From 8a503ffd481589f6bffeb5e48f9c288ca014729b Mon Sep 17 00:00:00 2001
From: Juan Marín Noguera <juan.marinn@um.es>
Date: Mon, 23 Mar 2020 17:34:50 +0100
Subject: ga/n2

---
 ga/n2.lyx | 4753 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 4753 insertions(+)
 create mode 100644 ga/n2.lyx

(limited to 'ga')

diff --git a/ga/n2.lyx b/ga/n2.lyx
new file mode 100644
index 0000000..d288366
--- /dev/null
+++ b/ga/n2.lyx
@@ -0,0 +1,4753 @@
+#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
+\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
+Un 
+\series bold
+dominio
+\series default
+ (
+\series bold
+de integridad
+\series default
+) es un anillo conmutativo en que todos los elementos no nulos son regulares,
+ y un 
+\series bold
+cuerpo
+\series default
+ es uno en que todos los elementos no nulos son invertibles.
+ Un 
+\series bold
+subdominio
+\series default
+ es un subanillo de un dominio que es dominio, y un 
+\series bold
+subcuerpo
+\series default
+ es un subanillo de un cuerpo que es cuerpo.
+ Todo cuerpo es un dominio.
+ Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un anillo conmutativo:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A$
+\end_inset
+
+ es un cuerpo si y sólo si los únicos ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+ son 0 y 
+\begin_inset Formula $A$
+\end_inset
+
+, si y sólo si todo homomorfismo de anillos 
+\begin_inset Formula $A\to B$
+\end_inset
+
+ con 
+\begin_inset Formula $B\neq0$
+\end_inset
+
+ es inyectivo.
+\end_layout
+
+\begin_deeper
+\begin_layout Description
+\begin_inset Formula $[1\implies2]$
+\end_inset
+
+ Sean 
+\begin_inset Formula $I\neq0$
+\end_inset
+
+ un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in I\setminus0$
+\end_inset
+
+, como 
+\begin_inset Formula $A$
+\end_inset
+
+ es cuerpo, 
+\begin_inset Formula $a$
+\end_inset
+
+ es invertible, luego 
+\begin_inset Formula $I=A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[2\implies1]$
+\end_inset
+
+ Si 
+\begin_inset Formula $A$
+\end_inset
+
+ no fuera cuerpo, habría 
+\begin_inset Formula $a\in A\setminus0$
+\end_inset
+
+ no invertible, luego 
+\begin_inset Formula $0\neq(a)\neq A\#$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[2\implies3]$
+\end_inset
+
+ Sea 
+\begin_inset Formula $f:A\to B\neq0$
+\end_inset
+
+ un homomorfismo de anillos, 
+\begin_inset Formula $f(1_{A})=1_{B}\neq0$
+\end_inset
+
+, luego 
+\begin_inset Formula $\ker f\neq A$
+\end_inset
+
+, pero como 
+\begin_inset Formula $\ker f$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $\ker f=0$
+\end_inset
+
+ y 
+\begin_inset Formula $f$
+\end_inset
+
+ es inyectivo.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[3\implies2]$
+\end_inset
+
+ Si hubiera un ideal 
+\begin_inset Formula $I$
+\end_inset
+
+ propio no nulo de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $A/I\neq0$
+\end_inset
+
+ y el homomorfismo proyección 
+\begin_inset Formula $A\to A/I$
+\end_inset
+
+ no es inyectivo porque su núcleo no es nulo.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Un elemento 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ es regular si y sólo si 
+\begin_inset Formula $\forall b\in A,(ab=0\implies b=0)$
+\end_inset
+
+.
+\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
+
+
+\begin_inset Formula $ab=0\implies ab=a0\implies b=0$
+\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
+
+Si 
+\begin_inset Formula $a$
+\end_inset
+
+ no fuera regular, existirían 
+\begin_inset Formula $b,c\in A$
+\end_inset
+
+, 
+\begin_inset Formula $b\neq c$
+\end_inset
+
+, con 
+\begin_inset Formula $ab=ac$
+\end_inset
+
+, luego 
+\begin_inset Formula $a(b-c)=ab-ac=0$
+\end_inset
+
+ pero 
+\begin_inset Formula $b-c\neq0\#$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:char-domain"
+
+\end_inset
+
+
+\begin_inset Formula $A$
+\end_inset
+
+ es un dominio si y sólo si 
+\begin_inset Formula $\forall a,b\in A\setminus\{0\},ab\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Consecuencia de lo anterior.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Todo subanillo de un dominio es un dominio.
+\end_layout
+
+\begin_layout Enumerate
+La característica de un dominio no trivial es 0 o un número primo.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si la característica de 
+\begin_inset Formula $A$
+\end_inset
+
+ es 1, 
+\begin_inset Formula $1=0$
+\end_inset
+
+ y 
+\begin_inset Formula $A=0\#$
+\end_inset
+
+.
+ Si es 
+\begin_inset Formula $n>1$
+\end_inset
+
+ no primo, pongamos 
+\begin_inset Formula $n=pq$
+\end_inset
+
+ con 
+\begin_inset Formula $0<p\neq n$
+\end_inset
+
+ primo, entonces 
+\begin_inset Formula $p1\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $q1\neq0$
+\end_inset
+
+, pero 
+\begin_inset Formula $(p1)(q1)=((p1)q)1=(pq)1=0$
+\end_inset
+
+, luego por 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:char-domain"
+plural "false"
+caps "false"
+noprefix "false"
+
+\end_inset
+
+, 
+\begin_inset Formula $A$
+\end_inset
+
+ no es un dominio.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Newpage pagebreak
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Algunos dominios:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+ es un dominio no cuerpo, y 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ son cuerpos.
+\end_layout
+
+\begin_layout Enumerate
+Para 
+\begin_inset Formula $n\geq2$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{Z}_{n}$
+\end_inset
+
+ es un dominio si y sólo si es un cuerpo, si y sólo si 
+\begin_inset Formula $n$
+\end_inset
+
+ es primo.
+\end_layout
+
+\begin_deeper
+\begin_layout Description
+\begin_inset Formula $[1\implies3]$
+\end_inset
+
+ 
+\begin_inset Formula $\mathbb{Z}_{n}$
+\end_inset
+
+ tiene característica 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[3\implies2]$
+\end_inset
+
+ Por el teorema de Euler en teoría de números, para 
+\begin_inset Formula $a\in\{0,\dots,n-1\}$
+\end_inset
+
+, como 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $n$
+\end_inset
+
+ son coprimos, 
+\begin_inset Formula $a^{\varphi(n)}\equiv1\pmod n$
+\end_inset
+
+, con 
+\begin_inset Formula $\varphi(n)=n-1$
+\end_inset
+
+, luego 
+\begin_inset Formula $a^{n-2}$
+\end_inset
+
+ es inverso de 
+\begin_inset Formula $[a]$
+\end_inset
+
+ en 
+\begin_inset Formula $\mathbb{Z}^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[2\implies1]$
+\end_inset
+
+ Obvio.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $m\in\mathbb{Z}$
+\end_inset
+
+ no es cuadrado de entero, 
+\begin_inset Formula $\mathbb{Z}[\sqrt{m}]$
+\end_inset
+
+ es un dominio no cuerpo y 
+\begin_inset Formula $\mathbb{Q}[\sqrt{m}]$
+\end_inset
+
+ es cuerpo.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Ambos son subanillos de 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ y por tanto subdominios.
+ 
+\begin_inset Formula $\mathbb{Z}[\sqrt{m}]$
+\end_inset
+
+ no es un cuerpo porque 
+\begin_inset Formula $\frac{1}{2}\notin\mathbb{Z}[\sqrt{m}]$
+\end_inset
+
+ y por tanto 2 no tiene inverso, pero 
+\begin_inset Formula $\mathbb{Q}[\sqrt{m}]$
+\end_inset
+
+ sí lo es porque, para 
+\begin_inset Formula $a,b\in\mathbb{Q}$
+\end_inset
+
+, 
+\begin_inset Formula $(a+b\sqrt{m})(a-b\sqrt{m})=a^{2}-b^{2}m=:q\in\mathbb{Z}\setminus\{0\}$
+\end_inset
+
+ y 
+\begin_inset Formula $aq^{-1}-bq^{-1}\sqrt{m}$
+\end_inset
+
+ es el inverso de 
+\begin_inset Formula $a+b\sqrt{m}$
+\end_inset
+
+.
+ Para ver que 
+\begin_inset Formula $q\neq0$
+\end_inset
+
+, vemos que si fuera 
+\begin_inset Formula $a^{2}-b^{2}m=0$
+\end_inset
+
+ sería 
+\begin_inset Formula $m=\frac{a^{2}}{b^{2}}=\left(\frac{a}{b}\right)^{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+ sería cuadrado de racional, pero si llamamos 
+\begin_inset Formula $\frac{p}{q}:=\frac{a}{b}$
+\end_inset
+
+ como fracción irreducible, 
+\begin_inset Formula $\frac{p^{2}}{q^{2}}$
+\end_inset
+
+ también es irreducible y debe ser 
+\begin_inset Formula $q^{2}=1$
+\end_inset
+
+, luego 
+\begin_inset Formula $\frac{a}{b}=\frac{p}{q}\in\mathbb{Z}$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+ es cuadrado de entero.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Un producto de anillos no triviales nunca es un dominio.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $(1,0)(0,1)=(0,0)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $A[X]$
+\end_inset
+
+ es un dominio si y sólo si lo es 
+\begin_inset Formula $A$
+\end_inset
+
+, pero no es un cuerpo.
