
+
+ con 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $|G|=n$
+\end_inset
+
+, 
+\begin_inset Formula $G\cong(\mathbb{Z}_{n},+)\cong C_{n}$
+\end_inset
+
+ y los subgrupos de 
+\begin_inset Formula $G$
+\end_inset
+
+ son exactamente uno de orden 
+\begin_inset Formula $d$
+\end_inset
+
+ por cada 
+\begin_inset Formula $d\mid n$
+\end_inset
+
+, 
+\begin_inset Formula $\langle a^{n/d}\rangle_{d}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Todos los subgrupos y grupos cociente de 
+\begin_inset Formula $G$
+\end_inset
+
+ son cíclicos.
+\end_layout
+
+\begin_layout Standard
+Así, si 
+\begin_inset Formula $p\in\mathbb{N}$
+\end_inset
+
+ es primo, todos los grupos de orden 
+\begin_inset Formula $p$
+\end_inset
+
+ son isomorfos a 
+\begin_inset Formula $(\mathbb{Z}_{p},+)$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $G=\langle g_{1},\dots,g_{n}\rangle$
+\end_inset
+
+ y 
+\begin_inset Formula $N\unlhd G$
+\end_inset
+
+, 
+\begin_inset Formula $G/N=\langle g_{1}N,\dots,g_{n}N\rangle$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema chino de los restos para grupos:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $G$
+\end_inset
+
+ y 
+\begin_inset Formula $H$
+\end_inset
+
+ son subgrupos cíclicos de órdenes respectivos 
+\begin_inset Formula $n$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+, 
+\begin_inset Formula $G\times H$
+\end_inset
+
+ es cíclico si y sólo si 
+\begin_inset Formula $n$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+ son coprimos.
+ [...]
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $g,h\in G$
+\end_inset
+
+ tienen órdenes respectivos 
+\begin_inset Formula $n$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+ coprimos y 
+\begin_inset Formula $gh=hg$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\langle g,h\rangle$
+\end_inset
+
+ es cíclico de orden 
+\begin_inset Formula $nm$
+\end_inset
+
+.
+ [...]
+\end_layout
+
+\begin_layout Standard
+Dados un grupo 
+\begin_inset Formula $G$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in G$
+\end_inset
+
+, llamamos 
+\series bold
+conjugado
+\series default
+ de 
+\begin_inset Formula $g\in G$
+\end_inset
+
+ por 
+\begin_inset Formula $a$
+\end_inset
+
+ a 
+\begin_inset Formula $g^{a}\coloneqq a^{-1}ga$
+\end_inset
+
+, y conjugado de 
+\begin_inset Formula $X\subseteq G$
+\end_inset
+
+ por 
+\begin_inset Formula $a$
+\end_inset
+
+ a 
+\begin_inset Formula $X^{a}\coloneqq\{x^{a}\}_{x\in X}$
+\end_inset
+
+.
+ Dos elementos 
+\begin_inset Formula $x,y\in G$
+\end_inset
+
+ o conjuntos 
+\begin_inset Formula $x,y\subseteq G$
+\end_inset
+
+ son 
+\series bold
+conjugados
+\series default
+ en 
+\begin_inset Formula $G$
+\end_inset
+
+ si existe 
+\begin_inset Formula $a\in G$
+\end_inset
+
+ con 
+\begin_inset Formula $x^{a}=y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $a\in G$
+\end_inset
+
+, llamamos 
+\series bold
+automorfismo interno
+\series default
+ definido por 
+\begin_inset Formula $a$
+\end_inset
+
+ al automorfismo 
+\begin_inset Formula $\iota_{a}:G\to G$
+\end_inset
+
+ dado por 
+\begin_inset Formula $\iota_{a}(x)\coloneqq x^{a}$
+\end_inset
+
+.
+ Su inverso es 
+\begin_inset Formula $\iota_{a^{-1}}$
+\end_inset
+
+.
+ El conjugado por 
+\begin_inset Formula $a$
+\end_inset
+
+ de un subgrupo de 
+\begin_inset Formula $G$
+\end_inset
+
+ es otro subgrupo de 
+\begin_inset Formula $G$
+\end_inset
+
+ del mismo orden.
+ [...]
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\forall g,a,b\in G,g^{ab}=(g^{a})^{b}$
+\end_inset
+
+, y [...] la relación de ser conjugados es de equivalencia.
+ Las clases de equivalencia se llaman 
+\series bold
+clases de conjugación
+\series default
+ de 
+\begin_inset Formula $G$
+\end_inset
+
+, y llamamos 
+\begin_inset Formula $a^{G}\coloneqq[a]=\{a^{g}\}_{g\in G}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $X$
+\end_inset
+
+ un conjunto.
+ Una 
+\series bold
+acción por la izquierda
+\series default
+ de 
+\begin_inset Formula $G$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+ es una función 
+\begin_inset Formula $\cdot:G\times X\to X$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall x\in X,(\forall g,h\in G,(gh)\cdot x=g\cdot(h\cdot x)\land1\cdot x=x)$
+\end_inset
+
+, y una 
+\series bold
+acción por la derecha
+\series default
+ de 
+\begin_inset Formula $G$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+ es una función 
+\begin_inset Formula $\cdot:X\times G\to X$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall x\in X,(\forall g,h\in G,x\cdot(gh)=(x\cdot g)\cdot h\land x\cdot1=x)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $\cdot:G\times X\to X$
+\end_inset
+
+ es una acción por la izquierda de 
+\begin_inset Formula $G$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+ y 
+\begin_inset Formula $x\in X$
+\end_inset
+
+, llamamos 
+\series bold
+órbita
+\series default
+ de 
+\begin_inset Formula $x$
+\end_inset
+
+ en 
+\begin_inset Formula $G$
+\end_inset
+
+ a 
+\begin_inset Formula $G\cdot x\coloneqq\{g\cdot x\}_{g\in G}$
+\end_inset
+
+ y 
+\series bold
+estabilizador
+\series default
+ de 
+\begin_inset Formula $x$
+\end_inset
+
+ en 
+\begin_inset Formula $G$
+\end_inset
+
+ a 
+\begin_inset Formula $\text{Estab}_{G}(x)\coloneqq\{g\in G\mid g\cdot x=x\}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $\cdot:X\times G\to X$
+\end_inset
+
+ es una acción por la derecha de 
+\begin_inset Formula $G$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+ y 
+\begin_inset Formula $x\in X$
+\end_inset
+
+, llamamos órbita de 
+\begin_inset Formula $x$
+\end_inset
+
+ en 
+\begin_inset Formula $G$
+\end_inset
+
+ a 
+\begin_inset Formula $x\cdot G\coloneqq\{x\cdot g\}_{g\in G}$
+\end_inset
+
+ y estabilizador de 
+\begin_inset Formula $x$
+\end_inset
+
+ en 
+\begin_inset Formula $G$
+\end_inset
+
+ a 
+\begin_inset Formula $\text{Estab}_{G}(x)\coloneqq\{g\in G\mid x\cdot g=x\}$
+\end_inset
+
+.
+ Las órbitas forman una partición de 
+\begin_inset Formula $G$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Llamamos 
+\series bold
+acción por traslación a la izquierda
+\series default
+ a la acción por la izquierda de 
+\begin_inset Formula $G$
+\end_inset
+
+ en 
+\begin_inset Formula $G/H$
+\end_inset
+
+ dada por 
+\begin_inset Formula $g\cdot xH=gxH$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $G\cdot xH=G/H$
+\end_inset
+
+ y 
+\begin_inset Formula 
+\[
+\text{Estab}_{G}(xH)=[...]=H^{x^{-1}}.
+\]
+
+\end_inset
+
+Análogamente llamamos 
+\series bold
+acción por traslación a la derecha
+\series default
+ a la acción por la derecha de 
+\begin_inset Formula $G$
+\end_inset
+
+ en 
+\begin_inset Formula $H\backslash G$
+\end_inset
+
+ dada por 
+\begin_inset Formula $Hx\cdot g=Hxg$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Cuando 
+\begin_inset Formula $H=1$
+\end_inset
+
+, la acción de traslación es de 
+\begin_inset Formula $G$
+\end_inset
+
+ en 
+\begin_inset Formula $G$
+\end_inset
+
+, con 
+\begin_inset Formula $G\cdot x=G$
+\end_inset
+
+ y 
+\begin_inset Formula $\text{Estab}_{G}(x)=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+La 
+\series bold
+acción por conjugación
+\series default
+ de 
+\begin_inset Formula $G$
+\end_inset
+
+ en 
+\begin_inset Formula $G$
+\end_inset
+
+ es la acción por la derecha 
+\begin_inset Formula $x\cdot g\coloneqq x^{g}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $x\cdot G=x^{G}$
+\end_inset
+
+ y 
+\begin_inset Formula $\text{Estab}_{G}(x)=C_{G}(x)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $S$
+\end_inset
+
+ es el conjunto de subgrupos de 
+\begin_inset Formula $G$
+\end_inset
+
+, la 
+\series bold
+acción por conjugación de 
+\begin_inset Formula $G$
+\end_inset
+
+ en sus subgrupos
+\series default
+ es la acción por la derecha de 
+\begin_inset Formula $G$
+\end_inset
+
+ en 
+\begin_inset Formula $S$
+\end_inset
+
+ 
+\begin_inset Formula $H\cdot g=H^{g}$
+\end_inset
+
+.
+ [...]
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ y 
+\begin_inset Formula $X$
+\end_inset
+
+ es un conjunto, 
+\begin_inset Formula $\cdot:S_{n}\times X^{n}\to X^{n}$
+\end_inset
+
+ dada por 
+\begin_inset Formula $\sigma\cdot(x_{1},\dots,x_{n})\coloneqq(x_{\sigma(1)},\dots,x_{\sigma(n)})$
+\end_inset
+
+ es una acción por la izquierda.
+\end_layout
+
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $\cdot:G\times X\to X$
+\end_inset
+
+ una acción por la izquierda, 
+\begin_inset Formula $H\leq G$
+\end_inset
+
+ e 
+\begin_inset Formula $Y\subseteq X$
+\end_inset
+
+, si 
+\begin_inset Formula $\forall h\in H,y\in Y,h\cdot y\in Y$
+\end_inset
+
+, 
+\begin_inset Formula $\cdot|_{H\times Y}$
+\end_inset
+
+ es una acción por la izquierda de 
+\begin_inset Formula $H$
+\end_inset
+
+ en 
+\begin_inset Formula $Y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $G$
+\end_inset
+
+ un grupo actuando sobre un conjunto 
+\begin_inset Formula $X$
+\end_inset
+
+, 
+\begin_inset Formula $x\in X$
+\end_inset
+
+ y 
+\begin_inset Formula $g\in G$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{Estab}_{G}(x)\leq G$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $[G:\text{Estab}_{G}(x)]=|G\cdot x|$
+\end_inset
+
+.
