asturbate

2 files changed, 5340 insertions, 0 deletions

diff --git a/ga/n.lyx b/ga/n.lyx
new file mode 100644
index 0000000..7e60de8
--- /dev/null
+++ b/ga/n.lyx
@@ -0,0 +1,161 @@
+#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 10
+\spacing single
+\use_hyperref false
+\papersize a5paper
+\use_geometry true
+\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
+\leftmargin 0.2cm
+\topmargin 0.7cm
+\rightmargin 0.2cm
+\bottommargin 0.7cm
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle empty
+\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 Title
+Grupos y Anillos
+\end_layout
+
+\begin_layout Date
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+def
+\backslash
+cryear{2020}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "../license.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Bibliografía:
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Anillos
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n1.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/ga/n1.lyx b/ga/n1.lyx
new file mode 100644
index 0000000..7f4aadf
--- /dev/null
+++ b/ga/n1.lyx
@@ -0,0 +1,5179 @@
+#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 Section
+Operaciones binarias
+\end_layout
+
+\begin_layout Standard
+Una 
+\series bold
+operación
+\series default
+ (
+\series bold
+binaria
+\series default
+) en un conjunto 
+\begin_inset Formula $X$
+\end_inset
+
+ es una aplicación 
+\begin_inset Formula $*:X\times X\to X$
+\end_inset
+
+, escribimos 
+\begin_inset Formula $a*b:=*(a,b)$
+\end_inset
+
+, y decimos que 
+\begin_inset Formula $*$
+\end_inset
+
+ es:
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Conmutativa
+\series default
+ si 
+\begin_inset Formula $\forall x,y\in X,x*y=y*x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Asociativa
+\series default
+ si 
+\begin_inset Formula $\forall x,y,z\in X,(x*y)*z=x*(y*z)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Un elemento 
+\begin_inset Formula $x\in X$
+\end_inset
+
+ es:
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Neutro por la izquierda
+\series default
+ de 
+\begin_inset Formula $X$
+\end_inset
+
+ con respecto a 
+\begin_inset Formula $*$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall y\in X,x*y=y$
+\end_inset
+
+, 
+\series bold
+por la derecha
+\series default
+ si 
+\begin_inset Formula $\forall y\in X,y*x=x$
+\end_inset
+
+ y 
+\series bold
+neutro
+\series default
+ si es neutro por la izquierda y por la derecha.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Cancelativo por la izquierda
+\series default
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+ respecto a 
+\begin_inset Formula $*$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall a,b\in X,(x*a=x*b\implies a=b)$
+\end_inset
+
+, 
+\series bold
+por la derecha
+\series default
+ si 
+\begin_inset Formula $\forall a,b\in X,(a*x=b*x\implies a=b)$
+\end_inset
+
+ y 
+\series bold
+cancelativo
+\series default
+ si es cancelativo por la izquierda y por la derecha.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Simétrico
+\series default
+ de 
+\begin_inset Formula $y\in X$
+\end_inset
+
+ 
+\series bold
+por la izquierda
+\series default
+ si existe un neutro 
+\begin_inset Formula $e$
+\end_inset
+
+ tal que 
+\begin_inset Formula $x*y=e$
+\end_inset
+
+, 
+\series bold
+por la derecha
+\series default
+ si 
+\begin_inset Formula $y*x=e$
+\end_inset
+
+ y 
+\series bold
+simétrico
+\series default
+ de 
+\begin_inset Formula $y$
+\end_inset
+
+ o 
+\series bold
+invertible
+\series default
+ si es simétrico por la izquierda y por la derecha.
+\end_layout
+
+\begin_layout Standard
+Dado un conjunto 
+\begin_inset Formula $X$
+\end_inset
+
+ y una operación 
+\begin_inset Formula $*$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+, 
+\begin_inset Formula $(X,*)$
+\end_inset
+
+ es:
+\end_layout
+
+\begin_layout Enumerate
+Un 
+\series bold
+semigrupo
+\series default
+ si 
+\begin_inset Formula $*$
+\end_inset
+
+ es asociativa.
+\end_layout
+
+\begin_layout Enumerate
+Un 
+\series bold
+monoide
+\series default
+ si además 
+\begin_inset Formula $X$
+\end_inset
+
+ tiene un elemento neutro respecto a 
+\begin_inset Formula $*$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Un 
+\series bold
+grupo
+\series default
+ si además todo elemento de 
+\begin_inset Formula $X$
+\end_inset
+
+ es invertible.
+\end_layout
+
+\begin_layout Enumerate
+Un 
+\series bold
+grupo abeliano
+\series default
+ si además 
+\begin_inset Formula $*$
+\end_inset
+
+ es conmutativa.
+\end_layout
+
+\begin_layout Standard
+Ejemplos:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $(\mathbb{N},+)$
+\end_inset
+
+ es un monoide conmutativo, y 
+\begin_inset Formula $(\mathbb{Z},+)$
+\end_inset
+
+, 
+\begin_inset Formula $(\mathbb{Q},+)$
+\end_inset
+
+, 
+\begin_inset Formula $(\mathbb{R},+)$
+\end_inset
+
+ y 
+\begin_inset Formula $(\mathbb{C},+)$
+\end_inset
+
+ son grupos abelianos.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+La suma es asociativa y conmutativa con elemento neutro 0, y todo elemento
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ tiene opuesto 
+\begin_inset Formula $-a$
+\end_inset
+
+, pero en 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+ solo el 0 tiene opuesto.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ son monoides conmutativos con el producto.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+El producto es asociativo y conmutativo con neutro 1, pero el 0 nunca tiene
+ opuesto.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Llamamos 
+\begin_inset Formula $Y^{X}$
+\end_inset
+
+ al conjunto de funciones de 
+\begin_inset Formula $X$
+\end_inset
+
+ a 
+\begin_inset Formula $Y$
+\end_inset
+
+.
+ Dado un conjunto 
+\begin_inset Formula $X$
+\end_inset
+
+, 
+\begin_inset Formula $(X^{X},\circ)$
+\end_inset
+
+ es un monoide, pero no es conmutativo si hay al menos dos elementos.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Claramente 
+\begin_inset Formula $\circ$
+\end_inset
+
+ es asociativa y tiene como neutro la identidad.
+ Si hay menos de dos elementos, claramente 
+\begin_inset Formula $\circ$
+\end_inset
+
+ es conmutativa porque 
+\begin_inset Formula $X^{X}$
+\end_inset
+
+ solo tiene un elemento, pero si tiene dos, por ejemplo 
+\begin_inset Formula $a,b\in X$
+\end_inset
+
+, sean 
+\begin_inset Formula $f(a):=a$
+\end_inset
+
+, 
+\begin_inset Formula $f(b):=a$
+\end_inset
+
+, 
+\begin_inset Formula $g(a):=b$
+\end_inset
+
+, 
+\begin_inset Formula $g(b):=a$
+\end_inset
+
+, entonces 
+\begin_inset Formula $(f\circ g)(a)=a$
+\end_inset
+
+ pero 
+\begin_inset Formula $(g\circ f)(a)=b$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Llamamos 
+\series bold
+grupo simétrico
+\series default
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+, 
+\begin_inset Formula $S_{X}$
+\end_inset
+
+, al conjunto de biyecciones de 
+\begin_inset Formula $X$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $(S_{X},\circ)$
+\end_inset
+
+ es un grupo.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Es asociativa, tiene como neutro la identidad y todo elemento es invertible.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $X$
+\end_inset
+
+ un conjunto, 
+\begin_inset Formula $(\mathbb{R}^{X},+)$
+\end_inset
+
+ con 
+\begin_inset Formula $(f+g)(a):=f(a)+g(a)$
+\end_inset
+
+ es un grupo abeliano, y 
+\begin_inset Formula $(\mathbb{R}^{X},\cdot)$
+\end_inset
+
+ con 
+\begin_inset Formula $(f\cdot g)(a):=f(a)g(a)$
+\end_inset
+
+ es un monoide conmutativo cuyos elementos invertibles son las funciones
+ que no se anulan.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Ambas operaciones son conmutativas y asociativas, la suma tiene como neutro
+ la función constante 0 y el producto la función constante 1.
+ El inverso de una función 
+\begin_inset Formula $f$
+\end_inset
+
+ respecto a la suma es 
+\begin_inset Formula $-f$
+\end_inset
+
+ dada por 
+\begin_inset Formula $(-f)(a):=-f(a)$
+\end_inset
+
+, pero respecto al producto es 
+\begin_inset Formula $g$
+\end_inset
+
+ dada por 
+\begin_inset Formula $g(a):=f(a)^{-1}$
+\end_inset
+
+, que solo existe si 
+\begin_inset Formula $f$
+\end_inset
+
+ no se anula.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Dada una operación 
+\begin_inset Formula $*$
+\end_inset
+
+ en un conjunto 
+\begin_inset Formula $X$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $*$
+\end_inset
+
+ es conmutativa, todo neutro por un lado es neutro, todo elemento cancelativo
+ por un lado es cancelativo y todo elemento con simétrico por un lado es
+ invertible.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $e$
+\end_inset
+
+ es neutro por la izquierda y 
+\begin_inset Formula $f$
+\end_inset
+
+ lo es por la derecha, entonces 
+\begin_inset Formula $e=f$
+\end_inset
+
+.
+ En particular, 
+\begin_inset Formula $X$
+\end_inset
+
+ tiene a lo sumo un neutro.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $f=e*f=e$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Dado un monoide 
+\begin_inset Formula $(X,*)$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in X$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $x$
+\end_inset
+
+ es simétrico por la izquierda de 
+\begin_inset Formula $a$
+\end_inset
+
+ e 
+\begin_inset Formula $y$
+\end_inset
+
+ es simétrico por la derecha de 
+\begin_inset Formula $a$
+\end_inset
+
+, entonces 
+\begin_inset Formula $x=y$
+\end_inset
+
+.
+ En particular, 
+\begin_inset Formula $a$
+\end_inset
+
+ tiene a lo sumo un simétrico.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $y=e*y=(x*a)*y=x*(a*y)=x*e=x$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $a$
+\end_inset
+
+ tiene simétrico por un lado, es cancelable por dicho lado.