+\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
+
+
+\begin_inset Formula $A$
+\end_inset
+
+ es subanillo de 
+\begin_inset Formula $A[X]$
+\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
+
+Sean 
+\begin_inset Formula $P,Q\in A[X]$
+\end_inset
+
+ no nulos, como los coeficientes principales respectivos de 
+\begin_inset Formula $P$
+\end_inset
+
+ y 
+\begin_inset Formula $Q$
+\end_inset
+
+ no son nulos y 
+\begin_inset Formula $A$
+\end_inset
+
+ es un dominio, el de 
+\begin_inset Formula $PQ$
+\end_inset
+
+ tampoco lo es.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $A[X]$
+\end_inset
+
+ no es un cuerpo porque 
+\begin_inset Formula $(X)$
+\end_inset
+
+ es un ideal propio no nulo.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Ideales maximales y primos
+\end_layout
+
+\begin_layout Standard
+Un ideal 
+\begin_inset Formula $I$
+\end_inset
+
+ del anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ es 
+\series bold
+maximal
+\series default
+ si no está contenido estrictamente en ningún ideal propio de 
+\begin_inset Formula $A$
+\end_inset
+
+, y es 
+\series bold
+primo
+\series default
+ si 
+\begin_inset Formula $\forall a,b\in A,(ab\in I\implies a\in I\lor b\in I)$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $I$
+\end_inset
+
+ un ideal propio de 
+\begin_inset Formula $A$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:char-maximal"
+
+\end_inset
+
+
+\begin_inset Formula $I$
+\end_inset
+
+ es maximal si y sólo si 
+\begin_inset Formula $A/I$
+\end_inset
+
+ es un cuerpo.
+\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
+
+Si 
+\begin_inset Formula $A/I$
+\end_inset
+
+ no fuera un cuerpo, existiría un ideal 
+\begin_inset Formula $K$
+\end_inset
+
+ de 
+\begin_inset Formula $A/I$
+\end_inset
+
+ con 
+\begin_inset Formula $0\subsetneq K\subsetneq A/I$
+\end_inset
+
+ y, por el teorema de la correspondencia, un ideal 
+\begin_inset Formula $J$
+\end_inset
+
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ con 
+\begin_inset Formula $J/I=K$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $I\subsetneq J\subsetneq A$
+\end_inset
+
+, luego 
+\begin_inset Formula $I$
+\end_inset
+
+ no es maximal.
+\begin_inset Formula $\#$
+\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
+
+Los únicos ideales de 
+\begin_inset Formula $A/I$
+\end_inset
+
+ son 0 y 
+\begin_inset Formula $A/I$
+\end_inset
+
+, y por el teorema de la correspondencia, los únicos ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contienen a 
+\begin_inset Formula $I$
+\end_inset
+
+ son 
+\begin_inset Formula $I$
+\end_inset
+
+ y 
+\begin_inset Formula $A$
+\end_inset
+
+, luego 
+\begin_inset Formula $I$
+\end_inset
+
+ es maximal.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:char-prime"
+
+\end_inset
+
+
+\begin_inset Formula $I$
+\end_inset
+
+ es primo si y sólo si 
+\begin_inset Formula $A/I$
+\end_inset
+
+ es un dominio.
+\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
+
+Sean 
+\begin_inset Formula $[a],[b]\in A/I\setminus0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a,b\notin I$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $ab\notin I$
+\end_inset
+
+, luego 
+\begin_inset Formula $[a][b]=[ab]\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $A/I$
+\end_inset
+
+ es un dominio.
+\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
+
+Sean 
+\begin_inset Formula $a,b\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $ab\in I$
+\end_inset
+
+, entonces 
+\begin_inset Formula $[ab]=[a][b]=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $[a]=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $a\in I$
+\end_inset
+
+ o 
+\begin_inset Formula $[b]=0$
+\end_inset
+
+ y 
+\begin_inset Formula $b\in I$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $I$
+\end_inset
+
+ es maximal, es primo.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si es maximal, 
+\begin_inset Formula $A/I$
+\end_inset
+
+ es cuerpo y por tanto dominio, luego 
+\begin_inset Formula $I$
+\end_inset
+
+ es primo.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $A$
+\end_inset
+
+ es cuerpo si y sólo si 0 es maximal.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $A\cong A/0$
+\end_inset
+
+ y se aplica el punto 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:char-maximal"
+plural "false"
+caps "false"
+noprefix "false"
+
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $A$
+\end_inset
+
+ es dominio si y sólo si 0 es primo.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $A\cong A/0$
+\end_inset
+
+ y se aplica el punto 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:char-prime"
+plural "false"
+caps "false"
+noprefix "false"
+
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Dado un conjunto 
+\begin_inset Formula $S$
+\end_inset
+
+ con un orden parcial, una 
+\series bold
+cadena
+\series default
+ de 
+\begin_inset Formula $S$
+\end_inset
+
+ es un subconjunto totalmente ordenado.
+ 
+\begin_inset Formula $S$
+\end_inset
+
+ es 
+\series bold
+inductivo
+\series default
+ si toda cadena suya tiene supremo.
+ Del axioma de elección se deduce el 
+\series bold
+lema de Zorn:
+\series default
+ Todo conjunto inductivo tiene un elemento maximal.
+\end_layout
+
+\begin_layout Standard
+Todo ideal propio de un anillo está contenido en un ideal maximal.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $I$
+\end_inset
+
+ un ideal propio de 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ el conjunto de los ideales propios de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ La unión de los elementos de una cadena 
+\begin_inset Formula $\{A_{i}\}_{i\in I}$
+\end_inset
+
+ del conjunto parcialmente ordenado 
+\begin_inset Formula $(\Omega,\subseteq)$
+\end_inset
+
+ es un ideal, pues para 
+\begin_inset Formula $x\in A_{i}$
+\end_inset
+
+ e 
+\begin_inset Formula $y\in A_{j}$
+\end_inset
+
+, bien 
+\begin_inset Formula $A_{i}\subseteq A_{j}$
+\end_inset
+
+ y 
+\begin_inset Formula $x\in A_{j}$
+\end_inset
+
+ o, análogamente, 
+\begin_inset Formula $y\in A_{i}$
+\end_inset
+
+, luego 
+\begin_inset Formula $x+y\in A_{i}\cup A_{j}$
+\end_inset
+
+, y para 
+\begin_inset Formula $x\in A_{i}$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, 
+\begin_inset Formula $ax\in A_{i}$
+\end_inset
+
+.
+ Además es propio, pues de lo contrario contendría al 1 y existiría 
+\begin_inset Formula $i\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $1\in A_{i}$
+\end_inset
+
+, pero 
+\begin_inset Formula $A_{i}$
+\end_inset
+
+ es propio.
+\begin_inset Formula $\#$
+\end_inset
+
+ Por tanto 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ es inductivo y, por el lema de Zorn, tiene un elemento maximal, que es
+ un ideal maximal de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contiene a 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Divisibilidad
+\end_layout
+
+\begin_layout Standard
+Dados un anillo conmutativo 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $a,b\in A$
+\end_inset
+
+, 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\series bold
+divide
+\series default
+ a 
+\begin_inset Formula $b$
+\end_inset
+
+, es un 
+\series bold
+divisor
+\series default
+ de 
+\begin_inset Formula $b$
+\end_inset
+
+ o 
+\begin_inset Formula $b$
+\end_inset
+
+ es un 
+\series bold
+múltiplo
+\series default
+ de 
+\begin_inset Formula $a$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $a\mid b$
+\end_inset
+
+, si existe 
+\begin_inset Formula $c\in A$
+\end_inset
+
+ tal que 
+\begin_inset Formula $b=ac$
+\end_inset
+
+.
+ Propiedades: 
+\begin_inset Formula $\forall a\in A$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+Reflexiva.
+\end_layout
+
+\begin_layout Enumerate
+Transitiva.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $1\mid a\mid0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $0\mid a\iff a=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a\mid1$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $a$
+\end_inset
+
+ es unidad, en cuyo caso 
+\begin_inset Formula $\forall x\in A,a\mid x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $a$
+\end_inset
+
+ divide a ciertos elementos, divide a cualquier combinación lineal de estos
+ con coeficientes en 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $c$
+\end_inset
+
+ es regular y 
+\begin_inset Formula $ac\mid bc$
+\end_inset
+
+, 
+\begin_inset Formula $a\mid b$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sea 
+\begin_inset Formula $x$
+\end_inset
+
+ tal que 
+\begin_inset Formula $xac=bc$
+\end_inset
+
+, como 
+\begin_inset Formula $c$
+\end_inset
+
+ es regular, 
+\begin_inset Formula $xa=b$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Dos elementos 
+\begin_inset Formula $a,b\in A$
+\end_inset
+
+ son 
+\series bold
+asociados
+\series default
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+ si 
+\begin_inset Formula $a\mid b\mid a$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+, si y sólo si tienen los mismos divisores, si y sólo si tienen los mismos
+ múltiplos.
+ Esta relación es de equivalencia.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{samepage}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $D$
+\end_inset
+
+ es un dominio, 
+\begin_inset Formula $a,b\in D$
+\end_inset
+
+ son asociados en 
+\begin_inset Formula $D$
+\end_inset
+
+ si y sólo si existe una unidad 
+\begin_inset Formula $u$
+\end_inset
+
+ de 
+\begin_inset Formula $D$
+\end_inset
+
+ con 
+\begin_inset Formula $b=au$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $a=0$
+\end_inset
+
+, 
+\begin_inset Formula $b=0$
+\end_inset
+
+ y es obvio.