+ En particular, si 
+\begin_inset Formula $G$
+\end_inset
+
+ es finito, 
+\begin_inset Formula $|G\cdot x|\mid|G|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si la acción es por la izquierda, 
+\begin_inset Formula $\text{Estab}_{G}(g\cdot x)=\text{Estab}_{G}(x)^{g^{-1}}$
+\end_inset
+
+, y si es por la derecha, 
+\begin_inset Formula $\text{Estab}_{G}(x\cdot g)=\text{Estab}_{G}(x)^{g}$
+\end_inset
+
+.
+ En particular, si 
+\begin_inset Formula $x,g\in G$
+\end_inset
+
+ y 
+\begin_inset Formula $H\leq G$
+\end_inset
+
+, 
+\begin_inset Formula $C_{G}(x^{g})=C_{G}(x)^{g}$
+\end_inset
+
+ y 
+\begin_inset Formula $N_{G}(H^{g})=N_{G}(H)^{g}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $R$
+\end_inset
+
+ es un conjunto irredundante de representantes de las órbitas, 
+\begin_inset Formula $|X|=\sum_{r\in R}|G\cdot r|=\sum_{r\in R}[G:\text{Estab}_{G}(r)]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así, si 
+\begin_inset Formula $G$
+\end_inset
+
+ es un grupo y 
+\begin_inset Formula $a\in G$
+\end_inset
+
+, 
+\begin_inset Formula $|a^{G}|=[G:C_{G}(a)]$
+\end_inset
+
+, y en particular 
+\begin_inset Formula $a^{G}$
+\end_inset
+
+ es unipuntual si y sólo si 
+\begin_inset Formula $a\in Z(G)$
+\end_inset
+
+.
+ 
+\series bold
+Ecuación de clases:
+\series default
+ Si 
+\begin_inset Formula $G$
+\end_inset
+
+ es finito y 
+\begin_inset Formula $X\subseteq G$
+\end_inset
+
+ contiene exactamente un elemento de cada clase de conjugación con al menos
+ dos elementos, entonces 
+\begin_inset Formula $|G|=|Z(G)|+\sum_{x\in X}[G:C_{G}(x)]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dado un número primo 
+\begin_inset Formula $p$
+\end_inset
+
+, un 
+\series bold
+
+\begin_inset Formula $p$
+\end_inset
+
+-grupo
+\series default
+ es un grupo en que todo elemento tiene orden potencia de 
+\begin_inset Formula $p$
+\end_inset
+
+, y un grupo finito es un 
+\begin_inset Formula $p$
+\end_inset
+
+-grupo si y sólo si su orden es potencia de 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ [...]
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de Cauchy:
+\series default
+ Si 
+\begin_inset Formula $G$
+\end_inset
+
+ es un grupo finito con orden múltiplo de un primo 
+\begin_inset Formula $p$
+\end_inset
+
+, 
+\begin_inset Formula $G$
+\end_inset
+
+ tiene un elemento de orden 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ [...]
+\end_layout
+
+\begin_layout Standard
+Dados un grupo finito 
+\begin_inset Formula $G$
+\end_inset
+
+ y un número primo 
+\begin_inset Formula $p$
+\end_inset
+
+, 
+\begin_inset Formula $H\leq G$
+\end_inset
+
+ es un 
+\series bold
+
+\begin_inset Formula $p$
+\end_inset
+
+-subgrupo de Sylow
+\series default
+ de 
+\begin_inset Formula $G$
+\end_inset
+
+ si es un 
+\begin_inset Formula $p$
+\end_inset
+
+-grupo y 
+\begin_inset Formula $[G:H]$
+\end_inset
+
+ es coprimo con 
+\begin_inset Formula $p$
+\end_inset
+
+, si y sólo si es un 
+\begin_inset Formula $p$
+\end_inset
+
+-grupo y 
+\begin_inset Formula $|H|$
+\end_inset
+
+ es la mayor potencia de 
+\begin_inset Formula $p$
+\end_inset
+
+ que divide a 
+\begin_inset Formula $|G|$
+\end_inset
+
+.
+ Llamamos 
+\begin_inset Formula $s_{p}(G)$
+\end_inset
+
+ al número de 
+\begin_inset Formula $p$
+\end_inset
+
+-subgrupos de Sylow de 
+\begin_inset Formula $G$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teoremas de Sylow:
+\series default
+ Sean 
+\begin_inset Formula $p$
+\end_inset
+
+ un número primo y 
+\begin_inset Formula $G$
+\end_inset
+
+ un grupo finito de orden 
+\begin_inset Formula $n\coloneqq p^{k}m$
+\end_inset
+
+ para ciertos 
+\begin_inset Formula $k,m\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $p\nmid m$
+\end_inset
+
+.
+ Entonces:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $G$
+\end_inset
+
+ tiene al menos un 
+\begin_inset Formula $p$
+\end_inset
+
+-subgrupo de Sylow, que tendrá orden 
+\begin_inset Formula $p^{k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $P$
+\end_inset
+
+ es un 
+\begin_inset Formula $p$
+\end_inset
+
+-subgrupo de Sylow de 
+\begin_inset Formula $G$
+\end_inset
+
+ y 
+\begin_inset Formula $Q$
+\end_inset
+
+ es un 
+\begin_inset Formula $p$
+\end_inset
+
+-subgrupo de 
+\begin_inset Formula $G$
+\end_inset
+
+, existe 
+\begin_inset Formula $g\in G$
+\end_inset
+
+ tal que 
+\begin_inset Formula $Q\subseteq P^{g}$
+\end_inset
+
+.
+ En particular, todos los 
+\begin_inset Formula $p$
+\end_inset
+
+-subgrupos de Sylow de 
+\begin_inset Formula $G$
+\end_inset
+
+ son conjugados en 
+\begin_inset Formula $G$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $s_{p}(G)\mid m$
+\end_inset
+
+ y 
+\begin_inset Formula $s_{p}(G)\equiv1\bmod p$
+\end_inset
+
+.
+ [...]
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{reminder}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Submódulos de torsión
+\end_layout
+
+\begin_layout Standard
+Un 
+\begin_inset Formula $x\in_{A}M$
+\end_inset
+
+ es un 
+\series bold
+elemento de torsión
+\series default
+ si 
+\begin_inset Formula $\text{ann}_{A}(x)\neq0$
+\end_inset
+
+, y es un 
+\series bold
+elemento de 
+\begin_inset Formula $p$
+\end_inset
+
+-torsión
+\series default
+ para cierto 
+\begin_inset Formula $p\in{\cal P}$
+\end_inset
+
+ si existe 
+\begin_inset Formula $t\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $\text{ann}_{A}(x)=(p^{t})$
+\end_inset
+
+, si y sólo si existe 
+\begin_inset Formula $s\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $p^{s}x=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+submódulo de torsión
+\series default
+ de 
+\begin_inset Formula $_{A}M$
+\end_inset
+
+ a 
+\begin_inset Formula $t(M)\coloneqq\{x\in M\mid x\text{ es de torsión}\}\leq_{A}M$
+\end_inset
+
+.
+ En efecto, para 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $x,y\in t(M)$
+\end_inset
+
+, sean 
+\begin_inset Formula $b\in\text{ann}_{A}(x)\setminus0$
+\end_inset
+
+ y 
+\begin_inset Formula $c\in\text{ann}_{A}(y)\setminus0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $bc(x-y)=bcx-bcy=0-0=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $0\neq bc\in\text{ann}_{A}(x-y)$
+\end_inset
+
+ y 
+\begin_inset Formula $x-y\in t(M)$
+\end_inset
+
+, y como 
+\begin_inset Formula $abx=0$
+\end_inset
+
+ y 
+\begin_inset Formula $ab\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $ax\in t(M)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Para 
+\begin_inset Formula $p\in{\cal P}$
+\end_inset
+
+, llamamos 
+\series bold
+subgrupo de 
+\begin_inset Formula $p$
+\end_inset
+
+-torsión
+\series default
+ de 
+\begin_inset Formula $_{A}M$
+\end_inset
+
+ a 
+\begin_inset Formula $M(p)\coloneqq\{x\in M\mid x\text{ es de }p\text{-torsión}\}\leq_{A}M$
+\end_inset
+
+.
+ En efecto, para 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $x,y\in M(p)$
+\end_inset
+
+, existe 
+\begin_inset Formula $s\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $p^{s}x=p^{s}y=0$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $ap^{s}x=0$
+\end_inset
+
+ y 
+\begin_inset Formula $p^{s}(x+y)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Para 
+\begin_inset Formula $_{A}M$
+\end_inset
+
+, 
+\begin_inset Formula $t(M)=\bigoplus_{p\in{\cal P}}M(p)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Claramente 
+\begin_inset Formula $\sum_{p\in{\cal P}}M(p)\leq_{A}t(M)$
+\end_inset
+
+.
+ Para ver que la suma es directa, sean 
+\begin_inset Formula $q\in{\cal P}$
+\end_inset
+
+ y 
+\begin_inset Formula $x\in M(q)\cap\sum_{p\in{\cal P}\setminus\{q\}}M(p)$
+\end_inset
+
+, existen 
+\begin_inset Formula $s\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $q^{s}x=0$
+\end_inset
+
+, una descomposición 
+\begin_inset Formula $x=x_{1}+\dots+x_{r}$
+\end_inset
+
+ con cada 
+\begin_inset Formula $x_{i}\in M(p_{i})$
+\end_inset
+
+ para cierto 
+\begin_inset Formula $p_{i}\in{\cal P}\setminus\{q\}$
+\end_inset
+
+ y, para cada 
+\begin_inset Formula $i$
+\end_inset
+
+, 
+\begin_inset Formula $t_{i}\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $p_{i}^{t_{i}}x_{i}=0$
+\end_inset
+
+, con lo que si 
+\begin_inset Formula $a\coloneqq\prod_{i=1}^{r}p_{i}^{t_{i}}$
+\end_inset
+
+, 
+\begin_inset Formula $ax=0$
+\end_inset
+
+, pero 
+\begin_inset Formula $\gcd\{q^{s},a\}=1$
+\end_inset
+
+, por lo que hay una identidad de Bézout 
+\begin_inset Formula $q^{s}b+ac=1$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $x=1x=q^{s}bx+acx=0$
+\end_inset
+
+.