+ En particular, todo elemento invertible es cancelable.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si, por ejemplo, 
+\begin_inset Formula $a$
+\end_inset
+
+ tiene simétrico por la izquierda, si 
+\begin_inset Formula $x,y\in X$
+\end_inset
+
+ cumplen 
+\begin_inset Formula $a*x=a*y$
+\end_inset
+
+, 
+\begin_inset Formula $x=e*x=(b*a)*x=b*(a*x)=b*(a*y)=(b*a)*y=e*y=y$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Anillos
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+anillo
+\series default
+ es una terna 
+\begin_inset Formula $(A,+,\cdot)$
+\end_inset
+
+ formada por un conjunto 
+\begin_inset Formula $A$
+\end_inset
+
+ y dos operaciones sobre 
+\begin_inset Formula $A$
+\end_inset
+
+ llamadas 
+\series bold
+suma
+\series default
+ y 
+\series bold
+producto
+\series default
+ tales que 
+\begin_inset Formula $(A,+)$
+\end_inset
+
+ es un grupo abeliano, 
+\begin_inset Formula $(A,\cdot)$
+\end_inset
+
+ es un monoide y el producto es 
+\series bold
+distributivo
+\series default
+ respecto de la suma, es decir, 
+\begin_inset Formula $\forall a,b,c\in A,(a\cdot(b+c)=(a\cdot b)+(a\cdot c)\land(a+b)\cdot c=(a\cdot c)+(b\cdot c))$
+\end_inset
+
+.
+ Si además 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ es conmutativo, decimos que 
+\begin_inset Formula $(A,+,\cdot)$
+\end_inset
+
+ es un 
+\series bold
+anillo conmutativo
+\series default
+.
+\end_layout
+
+\begin_layout Standard
+Asumimos que el producto tiene más preferencia que la suma, y escribimos
+ 
+\begin_inset Formula $ab:=a\cdot b$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+opuesto
+\series default
+ de 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, 
+\begin_inset Formula $-a$
+\end_inset
+
+, al simétrico de 
+\begin_inset Formula $a$
+\end_inset
+
+ respecto de la suma, y escribimos 
+\begin_inset Formula $a-b:=a+(-b)$
+\end_inset
+
+.
+ Si además 
+\begin_inset Formula $a$
+\end_inset
+
+ es invertible, llamamos 
+\series bold
+inverso
+\series default
+ de 
+\begin_inset Formula $a$
+\end_inset
+
+, 
+\begin_inset Formula $a^{-1}$
+\end_inset
+
+, al simétrico de 
+\begin_inset Formula $A$
+\end_inset
+
+ respecto del producto.
+ Si 
+\begin_inset Formula $b$
+\end_inset
+
+ es invertible en 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $A$
+\end_inset
+
+ es conmutativo, escribimos 
+\begin_inset Formula $a/b:=\frac{a}{b}:=ab^{-1}$
+\end_inset
+
+.
+ Decimos que 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ es 
+\series bold
+regular
+\series default
+ si es cancelable respecto del producto o 
+\series bold
+singular
+\series default
+ en caso contrario.
+ Una 
+\series bold
+unidad
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ es un elemento invertible, y llamamos 
+\begin_inset Formula $A^{*}$
+\end_inset
+
+ al conjunto de unidades de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ejemplos:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ son anillos conmutativos con la suma y el producto usuales.
+\end_layout
+
+\begin_layout Enumerate
+Dada una familia de anillos 
+\begin_inset Formula $(A_{i})_{i\in I}$
+\end_inset
+
+, el producto 
+\begin_inset Formula $\prod_{i\in I}A_{i}$
+\end_inset
+
+ es un anillos con las operaciones definidas componente a componente, esto
+ es, dados 
+\begin_inset Formula $a,b\in\prod_{i\in I}A_{i}$
+\end_inset
+
+, 
+\begin_inset Formula $a+b:=(a_{i}+b_{i})_{i\in I}$
+\end_inset
+
+ y 
+\begin_inset Formula $ab:=(a_{i}b_{i})_{i\in I}$
+\end_inset
+
+.
+ En particular, si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un anillo y 
+\begin_inset Formula $X$
+\end_inset
+
+ es un conjunto, el conjunto 
+\begin_inset Formula $A^{X}=\prod_{x\in X}A$
+\end_inset
+
+ es un anillo con la suma y el producto dados por 
+\begin_inset Formula $(f+g)(x):=f(x)+g(x)$
+\end_inset
+
+ y 
+\begin_inset Formula $(fg)(x):=f(x)g(x)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:series"
+
+\end_inset
+
+Llamamos 
+\begin_inset Formula $A[[X]]$
+\end_inset
+
+ al conjunto de las sucesiones de elementos del anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ entendidos como 
+\series bold
+series de potencias
+\series default
+ en una 
+\series bold
+indeterminada
+\series default
+ 
+\begin_inset Formula $X$
+\end_inset
+
+, 
+\begin_inset Formula $\sum_{n=0}^{\infty}a_{n}X^{n}$
+\end_inset
+
+.
+ Definiendo la suma como 
+\begin_inset Formula $(a_{n})_{n}+(b_{n})_{n}:=(a_{n}+b_{n})_{n}$
+\end_inset
+
+ y el producto como 
+\begin_inset Formula $(a_{n})_{n}(b_{n})_{n}:=(\sum_{i=0}^{n}a_{i}b_{n-i})_{n}$
+\end_inset
+
+, tenemos un anillo.
+\end_layout
+
+\begin_layout Enumerate
+Llamamos 
+\begin_inset Formula $A[X]$
+\end_inset
+
+ al conjunto de las sucesiones de elementos del anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ con un número finito de elementos no nulos, entendidas como 
+\series bold
+polinomios
+\series default
+ en una 
+\series bold
+indeterminada
+\series default
+ 
+\begin_inset Formula $X$
+\end_inset
+
+.
+ Este es un anillo con las mismas operaciones que en el punto 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:series"
+plural "false"
+caps "false"
+noprefix "false"
+
+\end_inset
+
+.
+ Dado un polinomio 
+\begin_inset Formula $P(X):=a_{0}+a_{1}X+\dots+a_{n}X^{n}$
+\end_inset
+
+, llamamos 
+\series bold
+coeficiente
+\series default
+ de 
+\series bold
+grado
+\series default
+ 
+\begin_inset Formula $i$
+\end_inset
+
+ de 
+\begin_inset Formula $P$
+\end_inset
+
+ a 
+\begin_inset Formula $a_{i}$
+\end_inset
+
+ y 
+\series bold
+coeficiente independiente
+\series default
+ de 
+\begin_inset Formula $P$
+\end_inset
+
+ a 
+\begin_inset Formula $a_{0}$
+\end_inset
+
+.
+ Además, si 
+\begin_inset Formula $a_{n}\neq0$
+\end_inset
+
+, decimos que 
+\begin_inset Formula $P$
+\end_inset
+
+ tiene 
+\series bold
+grado
+\series default
+ 
+\begin_inset Formula $n$
+\end_inset
+
+ y 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+ es su 
+\series bold
+coeficiente principal
+\series default
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un anillo y 
+\begin_inset Formula $n$
+\end_inset
+
+ es un entero positivo, entonces el conjunto 
+\begin_inset Formula ${\cal M}_{n}(A)$
+\end_inset
+
+ de matrices cuadradas en 
+\begin_inset Formula $A$
+\end_inset
+
+ de tamaño 
+\begin_inset Formula $n$
+\end_inset
+
+ es un anillo con la suma y el producto habituales de matrices.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $A$
+\end_inset
+
+ un anillo y 
+\begin_inset Formula $a,b,c\in A$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+Todo elemento es cancelable respecto de la suma.
+\end_layout
+
+\begin_layout Enumerate
+Todo elemento invertible es regular.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $b+a=a\implies b=0$
+\end_inset
+
+, 
+\begin_inset Formula $\forall a\in A,ba=a\implies b=1$
+\end_inset
+
+.
+ En particular, el 0 y el 1 son únicos.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $b+a=a\implies b=b+(a-a)=(b+a)-a=a-a=0$
+\end_inset
+
+, 
+\begin_inset Formula $\forall a\in A,ba=a\implies b=b1=1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+El opuesto de 
+\begin_inset Formula $a$
+\end_inset
+
+ es único, y si 
+\begin_inset Formula $a$
+\end_inset
+
+ es invertible, el inverso es único.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $0a=a0=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $0a+0a=(0+0)a=0a=0a+0\implies0a=0$
+\end_inset
+
+, y 
+\begin_inset Formula $a0=0$
+\end_inset
+
+ se prueba análogamente.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $a(-b)=(-a)b=-(ab)$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $a(-b)+ab=a(-b+b)=a0=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $a(-b)$
+\end_inset
+
+ es el opuesto de 
+\begin_inset Formula $ab$
+\end_inset
+
+ respecto de la suma y, por unicidad, 
+\begin_inset Formula $a(-b)=-(ab)$
+\end_inset
+
+.
+ Que 
+\begin_inset Formula $(-a)b=-(ab)$
+\end_inset
+
+ se prueba análogamente.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $a(b-c)=ab-ac$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $a(b-c)=a(b+(-c))=ab+a(-c)=ab+(-ac)=ab-ac$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ son invertibles si y sólo si lo son 
+\begin_inset Formula $ab$
+\end_inset
+
+ y 
+\begin_inset Formula $ba$
+\end_inset
+
+, en cuyo caso 
+\begin_inset Formula $(ab)^{-1}=b^{-1}a^{-1}$
+\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
+
+Basta ver que 
+\begin_inset Formula $abb^{-1}a^{-1},b^{-1}a^{-1}ab,baa^{-1}b^{-1},a^{-1}b^{-1}ba=1$
+\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
+
+Tenemos 
+\begin_inset Formula $ab(ab)^{-1}=1$
+\end_inset
+
+, luego 
+\begin_inset Formula $b(ab)^{-1}$
+\end_inset
+
+ es simétrico de 
+\begin_inset Formula $a$
+\end_inset
+
+ por la derecha, y 
+\begin_inset Formula $(ba)^{-1}ba=1$
+\end_inset
+
+, luego 
+\begin_inset Formula $(ba)^{-1}b$
+\end_inset
+
+ es simétrico de 
+\begin_inset Formula $a$
+\end_inset
+
+ por la izquierda, luego 
+\begin_inset Formula $a$
+\end_inset
+
+ es invertible.
+ Para 
+\begin_inset Formula $b$
+\end_inset
+
+, se hace de forma análoga.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $0=1$
+\end_inset
+
+, entonces 
+\begin_inset Formula $A=\{0\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $a\in A\implies a=a1=a0=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Dado un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+, llamamos 
+\begin_inset Formula $0_{A}$
+\end_inset
+
+ al cero de 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $1_{A}$
+\end_inset
+
+ al uno de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, definimos 
+\begin_inset Formula $0_{\mathbb{Z}}a:=0$
+\end_inset
+
+, y para 
+\begin_inset Formula $n\in\mathbb{Z}^{+}$
+\end_inset
+
+, 
+\begin_inset Formula $na:=(n-1)a+a$
+\end_inset
+
+ y 
+\begin_inset Formula $(-n)a:=-(na)$
+\end_inset
+
+.