+ De lo contrario, sean 
+\begin_inset Formula $x,y\in D$
+\end_inset
+
+ con 
+\begin_inset Formula $ax=b$
+\end_inset
+
+ y 
+\begin_inset Formula $by=a$
+\end_inset
+
+, 
+\begin_inset Formula $a=by=axy$
+\end_inset
+
+ y, como 
+\begin_inset Formula $a$
+\end_inset
+
+ es cancelable, 
+\begin_inset Formula $xy=1$
+\end_inset
+
+, luego 
+\begin_inset Formula $y$
+\end_inset
+
+ es inverso de 
+\begin_inset Formula $x$
+\end_inset
+
+ y en particular 
+\begin_inset Formula $x$
+\end_inset
+
+ es unidad con 
+\begin_inset Formula $b=ax$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Claramente 
+\begin_inset Formula $a\mid b$
+\end_inset
+
+ y, como 
+\begin_inset Formula $a=bu^{-1}$
+\end_inset
+
+, 
+\begin_inset Formula $b\mid a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{samepage}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $A$
+\end_inset
+
+ un anillo conmutativo y 
+\begin_inset Formula $a\in A\setminus(A^{*}\cup\{0\})$
+\end_inset
+
+, 
+\begin_inset Formula $a$
+\end_inset
+
+ es 
+\series bold
+irreducible
+\series default
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall b,c\in A,(a=bc\implies b\in A^{*}\lor c\in A^{*})$
+\end_inset
+
+, y es 
+\series bold
+primo
+\series default
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall b,c\in A,(a\mid bc\implies a\mid b\lor a\mid c)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un dominio, todo primo es irreducible.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $p$
+\end_inset
+
+ primo en 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $a,b\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $p=ab$
+\end_inset
+
+, entonces 
+\begin_inset Formula $p\mid a$
+\end_inset
+
+ o 
+\begin_inset Formula $p|b$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $p\mid a$
+\end_inset
+
+, como 
+\begin_inset Formula $a\mid p$
+\end_inset
+
+, 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $p$
+\end_inset
+
+ son asociados, luego existe una unidad 
+\begin_inset Formula $u$
+\end_inset
+
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ con 
+\begin_inset Formula $p=ab=au$
+\end_inset
+
+ y, como 
+\begin_inset Formula $a\neq0$
+\end_inset
+
+ porque de lo contrario sería 
+\begin_inset Formula $p=0$
+\end_inset
+
+, queda 
+\begin_inset Formula $b=u$
+\end_inset
+
+.
+ El caso en que 
+\begin_inset Formula $p\mid b$
+\end_inset
+
+ es análogo.
+\end_layout
+
+\begin_layout Standard
+Irreducible en un dominio no implica primo.
+ 
+\series bold
+Demostración:
+\series default
+ Primero vemos que 
+\begin_inset Formula $2$
+\end_inset
+
+ es irreducible en el dominio 
+\begin_inset Formula $\mathbb{Z}[\sqrt{-5}]$
+\end_inset
+
+.
+ El cuadrado del módulo de un elemento del dominio en 
+\begin_inset Formula $\mathbb{Z}[\sqrt{-5}]$
+\end_inset
+
+ es 
+\begin_inset Formula $|a+b\sqrt{-5}|^{2}=a^{2}+5b^{2}\in\mathbb{Z}$
+\end_inset
+
+.
+ Así, sean 
+\begin_inset Formula $x,y\in\mathbb{Z}[\sqrt{-5}]$
+\end_inset
+
+ con 
+\begin_inset Formula $xy=2$
+\end_inset
+
+, 
+\begin_inset Formula $|xy|^{2}=|x|^{2}|y|^{2}=4$
+\end_inset
+
+, luego o 
+\begin_inset Formula $|x|^{2}=1$
+\end_inset
+
+ o 
+\begin_inset Formula $|y|^{2}=1$
+\end_inset
+
+.
+ Si, por ejemplo, 
+\begin_inset Formula $|x|^{2}=1$
+\end_inset
+
+, sea 
+\begin_inset Formula $x=:a+b\sqrt{-5}$
+\end_inset
+
+, despejando, 
+\begin_inset Formula $0\leq a^{2}=1-5b^{2}$
+\end_inset
+
+, luego 
+\begin_inset Formula $b=0$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in\{1,-1\}$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $x\in\{-1,1\}$
+\end_inset
+
+ es unidad de 
+\begin_inset Formula $\mathbb{Z}[\sqrt{-5}]$
+\end_inset
+
+.
+ Pero 
+\begin_inset Formula $2$
+\end_inset
+
+ no es primo, pues 
+\begin_inset Formula $2|6=(1+\sqrt{-5})(1-\sqrt{-5})$
+\end_inset
+
+ y es claro que 
+\begin_inset Formula $2\mid6=(1+\sqrt{-5})(1-\sqrt{-5})$
+\end_inset
+
+ pero 
+\begin_inset Formula $2\nmid1+\sqrt{-5},1-\sqrt{-5}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $A$
+\end_inset
+
+ un anillo conmutativo y 
+\begin_inset Formula $a,b\in A$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a=0\iff(a)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a\in A^{*}\iff(a)=A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a\mid b\iff(b)\subseteq(a)\iff b\in(a)$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Description
+\begin_inset Formula $[1\implies2]$
+\end_inset
+
+ Sea 
+\begin_inset Formula $c\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $b=ac$
+\end_inset
+
+, para 
+\begin_inset Formula $d\in(b)$
+\end_inset
+
+, sea 
+\begin_inset Formula $t\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $d=bt$
+\end_inset
+
+, 
+\begin_inset Formula $d=act\in(a)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[2\implies3\implies1]$
+\end_inset
+
+ Obvio.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ son asociados si y sólo si 
+\begin_inset Formula $(a)=(b)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a$
+\end_inset
+
+ es primo si y sólo si 
+\begin_inset Formula $(a)$
+\end_inset
+
+ es un ideal primo no nulo de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $a$
+\end_inset
+
+ es primo si y sólo si 
+\begin_inset Formula $a\notin A^{*}\cup\{0\}$
+\end_inset
+
+ y 
+\begin_inset Formula $\forall b,c\in A,(a\mid bc\implies a\mid b\lor a\mid c)$
+\end_inset
+
+, pero 
+\begin_inset Formula $a\mid x$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $x\in(a)$
+\end_inset
+
+, luego 
+\begin_inset Quotes cld
+\end_inset
+
+traduciendo
+\begin_inset Quotes crd
+\end_inset
+
+, esto ocurre si y sólo si 
+\begin_inset Formula $(a)\neq A^{*},0$
+\end_inset
+
+ y 
+\begin_inset Formula $\forall b,c\in A,(bc\in(a)\implies b\in(a)\lor c\in(a))$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $(a)$
+\end_inset
+
+ es un ideal primo no nulo de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un dominio, 
+\begin_inset Formula $a$
+\end_inset
+
+ es irreducible si y sólo si 
+\begin_inset Formula $(a)$
+\end_inset
+
+ es maximal entre los ideales principales no nulos de 
+\begin_inset Formula $A$
+\end_inset
+
+, es decir, si 
+\begin_inset Formula $(a)\neq0,A$
+\end_inset
+
+ y 
+\begin_inset Formula $\forall b\in A,((a)\subseteq(b)\neq A\implies(a)=(b))$
+\end_inset
+
+.
+\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
+
+Como 
+\begin_inset Formula $a\notin\{0\},A^{*}$
+\end_inset
+
+, 
+\begin_inset Formula $(a)\neq0,A$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $b\in A$
+\end_inset
+
+ es tal que 
+\begin_inset Formula $(a)\subseteq(b)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $b\mid a$
+\end_inset
+
+, esto es, existe 
+\begin_inset Formula $c\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $a=bc$
+\end_inset
+
+, y como 
+\begin_inset Formula $a$
+\end_inset
+
+ es irreducible, o 
+\begin_inset Formula $b$
+\end_inset
+
+ es unidad y entonces 
+\begin_inset Formula $(b)\in A$
+\end_inset
+
+, o 
+\begin_inset Formula $c$
+\end_inset
+
+ es unidad y entonces 
+\begin_inset Formula $(a)=\{ax=bcx\}_{x\in A}=\{bx\}_{x\in(c)=A}=(b)$
+\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
+
+Como 
+\begin_inset Formula $(a)\neq0,A$
+\end_inset
+
+, 
+\begin_inset Formula $a\notin\{0\},A^{*}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $b,c\in A$
+\end_inset
+
+ cumplen 
+\begin_inset Formula $a=bc$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ no es unidad, como 
+\begin_inset Formula $b\mid a$
+\end_inset
+
+, 
+\begin_inset Formula $(a)\subseteq(b)\neq A$
+\end_inset
+
+, luego 
+\begin_inset Formula $(a)=(b)$
+\end_inset
+
+ y 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ son asociados.
+ Entonces, como 
+\begin_inset Formula $A$
+\end_inset
+
+ es dominio, existe una unidad 
+\begin_inset Formula $u$
+\end_inset
+
+ con 
+\begin_inset Formula $a=bc=bu$
+\end_inset
+
+ y, como 
+\begin_inset Formula $b\neq0$
+\end_inset
+
+ porque 
+\begin_inset Formula $a\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $c=u$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Dados un anillo conmutativo 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $S\subseteq A$
+\end_inset
+
+, 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ es un 
+\series bold
+máximo común divisor
+\series default
+ de 
+\begin_inset Formula $S$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $a=\text{mcm}S$
+\end_inset
+
+, si divide a cada elemento de 
+\begin_inset Formula $S$
+\end_inset
+
+ y es múltiplo de cada elemento que cumple esto, y es un 
+\series bold
+mínimo común múltiplo
+\series default
+ de 
+\begin_inset Formula $S$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $a=\text{mcd}S$
+\end_inset
+
+, si es múltiplo de cada elemento de 
+\begin_inset Formula $S$
+\end_inset
+
+ y divide a cada elemento que cumple esto.
+ Para 
+\begin_inset Formula $a,b\in A$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a=\text{mcd}S$
+\end_inset
+
+ si y solo si 
+\begin_inset Formula $(a)$
+\end_inset
+
+ es el menor ideal principal de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contiene a 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ En particular, si 
+\begin_inset Formula $(a)=(S)$
+\end_inset
+
+, 
+\begin_inset Formula $a=\text{mcd}S$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $a$
+\end_inset
+
+ divide a cada elemento de 
+\begin_inset Formula $S$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\forall x\in S,(x)\subseteq(a)$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $S\subseteq(a)$
+\end_inset
+
+.