+ Queda ver que 
+\begin_inset Formula $t(M)\subseteq\sum_{p\in{\cal P}}M(p)$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $x\in t(M)\setminus0$
+\end_inset
+
+, 
+\begin_inset Formula $\text{ann}_{A}(x)\triangleleft A$
+\end_inset
+
+, pero como 
+\begin_inset Formula $A$
+\end_inset
+
+ es un DIP existen 
+\begin_inset Formula $(b)\in A\setminus(A^{*}\cup\{0\})$
+\end_inset
+
+ con 
+\begin_inset Formula $\text{ann}_{A}(x)=(b)$
+\end_inset
+
+ y una factorización en irreducibles 
+\begin_inset Formula $b=up_{1}^{t_{1}}\cdots p_{r}^{t_{r}}$
+\end_inset
+
+ con 
+\begin_inset Formula $u\in A^{*}$
+\end_inset
+
+, los 
+\begin_inset Formula $p_{i}\in{\cal P}$
+\end_inset
+
+ irreducibles distintos y los 
+\begin_inset Formula $t_{i}>0$
+\end_inset
+
+, y queremos ver que 
+\begin_inset Formula $x\in\sum_{i=1}^{r}M(p_{i})\subseteq\sum_{p\in{\cal P}}M(p)$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $r=1$
+\end_inset
+
+, 
+\begin_inset Formula $x\in M(p_{1})$
+\end_inset
+
+ y hemos terminado.
+ Si 
+\begin_inset Formula $r>1$
+\end_inset
+
+, por inducción, como 
+\begin_inset Formula $\gcd\{p_{1}^{t_{1}}\cdots p_{r-1}^{t_{r-1}},p_{r}^{t_{r}}\}=1$
+\end_inset
+
+, existe una identidad de Bézout 
+\begin_inset Formula $p_{1}^{t_{1}}\cdots p_{r-1}^{t_{r-1}}b+p_{r}^{t_{r}}c=1$
+\end_inset
+
+ y 
+\begin_inset Formula $x=p_{1}^{t_{1}}\cdots p_{r-1}^{t_{r-1}}bx+p_{r}^{t_{r}}cx$
+\end_inset
+
+, donde el primer sumando es anulado por 
+\begin_inset Formula $p_{r}^{t_{r}}$
+\end_inset
+
+ y por tanto está en 
+\begin_inset Formula $M(p_{r})$
+\end_inset
+
+ y el segundo es anulado por 
+\begin_inset Formula $p_{1}^{t_{1}}\cdots p_{r-1}^{t_{r-1}}$
+\end_inset
+
+ y por tanto está en 
+\begin_inset Formula $\sum_{i=1}^{r}M(p_{i})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $_{A}M\neq0$
+\end_inset
+
+ es finitamente generado, existen 
+\begin_inset Formula $p_{1},\dots,p_{r}\in{\cal P}$
+\end_inset
+
+, unívocamente determinados salvo permutación, tales que 
+\begin_inset Formula $t(M)=\bigoplus_{i=1}^{r}M(p_{i})$
+\end_inset
+
+ y cada 
+\begin_inset Formula $M(p_{i})\neq0$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Como 
+\begin_inset Formula $A$
+\end_inset
+
+ es noetheriano, 
+\begin_inset Formula $M$
+\end_inset
+
+ es noetheriano y 
+\begin_inset Formula $t(M)$
+\end_inset
+
+ es finitamente generado.
+ Como 
+\begin_inset Formula $t(M)=\bigoplus_{p\in{\cal P}}M(p)$
+\end_inset
+
+ es finitamente generado, digamos por 
+\begin_inset Formula $\{x_{1},\dots,x_{s}\}$
+\end_inset
+
+, entendiendo la suma directa como externa, como cada 
+\begin_inset Formula $x_{i}$
+\end_inset
+
+ tiene una cantidad finita de elementos no nulos, 
+\begin_inset Formula $(x_{1},\dots,x_{s})$
+\end_inset
+
+ tiene una cantidad finita de índices no nulos y casi todo 
+\begin_inset Formula $M(p)=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $t(M)=\bigoplus_{i=1}^{r}M(p_{i})$
+\end_inset
+
+ para ciertos 
+\begin_inset Formula $p_{i}$
+\end_inset
+
+.
+ La unicidad se sigue de que los 
+\begin_inset Formula $p_{i}$
+\end_inset
+
+ deben ser justo aquellos con 
+\begin_inset Formula $M(p_{i})\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $_{A}M$
+\end_inset
+
+ es 
+\series bold
+de torsión
+\series default
+ si 
+\begin_inset Formula $M=t(M)$
+\end_inset
+
+, y es 
+\series bold
+de 
+\begin_inset Formula $p$
+\end_inset
+
+-torsión
+\series default
+ para un 
+\begin_inset Formula $p\in{\cal P}$
+\end_inset
+
+ si 
+\begin_inset Formula $M=M(p)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $G$
+\end_inset
+
+ es un grupo abeliano, es finitamente generado de torsión si y sólo si es
+ finito, y para 
+\begin_inset Formula $p$
+\end_inset
+
+ primo positivo, es de 
+\begin_inset Formula $p$
+\end_inset
+
+-torsión finitamente generado si y sólo si es finito y 
+\begin_inset Formula $p^{m}M=0$
+\end_inset
+
+ para cierto 
+\begin_inset Formula $m>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $K$
+\end_inset
+
+ un cuerpo y 
+\begin_inset Formula $M_{(V,f)}$
+\end_inset
+
+ el 
+\begin_inset Formula $K[X]$
+\end_inset
+
+-módulo asociado a un par 
+\begin_inset Formula $(V,f)$
+\end_inset
+
+ de un espacio vectorial y un 
+\begin_inset Formula $K$
+\end_inset
+
+-endomorfismo 
+\begin_inset Formula $V\to V$
+\end_inset
+
+, 
+\begin_inset Formula $M_{(V,f)}$
+\end_inset
+
+ es de torsión finitamente generado si y sólo si 
+\begin_inset Formula $_{K}V$
+\end_inset
+
+ es de dimensión finita, y si 
+\begin_inset Formula $p\in K[X]$
+\end_inset
+
+ es irreducible, 
+\begin_inset Formula $M_{(V,f)}$
+\end_inset
+
+ es finitamente generado de 
+\begin_inset Formula $p$
+\end_inset
+
+-torsión si y sólo si 
+\begin_inset Formula $_{K}V$
+\end_inset
+
+ es de dimensión finita y 
+\begin_inset Formula $p(f)^{m}=0\in\text{End}_{K}(V)$
+\end_inset
+
+ para cierto 
+\begin_inset Formula $m>0$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+¿Qué será 
+\begin_inset Formula $p(f)^{m}$
+\end_inset
+
+?
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Parte libre de torsión de un módulo finitamente generado
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $_{A}F$
+\end_inset
+
+ es 
+\series bold
+libre de torsión
+\series default
+ si 
+\begin_inset Formula $t(F)=0$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+parte libre de torsión
+\series default
+ de 
+\begin_inset Formula $_{A}M$
+\end_inset
+
+ a 
+\begin_inset Formula $\frac{M}{t(M)}$
+\end_inset
+
+, que es libre de torsión.
+ 
+\series bold
+Demostración:
+\series default
+ Queremos ver que, para 
+\begin_inset Formula $\overline{x}\in\frac{M}{t(M)}\setminus0$
+\end_inset
+
+ es 
+\begin_inset Formula $\text{ann}_{A}(\overline{x})=0$
+\end_inset
+
+.
+ Sean entonces 
+\begin_inset Formula $x\in M$
+\end_inset
+
+ con 
+\begin_inset Formula $\text{ann}_{A}(\overline{x})\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in A\setminus0$
+\end_inset
+
+ con 
+\begin_inset Formula $a\overline{x}=0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $ax\in t(M)$
+\end_inset
+
+ y existe 
+\begin_inset Formula $b\in A\setminus0$
+\end_inset
+
+ con 
+\begin_inset Formula $bax=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $x\in t(M)$
+\end_inset
+
+ y 
+\begin_inset Formula $\overline{x}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Todo 
+\begin_inset Formula $A$
+\end_inset
+
+-módulo libre es libre de torsión, pues es isomorfo a un 
+\begin_inset Formula $A^{(I)}$
+\end_inset
+
+ y, si hubiera un 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ y un 
+\begin_inset Formula $v\in A^{(I)}$
+\end_inset
+
+ con 
+\begin_inset Formula $av=0$
+\end_inset
+
+, como estamos en un dominio, 
+\begin_inset Formula $a=0$
+\end_inset
+
+ o 
+\begin_inset Formula $v=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+:
+\end_layout
+
+\begin_layout Enumerate
+Un 
+\begin_inset Formula $A$
+\end_inset
+
+-módulo es finitamente generado y libre de torsión si y sólo si es libre
+ de rango finito.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Para 
+\begin_inset Formula $_{A}F=0$
+\end_inset
+
+ es obvio.
+ Si 
+\begin_inset Formula $_{A}F\neq0$
+\end_inset
+
+, sean 
+\begin_inset Formula $X=\{x_{1},\dots,x_{n}\}$
+\end_inset
+
+ un generador finito de 
+\begin_inset Formula $F$
+\end_inset
+
+ y 
+\begin_inset Formula $S=\{x_{1},\dots,x_{k}\}$
+\end_inset
+
+ con 
+\begin_inset Formula $k\leq n$
+\end_inset
+
+ un subconjunto linealmente independiente maximal, 
+\begin_inset Formula $G\coloneqq(x_{1},\dots,x_{k})\leq_{A}F$
+\end_inset
+
+ es libre con base 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\frac{F}{G}$
+\end_inset
+
+ es finitamente generado, y queremos ver que es de torsión.
+ Para 
+\begin_inset Formula $x\in X\setminus S$
+\end_inset
+
+, 
+\begin_inset Formula $S\cup\{x\}$
+\end_inset
+
+ no es linealmente independiente, luego existen 
+\begin_inset Formula $a_{1},\dots,a_{k},a\in A$
+\end_inset
+
+ no todos nulos con 
+\begin_inset Formula $a_{1}x_{1}+\dots+a_{k}x_{k}=ax$
+\end_inset
+
+, lo que implica que 
+\begin_inset Formula $a\neq0$
+\end_inset
+
+ y que 
+\begin_inset Formula $ax\in(X)=G$
+\end_inset
+
+, luego 
+\begin_inset Formula $a\overline{x}=0$
+\end_inset
+
+ con 
+\begin_inset Formula $a\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $\overline{x}\in t(\frac{F}{G})$
+\end_inset
+
+.
+ Entonces, como 
+\begin_inset Formula $\frac{F}{G}=(\overline{X})=(\overline{x_{1}},\dots,\overline{x_{n}})=(\overline{0},\dots,\overline{0},\overline{x_{k+1}},\dots,\overline{x_{n}})=(\overline{X\setminus S})$
+\end_inset
+
+, 
+\begin_inset Formula $X\subseteq t(\frac{F}{G})$
+\end_inset
+
+ y 
+\begin_inset Formula $\frac{F}{G}=t(\frac{F}{G})$
+\end_inset
+
+.