+ Definimos 
+\begin_inset Formula $a^{0_{\mathbb{Z}}}:=1_{A}$
+\end_inset
+
+, para 
+\begin_inset Formula $n\in\mathbb{Z}^{+}$
+\end_inset
+
+, 
+\begin_inset Formula $a^{n}:=a^{n-1}a$
+\end_inset
+
+, y si 
+\begin_inset Formula $a$
+\end_inset
+
+ es invertible, 
+\begin_inset Formula $a^{-n}:=(a^{-1})^{n}$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Standard
+Propiedades: Dados un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $a,b\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $m,n\in\mathbb{Z}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $n(a+b)=na+nb$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para 
+\begin_inset Formula $n=0$
+\end_inset
+
+, 
+\begin_inset Formula $0(a+b)=0=0+0=0a+0b$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $n>0$
+\end_inset
+
+, por inducción, 
+\begin_inset Formula $n(a+b)=(n-1)(a+b)+(a+b)=(n-1)a+(n-1)b+a+b=(n-1)a+a+(n-1)b+b=na+nb$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $(-n)(a+b)=-(n(a+b))=-(na+nb)=-(na)-(nb)=(-n)a+(-n)b$
+\end_inset
+
+.
+ El penúltimo paso de la última igualdad se debe a que 
+\begin_inset Formula $a+b+(-a)+(-b)=a+(-a)+b+(-b)=0+0=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $-(a+b)=(-a)+(-b)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $(n+m)a=na+ma$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para 
+\begin_inset Formula $m=0$
+\end_inset
+
+, 
+\begin_inset Formula $(n+0)a=na=na+0=na+0a$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $m>0$
+\end_inset
+
+, por inducción, 
+\begin_inset Formula $(n+m)a=(n+m-1)a+a=na+(m-1)a+a=na+ma$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $(n-m)a+ma=(n-m+m)a=na$
+\end_inset
+
+, luego restando 
+\begin_inset Formula $ma$
+\end_inset
+
+ a ambos lados, 
+\begin_inset Formula $(n-m)a=na-ma$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $n(ma)=(nm)a$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para 
+\begin_inset Formula $n=0$
+\end_inset
+
+, 
+\begin_inset Formula $0(ma)=0=0a=(0m)a$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $n>0$
+\end_inset
+
+, por inducción, 
+\begin_inset Formula $n(ma)=(n-1)(ma)+ma=((n-1)m)a+ma=((n-1)m+m)a=(nm)a$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $(-n)(ma)=-(n(ma))=-((nm)a)=(-nm)a$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $n,m\geq0$
+\end_inset
+
+, 
+\begin_inset Formula $a^{n+m}=a^{n}a^{m}$
+\end_inset
+
+, y si 
+\begin_inset Formula $a$
+\end_inset
+
+ es invertible, esto también se cumple para 
+\begin_inset Formula $n$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+ enteros arbitrarios.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para 
+\begin_inset Formula $m=0$
+\end_inset
+
+, 
+\begin_inset Formula $a^{n+0}=a^{n}=a^{n}1=a^{n}a^{0}$
+\end_inset
+
+, y para 
+\begin_inset Formula $m>0$
+\end_inset
+
+, por inducción, 
+\begin_inset Formula $a^{n+m}=a^{n+m-1}a=a^{n}a^{m-1}a=a^{n}a^{m}$
+\end_inset
+
+.
+ Sea ahora 
+\begin_inset Formula $a$
+\end_inset
+
+ invertible.
+ La prueba anterior vale también para 
+\begin_inset Formula $n<0$
+\end_inset
+
+, luego queda ver el caso en que 
+\begin_inset Formula $m<0$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Standard
+Primero vemos que, para 
+\begin_inset Formula $m>0$
+\end_inset
+
+, 
+\begin_inset Formula $a^{m}=aa^{m-1}$
+\end_inset
+
+, pues para 
+\begin_inset Formula $m=1$
+\end_inset
+
+, 
+\begin_inset Formula $a^{1}=1a=a=aa^{0}=aa^{1-1}$
+\end_inset
+
+, y para 
+\begin_inset Formula $m>1$
+\end_inset
+
+, por inducción, 
+\begin_inset Formula $a^{m}=a^{m-1}a=aa^{m-2}a=aa^{m-1}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $(a^{m})^{-1}=(a^{-1})^{m}=a^{-m}$
+\end_inset
+
+, pues para 
+\begin_inset Formula $m=0$
+\end_inset
+
+, 
+\begin_inset Formula $(a^{0})^{-1}=1^{-1}=1=(a^{-1})^{0}$
+\end_inset
+
+, y para 
+\begin_inset Formula $m>0$
+\end_inset
+
+, por inducción, 
+\begin_inset Formula $(a^{m})^{-1}=(aa^{m-1})^{-1}=(a^{m-1})^{-1}a^{-1}=(a^{-1})^{m-1}(a^{-1})^{1}=(a^{-1})^{m}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Con esto, sea 
+\begin_inset Formula $m>0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a^{n-m}a^{m}=a^{n-m+m}=a^{n}$
+\end_inset
+
+, luego 
+\begin_inset Formula $a^{n-m}=a^{n-m}a^{m}(a^{m})^{-1}=a^{n}(a^{m})^{-1}=a^{n}a^{-m}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es conmutativo y 
+\begin_inset Formula $n\geq0$
+\end_inset
+
+, 
+\begin_inset Formula $(ab)^{n}=a^{n}b^{n}$
+\end_inset
+
+, y si además 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ son invertibles, esto también se cumple para todo entero 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para 
+\begin_inset Formula $n=0$
+\end_inset
+
+, 
+\begin_inset Formula $(ab)^{0}=1=1\cdot1=a^{0}b^{0}$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $n>0$
+\end_inset
+
+, por inducción, 
+\begin_inset Formula $(ab)^{n}=(ab)^{n-1}ab=a^{n-1}b^{n-1}ab=a^{n-1}ab^{n-1}b=a^{n}b^{n}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ son invertibles, entonces 
+\begin_inset Formula $ab$
+\end_inset
+
+ también lo es, y 
+\begin_inset Formula $(ab)^{-n}=((ab)^{-1})^{n}=(b^{-1}a^{-1})^{n}=(b^{-1})^{n}(a^{-1})^{n}=b^{-n}a^{-n}=a^{-n}b^{-n}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Subanillos
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $*$
+\end_inset
+
+ una operación sobre un conjunto 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B\subseteq A$
+\end_inset
+
+, decimos que 
+\begin_inset Formula $B$
+\end_inset
+
+ es 
+\series bold
+cerrado
+\series default
+ respecto a 
+\begin_inset Formula $*$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall a,b\in B,a*b\in B$
+\end_inset
+
+, en cuyo caso 
+\begin_inset Formula $\hat{*}:B\times B\to B$
+\end_inset
+
+ dada por 
+\begin_inset Formula $x\hat{*}y:=x*y$
+\end_inset
+
+ es la operación 
+\series bold
+inducida
+\series default
+ en 
+\begin_inset Formula $B$
+\end_inset
+
+ por 
+\begin_inset Formula $*$
+\end_inset
+
+, que identificamos con 
+\begin_inset Formula $*$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $B$
+\end_inset
+
+ es cerrado respecto a 
+\begin_inset Formula $*$
+\end_inset
+
+, si 
+\begin_inset Formula $(A,*)$
+\end_inset
+
+ y 
+\begin_inset Formula $(B,*)$
+\end_inset
+
+ son semigrupos, 
+\begin_inset Formula $B$
+\end_inset
+
+ es un 
+\series bold
+subsemigrupo
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+; si son monoides con el mismo neutro, es un 
+\series bold
+submonoide
+\series default
+, y si son grupos con el mismo neutro, es un 
+\series bold
+subgrupo
+\series default
+.
+ Si 
+\begin_inset Formula $B$
+\end_inset
+
+ es cerrado respecto a las operaciones 
+\begin_inset Formula $+$
+\end_inset
+
+ y 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $(A,+,\cdot)$
+\end_inset
+
+ y 
+\begin_inset Formula $(B,+,\cdot)$
+\end_inset
+
+ son anillos con el mismo uno, 
+\begin_inset Formula $B$
+\end_inset
+
+ es un 
+\series bold
+subanillo
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Para que 
+\begin_inset Formula $B\subseteq A$
+\end_inset
+
+ sea un subsemigrupo del semigrupo 
+\begin_inset Formula $(A,*)$
+\end_inset
+
+, basta con que 
+\begin_inset Formula $B$
+\end_inset
+
+ sea cerrado respecto a 
+\begin_inset Formula $*$
+\end_inset
+
+; para que sea un submonoide del monoide 
+\begin_inset Formula $(A,*)$
+\end_inset
+
+, también debe contener al neutro, y para que sea un subgrupo del grupo
+ 
+\begin_inset Formula $(A,*)$
+\end_inset
+
+, debe además ser cerrado respecto a inversos.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $B\subseteq A$
+\end_inset
+
+ es un subanillo de un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ si y sólo si contiene al 1 y es cerrado para sumas productos y opuestos,
+ si y sólo si contiene al 1 es cerrado para restas y productos, y en tal
+ caso el cero de 
+\begin_inset Formula $A$
+\end_inset
+
+ es el de 
+\begin_inset Formula $B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[1\implies2]$
+\end_inset
+
+ Por definición, 
+\begin_inset Formula $B$
+\end_inset
+
+ es cerrado para sumas y productos y 
+\begin_inset Formula $1\in B$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $B$
+\end_inset
+
+ es un anillo, tiene un cero 
+\begin_inset Formula $0_{B}$
+\end_inset
+
+ y cada 
+\begin_inset Formula $b\in B$
+\end_inset
+
+ tiene un opuesto en 
+\begin_inset Formula $B$
+\end_inset
+
+, pero 
+\begin_inset Formula $0_{B}+0_{B}=0_{B}=0+0_{B}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $0_{B}=0$
+\end_inset
+
+, luego el opuesto en 
+\begin_inset Formula $B$
+\end_inset
+
+ es el mismo que en 
+\begin_inset Formula $A$
+\end_inset
+
+ por unicidad y 
+\begin_inset Formula $B$
+\end_inset
+
+ es cerrado para opuestos.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[2\implies3]$
+\end_inset
+
+ Trivial.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[3\implies1]$
+\end_inset
+
+ Solo hay que ver que 
+\begin_inset Formula $B$
+\end_inset
+
+ es cerrado para sumas y tiene un 0 con el cual es cerrado para opuestos.