+ Además, 
+\begin_inset Formula $a$
+\end_inset
+
+ es múltiplo de cada 
+\begin_inset Formula $b\in A$
+\end_inset
+
+ que cumple esto, esto es, que cumple 
+\begin_inset Formula $S\subseteq(b)$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $\forall b\in A,(S\subseteq(b)\implies(a)\subseteq(b))$
+\end_inset
+
+.
+ Juntando ambos se obtiene el resultado.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $a=\text{mcm}S$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $(a)$
+\end_inset
+
+ es el mayor ideal principal de 
+\begin_inset Formula $A$
+\end_inset
+
+ contenido en 
+\begin_inset Formula $\bigcap_{s\in S}(s)$
+\end_inset
+
+.
+ En particular, si 
+\begin_inset Formula $(a)=\bigcap_{s\in S}(s)$
+\end_inset
+
+, 
+\begin_inset Formula $a=\text{mcm}S$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $a$
+\end_inset
+
+ es múltiplo de cada elemento de 
+\begin_inset Formula $S$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\forall s\in S,(a)\subseteq(s)$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $(a)\subseteq\bigcap_{s\in S}(s)$
+\end_inset
+
+.
+ Además, 
+\begin_inset Formula $a$
+\end_inset
+
+ divide a cada 
+\begin_inset Formula $b\in A$
+\end_inset
+
+ que cumple esto si y sólo si 
+\begin_inset Formula $\forall b\in A,((b)\subseteq\bigcap_{s\in S}(s)\implies(b)\subseteq(a))$
+\end_inset
+
+.
+ Juntando ambos se obtiene el resultado.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $a=\text{mcd}S$
+\end_inset
+
+, 
+\begin_inset Formula $b=\text{mcd}S$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ son asociados en 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Se obtiene de la caracterización de máximo común divisor y la de asociados
+ por ideales principales.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $a=\text{mcm}S$
+\end_inset
+
+, 
+\begin_inset Formula $b=\text{mcm}S$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ son asociados en 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Análogo.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:case-gcd"
+
+\end_inset
+
+Si 
+\begin_inset Formula $a$
+\end_inset
+
+ divide a todo elemento de 
+\begin_inset Formula $S$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in(S)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a=\text{mcd}S$
+\end_inset
+
+.
+ En tal caso llamamos 
+\series bold
+identidad de Bézout
+\series default
+ a una expresión de la forma 
+\begin_inset Formula $a=a_{1}s_{1}+\dots+a_{n}s_{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{1},\dots,a_{n}\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $s_{1},\dots,s_{n}\in S$
+\end_inset
+
+, que existe porque 
+\begin_inset Formula $a\in(S)$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si 
+\begin_inset Formula $b\in A$
+\end_inset
+
+ divide a todos los elementos de 
+\begin_inset Formula $S$
+\end_inset
+
+, 
+\begin_inset Formula $b\mid a_{1}s_{1}+\dots+a_{n}s_{n}=a$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\text{mcd}S=1$
+\end_inset
+
+ si y sólo si los únicos divisores comunes de los elementos de 
+\begin_inset Formula $S$
+\end_inset
+
+ son las unidades de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\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
+
+Si hubiera un divisor común 
+\begin_inset Formula $b\in A\setminus A^{*}$
+\end_inset
+
+, como 
+\begin_inset Formula $b\nmid1$
+\end_inset
+
+, 1 no sería múltiplo de todos los divisores comunes a los elementos de
+ 
+\begin_inset Formula $S\#$
+\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
+
+1 es divisor de todos y múltiplo de las unidades de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $1\in(S)$
+\end_inset
+
+, 
+\begin_inset Formula $\text{mcd}S=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Basta aplicar 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:case-gcd"
+plural "false"
+caps "false"
+noprefix "false"
+
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Dominios de factorización única
+\end_layout
+
+\begin_layout Standard
+Dado un dominio 
+\begin_inset Formula $D$
+\end_inset
+
+, una 
+\series bold
+factorización en producto de irreducibles
+\series default
+ de 
+\begin_inset Formula $a\in D$
+\end_inset
+
+ es una expresión de la forma 
+\begin_inset Formula $a=up_{1}\cdots p_{n}$
+\end_inset
+
+, donde 
+\begin_inset Formula $u$
+\end_inset
+
+ es una unidad de 
+\begin_inset Formula $D$
+\end_inset
+
+ y 
+\begin_inset Formula $p_{1},\dots,p_{n}$
+\end_inset
+
+ son irreducibles en 
+\begin_inset Formula $D$
+\end_inset
+
+.
+ Dos factorizaciones en producto de irreducibles de 
+\begin_inset Formula $a\in D$
+\end_inset
+
+, 
+\begin_inset Formula $a=up_{1}\cdots p_{n}$
+\end_inset
+
+ y 
+\begin_inset Formula $a=vq_{1}\cdots q_{n}$
+\end_inset
+
+, son 
+\series bold
+equivalentes
+\series default
+ si 
+\begin_inset Formula $m=n$
+\end_inset
+
+ y existe una permutación 
+\begin_inset Formula $\sigma$
+\end_inset
+
+ de 
+\begin_inset Formula $\mathbb{N}_{n}:=\{1,\dots,n\}$
+\end_inset
+
+ tal que para 
+\begin_inset Formula $k\in\mathbb{N}_{n}$
+\end_inset
+
+, 
+\begin_inset Formula $p_{k}$
+\end_inset
+
+ y 
+\begin_inset Formula $q_{\sigma(k)}$
+\end_inset
+
+ son asociados, en cuyo caso 
+\begin_inset Formula $u$
+\end_inset
+
+ y 
+\begin_inset Formula $v$
+\end_inset
+
+ también lo son.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $D$
+\end_inset
+
+ es un 
+\series bold
+dominio de factorización
+\series default
+ (
+\series bold
+DF
+\series default
+) si todo elemento no nulo de 
+\begin_inset Formula $D$
+\end_inset
+
+ admite una factorización en producto de irreducibles, y es un 
+\series bold
+dominio de factorización única
+\series default
+ (
+\series bold
+DFU
+\series default
+ o 
+\series bold
+UFD
+\series default
+) si, además, todas las factorizaciones de un mismo elemento son equivalentes.
+ Ejemplos:
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Teorema Fundamental de la Aritmética:
+\series default
+ 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+ es un DFU.
+\end_layout
+
+\begin_layout Enumerate
+Dado 
+\begin_inset Formula $m\in\mathbb{Z}^{+}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{Z}[\sqrt{m}]$
+\end_inset
+
+ es un DF.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si 
+\begin_inset Formula $m$
+\end_inset
+
+ es un cuadrado en 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{Z}[\sqrt{m}]=\mathbb{Z}$
+\end_inset
+
+ es un DF.
+ Supongamos que no lo es.
+ Consideramos la norma 
+\begin_inset Formula $N:\mathbb{Z}[\sqrt{m}]\to\mathbb{Z}$
+\end_inset
+
+ dada por 
+\begin_inset Formula $N(x)=x\overline{x}$
+\end_inset
+
+, siendo 
+\begin_inset Formula $\overline{x}$
+\end_inset
+
+ el conjugado de 
+\begin_inset Formula $x$
+\end_inset
+
+, y como la conjugación es un homomorfismo, 
+\begin_inset Formula $N(xy)=xy\overline{xy}=xy\overline{x}\,\overline{y}=x\overline{x}y\overline{y}=N(x)N(y)$
+\end_inset
+
+.
+ Veamos que 
+\begin_inset Formula $\mathbb{Z}[\sqrt{m}]^{*}=\{x:|N(x)|=1\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\subseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $x$
+\end_inset
+
+ es invertible, 
+\begin_inset Formula $1=N(1)=N(xx^{-1})$
+\end_inset
+
+, pero luego 
+\begin_inset Formula $1=N(xx^{-1})=N(x)N(x^{-1})$
+\end_inset
+
+ y 
+\begin_inset Formula $N(x)$
+\end_inset
+
+ es invertible en 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+, esto es, 
+\begin_inset Formula $N(x)\in\{-1,1\}$
+\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
+
+Si 
+\begin_inset Formula $N(x)=x\overline{x}\in\{-1,1\}$
+\end_inset
+
+, 
+\begin_inset Formula $x$
+\end_inset
+
+ es invertible con inverso 
+\begin_inset Formula $-\overline{x}$
+\end_inset
+
+ o 
+\begin_inset Formula $\overline{x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sea ahora 
+\begin_inset Formula $x=a+b\sqrt{m}\neq0$
+\end_inset
+
+ y queremos ver que 
+\begin_inset Formula $x$
+\end_inset
+
+ tiene una factorización en 
+\begin_inset Formula $\mathbb{Z}[\sqrt{m}]$
+\end_inset
+
+.
+ 
+\begin_inset Formula $|N(x)|\neq0$
+\end_inset
+
+, pues si fuera 0 sería 
+\begin_inset Formula $a^{2}=b^{2}m$
+\end_inset
+
+ y, como 
+\begin_inset Formula $x\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $m$
+\end_inset
+
+ sería un cuadrado en 
+\begin_inset Formula $\mathbb{Z}\#$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $|N(x)|=1$
+\end_inset
+
+, 
+\begin_inset Formula $x$
+\end_inset
+
+ es unidad y por tanto tiene una factorización.
+ Para 
+\begin_inset Formula $|N(x)|=k>1$
+\end_inset
+
+, supuesta probada la propiedad para 
+\begin_inset Formula $|N(x)|<k$
+\end_inset
+
+, podemos suponer que 
+\begin_inset Formula $x$
+\end_inset
+
+ no es irreducible, pues de serlo ya sabemos que tiene una factorización.