+ Por tanto, para 
+\begin_inset Formula $i\in\{k+1,\dots,n\}$
+\end_inset
+
+ existe 
+\begin_inset Formula $a_{i}\in A\setminus0$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{i}\overline{x_{i}}=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $r\coloneqq a_{k+1}\cdots a_{n}\neq0$
+\end_inset
+
+ cumple 
+\begin_inset Formula $rx\in G$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x\in X$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $rF\subseteq G$
+\end_inset
+
+, pero 
+\begin_inset Formula $F\to rF$
+\end_inset
+
+ dada por 
+\begin_inset Formula $z\mapsto rz$
+\end_inset
+
+ es un 
+\begin_inset Formula $A$
+\end_inset
+
+-isomorfismo ya que es un epimorfismo y, como 
+\begin_inset Formula $r$
+\end_inset
+
+ es libre de torsión, 
+\begin_inset Formula $z\neq0\implies rz\neq0$
+\end_inset
+
+.
+ Entonces, como 
+\begin_inset Formula $G$
+\end_inset
+
+ es libre y 
+\begin_inset Formula $rF\leq_{A}G$
+\end_inset
+
+, 
+\begin_inset Formula $rF$
+\end_inset
+
+ es libre y por tanto 
+\begin_inset Formula $F$
+\end_inset
+
+ también con 
+\begin_inset Formula $\text{rg}F\leq\text{rg}G=k$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Es libre de torsión por ser libre y es finitamente generado por ser de rango
+ finito.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Todo 
+\begin_inset Formula $_{A}M$
+\end_inset
+
+ finitamente generado admite una descomposición en suma directa interna
+ 
+\begin_inset Formula $M=t(M)\oplus L$
+\end_inset
+
+ con 
+\begin_inset Formula $_{A}L$
+\end_inset
+
+ libre de rango finito.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\frac{M}{t(M)}$
+\end_inset
+
+ es finitamente generado y libre de torsión, y por el apartado anterior
+ es libre de rango finito, luego la proyección canónica 
+\begin_inset Formula $p:M\twoheadrightarrow\frac{M}{t(M)}$
+\end_inset
+
+ tiene inversa por la derecha 
+\begin_inset Formula $\alpha:\frac{M}{t(M)}\to M$
+\end_inset
+
+ y si 
+\begin_inset Formula $L\coloneqq\text{Im}\alpha\cong\frac{M}{t(M)}$
+\end_inset
+
+, 
+\begin_inset Formula $M=\ker p\oplus\text{Im}\alpha=t(M)\oplus F$
+\end_inset
+
+ con 
+\begin_inset Formula $F$
+\end_inset
+
+ libre de rango finito.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $_{A}F$
+\end_inset
+
+ es libre de rango finito, 
+\begin_inset Formula $|S|=\text{rg}F$
+\end_inset
+
+ para todo 
+\begin_inset Formula $S\subseteq F$
+\end_inset
+
+ linealmente independiente maximal.
+ 
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $S$
+\end_inset
+
+ es finito porque, de no serlo, 
+\begin_inset Formula $G\coloneqq(S)$
+\end_inset
+
+ sería un submódulo libre de rango infinito de uno de rango finito, y como
+ 
+\begin_inset Formula $\frac{F}{G}$
+\end_inset
+
+ es finitamente generado con un cierto generador 
+\begin_inset Formula $\{\overline{y_{1}},\dots,\overline{y_{s}}\}$
+\end_inset
+
+, 
+\begin_inset Formula $X\coloneqq S\cup\{y_{1},\dots,y_{s}\}$
+\end_inset
+
+ es un generador finito de 
+\begin_inset Formula $F$
+\end_inset
+
+ en que 
+\begin_inset Formula $S$
+\end_inset
+
+ es linealmente independiente maximal.
+ Entonces, por un argumento como el del primer apartado, existe 
+\begin_inset Formula $r\in F$
+\end_inset
+
+ con 
+\begin_inset Formula $rF\leq_{A}G\leq_{A}F$
+\end_inset
+
+ y 
+\begin_inset Formula $rF\cong F$
+\end_inset
+
+, pero como 
+\begin_inset Formula $G$
+\end_inset
+
+ es libre, 
+\begin_inset Formula $rF\cong F$
+\end_inset
+
+ también y 
+\begin_inset Formula $\text{rg}F\leq\text{rg}G$
+\end_inset
+
+, y como ahora 
+\begin_inset Formula $F$
+\end_inset
+
+ es libre, 
+\begin_inset Formula $\text{rg}G\leq\text{rg}F$
+\end_inset
+
+, luego 
+\begin_inset Formula $\text{rg}F=\text{rg}G=|S|$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Módulos finitamente generados de 
+\begin_inset Formula $p$
+\end_inset
+
+-torsión sobre un DIP
+\end_layout
+
+\begin_layout Standard
+En un anillo conmutativo unitario 
+\begin_inset Formula $A$
+\end_inset
+
+ arbitrario, 
+\begin_inset Formula $N\leq_{A}M$
+\end_inset
+
+ es un 
+\series bold
+submódulo esencial
+\series default
+ de 
+\begin_inset Formula $M$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall L\leq_{A}M,(L\neq0\implies L\cap N\neq0)$
+\end_inset
+
+, y un monomorfismo de 
+\begin_inset Formula $A$
+\end_inset
+
+-módulos 
+\begin_inset Formula $f:L\rightarrowtail M$
+\end_inset
+
+ es un 
+\series bold
+monomorfismo esencial
+\series default
+ si su imagen es un submódulo esencial de 
+\begin_inset Formula $M$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un anillo conmutativo unitario arbitrario:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $N\leq_{A}M$
+\end_inset
+
+, un 
+\series bold
+pseudocomplemento
+\series default
+ de 
+\begin_inset Formula $N$
+\end_inset
+
+ en 
+\begin_inset Formula $M$
+\end_inset
+
+ es un elemento maximal 
+\begin_inset Formula $X$
+\end_inset
+
+ de 
+\begin_inset Formula ${\cal C}_{N}(M)\coloneqq\{L\leq_{A}M:L\cap N=0\}$
+\end_inset
+
+ por inclusión, que siempre existe.
+ El homomorfismo 
+\begin_inset Formula $f:N\hookrightarrow M\overset{\pi}{\twoheadrightarrow}\frac{M}{X}$
+\end_inset
+
+ es un monomorfismo esencial, y es un isomorfismo si y sólo si 
+\begin_inset Formula $X$
+\end_inset
+
+ es complemento directo de 
+\begin_inset Formula $N$
+\end_inset
+
+ en 
+\begin_inset Formula $M$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+La existencia es por el lema de Zorn.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un dominio, los ideales (
+\begin_inset Formula $A$
+\end_inset
+
+-submódulos) esenciales de 
+\begin_inset Formula $A$
+\end_inset
+
+ son precisamente los ideales no nulos.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un dominio, 
+\begin_inset Formula $_{A}F$
+\end_inset
+
+ es libre y 
+\begin_inset Formula $N\leq_{A}F$
+\end_inset
+
+ contiene un subconjunto linealmente independiente maximal de 
+\begin_inset Formula $F$
+\end_inset
+
+ entonces 
+\begin_inset Formula $N$
+\end_inset
+
+ es un submódulo esencial de 
+\begin_inset Formula $F$
+\end_inset
+
+.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $_{A}M$
+\end_inset
+
+ es finitamente generado de 
+\begin_inset Formula $p$
+\end_inset
+
+-torsión para cierto 
+\begin_inset Formula $p\in{\cal P}$
+\end_inset
+
+, existen 
+\begin_inset Formula $r>0$
+\end_inset
+
+ y 
+\begin_inset Formula $0<n_{1}\leq\dots\leq n_{r}$
+\end_inset
+
+ únicos tales que 
+\begin_inset Formula $M\cong\frac{A}{(p^{n_{1}})}\oplus\dots\oplus\frac{A}{(p^{n_{r}})}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ El enunciado se puede reescribir cambiando 
+\begin_inset Formula $0<n_{1}\leq\dots\leq n_{r}$
+\end_inset
+
+ por 
+\begin_inset Formula $n_{1}\geq\dots\geq n_{r}>0$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $X=\{x_{1},\dots,x_{k}\}$
+\end_inset
+
+ un generador de 
+\begin_inset Formula $M$
+\end_inset
+
+, para la existencia hacemos inducción en 
+\begin_inset Formula $k$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $k=1$
+\end_inset
+
+, 
+\begin_inset Formula $M=(x_{1})$
+\end_inset
+
+ y, como 
+\begin_inset Formula $x_{1}$
+\end_inset
+
+ es de 
+\begin_inset Formula $p$
+\end_inset
+
+-torsión, existe 
+\begin_inset Formula $n>0$
+\end_inset
+
+ con 
+\begin_inset Formula $\text{ann}_{A}(x_{1})=(p^{n})$
+\end_inset
+
+, luego 
+\begin_inset Formula $\frac{A}{(p^{n})}=\frac{A}{\text{ann}_{A}(x_{1})}\cong(x_{1})=M$
+\end_inset
+
+ por el primer teorema de isomorfía sobre 
+\begin_inset Formula $a\mapsto ax_{1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $k>1$
+\end_inset
+
+, sea 
+\begin_inset Formula $n_{1}>0$
+\end_inset
+
+ mínimo con 
+\begin_inset Formula $p^{n_{1}}x_{i}=0$
+\end_inset
+
+ para todo 
+\begin_inset Formula $i$
+\end_inset
+
+, entonces 
+\begin_inset Formula $p^{n_{1}}M=0\neq p^{n_{1}-1}M$
+\end_inset
+
+, y podemos suponer 
+\begin_inset Formula $p^{n_{1}-1}x_{1}\neq0$
+\end_inset
+
+.
+ Sean ahora 
+\begin_inset Formula $Z$
+\end_inset
+
+ un pseudocomplemento de 
+\begin_inset Formula $(x_{1})$
+\end_inset
+
+ en 
+\begin_inset Formula $M$
+\end_inset
+
+ y 
+\begin_inset Formula $f:(x_{1})\hookrightarrow M\overset{\pi}{\twoheadrightarrow}\frac{M}{Z}$
+\end_inset
+
+ un monomorfismo esencial, 
+\begin_inset Formula $p^{n_{1}}\frac{M}{Z}=\{\overline{p^{n_{1}}m}\}_{m\in\mathbb{Z}}=0$
+\end_inset
+
+ y 
+\begin_inset Formula $p^{n_{1}-1}f(x_{1})=f(p^{n_{1}-1}x_{1})\neq0$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Supongamos 
+\begin_inset Formula $(y_{1}\coloneqq f(x_{1}))\subsetneq\frac{M}{Z}$
+\end_inset
+
+.