+ Tenemos 
+\begin_inset Formula $0=1-1\in B$
+\end_inset
+
+ y, para 
+\begin_inset Formula $b\in B$
+\end_inset
+
+, 
+\begin_inset Formula $-b=0-b\in B$
+\end_inset
+
+, luego es cerrado para opuestos, y para 
+\begin_inset Formula $a,b\in B$
+\end_inset
+
+, 
+\begin_inset Formula $a+b=a-(-b)\in B$
+\end_inset
+
+, luego es cerrado para sumas.
+\end_layout
+
+\begin_layout Standard
+Algunos subanillos:
+\end_layout
+
+\begin_layout Enumerate
+Todo anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ es un subanillo de sí mismo, el 
+\series bold
+subanillo impropio
+\series default
+, y el resto de subanillos son 
+\series bold
+propios
+\series default
+.
+\end_layout
+
+\begin_layout Enumerate
+Cada uno de 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ es un subanillo de los posteriores.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\{0\}$
+\end_inset
+
+ es subanillo de 
+\begin_inset Formula $A$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $A=\{0\}$
+\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
+
+Si 
+\begin_inset Formula $\{0\}$
+\end_inset
+
+ es subanillo, 
+\begin_inset Formula $1\in\{0\}$
+\end_inset
+
+, luego 
+\begin_inset Formula $0=1$
+\end_inset
+
+ y 
+\begin_inset Formula $A=\{0\}$
+\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
+
+Obvio.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Llamamos 
+\series bold
+subanillo primo
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ a 
+\begin_inset Formula $\mathbb{Z}1:=\{n1_{A}\}_{n\in\mathbb{Z}}$
+\end_inset
+
+, el menor subanillo de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Claramente 
+\begin_inset Formula $\mathbb{Z}1$
+\end_inset
+
+ contiene al 1 y es cerrado para restas y productos.
+ Sea 
+\begin_inset Formula $B$
+\end_inset
+
+ un subanillo de 
+\begin_inset Formula $A$
+\end_inset
+
+, queremos ver que para 
+\begin_inset Formula $n\in\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $n1\in B$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $n=0$
+\end_inset
+
+, 
+\begin_inset Formula $0\cdot1=0\in B$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $n>0$
+\end_inset
+
+, por inducción, 
+\begin_inset Formula $n1=(n-1)1+1\in B$
+\end_inset
+
+ por ser suma de dos elementos de 
+\begin_inset Formula $B$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $(-n)1=-(n1)\in B$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ son anillos y 
+\begin_inset Formula $B\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $A\times\{0_{B}\}$
+\end_inset
+
+ no es cerrado para sumas y productos pero no es un subanillo de 
+\begin_inset Formula $A\times B$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+No contiene al 1.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Dado 
+\begin_inset Formula $z\in\mathbb{C}$
+\end_inset
+
+, llamamos 
+\begin_inset Formula $\mathbb{Z}[z]:=\{a+bz\}_{a,b\in\mathbb{Z}}$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{Q}[z]:=\{a+bz\}_{a,b\in\mathbb{Q}}$
+\end_inset
+
+.
+ Dado 
+\begin_inset Formula $m\in\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{Z}[\sqrt{m}]$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{Q}[\sqrt{m}]$
+\end_inset
+
+ son subanillos de 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+, y si además 
+\begin_inset Formula $m\geq0$
+\end_inset
+
+, lo son también de 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $m$
+\end_inset
+
+ es el cuadrado de un entero, entonces 
+\begin_inset Formula $\mathbb{Z}[\sqrt{m}]=\mathbb{Z}$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{Q}[\sqrt{m}]=\mathbb{Q}$
+\end_inset
+
+, y de lo contrario 
+\begin_inset Formula $a+b\sqrt{m}=c+d\sqrt{m}\implies a=c\land b=d$
+\end_inset
+
+.
+ Podemos ver 
+\begin_inset Formula $\mathbb{Z}[\sqrt{m}]$
+\end_inset
+
+ con 
+\begin_inset Formula $m<0$
+\end_inset
+
+ como el conjunto de vértices de un enlosado del plano complejo por losas
+ rectangulares con base 1 y altura 
+\begin_inset Formula $\sqrt{m}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Dado un espacio topológico 
+\begin_inset Formula $X$
+\end_inset
+
+, 
+\begin_inset Formula $\{f\in\mathbb{R}^{X}:f\text{ continua}\}$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $\mathbb{R}^{X}$
+\end_inset
+
+ con la suma y el producto por elementos.
+\end_layout
+
+\begin_layout Enumerate
+Dado un espacio vectorial 
+\begin_inset Formula $V$
+\end_inset
+
+, 
+\begin_inset Formula $\{f\in V^{V}:f\text{ lineal}\}$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $(V^{V},+,\circ)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Todo anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $A[X]$
+\end_inset
+
+ identificando los elementos de 
+\begin_inset Formula $A$
+\end_inset
+
+ con los 
+\series bold
+polinomios constantes
+\series default
+, de la forma 
+\begin_inset Formula $P(X)=a_{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Dado un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ y un conjunto 
+\begin_inset Formula $X$
+\end_inset
+
+, 
+\begin_inset Formula $\{f\in A^{X}:f\text{ constante}\}$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $A^{X}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Homomorfismos
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+homomorfismo
+\series default
+ entre dos anillos 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ es una aplicación 
+\begin_inset Formula $f:A\to B$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall x,y\in A,(f(x+y)=f(x)+f(y)\land f(xy)=f(x)f(y))$
+\end_inset
+
+ y 
+\begin_inset Formula $f(1)=1$
+\end_inset
+
+.
+ Un 
+\series bold
+automorfismo
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ es un homomorfismo de 
+\begin_inset Formula $A$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+, y un 
+\series bold
+isomorfismo de anillos
+\series default
+ es un homomorfismo de anillos biyectivo.
+ Dos anillos 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ son 
+\series bold
+isomorfos
+\series default
+ si existe un isomorfismo entre ellos.
+\end_layout
+
+\begin_layout Standard
+Propiedades: Sean 
+\begin_inset Formula $f:A\to B$
+\end_inset
+
+ un homomorfismo de anillos y 
+\begin_inset Formula $a,b,a_{1},\dots,a_{n}\in A$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f(0)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $0+f(0)=f(0)=f(0+0)=f(0)+f(0)\implies0=f(0)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f(-a)=-f(a)$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $f(a)+f(-a)=f(a+(-a))=f(0)=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f(a-b)=f(a)-f(b)$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $f(a-b)=f(a)+f(-b)=f(a)-f(b)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f(a_{1}+\dots+a_{n})=f(a_{1})+\dots+f(a_{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f(na)=nf(a)$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para 
+\begin_inset Formula $n=0$
+\end_inset
+
+, 
+\begin_inset Formula $f(0a)=f(0)=0=0f(a)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $a$
+\end_inset
+
+ es invertible, 
+\begin_inset Formula $f(a)$
+\end_inset
+
+ también lo es y 
+\begin_inset Formula $f(a)^{-1}=f(a^{-1})$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $f(a)f(a^{-1})=f(aa^{-1})=f(1)=1$
+\end_inset
+
+, 
+\begin_inset Formula $f(a^{-1})f(a)=f(a^{-1}a)=f(1)=1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f(a_{1}\cdots a_{n})=f(a_{1})\cdots f(a_{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A'$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $f(A')$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $B$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $1=f(1)\in f(A')$
+\end_inset
+
+.
+ Sean 
+\begin_inset Formula $a,b\in f(A')$
+\end_inset
+
+, existen 
+\begin_inset Formula $x,y\in A'$
+\end_inset
+
+ con 
+\begin_inset Formula $f(x)=a$
+\end_inset
+
+ y 
+\begin_inset Formula $f(y)=b$
+\end_inset
+
+, luego 
+\begin_inset Formula $a-b=f(x)-f(y)=f(x-y)\in f(A')$
+\end_inset
+
+, y 
+\begin_inset Formula $ab=f(x)f(y)=f(xy)\in f(A')$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $B'$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $B$
+\end_inset
+
+, 
+\begin_inset Formula $f^{-1}(B')$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $1\in f^{-1}(1)\in f^{-1}(B')$
+\end_inset
+
+.
+ Sean 
+\begin_inset Formula $a,b\in f^{-1}(B')$
+\end_inset
+
+, 
+\begin_inset Formula $f(a),f(b)\in B'$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $f(a-b)=f(a)-f(b)\in B'$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $a-b\in f^{-1}(B')$
+\end_inset
+
+ y 
+\begin_inset Formula $f(ab)=f(a)f(b)\in B'$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $ab\in f^{-1}(B')$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es un isomorfismo de anillos, 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ también.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $f^{-1}(1)=1$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $a,b\in B$
+\end_inset
+
+, sean 
+\begin_inset Formula $x,y\in A$
+\end_inset
+
+ tales que 
+\begin_inset Formula $f(x)=a$
+\end_inset
+
+ y 
+\begin_inset Formula $f(y)=b$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f(x+y)=f(x)+f(y)=a+b$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $f^{-1}(a+b)=x+y=f^{-1}(a)+f^{-1}(b)$
+\end_inset
+
+, y 
+\begin_inset Formula $f(xy)=f(x)f(y)=ab$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $f^{-1}(ab)=xy=f^{-1}(a)f^{-1}(b)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Ejemplos:
+\end_layout
+
+\begin_layout Enumerate
+Dados anillos 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+, 
+\begin_inset Formula $f:A\to B$
+\end_inset
+
+ dada por 
+\begin_inset Formula $f(a)=0$
+\end_inset
+
+ es un homomorfismo si y sólo si 
+\begin_inset Formula $B=0$
+\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 $1=f(1)=0$
+\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
+
+
+\begin_inset Formula $f(1)=0=1$
+\end_inset
+
+, 
+\begin_inset Formula $f(a+b)=0=f(a)+f(b)$
+\end_inset
+
+, 
+\begin_inset Formula $f(ab)=0=f(a)f(b)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $B$
+\end_inset
+
+ un subanillo de 
+\begin_inset Formula $A$
+\end_inset
+
+, la inclusión 
+\begin_inset Formula $i:B\to A$
+\end_inset
+
+ es un homomorfismo.