+ Entonces existen 
+\begin_inset Formula $p,q\in\mathbb{Z}[\sqrt{m}]$
+\end_inset
+
+ con 
+\begin_inset Formula $x=pq$
+\end_inset
+
+ y 
+\begin_inset Formula $p$
+\end_inset
+
+ y 
+\begin_inset Formula $q$
+\end_inset
+
+ no unidades, con lo que 
+\begin_inset Formula $|N(p)|$
+\end_inset
+
+ y 
+\begin_inset Formula $|N(q)|$
+\end_inset
+
+ son divisores propios de 
+\begin_inset Formula $|N(x)|$
+\end_inset
+
+ y, por la hipótesis de inducción, 
+\begin_inset Formula $p$
+\end_inset
+
+ y 
+\begin_inset Formula $q$
+\end_inset
+
+ son producto de irreducibles por alguna unidad, luego 
+\begin_inset Formula $x$
+\end_inset
+
+ también lo es.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Un dominio 
+\begin_inset Formula $D$
+\end_inset
+
+ es un DFU si y sólo si todo elemento no nulo de 
+\begin_inset Formula $D$
+\end_inset
+
+ es producto de una unidad por primos, si y sólo si 
+\begin_inset Formula $D$
+\end_inset
+
+ es un dominio de factorización en el que todo elemento irreducible es primo.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Todo elemento no nulo de 
+\begin_inset Formula $D$
+\end_inset
+
+ puede expresarse como producto de una unidad por irreducibles.
+ Ahora bien, sea 
+\begin_inset Formula $p\in D$
+\end_inset
+
+ irreducible, queremos ver que 
+\begin_inset Formula $p$
+\end_inset
+
+ es primo, esto es, que para 
+\begin_inset Formula $a,b\in D$
+\end_inset
+
+ con 
+\begin_inset Formula $p\mid ab$
+\end_inset
+
+, 
+\begin_inset Formula $p\mid a$
+\end_inset
+
+ o 
+\begin_inset Formula $p\mid b$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $a=0$
+\end_inset
+
+ o 
+\begin_inset Formula $b=0$
+\end_inset
+
+ esto es claro, por lo que suponemos 
+\begin_inset Formula $a,b\neq0$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $t\in D$
+\end_inset
+
+ con 
+\begin_inset Formula $pt=ab$
+\end_inset
+
+, si 
+\begin_inset Formula $t=up_{1}\cdots p_{n}$
+\end_inset
+
+, 
+\begin_inset Formula $a=vq_{1}\cdots q_{m}$
+\end_inset
+
+ y 
+\begin_inset Formula $b=wr_{1}\cdots r_{k}$
+\end_inset
+
+ son las factorizaciones en irreducibles de 
+\begin_inset Formula $t$
+\end_inset
+
+, 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+, entonces 
+\begin_inset Formula $upp_{1}\cdots p_{n}=(vw)q_{1}\cdots q_{m}r_{1}\cdots r_{k}$
+\end_inset
+
+, y por la unicidad de la factorización, 
+\begin_inset Formula $p$
+\end_inset
+
+ es asociado de algún 
+\begin_inset Formula $q_{i}$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $p\mid a$
+\end_inset
+
+, o de algún 
+\begin_inset Formula $r_{i}$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $p\mid b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\implies3]$
+\end_inset
+
+ Como los primos en un dominio son irreducibles, 
+\begin_inset Formula $D$
+\end_inset
+
+ es un DF.
+ Sean 
+\begin_inset Formula $p\in D$
+\end_inset
+
+ irreducible y sean 
+\begin_inset Formula $u\in D^{*}$
+\end_inset
+
+ y 
+\begin_inset Formula $q_{1},\dots,q_{k}\in D$
+\end_inset
+
+ primos con 
+\begin_inset Formula $p=uq_{1}\cdots q_{k}$
+\end_inset
+
+.
+ Entonces o 
+\begin_inset Formula $q_{1}$
+\end_inset
+
+ es unidad o lo es 
+\begin_inset Formula $uq_{2}\cdots q_{k}$
+\end_inset
+
+, pero como 
+\begin_inset Formula $q_{1}$
+\end_inset
+
+ no lo es, debe ser 
+\begin_inset Formula $k=1$
+\end_inset
+
+.
+ De aquí, 
+\begin_inset Formula $p$
+\end_inset
+
+ y 
+\begin_inset Formula $q_{1}$
+\end_inset
+
+ son asociados, y como 
+\begin_inset Formula $q_{1}$
+\end_inset
+
+ es primo, 
+\begin_inset Formula $p$
+\end_inset
+
+ también.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $3\implies1]$
+\end_inset
+
+ Sean 
+\begin_inset Formula $d=up_{1}\cdots p_{n}=vq_{1}\cdots q_{m}$
+\end_inset
+
+ factorizaciones en producto de irreducibles de un cierto 
+\begin_inset Formula $d\neq0$
+\end_inset
+
+ y podemos suponer 
+\begin_inset Formula $n\leq m$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $n=0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $d$
+\end_inset
+
+ es unidad y, como los divisores de unidades son unidades, 
+\begin_inset Formula $m=0$
+\end_inset
+
+ y las factorizaciones son equivalentes.
+ Si 
+\begin_inset Formula $n>0$
+\end_inset
+
+, supuesto esto probado para 
+\begin_inset Formula $n-1$
+\end_inset
+
+, como 
+\begin_inset Formula $p_{n}$
+\end_inset
+
+ es primo, divide a algún 
+\begin_inset Formula $q_{i}$
+\end_inset
+
+ y, como 
+\begin_inset Formula $q_{i}$
+\end_inset
+
+ es irreducible por ser primo, existe 
+\begin_inset Formula $w\in D^{*}$
+\end_inset
+
+ con 
+\begin_inset Formula $p_{n}w=q_{i}$
+\end_inset
+
+ y ambos son asociados.
+ Podemos suponer 
+\begin_inset Formula $i=m$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $up_{1}\cdots p_{n}=vq_{1}\cdots q_{m-1}wp_{n}$
+\end_inset
+
+ y 
+\begin_inset Formula $up_{1}\cdots p_{n-1}=(vw)q_{1}\cdots q_{m-1}$
+\end_inset
+
+.
+ Por la hipótesis de inducción, 
+\begin_inset Formula $n-1=m-1$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $n=m$
+\end_inset
+
+, y existe una permutación 
+\begin_inset Formula $\tau$
+\end_inset
+
+ en 
+\begin_inset Formula $\mathbb{N}_{n-1}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $p_{i}$
+\end_inset
+
+ y 
+\begin_inset Formula $q_{\tau(i)}$
+\end_inset
+
+ son asociados para cada 
+\begin_inset Formula $i$
+\end_inset
+
+, y obviamente 
+\begin_inset Formula $\tau$
+\end_inset
+
+ se extiende a una permutación 
+\begin_inset Formula $\sigma$
+\end_inset
+
+ de 
+\begin_inset Formula $\mathbb{N}_{n}$
+\end_inset
+
+ con esta propiedad.
+ Por tanto las factorizaciones iniciales son equivalentes.
+\end_layout
+
+\begin_layout Section
+Dominios de ideales principales
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+dominio de ideales principales
+\series default
+ (
+\series bold
+DIP
+\series default
+ o 
+\series bold
+PID
+\series default
+) es un dominio en que todos los ideales son principales.
+ Si 
+\begin_inset Formula $D$
+\end_inset
+
+ es un DIP y 
+\begin_inset Formula $a\in D\setminus(D^{*}\cup\{0\})$
+\end_inset
+
+, 
+\begin_inset Formula $a$
+\end_inset
+
+ es irreducible si y sólo si 
+\begin_inset Formula $(a)$
+\end_inset
+
+ es un ideal maximal, si y sólo si 
+\begin_inset Formula $\frac{A}{(a)}$
+\end_inset
+
+ es un cuerpo, si y sólo si 
+\begin_inset Formula $a$
+\end_inset
+
+ es primo, si y sólo si 
+\begin_inset Formula $(a)$
+\end_inset
+
+ es un ideal primo, si y sólo si 
+\begin_inset Formula $\frac{A}{(a)}$
+\end_inset
+
+ es un dominio.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\iff2]$
+\end_inset
+
+ Sabemos que 
+\begin_inset Formula $a$
+\end_inset
+
+ es irreducible si y sólo si 
+\begin_inset Formula $(a)$
+\end_inset
+
+ es maximal entre los ideales principales no nulos de 
+\begin_inset Formula $D$
+\end_inset
+
+, pero en un DIP estos son precisamente todos los ideales no nulos de 
+\begin_inset Formula $D$
+\end_inset
+
+, y como 
+\begin_inset Formula $a\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $(a)\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\iff3]$
+\end_inset
+
+ Visto.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\impliedby4\iff5\iff6]$
+\end_inset
+
+ Visto.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $3\implies6]$
+\end_inset
+
+ Obvio.
+\end_layout
+
+\begin_layout Standard
+Todo DIP es un DFU.
+ 
+\series bold
+Demostración:
+\series default
+ Supongamos que existe 
+\begin_inset Formula $a_{0}\in D\setminus\{0\}$
+\end_inset
+
+ que no admite factorización en irreducibles.
+ Si 
+\begin_inset Formula $a\in D\setminus\{0\}$
+\end_inset
+
+ no admite factorización, como 
+\begin_inset Formula $a$
+\end_inset
+
+ no es irreducible ni unidad, existen 
+\begin_inset Formula $x,y\in D\setminus(D^{*}\cup\{0\})$
+\end_inset
+
+ con 
+\begin_inset Formula $a=xy$
+\end_inset
+
+, y al menos 
+\begin_inset Formula $x$
+\end_inset
+
+ o 
+\begin_inset Formula $y$
+\end_inset
+
+ (por ejemplo 
+\begin_inset Formula $x$
+\end_inset
+
+) no admite factorización, con lo que 
+\begin_inset Formula $(a)\subsetneq(x)$
+\end_inset
+
+.