+ Sean 
+\begin_inset Formula $\xi\in M\setminus(y_{1})$
+\end_inset
+
+ y 
+\begin_inset Formula $k\coloneqq\min\{i\in\mathbb{N}^{*}\mid p^{i}\xi\in(y_{1})\}$
+\end_inset
+
+, que existe porque 
+\begin_inset Formula $p^{n_{1}}\xi=0\in(y_{1})$
+\end_inset
+
+, entonces 
+\begin_inset Formula $p^{k}\xi\in(y_{1})$
+\end_inset
+
+ y 
+\begin_inset Formula $p^{k-1}\xi\notin(y_{1})$
+\end_inset
+
+, luego 
+\begin_inset Formula $z\coloneqq p^{k-1}\xi\in M\setminus(y_{1})$
+\end_inset
+
+ cumple 
+\begin_inset Formula $pz\in(y_{1})$
+\end_inset
+
+ y existe 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $pz=ay_{1}$
+\end_inset
+
+.
+ En la factorización 
+\begin_inset Formula $a\eqqcolon p^{t}b$
+\end_inset
+
+ con 
+\begin_inset Formula $t\in\mathbb{N}$
+\end_inset
+
+ y 
+\begin_inset Formula $p\nmid b$
+\end_inset
+
+, se tiene 
+\begin_inset Formula $t>0$
+\end_inset
+
+.
+ En efecto, si no lo fuera sería 
+\begin_inset Formula $pz=bx$
+\end_inset
+
+ con 
+\begin_inset Formula $b$
+\end_inset
+
+ y 
+\begin_inset Formula $p$
+\end_inset
+
+ coprimos, pero como 
+\begin_inset Formula $p^{n_{1}}\frac{M}{Z}=0$
+\end_inset
+
+, 
+\begin_inset Formula $\frac{M}{Z}$
+\end_inset
+
+ se puede ver como un 
+\begin_inset Formula $\frac{A}{(p^{n_{1}})}$
+\end_inset
+
+-módulo y entonces 
+\begin_inset Formula $\overline{b}=b+(p^{n_{1}})$
+\end_inset
+
+ es unidad de 
+\begin_inset Formula $\frac{A}{(p^{n_{1}})}$
+\end_inset
+
+ y 
+\begin_inset Formula $(\overline{y_{1}})=(\overline{by_{1}})$
+\end_inset
+
+, y por la correspondencia, 
+\begin_inset Formula $(y_{1})=(by_{1})$
+\end_inset
+
+, con lo que existe 
+\begin_inset Formula $a$
+\end_inset
+
+ tal que 
+\begin_inset Formula $aby_{1}=y_{1}$
+\end_inset
+
+ y como 
+\begin_inset Formula $p^{n-1}y_{1}=p^{n-1}aby_{1}\neq0$
+\end_inset
+
+ es 
+\begin_inset Formula $p^{n}z=p^{n-1}by_{1}\neq0$
+\end_inset
+
+, pero 
+\begin_inset Formula $p^{n}M=0\#$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $t>0$
+\end_inset
+
+, luego 
+\begin_inset Formula $pz=p^{t}by_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $p(z-p^{t-1}by_{1})=0$
+\end_inset
+
+.
+ Sea entonces 
+\begin_inset Formula $y'\coloneqq z-p^{t-1}by_{1}$
+\end_inset
+
+, 
+\begin_inset Formula $y'\notin(y_{1})$
+\end_inset
+
+ porque 
+\begin_inset Formula $z\notin(y_{1})$
+\end_inset
+
+ y 
+\begin_inset Formula $py'=0$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $\frac{A}{(p)}\to(y')$
+\end_inset
+
+ dado por 
+\begin_inset Formula $a+(p)\mapsto ay'$
+\end_inset
+
+ es un isomorfismo de 
+\begin_inset Formula $A$
+\end_inset
+
+-módulos y los únicos submódulos de 
+\begin_inset Formula $(y')$
+\end_inset
+
+ son 0 e 
+\begin_inset Formula $(y')$
+\end_inset
+
+, pero 
+\begin_inset Formula $(y_{1})$
+\end_inset
+
+ es esencial por ser la imagen de un monomorfismo esencial, luego 
+\begin_inset Formula $(y_{1})\cap(y')\neq0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $(y_{1})\cap(y')=(y')$
+\end_inset
+
+ e 
+\begin_inset Formula $(y')\subseteq(y_{1})$
+\end_inset
+
+, pero 
+\begin_inset Formula $y'\notin(y_{1})\#$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $\frac{M}{Z}=(y_{1})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Con esto 
+\begin_inset Formula $f$
+\end_inset
+
+ es un isomorfismo y 
+\begin_inset Formula $M=(x_{1})\oplus Z\cong\frac{A}{(p^{n_{1}})}\oplus Z$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $Z\cong\frac{(x_{1})\oplus Z}{(x_{1})}=\frac{M}{(x_{1})}$
+\end_inset
+
+, que es un 
+\begin_inset Formula $A$
+\end_inset
+
+-módulo generado por 
+\begin_inset Formula $\{\overline{x_{2}},\dots,\overline{x_{k}}\}$
+\end_inset
+
+, y por hipótesis de inducción, 
+\begin_inset Formula $Z\cong\frac{A}{(p^{n_{2}})}\oplus\dots\oplus\frac{A}{(p^{n_{r}})}$
+\end_inset
+
+ con 
+\begin_inset Formula $n_{2}\geq\dots\geq n_{r}>0$
+\end_inset
+
+ y se tiene 
+\begin_inset Formula $n_{2}=\min\{s\in\mathbb{N}^{*}\mid p^{s}Z=0\}$
+\end_inset
+
+, pero 
+\begin_inset Formula $p^{n_{1}}Z\subseteq p^{n_{1}}M=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $n_{1}\geq n_{2}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Para la unicidad, si 
+\begin_inset Formula $M\cong\frac{A}{(p^{n_{1}})}\oplus\dots\oplus\frac{A}{(p^{n_{r}})}\cong\frac{A}{(p^{m_{1}})}\oplus\dots\oplus\frac{A}{(p^{m_{s}})}$
+\end_inset
+
+ con 
+\begin_inset Formula $r,s>0$
+\end_inset
+
+, 
+\begin_inset Formula $0<n_{1}\leq\dots\leq n_{r}$
+\end_inset
+
+ y 
+\begin_inset Formula $0<m_{1}\leq\dots\leq m_{s}$
+\end_inset
+
+, 
+\begin_inset Formula $M\cong\frac{A}{(p^{n_{1}})}\times\dots\times\frac{A}{(p^{n_{r}})}$
+\end_inset
+
+ y 
+\begin_inset Formula $\frac{M}{pM}\cong\bigoplus_{i=1}^{r}\frac{A}{(p)}\cong\left(\frac{A}{(p)}\right)^{r}$
+\end_inset
+
+, y análogamente 
+\begin_inset Formula $\frac{M}{pM}\cong\left(\frac{A}{(p)}\right)^{s}$
+\end_inset
+
+, pero como 
+\begin_inset Formula $(p)$
+\end_inset
+
+ es maximal, 
+\begin_inset Formula $\frac{A}{(p)}$
+\end_inset
+
+ es un cuerpo y los isomorfismos son entre 
+\begin_inset Formula $\frac{A}{(p)}$
+\end_inset
+
+-espacios vectoriales, luego 
+\begin_inset Formula $r=s$
+\end_inset
+
+ por la unicidad de la dimensión entre espacios vectoriales.
+ Entonces 
+\begin_inset Formula $n_{1}$
+\end_inset
+
+ es el mínimo 
+\begin_inset Formula $j>0$
+\end_inset
+
+ tal que 
+\begin_inset Formula $p^{j}M$
+\end_inset
+
+ admite una descomposición en menos de 
+\begin_inset Formula $r$
+\end_inset
+
+ sumandos y 
+\begin_inset Formula $m_{1}$
+\end_inset
+
+ también, luego 
+\begin_inset Formula $n_{1}=m_{1}$
+\end_inset
+
+.
+ Por inducción en 
+\begin_inset Formula $r$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $r=1$
+\end_inset
+
+ hemos terminado.
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $r>1$
+\end_inset
+
+, 
+\begin_inset Formula $p^{n_{1}}M\cong\bigoplus_{i\geq2}\frac{(p^{n_{1}})}{(p^{n_{i}})}\cong\bigoplus_{i}\frac{A}{(p^{n_{i}-n_{1}})}$
+\end_inset
+
+ y del mismo modo 
+\begin_inset Formula $p^{n_{1}}M\cong\bigoplus_{i\geq2}\frac{A}{(p^{m_{i}-n_{1}})}$
+\end_inset
+
+, y tomando el mínimo 
+\begin_inset Formula $t$
+\end_inset
+
+ con 
+\begin_inset Formula $n_{t}>n_{1}$
+\end_inset
+
+ y el mínimo 
+\begin_inset Formula $t'$
+\end_inset
+
+ con 
+\begin_inset Formula $n_{t'}>n_{1}$
+\end_inset
+
+, por hipótesis de inducción, 
+\begin_inset Formula $(n_{i}-n_{1})_{i\geq t}=(m_{i}-m_{1})_{i\geq t'}$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $t=t'$
+\end_inset
+
+ y hay exactamente 
+\begin_inset Formula $t-1$
+\end_inset
+
+ apariciones de 
+\begin_inset Formula $\frac{A}{(p^{n_{1}})}$
+\end_inset
+
+ y de 
+\begin_inset Formula $\frac{A}{(p^{m_{1}})}$
+\end_inset
+
+, pero 
+\begin_inset Formula $n_{1}=m_{1}$
+\end_inset
+
+, luego al final cada 
+\begin_inset Formula $n_{i}=m_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Módulos finitamente generados sobre un DIP
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, si 
+\begin_inset Formula $_{A}M\neq0$
+\end_inset
+
+ es finitamente generado, existen un único 
+\begin_inset Formula $r\in\mathbb{N}$
+\end_inset
+
+, el 
+\series bold
+rango libre de torsión
+\series default
+ de 
+\begin_inset Formula $M$
+\end_inset
+
+, y una familia 
+\begin_inset Formula $p_{1}^{n_{11}},\dots,p_{1}^{n_{1r_{1}}},\dots,p_{k}^{n_{k1}},\dots p_{k}^{n_{kr_{k}}}$
+\end_inset
+
+ de 
+\series bold
+divisores elementales
+\series default
+ de 
+\begin_inset Formula $M$
+\end_inset
+
+, única salvo asociados y orden de los 
+\begin_inset Formula $p_{i}$
+\end_inset
+
+, con los 
+\begin_inset Formula $p_{i}$
+\end_inset
+
+ irreducibles no asociados dos a dos y 
+\begin_inset Formula $0<n_{i1}\leq\dots\leq n_{ir_{i}}$
+\end_inset
+
+ para cada 
+\begin_inset Formula $i$
+\end_inset
+
+, de forma que 
+\begin_inset Formula 
+\[
+M\cong A^{r}\oplus\bigoplus_{i=1}^{k}\bigoplus_{j=1}^{r_{i}}\frac{A}{(p_{i}^{n_{ij}})},
+\]
+
+\end_inset
+
+lo que llamamos la 
+\series bold
+descomposición indescomponible
+\series default
+ de 
+\begin_inset Formula $M$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $M\cong t(M)\oplus\frac{M}{t(M)}$
+\end_inset
+
+ con 
+\begin_inset Formula $\frac{M}{t(M)}$
+\end_inset
+
+ libre de rango finito 
+\begin_inset Formula $r$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $M\cong A^{r}\oplus t(M)$
+\end_inset
+
+.