+\end_layout
+
+\begin_layout Enumerate
+Dado un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $\mu:\mathbb{Z}\to A$
+\end_inset
+
+ dada por 
+\begin_inset Formula $\mu(n):=n1$
+\end_inset
+
+ es el único homomorfismo de anillos 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\mu(1)=1$
+\end_inset
+
+, 
+\begin_inset Formula $\mu(a+b)=(a+b)1=a1+b1$
+\end_inset
+
+, 
+\begin_inset Formula $\mu(ab)=(ab)1=a(b1)=(a1)(b1)$
+\end_inset
+
+.
+ Para este último paso si 
+\begin_inset Formula $a=0$
+\end_inset
+
+, 
+\begin_inset Formula $a(b1)=0(b1)=(0\cdot1)(b1)$
+\end_inset
+
+, y si 
+\begin_inset Formula $a>0$
+\end_inset
+
+, por inducción, 
+\begin_inset Formula $a(b1)=(a-1)(b1)+b1=((a-1)1)(b1)+1(b1)=((a-1)1+1\cdot1)(b1)=(a1)(b1)$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $(-a)(b1)=-(a(b1))=-((a1)(b1))=((-a)1)(b1)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Dada una familia de anillos 
+\begin_inset Formula $(A_{i})_{i\in I}$
+\end_inset
+
+ y 
+\begin_inset Formula $j\in I$
+\end_inset
+
+, la 
+\series bold
+proyección
+\series default
+ 
+\begin_inset Formula $p_{j}:\prod_{i\in I}A_{i}\to A_{j}$
+\end_inset
+
+ dada por 
+\begin_inset Formula $p_{j}(a):=a_{j}$
+\end_inset
+
+ es un homomorfismo.
+\end_layout
+
+\begin_layout Enumerate
+La 
+\series bold
+conjugación
+\series default
+ de complejos, dada por 
+\begin_inset Formula $\overline{a+bi}:=a-bi$
+\end_inset
+
+ para 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+, es un automorfismo en 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+.
+ Del mismo modo, si 
+\begin_inset Formula $d$
+\end_inset
+
+ es un entero que no es un cuadrado, definiendo el conjugado de 
+\begin_inset Formula $a+b\sqrt{d}$
+\end_inset
+
+ como 
+\begin_inset Formula $a-b\sqrt{d}$
+\end_inset
+
+ en 
+\begin_inset Formula $\mathbb{Z}[\sqrt{d}]$
+\end_inset
+
+ o en 
+\begin_inset Formula $\mathbb{Q}[\sqrt{d}]$
+\end_inset
+
+ tenemos un automorfismo.
+\end_layout
+
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $A$
+\end_inset
+
+ un anillo y 
+\begin_inset Formula $b\in A$
+\end_inset
+
+, definimos el 
+\series bold
+homomorfismo de sustitución
+\series default
+ en 
+\begin_inset Formula $b$
+\end_inset
+
+ como la función 
+\begin_inset Formula $S_{b}:A[X]\to A$
+\end_inset
+
+ dada por 
+\begin_inset Formula $S_{b}(a_{0}+a_{1}X+\dots+a_{n}X^{n}):=a_{0}+a_{1}b+\dots+a_{n}b^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Ideales
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+ideal
+\series default
+ de un anillo conmutativo 
+\begin_inset Formula $A$
+\end_inset
+
+ es un subconjunto 
+\begin_inset Formula $I\subseteq A$
+\end_inset
+
+ no vacío tal que 
+\begin_inset Formula $\forall x,y\in I,x+y\in I$
+\end_inset
+
+ y 
+\begin_inset Formula $\forall x\in I,\forall a\in A,ax\in I$
+\end_inset
+
+.
+ Todo ideal contiene al 0, pues tomando 
+\begin_inset Formula $a\in I$
+\end_inset
+
+, 
+\begin_inset Formula $0=a+(-1)a\in I$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Standard
+\begin_inset Newpage pagebreak
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Ejemplos:
+\end_layout
+
+\begin_layout Enumerate
+Dado un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $0:=\{0\}$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+ llamado 
+\series bold
+ideal cero
+\series default
+, y 
+\begin_inset Formula $A$
+\end_inset
+
+ es un ideal llamado 
+\series bold
+ideal impropio
+\series default
+, en oposición al resto que son 
+\series bold
+ideales propios
+\series default
+.
+\end_layout
+
+\begin_layout Enumerate
+Dado un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $T\subseteq A$
+\end_inset
+
+, llamamos 
+\series bold
+ideal generado
+\series default
+ por 
+\begin_inset Formula $T$
+\end_inset
+
+ a
+\begin_inset Formula 
+\[
+TA:=(T):=\left\{ \sum_{k=1}^{n}a_{k}t_{k}\right\} _{n\in\mathbb{N},a_{k}\in A,t_{k}\in T}.
+\]
+
+\end_inset
+
+En particular, dado 
+\begin_inset Formula $b\in A$
+\end_inset
+
+, llamamos 
+\series bold
+ideal principal
+\series default
+ generado por 
+\begin_inset Formula $b$
+\end_inset
+
+ a 
+\begin_inset Formula $(b):=bA:=\{b\}A$
+\end_inset
+
+.
+ Todos los ideales de 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+ son de esta forma.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sea 
+\begin_inset Formula $I$
+\end_inset
+
+ un ideal de 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $I=0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $I=(0)$
+\end_inset
+
+, por lo que suponemos 
+\begin_inset Formula $I\neq0$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $n\in I\setminus\{0\}$
+\end_inset
+
+, o 
+\begin_inset Formula $n$
+\end_inset
+
+ o 
+\begin_inset Formula $-n$
+\end_inset
+
+ es positivo, luego 
+\begin_inset Formula $I$
+\end_inset
+
+ contiene al menos un positivo y podemos definir 
+\begin_inset Formula $a:=\min(I\cap\mathbb{Z}^{+})$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $b\in I$
+\end_inset
+
+, existen 
+\begin_inset Formula $q,r\in\mathbb{Z}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $b=aq+r$
+\end_inset
+
+ y 
+\begin_inset Formula $0\leq r<a$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $r=b+(-q)a\in I$
+\end_inset
+
+, luego debe ser 
+\begin_inset Formula $r=0$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $b=aq\in(a)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $I$
+\end_inset
+
+ un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $J$
+\end_inset
+
+ un ideal de 
+\begin_inset Formula $B$
+\end_inset
+
+, 
+\begin_inset Formula $I\times J$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A\times B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Dado un ideal 
+\begin_inset Formula $I$
+\end_inset
+
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $\{a_{0}+a_{1}X+\dots+a_{n}X^{n}\in A[X]:a_{0}\in I\}$
+\end_inset
+
+ y 
+\begin_inset Formula $\{a_{0}+a_{1}X+\dots+a_{n}X^{n}\in A[X]:a_{0},\dots,a_{n}\in I\}$
+\end_inset
+
+ son ideales de 
+\begin_inset Formula $A[X]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dado un ideal 
+\begin_inset Formula $I$
+\end_inset
+
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $a,b\in A$
+\end_inset
+
+ son 
+\series bold
+congruentes módulo 
+\begin_inset Formula $I$
+\end_inset
+
+
+\series default
+, 
+\begin_inset Formula $a\equiv b$
+\end_inset
+
+, si 
+\begin_inset Formula $b-a\in I$
+\end_inset
+
+, y esta es una relación de equivalencia en 
+\begin_inset Formula $A$
+\end_inset
+
+ con clases de equivalencia de la forma 
+\begin_inset Formula $[a]:=a+I:=\{a+x\}_{x\in I}$
+\end_inset
+
+ y conjunto cociente 
+\begin_inset Formula $A/I=\{[a]\}_{a\in A}$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Standard
+Las operaciones 
+\begin_inset Formula $[a]+[b]:=[a+b]$
+\end_inset
+
+ y 
+\begin_inset Formula $[a][b]:=[ab]$
+\end_inset
+
+ están bien definidas y dotan a 
+\begin_inset Formula $A/I$
+\end_inset
+
+ de una estructura de anillo conmutativo con cero 
+\begin_inset Formula $[0]$
+\end_inset
+
+ y uno 
+\begin_inset Formula $[1]$
+\end_inset
+
+, y llamamos a este anillo 
+\series bold
+anillo cociente de 
+\begin_inset Formula $A$
+\end_inset
+
+ módulo 
+\begin_inset Formula $I$
+\end_inset
+
+
+\series default
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $a\equiv a'$
+\end_inset
+
+ y 
+\begin_inset Formula $b\equiv b'$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a-a',b-b'\in I$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $(a+b)-(a'+b')=(a-a')+(b-b')\in I$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $a+b\equiv a'+b'$
+\end_inset
+
+, 
+\begin_inset Formula $[a+b]=[a'+b']$
+\end_inset
+
+ y la suma está bien definida.
+ Para el producto, 
+\begin_inset Formula $a'b'-ab=a'b'-a'b+a'b-ab=a'(b'-b)+(a'-a)b\in I$
+\end_inset
+
+, luego 
+\begin_inset Formula $[ab]=[a'b']$
+\end_inset
+
+.
+ Es claro que la suma y el producto definidos son asociativos y conmutativos,
+ pues 
+\begin_inset Formula $A$
+\end_inset
+
+ es conmutativo, y que el producto es distributivo respecto de la suma.
+ Además, 
+\begin_inset Formula $[0]+[a]=[0+a]=[a]$
+\end_inset
+
+ y 
+\begin_inset Formula $[1][b]=[1b]=[b]$
+\end_inset
+
+, luego 
+\begin_inset Formula $[0]$
+\end_inset
+
+ y 
+\begin_inset Formula $[1]$
+\end_inset
+
+ son respectivamente el 0 y el 1.