+ Por inducción existe una sucesión 
+\begin_inset Formula $(a_{n})_{n\in\mathbb{N}}$
+\end_inset
+
+ en 
+\begin_inset Formula $D$
+\end_inset
+
+ con 
+\begin_inset Formula $(a_{0})\subsetneq(a_{1})\subsetneq\dots$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $I:=(a_{1},a_{2},\dots)=\bigcup_{n\in\mathbb{N}}(a_{n})$
+\end_inset
+
+, como 
+\begin_inset Formula $D$
+\end_inset
+
+ es un DIP, existe 
+\begin_inset Formula $x\in D$
+\end_inset
+
+ con 
+\begin_inset Formula $I=(x)$
+\end_inset
+
+, luego 
+\begin_inset Formula $x\in\bigcup_{n\in\mathbb{N}}(a_{n})$
+\end_inset
+
+ y existe 
+\begin_inset Formula $n$
+\end_inset
+
+ con 
+\begin_inset Formula $x\in(a_{n})$
+\end_inset
+
+.
+ Como además 
+\begin_inset Formula $a_{n}\in I=(x)$
+\end_inset
+
+, 
+\begin_inset Formula $(a_{n})=(x)=I$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $(a_{n})=(a_{n+1})\#$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Dominios euclídeos
+\end_layout
+
+\begin_layout Standard
+Dado un dominio 
+\begin_inset Formula $D\neq0$
+\end_inset
+
+, una función 
+\begin_inset Formula $\delta:D\setminus\{0\}\to\mathbb{N}$
+\end_inset
+
+ es 
+\series bold
+euclídea
+\series default
+ si cumple:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall a,b\in D\setminus\{0\},(a\mid b\implies\delta(a)\leq\delta(b))$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall a\in D,b\in D\setminus\{0\},\exists q,r\in D:(a=bq+r\land(r=0\lor\delta(r)<\delta(b)))$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+dominio euclídeo
+\series default
+ es uno que admite una función euclídea.
+ Ejemplos:
+\end_layout
+
+\begin_layout Enumerate
+El valor absoluto es una función euclídea en 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+ es un cuerpo, el grado de un polinomio en 
+\begin_inset Formula $\mathbb{K}[X]$
+\end_inset
+
+ es una función euclídea.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+La primera condición es clara.
+ Para la segunda, sean 
+\begin_inset Formula $D:=\mathbb{K}[X]$
+\end_inset
+
+, 
+\begin_inset Formula $a,b\in D$
+\end_inset
+
+ y 
+\begin_inset Formula $b\neq0$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $a=0$
+\end_inset
+
+, tomando 
+\begin_inset Formula $q=r=0$
+\end_inset
+
+ se tiene el resultado.
+ Sean 
+\begin_inset Formula $a\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $n$
+\end_inset
+
+ el grado de 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+ el de 
+\begin_inset Formula $b$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $n<m$
+\end_inset
+
+, tomamos 
+\begin_inset Formula $q=0$
+\end_inset
+
+ y 
+\begin_inset Formula $r=a$
+\end_inset
+
+, y si 
+\begin_inset Formula $n=m=0$
+\end_inset
+
+, tomamos 
+\begin_inset Formula $q=ab^{-1}$
+\end_inset
+
+ y 
+\begin_inset Formula $r=0$
+\end_inset
+
+, lo que prueba la condición para 
+\begin_inset Formula $n=0$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $n>0$
+\end_inset
+
+, supuesto esto probado para grado menor que 
+\begin_inset Formula $n$
+\end_inset
+
+, sean 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ el término principal de 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $\beta$
+\end_inset
+
+ el término principal de 
+\begin_inset Formula $b$
+\end_inset
+
+, 
+\begin_inset Formula $c:=a-\alpha\beta^{-1}X^{n-m}b$
+\end_inset
+
+ tiene grado 
+\begin_inset Formula $n-m<n$
+\end_inset
+
+ si 
+\begin_inset Formula $m>0$
+\end_inset
+
+ o 
+\begin_inset Formula $n-1$
+\end_inset
+
+ si 
+\begin_inset Formula $m=0$
+\end_inset
+
+, luego existen 
+\begin_inset Formula $q',r'\in D$
+\end_inset
+
+ con 
+\begin_inset Formula $c:=q'b+r$
+\end_inset
+
+ y o bien 
+\begin_inset Formula $r=0$
+\end_inset
+
+ o 
+\begin_inset Formula $r$
+\end_inset
+
+ tiene grado menor que 
+\begin_inset Formula $b$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $a+\alpha\beta^{-1}X^{n-m}b=q'b+r$
+\end_inset
+
+, y tomando 
+\begin_inset Formula $q:=q'-\alpha\beta'X^{n-m}$
+\end_inset
+
+ se tiene 
+\begin_inset Formula $a=qb+r$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+El cuadrado del módulo complejo es una función euclídea en 
+\begin_inset Formula $\mathbb{Z}[i]$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si 
+\begin_inset Formula $x:=a+bi$
+\end_inset
+
+ con 
+\begin_inset Formula $a,b\in\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $\delta(x):=|x|^{2}=a^{2}+b^{2}\in\mathbb{N}$
+\end_inset
+
+.
+ Además, 
+\begin_inset Formula $\delta(x)=0\iff x=0$
+\end_inset
+
+ y 
+\begin_inset Formula $\delta(xy)=\delta(x)\delta(y)$
+\end_inset
+
+, de donde se obtiene la primera condición.
+ Sean ahora 
+\begin_inset Formula $a:=a_{1}+a_{2}i$
+\end_inset
+
+ y 
+\begin_inset Formula $b:=b_{1}+b_{2}i\neq0$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{1},a_{2},b_{1},b_{2}\in\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $x:=x_{1}+x_{2}i:=\frac{a}{b}$
+\end_inset
+
+ con 
+\begin_inset Formula $x_{1},x_{2}\in\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $q_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $q_{2}$
+\end_inset
+
+ los enteros más próximos a 
+\begin_inset Formula $x_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $x_{2}$
+\end_inset
+
+ respectivamente, 
+\begin_inset Formula $q:=q_{1}+q_{2}i$
+\end_inset
+
+ y 
+\begin_inset Formula $r:=a-bq$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $a=bq+r$
+\end_inset
+
+ y, como 
+\begin_inset Formula $|x_{i}-q_{i}|\leq\frac{1}{2}$
+\end_inset
+
+, 
+\begin_inset Formula 
+\begin{multline*}
+\delta(r):=|a-bq|^{2}=|b(\frac{a}{b}-q)|^{2}=|b|^{2}|x-q|^{2}=\\
+=\delta(b)((x_{1}-q_{1})^{2}+(x_{2}-q_{2})^{2})\leq\delta(b)\left(\frac{1}{4}+\frac{1}{4}\right)=\frac{\delta(b)}{2}<\delta(b).
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $\delta$
+\end_inset
+
+ una función euclídea en 
+\begin_inset Formula $D$
+\end_inset
+
+, 
+\begin_inset Formula $I$
+\end_inset
+
+ un ideal de 
+\begin_inset Formula $D$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in I\setminus\{0\}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $I=(a)\iff\forall x\in I\setminus\{0\},\delta(a)\leq\delta(x)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Dado 
+\begin_inset Formula $0\neq x\in I=(a)$
+\end_inset
+
+, como 
+\begin_inset Formula $a\mid x$
+\end_inset
+
+, 
+\begin_inset Formula $\delta(a)\leq\delta(x)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\supseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $a\in I\implies(a)\subseteq I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\subseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $x\in I$
+\end_inset
+
+, si 
+\begin_inset Formula $x=0$
+\end_inset
+
+, 
+\begin_inset Formula $x\in(a)$
+\end_inset
+
+.
+ De lo contrario existen 
+\begin_inset Formula $q,r\in D$
+\end_inset
+
+ con 
+\begin_inset Formula $x=aq+r$
+\end_inset
+
+ y o 
+\begin_inset Formula $r=0$
+\end_inset
+
+ o 
+\begin_inset Formula $\delta(r)<\delta(a)$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $r=x-aq\in I$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\delta(a)\leq\delta(x)$
+\end_inset
+
+, luego necesariamente 
+\begin_inset Formula $r=0$
+\end_inset
+
+ y 
+\begin_inset Formula $x\in(a)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, todo dominio euclídeo es DIP, pues si 
+\begin_inset Formula $\delta$
+\end_inset
+
+ es una función euclídea en el dominio e 
+\begin_inset Formula $I$
+\end_inset
+
+ es un ideal no nulo (el ideal nulo es 
+\begin_inset Formula $(0)$
+\end_inset
+
+), 
+\begin_inset Formula $\delta(I\setminus\{0\})$
+\end_inset
+
+ tendrá un mínimo que se alcanzará para algún 
+\begin_inset Formula $a\in I\setminus\{0\}$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $I=(a)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $\delta$
+\end_inset
+
+ es una función euclídea en 
+\begin_inset Formula $D$
+\end_inset
+
+, un elemento 
+\begin_inset Formula $a\in D$
+\end_inset
+
+ es una unidad si y sólo si 
+\begin_inset Formula $\delta(a)=\delta(1)$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $\forall x\in D\setminus\{0\},\delta(a)\leq\delta(x)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[1\iff3]$
+\end_inset
+
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ es unidad si y sólo si 
+\begin_inset Formula $(a)=D$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $\forall x\in D\setminus\{0\},\delta(a)\leq\delta(x)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[2\implies3]$
+\end_inset
+
+ Dado 
+\begin_inset Formula $x\in D\setminus\{0\}$
+\end_inset
+
+, como 
+\begin_inset Formula $1\mid x$
+\end_inset
+
+, 
+\begin_inset Formula $\delta(a)=\delta(1)\leq\delta(x)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[3\implies2]$
+\end_inset
+
+ Obvio.