+ Ahora bien, existen irreducibles distintos 
+\begin_inset Formula $p_{1},\dots,p_{k}\in{\cal P}$
+\end_inset
+
+ con 
+\begin_inset Formula $t(M)=M(p_{1})\oplus\dots\oplus M(p_{k})$
+\end_inset
+
+ y cada 
+\begin_inset Formula $M(p_{i})\neq0$
+\end_inset
+
+, pero como cada 
+\begin_inset Formula $M(p_{i})$
+\end_inset
+
+ es finitamente generado de 
+\begin_inset Formula $p_{i}$
+\end_inset
+
+-torsión existen 
+\begin_inset Formula $0<n_{i1}\leq\dots\leq n_{ir_{i}}$
+\end_inset
+
+ con 
+\begin_inset Formula $M(p_{i})\cong\bigoplus_{j=1}^{r_{i}}\frac{A}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+.
+ Para la unicidad, si hay otra descomposición 
+\begin_inset Formula $M\cong A^{s}\oplus\bigoplus_{i=1}^{l}\bigoplus_{j=1}^{s_{i}}\frac{A}{(q_{i}^{m_{ij}})}$
+\end_inset
+
+ de la misma forma, donde podemos suponer que los 
+\begin_inset Formula $q_{i}\in{\cal P}$
+\end_inset
+
+, la parte libre de torsión de la suma es isomorfa a 
+\begin_inset Formula $A^{s}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\frac{M}{t(M)}\cong A^{s}$
+\end_inset
+
+, y la parte de torsión isomorfa al sumando derecho y a 
+\begin_inset Formula $t(M)=M(p_{1})\oplus\dots\oplus M(p_{k})=M(q_{1})\oplus\dots\oplus M(q_{l})$
+\end_inset
+
+, luego por unicidad queda 
+\begin_inset Formula $\{p_{1},\dots,p_{k}\}=\{q_{1},\dots,q_{l}\}$
+\end_inset
+
+, 
+\begin_inset Formula $k=l$
+\end_inset
+
+ y, reordenando, cada 
+\begin_inset Formula $p_{i}=q_{i}$
+\end_inset
+
+.Entonces, como cada 
+\begin_inset Formula $M(p_{i})$
+\end_inset
+
+ es de 
+\begin_inset Formula $p_{i}$
+\end_inset
+
+-torsión, por la proposición anterior es 
+\begin_inset Formula $(n_{i1},\dots,n_{ir_{i}})=(m_{i1},\dots,m_{ir_{i}})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $_{A}M$
+\end_inset
+
+ es finitamente generado, existe una descomposición 
+\begin_inset Formula $M=L\oplus\bigoplus_{i=1}^{k}\bigoplus_{j=1}^{r_{i}}M_{ij}$
+\end_inset
+
+ como suma directa interna con 
+\begin_inset Formula $L$
+\end_inset
+
+ libre de rango igual al rango libre de torsión de 
+\begin_inset Formula $M$
+\end_inset
+
+ y cada 
+\begin_inset Formula $M_{ij}\cong\frac{A}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+, siendo los 
+\begin_inset Formula $p_{i}^{n_{ij}}$
+\end_inset
+
+ los divisores elementales de 
+\begin_inset Formula $M$
+\end_inset
+
+.
+ En efecto, por el teorema hay un isomorfismo 
+\begin_inset Formula $\phi:A^{r}\oplus\bigoplus_{i=1}^{k}\bigoplus_{j=1}^{r_{i}}\frac{A}{(p_{i}^{n_{ij}})}\to M$
+\end_inset
+
+ y basta tomar 
+\begin_inset Formula $L\coloneqq\phi(A^{r})$
+\end_inset
+
+ y 
+\begin_inset Formula $M_{ij}\coloneqq\phi(\frac{A}{(p_{i}^{n_{ij}})})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, si 
+\begin_inset Formula $_{A}M\neq0$
+\end_inset
+
+ es finitamente generado, existe un único 
+\begin_inset Formula $r\in\mathbb{N}^{*}$
+\end_inset
+
+ y una única secuencia 
+\begin_inset Formula $d_{1},\dots,d_{t}\in A\setminus(A^{*}\cup\{0\})$
+\end_inset
+
+ de 
+\series bold
+factores invariantes
+\series default
+ de 
+\begin_inset Formula $M$
+\end_inset
+
+, única salvo asociados, tal que 
+\begin_inset Formula $d_{1}\mid\dots\mid d_{t}$
+\end_inset
+
+ y 
+\begin_inset Formula 
+\[
+M\cong A^{r}\oplus\bigoplus_{j=1}^{t}\frac{A}{(d_{j})},
+\]
+
+\end_inset
+
+y de hecho 
+\begin_inset Formula $r$
+\end_inset
+
+ es el rango libre de torsión de 
+\begin_inset Formula $A$
+\end_inset
+
+ y si 
+\begin_inset Formula $(p_{i}^{n_{ij}})_{1\leq i\leq k}^{1\leq j\leq r_{i}}$
+\end_inset
+
+ son los divisores elementales de 
+\begin_inset Formula $M$
+\end_inset
+
+, 
+\begin_inset Formula $t\coloneqq\max_{i}r_{i}$
+\end_inset
+
+ y cada 
+\begin_inset Formula $d_{j}=\prod_{i}p_{i}^{n_{i,r_{i}-t+j}}$
+\end_inset
+
+, tomando 
+\begin_inset Formula $n_{ij}\coloneqq0$
+\end_inset
+
+ para 
+\begin_inset Formula $j<0$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Claramente 
+\begin_inset Formula $d_{1}\mid\dots\mid d_{t}$
+\end_inset
+
+ y, como 
+\begin_inset Formula $\frac{A}{(d_{j})}\cong\bigoplus_{i}\frac{A}{(p_{i}^{n_{i,r_{i}-t+j}})}$
+\end_inset
+
+, 
+\begin_inset Formula $\bigoplus_{j=1}^{t}\frac{A}{(d_{j})}\cong\bigoplus_{i=1}^{k}\bigoplus_{j=1}^{r_{i}}\frac{A}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+.
+ Para la unicidad, la de 
+\begin_inset Formula $r$
+\end_inset
+
+ es como en el teorema anterior, y para la del sumando derecho queremos
+ ver que toda descomposición 
+\begin_inset Formula $t(M)\cong\bigoplus_{j=1}^{u}\frac{A}{(\delta_{j})}$
+\end_inset
+
+ con los 
+\begin_inset Formula $\delta_{j}\notin A^{*}\cup\{0\}$
+\end_inset
+
+ y 
+\begin_inset Formula $\delta_{1}\mid\dots\mid\delta_{u}$
+\end_inset
+
+ cumple 
+\begin_inset Formula $t=u$
+\end_inset
+
+ y 
+\begin_inset Formula $\frac{A}{(\delta_{j})}\cong\frac{A}{(d_{j})}$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $\delta_{j}$
+\end_inset
+
+ y 
+\begin_inset Formula $d_{j}$
+\end_inset
+
+ serán asociados.
+ Cada 
+\begin_inset Formula $\delta_{k}$
+\end_inset
+
+ debe ser suma de submódulos de los 
+\begin_inset Formula $\frac{A}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+, pero estos submódulos son de la forma 
+\begin_inset Formula $\frac{A}{(p_{i}^{z})}$
+\end_inset
+
+ para ciertos 
+\begin_inset Formula $z$
+\end_inset
+
+, por lo que finalmente 
+\begin_inset Formula $\delta_{j}$
+\end_inset
+
+ debe ser de la forma 
+\begin_inset Formula $p_{1}^{m_{1j}}\cdots p_{k}^{m_{kj}}$
+\end_inset
+
+ para ciertos 
+\begin_inset Formula $m_{kj}$
+\end_inset
+
+, y claramente para cada 
+\begin_inset Formula $i$
+\end_inset
+
+ es 
+\begin_inset Formula $0\leq m_{i1}\leq\dots\leq m_{iu}$
+\end_inset
+
+.
+ Por el teorema chino de los restos, 
+\begin_inset Formula $\frac{A}{(\delta_{j})}\cong\bigoplus_{i=1}^{k}\frac{A}{(p_{i}^{m_{ij}})}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $t(M)\cong\bigoplus_{i,j}\frac{A}{(p_{i}^{m_{ij}})}$
+\end_inset
+
+, lo que tras eliminar los sumandos nulos y reordenar debe coincidir con
+ la descomposición indescomponible de 
+\begin_inset Formula $t(M)$
+\end_inset
+
+, lo que junto a que 
+\begin_inset Formula $0\leq m_{i1}\leq\dots\leq m_{iu}$
+\end_inset
+
+ determina los 
+\begin_inset Formula $m_{ij}$
+\end_inset
+
+ y por tanto los 
+\begin_inset Formula $\delta_{j}$
+\end_inset
+
+ salvo asociados.