+ Finalmente, para 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, 
+\begin_inset Formula $[-a]+[a]=[-a+a]=[0]$
+\end_inset
+
+, luego todo elemento 
+\begin_inset Formula $[a]$
+\end_inset
+
+ tiene opuesto 
+\begin_inset Formula $[-a]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Algunos anillos cociente:
+\end_layout
+
+\begin_layout Enumerate
+Dado 
+\begin_inset Formula $n\in\mathbb{Z}^{+}$
+\end_inset
+
+, llamamos 
+\begin_inset Formula $\mathbb{Z}_{n}:=\frac{\mathbb{Z}}{n\mathbb{Z}}=\{0+n\mathbb{Z},\dots,(n-1)+n\mathbb{Z}\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Dado 
+\begin_inset Formula $a\in\mathbb{Z}$
+\end_inset
+
+, si 
+\begin_inset Formula $r$
+\end_inset
+
+ es el resto de 
+\begin_inset Formula $a$
+\end_inset
+
+ entre 
+\begin_inset Formula $n$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a\equiv r\bmod n$
+\end_inset
+
+, pero para 
+\begin_inset Formula $0\leq a,b<n$
+\end_inset
+
+, 
+\begin_inset Formula $a\equiv b\iff a-b\in n\mathbb{Z}\iff n|a-b\overset{|a-b|<n}{\iff}a=b$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $A/0\cong A$
+\end_inset
+
+ y 
+\begin_inset Formula $A/A\cong0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Dado un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $A[X]/(X)\cong A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $P,Q\in A[X]$
+\end_inset
+
+ son congruentes módulo 
+\begin_inset Formula $(X)$
+\end_inset
+
+ si y sólo si el coeficiente independiente de 
+\begin_inset Formula $P-Q$
+\end_inset
+
+ es 0, si y sólo si 
+\begin_inset Formula $P$
+\end_inset
+
+ y 
+\begin_inset Formula $Q$
+\end_inset
+
+ tienen igual coeficiente independiente, y es claro que entonces la composición
+ de la inclusión con la proyección, 
+\begin_inset Formula $A\overset{i}{\to}A[X]\overset{\pi}{\to}A[X]/(X)$
+\end_inset
+
+ es un isomorfismo.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Dado un anillo conmutativo 
+\begin_inset Formula $A$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $b\in A$
+\end_inset
+
+ es invertible si y sólo si 
+\begin_inset Formula $(b)=A$
+\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
+
+Dado 
+\begin_inset Formula $x\in A$
+\end_inset
+
+, 
+\begin_inset Formula $x=(bb^{-1})x=b(b^{-1}x)\in(b)$
+\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
+
+En particular 
+\begin_inset Formula $1\in(b)$
+\end_inset
+
+, luego existe 
+\begin_inset Formula $a$
+\end_inset
+
+ tal que 
+\begin_inset Formula $ba=1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Un ideal 
+\begin_inset Formula $I$
+\end_inset
+
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ es impropio si y sólo si 
+\begin_inset Formula $1\in I$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $I$
+\end_inset
+
+ contiene una unidad de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Description
+\begin_inset Formula $[1\implies2\implies3]$
+\end_inset
+
+ Obvio.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[3\implies1]$
+\end_inset
+
+ Si 
+\begin_inset Formula $a\in I\cap A^{*}$
+\end_inset
+
+, 
+\begin_inset Formula $(a)\subseteq I$
+\end_inset
+
+, pero 
+\begin_inset Formula $a$
+\end_inset
+
+ es invertible, luego 
+\begin_inset Formula $(a)=A$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $I=A$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f:A\to B$
+\end_inset
+
+ un homomorfismo de anillos, llamamos 
+\series bold
+núcleo
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ a 
+\begin_inset Formula $\ker f:=f^{-1}(0)$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\text{Im}f$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $B$
+\end_inset
+
+ y 
+\begin_inset Formula $\ker f$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Para lo primero, 
+\begin_inset Formula $f(1)=1\in\text{Im}f$
+\end_inset
+
+, y para 
+\begin_inset Formula $a,b\in\text{Im}f$
+\end_inset
+
+, existen 
+\begin_inset Formula $x,y\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $f(x)=a$
+\end_inset
+
+ y 
+\begin_inset Formula $f(y)=b$
+\end_inset
+
+, luego 
+\begin_inset Formula $a-b=f(x)-f(y)=f(x-y)\in\text{Im}f$
+\end_inset
+
+ y 
+\begin_inset Formula $ab=f(x)f(y)=f(xy)\in\text{Im}f$
+\end_inset
+
+, luego 
+\begin_inset Formula $\text{Im}f$
+\end_inset
+
+ contiene al 1 y es cerrado para sumas y productos y por tanto es un subanillo.
+ Para lo segundo, sabemos que 
+\begin_inset Formula $0\in\ker f$
+\end_inset
+
+, luego 
+\begin_inset Formula $\ker f\neq\emptyset$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $x,y\in\ker f$
+\end_inset
+
+, 
+\begin_inset Formula $f(x+y)=f(x)+f(y)=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $x+y\in\ker f$
+\end_inset
+
+, y para 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, 
+\begin_inset Formula $f(ax)=f(a)f(x)=f(a)0=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $ax\in\ker f$
+\end_inset
+
+, lo que prueba que 
+\begin_inset Formula $\ker f$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Un homomorfismo de anillos 
+\begin_inset Formula $f:A\to B$
+\end_inset
+
+ es inyectivo si y sólo si 
+\begin_inset Formula $\ker f=0$
+\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 $f(a)=0=f(0)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $\ker f=0$
+\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
+
+Sean 
+\begin_inset Formula $a,b\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $a\neq b$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f(a)-f(b)=f(a-b)\neq0$
+\end_inset
+
+, luego 
+\begin_inset Formula $f(a)\neq f(b)$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $f$
+\end_inset
+
+ es inyectiva.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de la correspondencia:
+\series default
+ Si 
+\begin_inset Formula $I$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $J\overset{\pi}{\mapsto}J/I:=\{[a]\}_{a\in J}$
+\end_inset
+
+ es una biyección entre el conjunto de los ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contienen a 
+\begin_inset Formula $I$
+\end_inset
+
+ y el conjunto de los ideales de 
+\begin_inset Formula $A/I$
+\end_inset
+
+ y tanto 
+\begin_inset Formula $\pi$
+\end_inset
+
+ como 
+\begin_inset Formula $\pi^{-1}$
+\end_inset
+
+ preservan la inclusión.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Empezamos viendo que, si 
+\begin_inset Formula $J\supseteq I$
+\end_inset
+
+ es ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+ , 
+\begin_inset Formula $J/I$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A/I$
+\end_inset
+
+.
+ En efecto, como 
+\begin_inset Formula $0\in J$
+\end_inset
+
+, 
+\begin_inset Formula $[0]\in J/I\neq\emptyset$
+\end_inset
+
+; para 
+\begin_inset Formula $[x],[y]\in J/I$
+\end_inset
+
+, 
+\begin_inset Formula $[x]+[y]=[x+y]\in J/I$
+\end_inset
+
+, y para 
+\begin_inset Formula $[x]\in J/I$
+\end_inset
+
+ y 
+\begin_inset Formula $[a]\in A/I$
+\end_inset
+
+, 
+\begin_inset Formula $[a][x]=[ax]\in J/I$
+\end_inset
+
+.
+ Además, 
+\begin_inset Formula $\pi^{-1}(J/I)=J$
+\end_inset
+
+, pues 
+\begin_inset Formula $\pi^{-1}(J/I)=\{x:\pi(x)=[x]\in J/I\}$
+\end_inset
+
+, pero si 
+\begin_inset Formula $x\in J$
+\end_inset
+
+, 
+\begin_inset Formula $[x]\in J/I$
+\end_inset
+
+, y si 
+\begin_inset Formula $[x]\in J/I$
+\end_inset
+
+, existe 
+\begin_inset Formula $a\in I\subseteq J$
+\end_inset
+
+ con 
+\begin_inset Formula $x+a\in J$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $-a\in J$
+\end_inset
+
+ y 
+\begin_inset Formula $(x+a)-a=x\in J$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ahora vemos que, dado un ideal 
+\begin_inset Formula $X$
+\end_inset
+
+ de 
+\begin_inset Formula $A/I$
+\end_inset
+
+, 
+\begin_inset Formula $\pi^{-1}(X)$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contiene a 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ En efecto, como 
+\begin_inset Formula $[0]=I\in X$
+\end_inset
+
+, 
+\begin_inset Formula $\pi^{-1}(X)=\{x:[x]\in X\}\ni0$
+\end_inset
+
+; para 
+\begin_inset Formula $x,y\in\pi^{-1}(X)$
+\end_inset
+
+, 
+\begin_inset Formula $[x],[y]\in X$
+\end_inset
+
+, luego 
+\begin_inset Formula $[x+y]=[x]+[y]\in X$
+\end_inset
+
+ y 
+\begin_inset Formula $x+y\in\pi^{-1}(X)$
+\end_inset
+
+; para 
+\begin_inset Formula $x\in\pi^{-1}(X)$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, 
+\begin_inset Formula $[ax]=[a][x]\in X$
+\end_inset
+
+ y 
+\begin_inset Formula $ax\in\pi^{-1}(X)$
+\end_inset
+
+, y para 
+\begin_inset Formula $x\in I$
+\end_inset
+
+, 
+\begin_inset Formula $[x]\in I/I=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $x\in\pi^{-1}(0)\subseteq\pi^{-1}(X)$
+\end_inset
+
+ e 
+\begin_inset Formula $I\subseteq\pi^{-1}(X)$
+\end_inset
+
+.
+ Además, 
+\begin_inset Formula $\pi^{-1}(X)/I=\{x:[x]\in X\}/I=\{[x]:[x]\in X\}=X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Para ver que 
+\begin_inset Formula $\pi$
+\end_inset
+
+ preserva la inclusión, sean 
+\begin_inset Formula $I\subseteq J\subseteq K$
+\end_inset
+
+ ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+, queremos ver que 
+\begin_inset Formula $J/I\subseteq K/I$
+\end_inset
+
+, pero para 
+\begin_inset Formula $[x]\in J/I$
+\end_inset
+
+, existe 
+\begin_inset Formula $a\in I\subseteq J$
+\end_inset
+
+ con 
+\begin_inset Formula $x+a\in J$
+\end_inset
+
+, luego 
+\begin_inset Formula $x=(x+a)-a\in J\subseteq K$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $[x]\in K/I$
+\end_inset
+
+.
+ Queda ver que 
+\begin_inset Formula $\pi^{-1}$
+\end_inset
+
+ también preserva la inclusión.
+ Sean 
+\begin_inset Formula $X\subseteq Y$
+\end_inset
+
+ ideales de 
+\begin_inset Formula $A/I$
+\end_inset
+
+, queremos ver que 
+\begin_inset Formula $\pi^{-1}(X)\subseteq\pi^{-1}(Y)$
+\end_inset
+
+.
+ En efecto, sea 
+\begin_inset Formula $x\in\pi^{-1}(X)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $[x]\in X\subseteq Y$
+\end_inset
+
+, luego existe 
+\begin_inset Formula $a\in I\subseteq\pi^{-1}(Y)$
+\end_inset
+
+ con 
+\begin_inset Formula $x+a\in\pi^{-1}(Y)$
+\end_inset
+
+, pero como 
+\begin_inset Formula $\pi^{-1}(Y)$
+\end_inset
+
+ es un ideal, 
+\begin_inset Formula $x=(x+a)-a\in\pi^{-1}(Y)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Operaciones con ideales
+\end_layout
+
+\begin_layout Standard
+La intersección de una familia de ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $(X)$
+\end_inset
+
+ es la intersección de todos los ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contienen a 
+\begin_inset Formula $X$
+\end_inset
+
+.