+\end_layout
+
+\begin_layout Section
+Cuerpos de fracciones
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $D\neq0$
+\end_inset
+
+ un dominio y 
+\begin_inset Formula $X:=D\times(D\setminus\{0\})$
+\end_inset
+
+, definimos la relación binaria 
+\begin_inset Formula $(a_{1},s_{1})\sim(a_{2},s_{2}):\iff a_{1}s_{2}=a_{2}s_{1}$
+\end_inset
+
+.
+ Esta relación es de equivalencia; en efecto, las propiedades reflexiva
+ y transitiva son claras, y si 
+\begin_inset Formula $a_{1}s_{2}=a_{2}s_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $a_{2}s_{3}=a_{3}s_{2}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a_{1}s_{2}a_{2}s_{3}=a_{2}s_{1}a_{3}s_{2}$
+\end_inset
+
+ y, o bien 
+\begin_inset Formula $a_{2}=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $a_{1}=a_{3}=0$
+\end_inset
+
+, o podemos cancelar 
+\begin_inset Formula $s_{2}a_{2}$
+\end_inset
+
+, y en cualquier caso 
+\begin_inset Formula $a_{1}s_{3}=s_{1}a_{3}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\begin_inset Formula $a/s:=\frac{a}{s}:=[(a,s)]\in Q(D):=X/\sim$
+\end_inset
+
+, y se tiene que las operaciones
+\begin_inset Formula 
+\begin{align*}
+\frac{a_{1}}{s_{1}}+\frac{a_{2}}{s_{2}} & :=\frac{a_{1}s_{2}+a_{2}s_{1}}{s_{1}s_{2}}, & \frac{a_{1}}{s_{1}}\cdot\frac{a_{2}}{s_{2}} & :=\frac{a_{1}a_{2}}{s_{1}s_{2}},
+\end{align*}
+
+\end_inset
+
+están bien definidas.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $a_{1}/s_{1}=b_{1}/t_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $a_{2}/s_{2}=b_{2}/t_{2}$
+\end_inset
+
+, esto es, 
+\begin_inset Formula $a_{1}t_{1}=b_{1}s_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $a_{2}t_{2}=b_{2}s_{2}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $(a_{1}s_{2}+a_{2}s_{1})t_{1}t_{2}=a_{1}s_{2}t_{1}t_{2}+a_{2}s_{1}t_{1}t_{2}=b_{1}s_{2}s_{1}t_{2}+b_{2}s_{1}s_{2}t_{2}=(b_{1}t_{2}+b_{2}t_{1})s_{1}s_{2}$
+\end_inset
+
+, luego 
+\begin_inset Formula $\frac{a_{1}s_{2}+a_{2}s_{1}}{s_{1}s_{2}}=\frac{b_{1}t_{2}+b_{2}t_{1}}{t_{1}t_{2}}$
+\end_inset
+
+.
+ Por otro lado, 
+\begin_inset Formula $a_{1}a_{2}t_{1}t_{2}=b_{1}b_{2}s_{1}s_{2}$
+\end_inset
+
+, luego 
+\begin_inset Formula $\frac{a_{1}a_{2}}{s_{1}s_{2}}=\frac{b_{1}b_{2}}{t_{1}t_{2}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Propiedades: 
+\begin_inset Formula $\forall a,b\in D;s,t\in D\setminus\{0\}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}=\frac{0}{1}\iff a=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\frac{a}{s}=\frac{0}{1}\iff a=a1=s0=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}=\frac{1}{1}\iff a=s$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\frac{a}{s}=\frac{1}{1}\iff a=a1=s1=s$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\frac{at}{st}=\frac{a}{s}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $ats=ast$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}=\frac{b}{s}\iff a=b$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\frac{a}{s}=\frac{b}{s}\iff as=bs\overset{s\neq0}{\iff}a=b$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}+\frac{b}{s}=\frac{a+b}{s}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\frac{a}{s}+\frac{b}{s}=\frac{as+bs}{ss}=\frac{(a+b)s}{ss}=\frac{a+b}{s}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+De aquí, 
+\begin_inset Formula $(Q(D),+,\cdot)$
+\end_inset
+
+ es un cuerpo llamado 
+\series bold
+cuerpo de fracciones
+\series default
+ o 
+\series bold
+de cocientes
+\series default
+ de 
+\begin_inset Formula $D$
+\end_inset
+
+ cuyo cero es 
+\begin_inset Formula $\frac{0}{1}$
+\end_inset
+
+ y cuyo uno es 
+\begin_inset Formula $\frac{1}{1}$
+\end_inset
+
+ .
+ 
+\series bold
+Demostración:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}+\frac{b}{t}=\frac{at+bs}{st}=\frac{bs+at}{ts}=\frac{b}{t}+\frac{a}{s}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(\frac{a}{s}+\frac{b}{t}\right)+\frac{c}{u}=\frac{at+bs}{st}+\frac{c}{u}=\frac{atu+bsu+cst}{stu}=\frac{a}{s}+\frac{bu+ct}{tu}=\frac{a}{s}+\left(\frac{b}{t}+\frac{c}{u}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}+\frac{0}{1}=\frac{a1+s0}{s1}=\frac{a+0}{s}=\frac{a}{s}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}+\frac{-a}{s}=\frac{a+(-a)}{s}=\frac{0}{s}=\frac{0}{1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}\frac{b}{t}=\frac{ab}{st}=\frac{ba}{ts}=\frac{b}{t}\frac{a}{s}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}\left(\frac{b}{t}\frac{c}{u}\right)=\frac{a}{s}\frac{bc}{tu}=\frac{abc}{stu}=\frac{ab}{st}\frac{c}{u}=\left(\frac{a}{s}\frac{b}{t}\right)\frac{c}{u}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}\frac{1}{1}=\frac{a1}{s1}=\frac{a}{s}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}\frac{s}{a}=\frac{as}{sa}=\frac{as}{as}=\frac{1}{1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\frac{a}{s}\left(\frac{b}{t}+\frac{c}{u}\right)=\frac{a}{s}\frac{bu+ct}{tu}=\frac{abu+act}{stu}=\frac{abu}{stu}+\frac{act}{stu}=\frac{ab}{st}+\frac{ac}{su}=\frac{a}{s}\frac{b}{t}+\frac{a}{s}\frac{c}{u}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así, 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+ es el cuerpo de fracciones de 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+, y si 
+\begin_inset Formula $A[X]$
+\end_inset
+
+ es un dominio llamamos 
+\series bold
+cuerpo de las funciones racionales
+\series default
+ sobre 
+\begin_inset Formula $A$
+\end_inset
+
+ a 
+\begin_inset Formula $A(X):=D(A[X])$
+\end_inset
+
+.
+ Es fácil ver que función 
+\begin_inset Formula $u:D\to Q(D)$
+\end_inset
+
+ dada por 
+\begin_inset Formula $u(a):=a/1$
+\end_inset
+
+ es un homomorfismo inyectivo, por lo que podemos ver a 
+\begin_inset Formula $D$
+\end_inset
+
+ como un subdominio de 
+\begin_inset Formula $Q(D)$
+\end_inset
+
+ identificando a cada 
+\begin_inset Formula $a\in D$
+\end_inset
+
+ con 
+\begin_inset Formula $a/1\in Q(D)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así, dados un dominio 
+\begin_inset Formula $D$
+\end_inset
+
+ y 
+\begin_inset Formula $u:D\to Q(D)$
+\end_inset
+
+ dada por 
+\begin_inset Formula $u(a):=a/1$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Propiedad universal del cuerpo de fracciones:
+\series default
+ Sean 
+\begin_inset Formula $K$
+\end_inset
+
+ un cuerpo y 
+\begin_inset Formula $f:D\to K$
+\end_inset
+
+ un homomorfismo inyectivo, el único homomorfismo de cuerpos 
+\begin_inset Formula $\tilde{f}:Q(D)\to K$
+\end_inset
+
+ con 
+\begin_inset Formula $\tilde{f}\circ u=f$
+\end_inset
+
+ viene dado por 
+\begin_inset Formula $\tilde{f}(\frac{a}{s})=f(a)f(s)^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si 
+\begin_inset Formula $\hat{f}:Q(D)\to K$
+\end_inset
+
+ es un homomorfismo de cuerpos con 
+\begin_inset Formula $\hat{f}\circ u=f$
+\end_inset
+
+, para todo 
+\begin_inset Formula $\frac{a}{s}\in Q(D)$
+\end_inset
+
+, 
+\begin_inset Formula $\hat{f}(\frac{a}{s})=\hat{f}(\frac{a}{1}\frac{s^{-1}}{1})=\hat{f}(u(a)u(s)^{-1})=\hat{f}(u(a))\hat{f}(u(s))^{-1}=f(a)f(s)^{-1}$
+\end_inset
+
+, luego el homomorfismo es único.
+ Para ver que está bien definido, si 
+\begin_inset Formula $\frac{a}{s}=\frac{b}{t}$
+\end_inset
+
+, 
+\begin_inset Formula $at=bs$
+\end_inset
+
+, luego 
+\begin_inset Formula $f(a)f(t)=f(b)f(s)$
+\end_inset
+
+ y 
+\begin_inset Formula $f(a)f(s)^{-1}=f(b)f(t)^{-1}$
+\end_inset
+
+.