+\end_layout
+
+\begin_layout Standard
+Así, si 
+\begin_inset Formula $_{A}M\neq0$
+\end_inset
+
+ es finitamente generado, se puede expresar como suma directa interna de
+ la forma 
+\begin_inset Formula $L\oplus\bigoplus_{i=1}^{t}M_{i}$
+\end_inset
+
+ con 
+\begin_inset Formula $L$
+\end_inset
+
+ libre de rango igual al rango libre de torsión de 
+\begin_inset Formula $M$
+\end_inset
+
+ y cada 
+\begin_inset Formula $M_{i}\cong\frac{A}{(d_{i})}$
+\end_inset
+
+, siendo los 
+\begin_inset Formula $d_{i}$
+\end_inset
+
+ los factores invariantes de 
+\begin_inset Formula $M$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teoremas de estructura de los grupos abelianos finitos:
+\series default
+ Si 
+\begin_inset Formula $M$
+\end_inset
+
+ es un grupo abeliano finito:
+\end_layout
+
+\begin_layout Enumerate
+Existen números primos 
+\begin_inset Formula $1<p_{1}<\dots<p_{k}$
+\end_inset
+
+ y enteros 
+\begin_inset Formula $0<n_{i1}\leq\dots\leq n_{ir_{i}}$
+\end_inset
+
+ para 
+\begin_inset Formula $i\in\{1,\dots,k\}$
+\end_inset
+
+, únicos, con 
+\begin_inset Formula $M\cong\bigoplus_{i=1}^{k}\bigoplus_{j=1}^{r_{i}}\mathbb{Z}_{p_{i}^{n_{ij}}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Existen enteros 
+\begin_inset Formula $1<d_{1}\mid\dots\mid d_{t}$
+\end_inset
+
+ únicos con 
+\begin_inset Formula $M\cong\bigoplus_{j=1}^{t}\mathbb{Z}_{d_{j}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teoremas de clasificación de endomorfismos de espacios vectoriales:
+\series default
+ Sean 
+\begin_inset Formula $V$
+\end_inset
+
+ un 
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial de dimensión finita y 
+\begin_inset Formula $f:V\to V$
+\end_inset
+
+ un 
+\begin_inset Formula $K$
+\end_inset
+
+-endomorfismo:
+\end_layout
+
+\begin_layout Enumerate
+Existen 
+\begin_inset Formula $p_{1},\dots,p_{k}\in K[X]$
+\end_inset
+
+ mónicos irreducibles distintos y 
+\begin_inset Formula $n_{ij}\in\mathbb{N}^{*}$
+\end_inset
+
+ para 
+\begin_inset Formula $i\in\{1,\dots,k\}$
+\end_inset
+
+ y 
+\begin_inset Formula $j\in\{1,\dots,r_{i}\}$
+\end_inset
+
+, unívocamente determinados, y vectores 
+\begin_inset Formula $v_{ij}\in V$
+\end_inset
+
+, tales que 
+\begin_inset Formula $\bigoplus_{i=1}^{k}\bigoplus_{j=1}^{r_{i}}K\{f^{s}(v_{ij})\}_{s\geq0}$
+\end_inset
+
+ es una descomposición de 
+\begin_inset Formula $V$
+\end_inset
+
+ en suma directa interna de subespacios vectoriales 
+\begin_inset Formula $f$
+\end_inset
+
+-invariantes y cada 
+\begin_inset Formula $p_{i}(f)^{n_{ij}}(v_{ij})=0\neq p_{i}(f)^{n_{ij}-1}(v_{ij})$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $M$
+\end_inset
+
+ el 
+\begin_inset Formula $K[X]$
+\end_inset
+
+-módulo asociado a 
+\begin_inset Formula $(V,f)$
+\end_inset
+
+, 
+\begin_inset Formula $W\leq V$
+\end_inset
+
+ y 
+\begin_inset Formula $N$
+\end_inset
+
+ el 
+\begin_inset Formula $K[X]$
+\end_inset
+
+-submódulo de 
+\begin_inset Formula $M$
+\end_inset
+
+ asociado a 
+\begin_inset Formula $(W,f|_{W})$
+\end_inset
+
+, basta ver que 
+\begin_inset Formula $N\cong\frac{K[X]}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+ si y sólo si existe 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ tal que 
+\begin_inset Formula $W=K\{f^{s}(v)_{s\geq0}\}$
+\end_inset
+
+ y 
+\begin_inset Formula $p_{i}(f)^{n_{ij}}(v)=0\neq p_{i}(f)^{n_{ij}-1}(v)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sean 
+\begin_inset Formula $\phi:\frac{K[X]}{(p_{i}^{n_{ij}})}\to N$
+\end_inset
+
+ el isomorfismo y 
+\begin_inset Formula $v\coloneqq\phi(\overline{1})$
+\end_inset
+
+, 
+\begin_inset Formula $p_{i}^{n_{ij}}\overline{1}=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $0=p_{i}^{n_{ij}}\phi(\overline{1})=p_{i}^{n_{ij}}v=p_{i}(f)^{n_{ij}}(v)$
+\end_inset
+
+ por la definición del 
+\begin_inset Formula $K[X]$
+\end_inset
+
+-módulo, pero 
+\begin_inset Formula $p_{i}^{n_{ij}-1}\overline{1}\neq0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $p_{i}(f)^{n_{ij}-1}(v_{ij})\neq0$
+\end_inset
+
+.
+ Finalmente, como 
+\begin_inset Formula $\frac{K[X]}{(p_{i}^{n_{ij}})}=K\{\overline{1},X\overline{1},\dots,X^{s}\overline{1},\dots\}$
+\end_inset
+
+, 
+\begin_inset Formula $M=K\{f^{s}(v)\}_{s\geq0}$
+\end_inset
+
+ ya que 
+\begin_inset Formula $\phi(X^{s}\overline{1})=X^{s}\phi(\overline{1})=f^{s}(v)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Por la hipótesis y la definición de 
+\begin_inset Formula $N$
+\end_inset
+
+, 
+\begin_inset Formula $N=(v)$
+\end_inset
+
+, pero 
+\begin_inset Formula $v$
+\end_inset
+
+ es anulado por 
+\begin_inset Formula $p_{i}(f)^{n_{ij}}$
+\end_inset
+
+ y por tanto hay un epimorfismo 
+\begin_inset Formula $\psi:\frac{K[X]}{(p_{i}^{n_{ij}})}\twoheadrightarrow K[X]v=N$
+\end_inset
+
+ con 
+\begin_inset Formula $\ker\psi\trianglelefteq\frac{K[X]}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+, pero los únicos ideales de 
+\begin_inset Formula $\frac{K[X]}{(p_{i}^{n_{ij}})}$
+\end_inset
+
+ son 
+\begin_inset Formula $(\overline{p_{i}}^{k})$
+\end_inset
+
+ con 
+\begin_inset Formula $k\in\{0,\dots,n_{ij}\}$
+\end_inset
+
+, y como 
+\begin_inset Formula $p_{i}(f)^{n_{ij}-1}(v)\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $\overline{p_{i}}^{n_{ij}-1}\notin\ker\psi$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $\ker\psi=0$
+\end_inset
+
+ y 
+\begin_inset Formula $\psi$
+\end_inset
+
+ es un isomorfismo.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Existen polinomios mónicos no constantes 
+\begin_inset Formula $d_{1}\mid\dots\mid d_{t}$
+\end_inset
+
+ unívocamente determinados y vectores 
+\begin_inset Formula $v_{j}\in V$
+\end_inset
+
+ tales que 
+\begin_inset Formula $\bigoplus_{i=1}^{t}\text{span}\{f^{s}(v_{j})\}_{s\in\mathbb{N}_{\text{gr}(d_{j})}}$
+\end_inset
+
+ es una descomposición de 
+\begin_inset Formula $V$
+\end_inset
+
+ en subespacios 
+\begin_inset Formula $f$
+\end_inset
+
+-invariantes y cada 
+\begin_inset Formula $d_{j}(f)(v_{j})=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $M$
+\end_inset
+
+ el 
+\begin_inset Formula $K[X]$
+\end_inset
+
+-módulo asociado a 
+\begin_inset Formula $(V,f)$
+\end_inset
+
+, 
+\begin_inset Formula $W\leq V$
+\end_inset
+
+ y 
+\begin_inset Formula $N$
+\end_inset
+
+ el 
+\begin_inset Formula $K[X]$
+\end_inset
+
+-submódulo de 
+\begin_inset Formula $M$
+\end_inset
+
+ asociado a 
+\begin_inset Formula $(W,f|_{W})$
+\end_inset
+
+, basta ver que 
+\begin_inset Formula $N\cong\frac{K[X]}{(d_{j})}$
+\end_inset
+
+ si y sólo si existe 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\{f^{s}(v)\}{}_{s\in\mathbb{N}_{\text{gr}(d_{j})}}$
+\end_inset
+
+ es base de 
+\begin_inset Formula $W$
+\end_inset
+
+ como 
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial y 
+\begin_inset Formula $d_{j}(f)(v)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sean 
+\begin_inset Formula $\phi:\frac{K[X]}{(p_{i}^{n_{ij}})}\to N$
+\end_inset
+
+ el isomorfismo y 
+\begin_inset Formula $v\coloneqq\phi(\overline{1})$
+\end_inset
+
+, 
+\begin_inset Formula $d_{j}\overline{1}=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $0=d_{j}\phi(\overline{1})=d_{j}v=d_{j}(f)(v)$
+\end_inset
+
+, y como 
+\begin_inset Formula $\frac{K[X]}{(d_{j})}=K\{\overline{1},X\overline{1},\dots,X^{\text{gr}d_{j}-1}\overline{1}\}$
+\end_inset
+
+ con 
+\begin_inset Formula $(X^{s}\overline{1})_{s\in\mathbb{N}_{\text{gr}(d_{j})}}$
+\end_inset
+
+ linealmente independiente, 
+\begin_inset Formula $N=K\{f^{s}(v)\}_{s\in\mathbb{N}_{\text{gr}(d_{j})}}$
+\end_inset
+
+ con 
+\begin_inset Formula $(f^{s}(v))_{s\in\mathbb{N}_{\text{gr}(d_{j})}}$
+\end_inset
+
+ linealmente independiente.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $v$
+\end_inset
+
+ es anulado por 
+\begin_inset Formula $p_{i}(f)^{n_{ij}}$
+\end_inset
+
+ y por tanto hay un epimorfismo 
+\begin_inset Formula $\psi:\frac{K[X]}{(d_{j})}\twoheadrightarrow K[X]v=K\{f^{s}(v)\}_{s\in\mathbb{N}}=K\{f^{s}(v)\}_{s\in\mathbb{N}_{\text{gr}(d_{j})}}=N$
+\end_inset
+
+, pero si 
+\begin_inset Formula $p\in K[X]$
+\end_inset
+
+ con 
+\begin_inset Formula $\text{gr}p<\text{gr}d_{j}$
+\end_inset
+
+ cumple 
+\begin_inset Formula $\psi(\overline{p})=p(f)(v)=\sum_{i}p_{i}f^{i}(v)=0$
+\end_inset
+
+, como los 
+\begin_inset Formula $f^{i}(v)$
+\end_inset
+
+ son linealmente independiente, cada 
+\begin_inset Formula $p_{i}=0$
+\end_inset
+
+ y 
+\begin_inset Formula $p=0$
+\end_inset
+
+, y como cada elemento de 
+\begin_inset Formula $\frac{K[X]}{(d_{j})}$
+\end_inset
+
+ tiene un representante de grado menor que el de 
+\begin_inset Formula $d_{j}$
+\end_inset
+
+, 
+\begin_inset Formula $\ker\psi=0$
+\end_inset
+
+ y 
+\begin_inset Formula $\psi$
+\end_inset
+
+ es un isomorfismo.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{reminder}{GyA}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Un grupo cíclico 
+\begin_inset Formula $\langle a\rangle_{n}$
+\end_inset
+
+ es indescomponible si y sólo si tiene orden potencia de primo.