+ Por otro lado, si 
+\begin_inset Formula $\{I_{x}\}_{x\in X}$
+\end_inset
+
+ es una familia de ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+, definimos 
+\begin_inset Formula 
+\begin{eqnarray*}
+\sum_{x\in X}I_{x} & := & \left\{ \sum_{x\in S}a_{x}:S\subseteq X\text{ finito},a_{x}\in I_{x}\right\} ,\\
+\prod_{x\in X}I_{x} & := & \left\{ \sum_{k=1}^{n}\prod_{x\in S}a_{kx}:n\in\mathbb{N},S\subseteq X\text{ finito},a_{kx}\in I_{x}\right\} .
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Entonces:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $\{I_{x}\}_{x\in X}$
+\end_inset
+
+ es una familia de ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $\sum_{x\in X}I_{x}=\left(\bigcup_{x\in X}I_{x}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $I_{1},\dots,I_{n}$
+\end_inset
+
+ son ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $I_{1}\cdots I_{n}=\left(\left\{ x_{1}\cdots x_{n}\right\} _{x_{k}\in I_{k}}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ejemplos:
+\end_layout
+
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $n,m\in\mathbb{Z}$
+\end_inset
+
+ coprimos, 
+\begin_inset Formula $(n)(m)=(nm)$
+\end_inset
+
+, 
+\begin_inset Formula $(n)\cap(m)=(\text{mcm}(n,m))$
+\end_inset
+
+, 
+\begin_inset Formula $(n)+(m)=(\text{mcd}(n,m))$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $(n)(m)=(\{ab\}_{a\in(n),b\in(m)})=(\{pnqm\}_{p,q\in\mathbb{Z}})=(\{knm\})_{k\in\mathbb{Z}}=(nm)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $(n)\cap(m)=\{k\in\mathbb{Z}:n,m|k\}=\{k:\text{mcm}(n,m)|k\}=(\text{mcm}(n,m))$
+\end_inset
+
+.
+ 
+\begin_inset Formula $(n)+(m)=\{a+b\}_{a\in(n),b\in(m)}=\{pn+qm\}_{p,q\in\mathbb{Z}}=\{k\text{mcd}(n,m)\}_{k\in\mathbb{Z}}=(\text{mcd}(n,m))$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+En 
+\begin_inset Formula $\mathbb{Z}[X]$
+\end_inset
+
+, 
+\begin_inset Formula $(2)+(X)$
+\end_inset
+
+, formado por los polinomios cuyo término principal es par, no es principal.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Supongamos que existe 
+\begin_inset Formula $a\in\mathbb{Z}[X]$
+\end_inset
+
+ con 
+\begin_inset Formula $(2)+(X)=(a)$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $ba=2\in(2)$
+\end_inset
+
+ para algún polinomio 
+\begin_inset Formula $b$
+\end_inset
+
+, luego 
+\begin_inset Formula $a\in\mathbb{Z}$
+\end_inset
+
+, y como 
+\begin_inset Formula $a\in(2,X)$
+\end_inset
+
+, 
+\begin_inset Formula $a$
+\end_inset
+
+ es par, luego 
+\begin_inset Formula $X\notin(a)=(2)+(X)\#$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Teoremas de isomorfía
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Primer teorema de isomorfía:
+\series default
+ Dado un homomorfismo de anillos conmutativos 
+\begin_inset Formula $f:A\to B$
+\end_inset
+
+, existe un único isomorfismo de anillos 
+\begin_inset Formula $\tilde{f}:A/\ker f\to\text{Im}f$
+\end_inset
+
+ tal que 
+\begin_inset Formula $i\circ\tilde{f}\circ p=f$
+\end_inset
+
+, donde 
+\begin_inset Formula $i:\text{Im}f\to B$
+\end_inset
+
+ es la inclusión y 
+\begin_inset Formula $p:A\to A/\ker f$
+\end_inset
+
+ es la proyección.
+ En particular, 
+\begin_inset Formula 
+\[
+A/\ker f\cong\text{Im}f.
+\]
+
+\end_inset
+
+
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $K:=\ker f$
+\end_inset
+
+ e 
+\begin_inset Formula $I:=\text{Im}f$
+\end_inset
+
+.
+ La función 
+\begin_inset Formula $\tilde{f}:A/K\to I$
+\end_inset
+
+ dada por 
+\begin_inset Formula $\tilde{f}(x+K):=f(x)$
+\end_inset
+
+ está bien definida, pues si 
+\begin_inset Formula $x+K=y+K$
+\end_inset
+
+, 
+\begin_inset Formula $x-y\in F$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $f(x)-f(y)=f(x-y)=0$
+\end_inset
+
+ y 
+\begin_inset Formula $f(x)=f(y)$
+\end_inset
+
+.
+ Es claro que 
+\begin_inset Formula $\tilde{f}$
+\end_inset
+
+ es un homomorfismo de anillos suprayectivo.
+ Para ver que es inyectivo, sea 
+\begin_inset Formula $x+K\in\ker\tilde{f}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $0=\tilde{f}(x+K)=f(x)$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $x\in K$
+\end_inset
+
+ y 
+\begin_inset Formula $x+K=0+K=0$
+\end_inset
+
+.
+ De aquí que 
+\begin_inset Formula $\tilde{f}$
+\end_inset
+
+ es un homomorfismo.
+ Para ver que 
+\begin_inset Formula $i\circ\tilde{f}\circ p=f$
+\end_inset
+
+, 
+\begin_inset Formula $(i\circ\tilde{f}\circ p)(x)=\tilde{f}(x+K)=f(x)$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x\in A$
+\end_inset
+
+.
+ Para la unicidad, sea 
+\begin_inset Formula $\tilde{f}:A/K\to I$
+\end_inset
+
+ otro isomorfismo con 
+\begin_inset Formula $i\circ\hat{f}\circ p=f$
+\end_inset
+
+, para 
+\begin_inset Formula $x\in A$
+\end_inset
+
+, 
+\begin_inset Formula $\hat{f}(x+K)=i(\hat{f}(p(x))=f(x)=\tilde{f}(x+K)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Newpage pagebreak
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Así:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ son anillos conmutativos, 
+\begin_inset Formula $\frac{A\times B}{0\times B}\cong A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+El homomorfismo de proyección 
+\begin_inset Formula $f:A\times B\to A$
+\end_inset
+
+, 
+\begin_inset Formula $f(a,b):=a$
+\end_inset
+
+, es suprayectivo con núcleo 
+\begin_inset Formula $0\times B$
+\end_inset
+
+, de donde se obtiene el resultado por el primer teorema de isomorfía.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un anillo conmutativo, 
+\begin_inset Formula $\frac{A[X]}{(X)}\cong A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+El homomorfismo de sustitución en el 0 
+\begin_inset Formula $f:A[X]\to A$
+\end_inset
+
+, 
+\begin_inset Formula $f(\sum_{i=0}^{n}a_{i}X^{i})=a_{0}$
+\end_inset
+
+, es suprayectivo con núcleo 
+\begin_inset Formula $(X)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $I$
+\end_inset
+
+ un ideal de de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $\frac{A[X]}{I[X]}\cong(A/I)[X]$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+La función 
+\begin_inset Formula $f:A[X]\to(A/I)[X]$
+\end_inset
+
+, 
+\begin_inset Formula $f(\sum_{i=0}^{n}a_{i}X^{i})=\sum_{i=0}^{n}[a_{i}]X^{i}$
+\end_inset
+
+, es un homomorfismo suprayectivo con núcleo 
+\begin_inset Formula $I[X]=\{\sum_{i=0}^{n}a_{i}X^{i}:a_{i}\in I\}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+
+\series bold
+Segundo teorema de isomorfía:
+\series default
+ Dados dos ideales 
+\begin_inset Formula $I\subseteq J$
+\end_inset
+
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+, el ideal 
+\begin_inset Formula $J/I$
+\end_inset
+
+ de 
+\begin_inset Formula $A/I$
+\end_inset
+
+ cumple 
+\begin_inset Formula 
+\[
+\frac{A/I}{J/I}\cong\frac{A}{J}.
+\]
+
+\end_inset
+
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $J/I$
+\end_inset
+
+ es ideal de 
+\begin_inset Formula $A/I$
+\end_inset
+
+ por el teorema de la correspondencia.
+ Sea 
+\begin_inset Formula $f:A/I\to A/J$
+\end_inset
+
+ dada por 
+\begin_inset Formula $f(a+I):=a+J$
+\end_inset
+
+, es fácil ver que 
+\begin_inset Formula $f$
+\end_inset
+
+ es un homomorfismo de anillos suprayectivo y que 
+\begin_inset Formula $\ker f=J/I$
+\end_inset
+
+, y entonces basta aplicar el primer teorema de isomorfía.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Tercer teorema de isomorfía:
+\series default
+ Sea 
+\begin_inset Formula $A$
+\end_inset
+
+ un anillo con un subanillo 
+\begin_inset Formula $B$
+\end_inset
+
+ y un ideal 
+\begin_inset Formula $I$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $B\cap I$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $B$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $x,y\in B\cap I$
+\end_inset
+
+, 
+\begin_inset Formula $x+y\in B$
+\end_inset
+
+ y 
+\begin_inset Formula $x+y\in I$
+\end_inset
+
+, y sean 
+\begin_inset Formula $x\in B\cap I$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in B$
+\end_inset
+
+, 
+\begin_inset Formula $ax\in I$
+\end_inset
+
+ y 
+\begin_inset Formula $ax\in B$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $B+I$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contiene a 
+\begin_inset Formula $I$
+\end_inset
+
+ como ideal.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $a,b\in B$
+\end_inset
+
+ y 
+\begin_inset Formula $x,y\in I$
+\end_inset
+
+, 
+\begin_inset Formula $(a+x)-(b+y)=(a-b)+(x-y)\in B+I$
+\end_inset
+
+ y 
+\begin_inset Formula $(a+x)(b+y)=ab+ay+xb+xy=ab+(ay+bx+xy)\in B+I$
+\end_inset
+
+, y como 
+\begin_inset Formula $1\in B\subseteq B+I$
+\end_inset
+
+, 
+\begin_inset Formula $B+I$
+\end_inset
+
+ es un subanillo.