+ Para ver que es un homomorfismo, 
+\begin_inset Formula $\tilde{f}(1/1)=f(1)f(1)^{-1}=1$
+\end_inset
+
+, 
+\begin_inset Formula $\tilde{f}(\frac{a}{s}+\frac{b}{t})=\tilde{f}(\frac{at+bs}{st})=f(at+bs)f(st)^{-1}=(f(a)f(t)+f(b)f(s))f(s)^{-1}f(t)^{-1}=f(a)f(s)^{-1}+f(b)f(t)^{-1}=\tilde{f}(\frac{a}{s})+\tilde{f}(\frac{b}{t})$
+\end_inset
+
+ y 
+\begin_inset Formula $\tilde{f}(\frac{a}{s}\frac{b}{t})=\tilde{f}(\frac{ab}{st})=f(ab)f(st)^{-1}=f(a)f(s)^{-1}f(b)f(t)^{-1}=\tilde{f}(\frac{a}{s})\tilde{f}(\frac{b}{t})$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $K$
+\end_inset
+
+ un cuerpo no trivial y 
+\begin_inset Formula $g,h:Q(D)\to K$
+\end_inset
+
+ homomorfismos que coinciden en 
+\begin_inset Formula $D$
+\end_inset
+
+, entonces 
+\begin_inset Formula $g=h$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+El homomorfismo 
+\begin_inset Formula $f:=g\circ u=h\circ u$
+\end_inset
+
+ es inyectivo por serlo 
+\begin_inset Formula $g$
+\end_inset
+
+, al ser homomorfismo de cuerpos, y 
+\begin_inset Formula $u$
+\end_inset
+
+, y por la Propiedad Universal, existe un único 
+\begin_inset Formula $\tilde{f}$
+\end_inset
+
+ con 
+\begin_inset Formula $\tilde{f}\circ u=f$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\tilde{f}=g=h$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $F$
+\end_inset
+
+ un cuerpo no trivial y 
+\begin_inset Formula $v:D\to F$
+\end_inset
+
+ un homomorfismo inyectivo tal que para todo cuerpo 
+\begin_inset Formula $K$
+\end_inset
+
+ y homomorfismo inyectivo 
+\begin_inset Formula $f:D\to K$
+\end_inset
+
+ existe un único homomorfismo 
+\begin_inset Formula $\tilde{f}:F\to K$
+\end_inset
+
+ con 
+\begin_inset Formula $\tilde{f}\circ v=f$
+\end_inset
+
+, entonces existe un isomorfismo 
+\begin_inset Formula $\phi:F\to Q(D)$
+\end_inset
+
+ con 
+\begin_inset Formula $\phi\circ v=u$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Por la Propiedad Universal de 
+\begin_inset Formula $Q(D)$
+\end_inset
+
+, existe un único homomorfismo 
+\begin_inset Formula $\tilde{v}:Q(D)\to F$
+\end_inset
+
+ con 
+\begin_inset Formula $\tilde{v}\circ u=v$
+\end_inset
+
+, y por la hipótesis existe un único homomorfismo 
+\begin_inset Formula $\tilde{u}:F\to Q(D)$
+\end_inset
+
+ con 
+\begin_inset Formula $\tilde{u}\circ v=u$
+\end_inset
+
+, luego 
+\begin_inset Formula $\tilde{u}\circ\tilde{v}:Q(D)\to Q(D)$
+\end_inset
+
+ verifica 
+\begin_inset Formula $(\tilde{u}\circ\tilde{v})\circ u=\tilde{u}\circ v=u$
+\end_inset
+
+ y, por el punto anterior tomando como cuerpo 
+\begin_inset Formula $Q(D)$
+\end_inset
+
+, 
+\begin_inset Formula $\tilde{u}\circ\tilde{v}=id$
+\end_inset
+
+.
+ En particular 
+\begin_inset Formula $\tilde{u}$
+\end_inset
+
+ es suprayectiva y, como es inyectivo por ser homomorfismo de cuerpos no
+ triviales, es el isomorfismo buscado.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $D$
+\end_inset
+
+ un dominio, 
+\begin_inset Formula $K$
+\end_inset
+
+ un cuerpo no trivial y 
+\begin_inset Formula $f:D\to K$
+\end_inset
+
+ un homomorfismo inyectivo, 
+\begin_inset Formula $K$
+\end_inset
+
+ contiene un subcuerpo isomorfo a 
+\begin_inset Formula $Q(D)$
+\end_inset
+
+.
+ En efecto, por la propiedad universal, existe un homomorfismo 
+\begin_inset Formula $\tilde{f}:Q(D)\to K$
+\end_inset
+
+, y como este es inyectivo, 
+\begin_inset Formula $\text{Im}\tilde{f}$
+\end_inset
+
+ es un subcuerpo de 
+\begin_inset Formula $K$
+\end_inset
+
+ isomorfo a 
+\begin_inset Formula $Q(D)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+De aquí, para 
+\begin_inset Formula $m\in\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $Q(\mathbb{Z}[\sqrt{m}])\cong\mathbb{Q}[\sqrt{m}]$
+\end_inset
+
+, lo que nos permite identificar los elementos de 
+\begin_inset Formula $Q(\mathbb{Z}[\sqrt{m}])$
+\end_inset
+
+ con los de 
+\begin_inset Formula $\mathbb{Q}[\sqrt{m}]$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $m$
+\end_inset
+
+ es un cuadrado, esto significa 
+\begin_inset Formula $Q(\mathbb{Z})\cong\mathbb{Q}$
+\end_inset
+
+, lo que ya sabemos.
+ De lo contrario, sean 
+\begin_inset Formula $f:\mathbb{Z}[\sqrt{m}]\to\mathbb{C}$
+\end_inset
+
+ el homomorfismo inclusión y 
+\begin_inset Formula $\tilde{f}:Q(\mathbb{Z}[\sqrt{m}])\to\mathbb{C}$
+\end_inset
+
+ el que nos da la propiedad universal, dado por 
+\begin_inset Formula $\tilde{f}(\frac{a+b\sqrt{m}}{c+d\sqrt{m}})=\frac{a+b\sqrt{m}}{c+d\sqrt{m}}$
+\end_inset
+
+.
+ Queda ver que 
+\begin_inset Formula $\text{Im}\tilde{f}=\mathbb{Q}[\sqrt{m}]$
+\end_inset
+
+, pero esto es claro, pues para 
+\begin_inset Formula $a,b,c,d\in\mathbb{Z}$
+\end_inset
+
+ con 
+\begin_inset Formula $c,d\neq0$
+\end_inset
+
+, sea 
+\begin_inset Formula $t:=(c+d\sqrt{m})(c-d\sqrt{m})$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+\frac{a+b\sqrt{m}}{c+d\sqrt{m}}=\frac{(a+b\sqrt{m})(c-d\sqrt{m})}{(c+d\sqrt{m})(c-d\sqrt{m})}=\frac{ac-bdm}{t}+\frac{bc-ad}{t}\sqrt{m}\in\mathbb{Q}[\sqrt{m}],
+\]
+
+\end_inset
+
+y recíprocamente,
+\begin_inset Formula 
+\[
+\frac{a}{c}+\frac{b}{d}\sqrt{m}=\frac{ad+bc\sqrt{m}}{cd}\in Q(\mathbb{Z}[\sqrt{m}]).
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $K$
+\end_inset
+
+ un cuerpo no trivial, existe un subcuerpo 
+\begin_inset Formula $K'$
+\end_inset
+
+ de 
+\begin_inset Formula $K$
+\end_inset
+
+ llamado 
+\series bold
+subcuerpo primo
+\series default
+ de 
+\begin_inset Formula $K$
+\end_inset
+
+ contenido en cualquier subcuerpo de 
+\begin_inset Formula $K$
+\end_inset
+
+, y este es isomorfo a 
+\begin_inset Formula $\mathbb{Z}_{p}$
+\end_inset
+
+ si la característica de 
+\begin_inset Formula $K$
+\end_inset
+
+ es un entero primo 
+\begin_inset Formula $p$
+\end_inset
+
+ o a 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+ en caso contrario.
+ 
+\series bold
+Demostración:
+\series default
+ Si la característica es un primo 
+\begin_inset Formula $p$
+\end_inset
+
+, el subanillo primo de 
+\begin_inset Formula $K$
+\end_inset
+
+, isomorfo a 
+\begin_inset Formula $\mathbb{Z}_{p}$
+\end_inset
+
+, es un cuerpo y contiene a cualquier subanillo de 
+\begin_inset Formula $K$
+\end_inset
+
+, y por tanto a cualquier subcuerpo.
+ En otro caso, al ser 
+\begin_inset Formula $K$
+\end_inset
+
+ un cuerpo, la característica es 0, por lo que 
+\begin_inset Formula $f:\mathbb{Z}\to K$
+\end_inset
+
+ dado por 
+\begin_inset Formula $f(n):=n1$
+\end_inset
+
+ es un homomorfismo inyectivo y la propiedad universal nos da un homomorfismo
+ 
+\begin_inset Formula $\tilde{f}:Q(\mathbb{Z})=\mathbb{Q}\to K$
+\end_inset
+
+ dado por 
+\begin_inset Formula $f(\frac{n}{m})=f(n)f(m)^{-1}$
+\end_inset
+
+.
+ Es claro entonces que 
+\begin_inset Formula $K':=\tilde{f}(\mathbb{Q})$
+\end_inset
+
+ es isomorfo a 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+, y queda ver que está contenido en cualquier subcuerpo de 
+\begin_inset Formula $K$
+\end_inset
+
+.
+ Dado un tal 
+\begin_inset Formula $F$
+\end_inset
+
+, para 
+\begin_inset Formula $m\in\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $f(m)=m1\in F$
+\end_inset
+
+, y para 
+\begin_inset Formula $n\in\mathbb{Z}\setminus\{0\}$
+\end_inset
+
+, 
+\begin_inset Formula $f(n)\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $f(n)^{-1}\in F$
+\end_inset
+
+, luego 
+\begin_inset Formula $\tilde{f}(\frac{m}{n})=f(m)f(n)^{-1}\in F$
+\end_inset
+
+, y en resumen 
+\begin_inset Formula $\tilde{f}(\mathbb{Q})\subseteq F$
+\end_inset
+
+.
+ 
+\end_layout
+
+\end_body
+\end_document
-- 
cgit v1.2.3