+\end_layout
+
+\begin_layout Standard
+Dado un grupo 
+\begin_inset Formula $G$
+\end_inset
+
+, llamamos 
+\series bold
+exponente
+\series default
+ o 
+\series bold
+periodo
+\series default
+ de 
+\begin_inset Formula $G$
+\end_inset
+
+, 
+\begin_inset Formula $\text{Exp}(G)$
+\end_inset
+
+, al menor 
+\begin_inset Formula $n\in\mathbb{N}^{*}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall g\in G,g^{n}=1$
+\end_inset
+
+, o a 
+\begin_inset Formula $\infty$
+\end_inset
+
+ si este no existe.
+ [...]
+\end_layout
+
+\begin_layout Standard
+Si un grupo es finito tiene periodo finito, y si tiene periodo finito es
+ periódico.
+ Los recíprocos no se cumplen.
+ Todo 
+\begin_inset Formula $p$
+\end_inset
+
+-grupo es periódico, pero no necesariamente finito.
+ [...]
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un grupo abeliano, 
+\begin_inset Formula $B\leq A$
+\end_inset
+
+, 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ y 
+\begin_inset Formula $na=0$
+\end_inset
+
+, en 
+\begin_inset Formula $A/B$
+\end_inset
+
+ es 
+\begin_inset Formula $|a+B|\mid|a|$
+\end_inset
+
+.
+ En general estos órdenes no coinciden.
+ [...]
+\end_layout
+
+\begin_layout Standard
+Dados dos grupos abelianos finitos 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+, una descomposición por suma directa de 
+\begin_inset Formula $A$
+\end_inset
+
+ y una de 
+\begin_inset Formula $B$
+\end_inset
+
+ son 
+\series bold
+semejantes
+\series default
+ si existe una biyección entre los subgrupos en la descomposición de 
+\begin_inset Formula $A$
+\end_inset
+
+ y la de 
+\begin_inset Formula $B$
+\end_inset
+
+ que a cada subgrupo de 
+\begin_inset Formula $A$
+\end_inset
+
+ le asocia uno de 
+\begin_inset Formula $B$
+\end_inset
+
+ isomorfo.
+ [...]
+\end_layout
+
+\begin_layout Standard
+Dos grupos abelianos finitos son isomorfos si y sólo si tienen descomposiciones
+ primarias semejantes, si y sólo si tienen descomposiciones invariantes
+ semejantes, si y sólo si tienen la misma lista de divisores elementales,
+ si y sólo si tienen la misma lista de factores invariantes.
+ [...]
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{reminder}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Determinación de descomposiciones de módulos de torsión finitamente generados
+\end_layout
+
+\begin_layout Standard
+En esta sección, salvo que se indique lo contrario, 
+\begin_inset Formula $M$
+\end_inset
+
+ es un 
+\begin_inset Formula $A$
+\end_inset
+
+-módulo finitamente generado de torsión y 
+\begin_inset Formula $\{p_{1},\dots,p_{k}\}\coloneqq\{p\in{\cal P}\mid M(p)\neq0\}$
+\end_inset
+
+ son sus 
+\series bold
+divisores irreducibles
+\series default
+.
+ 
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $M\neq0$
+\end_inset
+
+ tiene factores invariantes 
+\begin_inset Formula $d_{1}\mid\dots\mid d_{t}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{ann}_{A}(M)=(d_{t})$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\subseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Para 
+\begin_inset Formula $a\in\text{ann}_{A}(M)$
+\end_inset
+
+, como 
+\begin_inset Formula $\frac{A}{(d_{t})}$
+\end_inset
+
+ es isomorfo a un sumando directo de 
+\begin_inset Formula $M$
+\end_inset
+
+, 
+\begin_inset Formula $a\frac{A}{(d_{t})}=0$
+\end_inset
+
+, pero 
+\begin_inset Formula $a\frac{A}{(d_{t})}=\frac{(a)+(d_{t})}{(d_{t})}=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $(a)+(d_{t})\subseteq(d_{t})$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in(d_{t})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\supseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $M\cong\bigoplus_{j=1}^{t}\frac{A}{(d_{j})}$
+\end_inset
+
+ y, como cada 
+\begin_inset Formula $d_{j}\mid d_{t}$
+\end_inset
+
+, 
+\begin_inset Formula $d_{t}M=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $(d_{t})\subseteq\text{ann}_{A}(M)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Un 
+\begin_inset Formula $p\in{\cal P}$
+\end_inset
+
+ es divisor irreducible de 
+\begin_inset Formula $M$
+\end_inset
+
+ si y sólo si lo es de 
+\begin_inset Formula $d_{t}$
+\end_inset
+
+, si y sólo si existe 
+\begin_inset Formula $x\in M\setminus\{0\}$
+\end_inset
+
+ con 
+\begin_inset Formula $px=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Description
+\begin_inset Formula $1\iff2]$
+\end_inset
+
+ Si 
+\begin_inset Formula $(p_{ij})_{1\leq i\leq k}^{1\leq j\leq r_{i}}$
+\end_inset
+
+ son los divisores elementales de 
+\begin_inset Formula $M$
+\end_inset
+
+, 
+\begin_inset Formula $d_{t}=p_{1}^{n_{1r_{1}}}\cdots p_{k}^{n_{kr_{k}}}$
+\end_inset
+
+, luego los divisores irreducibles son los irreducibles de la factorización
+ irreducible de 
+\begin_inset Formula $d_{t}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\implies3]$
+\end_inset
+
+ Si 
+\begin_inset Formula $M(p)\neq0$
+\end_inset
+
+, sea 
+\begin_inset Formula $z\in M(p)\setminus\{0\}$
+\end_inset
+
+ con 
+\begin_inset Formula $\text{ann}_{A}(z)=(p^{s})$
+\end_inset
+
+ y 
+\begin_inset Formula $s$
+\end_inset
+
+ mínimo, 
+\begin_inset Formula $s>0$
+\end_inset
+
+ ya que de lo contrario sería 
+\begin_inset Formula $(p^{s})=A$
+\end_inset
+
+ y 
+\begin_inset Formula $z=1z=0$
+\end_inset
+
+, y 
+\begin_inset Formula $x\coloneqq p^{s-1}z\in M\setminus\{0\}$
+\end_inset
+
+ cumple 
+\begin_inset Formula $px=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $3\implies1]$
+\end_inset
+
+ 
+\begin_inset Formula $x\in M(p)\neq0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Así, si 
+\begin_inset Formula $M$
+\end_inset
+
+ es un grupo abeliano finito, los divisores irreducibles de 
+\begin_inset Formula $M$
+\end_inset
+
+ son los 
+\begin_inset Formula $p>0$
+\end_inset
+
+ que dividen a 
+\begin_inset Formula $|M|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $V\in_{K}\text{Vect}$
+\end_inset
+
+ de dimensión finita y 
+\begin_inset Formula $f\in\text{End}_{K}(V)$
+\end_inset
+
+ con polinomio característico 
+\begin_inset Formula $\varphi\in K[X]$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Teorema de Cayley-Hamilton:
+\series default
+ 
+\begin_inset Formula $\varphi_{f}(f)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $C\in{\cal M}_{n}(K)$
+\end_inset
+
+ la matriz asociada a 
+\begin_inset Formula $f$
+\end_inset
+
+ bajo cualquier base de 
+\begin_inset Formula $V$
+\end_inset
+
+ e 
+\begin_inset Formula $I\coloneqq I_{n}$
+\end_inset
+
+, queremos ver que 
+\begin_inset Formula $\varphi=\det(XI-C)$
+\end_inset
+
+ cumple 
+\begin_inset Formula $\sum_{i=0}^{n}\varphi_{i}C^{i}=0$
+\end_inset
+
+.
+ Por la prueba de la fórmula de la matriz inversa, para toda matriz 
+\begin_inset Formula $A$
+\end_inset
+
+ es 
+\begin_inset Formula $A\cdot\text{adj}(A)^{\intercal}=|A|I$
+\end_inset
+
+, por lo que viendo 
+\begin_inset Formula $XI-C\in{\cal M}_{n}(K[X])$
+\end_inset
+
+ es 
+\begin_inset Formula $(XI-C)\text{adj}(XI-C)^{\intercal}=\varphi I$
+\end_inset
+
+.
+ Como las entradas de 
+\begin_inset Formula $\text{adj}(XI-C)^{\intercal}$
+\end_inset
+
+ son polinomios de grado máximo 
+\begin_inset Formula $n-1$
+\end_inset
+
+, podemos escribir 
+\begin_inset Formula $\text{adj}(XI-C)^{t}\eqqcolon\sum_{i=0}^{n-1}B_{i}X^{i}$
+\end_inset
+
+ con cada 
+\begin_inset Formula $B_{i}\in{\cal M}_{n}(K)$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $(XI-C)\sum_{i=0}^{n-1}B_{i}X^{i}=\sum_{i=0}^{n}\varphi_{i}I$
+\end_inset
+
+.
+ Viendo esta igualdad en 
+\begin_inset Formula ${\cal M}_{n}(K)[X]$
+\end_inset
+
+, igualando coeficientes,
+\begin_inset Formula 
+\begin{align*}
+B_{n-1} & =\varphi_{n}I, & B_{n-2}-CB_{n-1} & =\varphi_{n-1}I, &  & \cdots, & B_{0}-B_{1}C & =\varphi_{1}I, & -B_{0}C & =\varphi_{0}I,
+\end{align*}
+
+\end_inset
+
+y multiplicando la primera igualdad por 
+\begin_inset Formula $C^{n}$
+\end_inset
+
+, la segunda por 
+\begin_inset Formula $C^{n-1}$
+\end_inset
+
+, etc.,
+\begin_inset Formula 
+\begin{align*}
+C^{n}B_{n-1} & =\varphi_{n}I, & C^{n-1}B_{n-2}-C^{n}B_{n-1} & =\varphi_{n-1}I, &  & \dots,
+\end{align*}
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\end_body
+\end_document