+ Además, para 
+\begin_inset Formula $x,y\in I$
+\end_inset
+
+, 
+\begin_inset Formula $x+y\in I\subseteq B+I$
+\end_inset
+
+, y para 
+\begin_inset Formula $a\in B$
+\end_inset
+
+ y 
+\begin_inset Formula $x,y\in I$
+\end_inset
+
+, 
+\begin_inset Formula $(a+x)y\overset{a+x\in A}{\in}I\subseteq B+I$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula 
+\[
+\frac{B}{B\cap I}\cong\frac{B+I}{I}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f:B\to A/I$
+\end_inset
+
+ dada por 
+\begin_inset Formula $f(x):=x+I$
+\end_inset
+
+, es claro que 
+\begin_inset Formula $\ker f=B\cap I$
+\end_inset
+
+ e 
+\begin_inset Formula $\text{Im}f=(B+I)/I$
+\end_inset
+
+, y entonces basta aplicar el primer teorema de isomorfía.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Decimos que un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ tiene 
+\series bold
+característica
+\series default
+ 
+\begin_inset Formula $n\in\mathbb{Z}^{\geq0}$
+\end_inset
+
+ si 
+\begin_inset Formula $n$
+\end_inset
+
+ es el menor entero positivo con 
+\begin_inset Formula $n1_{A}=0_{A}$
+\end_inset
+
+, o 0 si no existe tal 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Sean 
+\begin_inset Formula $A$
+\end_inset
+
+ un anillo conmutativo, 
+\begin_inset Formula $f:\mathbb{Z}\to A$
+\end_inset
+
+ el único homomorfismo de anillos (
+\begin_inset Formula $f(n)=n1$
+\end_inset
+
+) y 
+\begin_inset Formula $n\geq0$
+\end_inset
+
+, 
+\begin_inset Formula $A$
+\end_inset
+
+ tiene característica 
+\begin_inset Formula $n$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\ker f=n\mathbb{Z}$
+\end_inset
+
+, si y sólo si el subanillo primo de 
+\begin_inset Formula $A$
+\end_inset
+
+ es isomorfo a 
+\begin_inset Formula $\mathbb{Z}_{n}$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $A$
+\end_inset
+
+ contiene un subanillo isomorfo a 
+\begin_inset Formula $\mathbb{Z}_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[1\implies2]$
+\end_inset
+
+ Si 
+\begin_inset Formula $n>0$
+\end_inset
+
+, para un cierto 
+\begin_inset Formula $m=qn+r$
+\end_inset
+
+ con 
+\begin_inset Formula $q,r\in\mathbb{Z}$
+\end_inset
+
+ y 
+\begin_inset Formula $0\leq r<n$
+\end_inset
+
+, 
+\begin_inset Formula $f(qn+r)=f(q)f(n)+f(r)\overset{n1=0}{=}f(r)$
+\end_inset
+
+, que es 0 si y sólo si 
+\begin_inset Formula $r=0$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $\ker f=\{qn\}_{q\in\mathbb{Z}}=n\mathbb{Z}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $n=0$
+\end_inset
+
+, 
+\begin_inset Formula $f(m)=0\iff m=0$
+\end_inset
+
+, pues para 
+\begin_inset Formula $m>0$
+\end_inset
+
+ es 
+\begin_inset Formula $m1\neq0$
+\end_inset
+
+ y para 
+\begin_inset Formula $m<0$
+\end_inset
+
+ es 
+\begin_inset Formula $f(m)=f((-1)(-m))=f(-1)f(-m)=(-1)f(-m)=-f(-m)\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[2\implies1]$
+\end_inset
+
+ Si 
+\begin_inset Formula $\ker f=n\mathbb{Z}$
+\end_inset
+
+ con 
+\begin_inset Formula $n>0$
+\end_inset
+
+, el menor entero positivo con 
+\begin_inset Formula $f(n)=n1=0$
+\end_inset
+
+ es 
+\begin_inset Formula $n1_{\mathbb{Z}}=n$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $\ker f=0\mathbb{Z}=0$
+\end_inset
+
+, 
+\begin_inset Formula $f(n)=n1=0\iff n=0$
+\end_inset
+
+ y 
+\begin_inset Formula $A$
+\end_inset
+
+ tiene característica 0.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[2\implies3]$
+\end_inset
+
+ 
+\begin_inset Formula $\text{Im}f=\{f(n)=n1\}_{n\in\mathbb{Z}}$
+\end_inset
+
+ es el subanillo primo de 
+\begin_inset Formula $A$
+\end_inset
+
+, y por el primer teorema de isomorfía, 
+\begin_inset Formula $\mathbb{Z}_{n}=\frac{\mathbb{Z}}{n\mathbb{Z}}\cong\text{Im}f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[3\implies4]$
+\end_inset
+
+ Obvio.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $[4\implies2]$
+\end_inset
+
+ Sean 
+\begin_inset Formula $g:\mathbb{Z}_{n}\to B$
+\end_inset
+
+ un isomorfismo de 
+\begin_inset Formula $\mathbb{Z}_{n}$
+\end_inset
+
+ con un subanillo 
+\begin_inset Formula $B$
+\end_inset
+
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $\pi:\mathbb{Z}\to\mathbb{Z}_{n}$
+\end_inset
+
+ la proyección y 
+\begin_inset Formula $u:B\to A$
+\end_inset
+
+ la inclusión, 
+\begin_inset Formula $u\circ g\circ\pi:\mathbb{Z}\to A$
+\end_inset
+
+ es un homomorfismo de anillos que debe coincidir con 
+\begin_inset Formula $f$
+\end_inset
+
+ por unicidad, y como 
+\begin_inset Formula $u\circ g$
+\end_inset
+
+ es inyectiva, 
+\begin_inset Formula $\ker f=\ker\pi=n\mathbb{Z}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema chino de los restos:
+\series default
+ Sean 
+\begin_inset Formula $A$
+\end_inset
+
+ un anillo conmutativo e 
+\begin_inset Formula $I_{1},\dots,I_{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $n\geq1$
+\end_inset
+
+ ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+ con 
+\begin_inset Formula $I_{i}+I_{j}=A$
+\end_inset
+
+ para 
+\begin_inset Formula $i\neq j$
+\end_inset
+
+, entonces 
+\begin_inset Formula $I_{1}\cap\dots\cap I_{n}=I_{1}\cdots I_{n}$
+\end_inset
+
+ y 
+\begin_inset Formula 
+\[
+\frac{A}{I_{1}\cap\dots\cap I_{n}}\cong\frac{A}{I_{1}}\times\cdots\times\frac{A}{I_{n}}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Supongamos primero 
+\begin_inset Formula $n=2$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $I_{1}+I_{2}=A$
+\end_inset
+
+, sean 
+\begin_inset Formula $x_{1}\in I_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $x_{2}\in I_{2}$
+\end_inset
+
+ con 
+\begin_inset Formula $x_{1}+x_{2}=1$
+\end_inset
+
+, para 
+\begin_inset Formula $a\in I_{1}\cap I_{2}$
+\end_inset
+
+, 
+\begin_inset Formula $a=x_{1}a+ax_{2}\in I_{1}I_{2}$
+\end_inset
+
+, luego 
+\begin_inset Formula $I_{1}\cap I_{2}\subseteq I_{1}I_{2}$
+\end_inset
+
+, y la otra inclusión es clara.
+ Por otro lado, 
+\begin_inset Formula $f:A\to\frac{A}{I_{1}}\times\frac{A}{I_{2}}$
+\end_inset
+
+ dada por 
+\begin_inset Formula $f(a):=(a+I_{1},a+I_{2})$
+\end_inset
+
+ es un homomorfismo de anillos con núcleo 
+\begin_inset Formula $I_{1}\cap I_{2}$
+\end_inset
+
+, y es suprayectiva porque para 
+\begin_inset Formula $(a_{1}+I_{1},a_{2}+I_{2})\in\frac{A}{I_{1}}\times\frac{A}{I_{2}}$
+\end_inset
+
+, 
+\begin_inset Formula 
+\begin{multline*}
+f(a_{1}x_{2}+a_{2}x_{1})=(a_{1}x_{2}+a_{2}x_{1}+I_{1},a_{1}x_{2}+a_{2}x_{1}+I_{2})=\\
+=(a_{1}x_{2}+I_{1},a_{2}x_{1}+I_{2})\stackrel[x_{1}\equiv1\bmod I_{2}]{x_{2}\equiv1\bmod I_{1}}{=}(a_{1}+I_{1},a_{2}+I_{2}).
+\end{multline*}
+
+\end_inset
+
+ El resultado se obtiene por el primer teorema de isomorfía.
+\end_layout
+
+\begin_layout Standard
+Para 
+\begin_inset Formula $n>2$
+\end_inset
+
+, supongamos que esto se cumple para 
+\begin_inset Formula $n-1$
+\end_inset
+
+.
+ Entonces, por la hipótesis de inducción, 
+\begin_inset Formula $I_{1}\cap\cdots\cap I_{n-1}\cap I_{n}=(I_{1}\cap\cdots\cap I_{n-1})I_{n}=I_{1}\cdots I_{n-1}I_{n}$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $k\leq n-1$
+\end_inset
+
+, existen 
+\begin_inset Formula $a_{k}\in I_{k}$
+\end_inset
+
+ y 
+\begin_inset Formula $b_{k}\in I_{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{k}+b_{k}=1$
+\end_inset
+
+ y, multiplicando, 
+\begin_inset Formula 
+\[
+1=\prod_{k=1}^{n-1}(a_{k}+b_{k})=:a_{1}\cdots a_{n-1}+b,
+\]
+
+\end_inset
+
+ con 
+\begin_inset Formula $b\in I_{n}$
+\end_inset
+
+ porque en cada sumando que incluye hay al menos un factor en 
+\begin_inset Formula $I_{n}$
+\end_inset
+
+, y como 
+\begin_inset Formula $a_{1}\cdots a_{n-1}\in I_{1}\cap\cdots\cap I_{n-1}$
+\end_inset
+
+, 
+\begin_inset Formula $1\in(I_{1}\cap\cdots\cap I_{n-1})+I_{n}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $(I_{1}\cap\cdots\cap I_{n-1})+I_{n}=A$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula 
+\[
+\frac{A}{I_{1}\cap\cdots\cap I_{n}}\cong\frac{A}{I_{1}\cap\cdots\cap I_{n-1}}\times\frac{A}{I_{n}}\cong\frac{A}{I_{1}}\times\cdots\times\frac{A}{I_{n-1}}\times\frac{A}{I_{n}}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document