AC tema 1

2 files changed, 3422 insertions, 85 deletions

diff --git a/ac/n.lyx b/ac/n.lyx
index cfda71e..7590366 100644
--- a/ac/n.lyx
+++ b/ac/n.lyx
@@ -151,8 +151,6 @@ Alberto del Valle Robles.
  Cuarto curso del Grado en Matemáticas.
  Departamento de Matemáticas, Universidad de Murcia.
  Basado en apuntes previos de José Luis García Hernández.
-\backslash
-
 \end_layout
 
 \begin_layout Itemize
diff --git a/ac/n1.lyx b/ac/n1.lyx
index 1f8cc1c..0e8bd67 100644
--- a/ac/n1.lyx
+++ b/ac/n1.lyx
@@ -331,154 +331,101 @@ Dados un anillo
 \begin_inset Formula $a,b,c\in A$
 \end_inset
 
-:
-\end_layout
-
-\begin_layout Enumerate
+, 
 \begin_inset Formula $a0=0$
 \end_inset
 
-.
+, 
 \begin_inset Note Comment
 status open
 
 \begin_layout Plain Layout
-\begin_inset Formula $a0+0=a0=a(0+0)=a0+a0\implies0=a0.$
+pues 
+\begin_inset Formula $a0+0=a0=a(0+0)=a0+a0\implies0=a0$
 \end_inset
 
-
+;
 \end_layout
 
 \end_inset
 
-
-\end_layout
-
-\begin_layout Enumerate
+ 
 \begin_inset Formula $-(-a)=a$
 \end_inset
 
-.
+, 
 \begin_inset Note Comment
 status open
 
 \begin_layout Plain Layout
-\begin_inset Formula $x=-(-a)\implies0=x+(-a)\implies a=x+(-a)+a=x.$
+pues 
+\begin_inset Formula $x=-(-a)\implies0=x+(-a)\implies a=x+(-a)+a=x$
 \end_inset
 
-
+;
 \end_layout
 
 \end_inset
 
-
-\end_layout
-
-\begin_layout Enumerate
+ 
 \begin_inset Formula $a-b=c\iff b+c=a$
 \end_inset
 
-.
+, 
 \begin_inset Note Comment
 status open
 
-\begin_layout Itemize
-\begin_inset Argument item:1
-status open
-
 \begin_layout Plain Layout
-\begin_inset Formula $\implies]$
-\end_inset
-
-
-\end_layout
-
-\end_inset
-
-
+pues 
 \begin_inset Formula $a-b=c\implies a=a-b+b=c+b=b+c$
 \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
-
-
+ y 
 \begin_inset Formula $b+c=a\implies c=-b+b+c=-b+a=a-b$
 \end_inset
 
-.
+;
 \end_layout
 
 \end_inset
 
-
-\end_layout
-
-\begin_layout Enumerate
+ 
 \begin_inset Formula $(a-b)c=ac-bc$
 \end_inset
 
-.
+
 \begin_inset Note Comment
 status open
 
 \begin_layout Plain Layout
-\begin_inset Formula $(a-b)c+bc=ac-bc+bc=ac\implies ac-bc=(a-b)c.$
+, pues 
+\begin_inset Formula $(a-b)c+bc=ac-bc+bc=ac\implies ac-bc=(a-b)c$
 \end_inset
 
-
+,
 \end_layout
 
 \end_inset
 
-
-\end_layout
-
-\begin_layout Enumerate
+ y 
 \begin_inset Formula $(-a)b=-(ab)$
 \end_inset
 
-.
+
 \begin_inset Note Comment
 status open
 
 \begin_layout Plain Layout
-\begin_inset Formula $(-a)b=(0-a)b=0b-ab=0-ab=-ab.$
-\end_inset
-
-
-\end_layout
-
+, pues 
+\begin_inset Formula $(-a)b=(0-a)b=0b-ab=0-ab=-ab$
 \end_inset
 
 
-\begin_inset Note Note
-status open
-
-\begin_layout Plain Layout
-Anillo trivial si y solo si 
-\begin_inset Formula $1=0$
-\end_inset
-
-, salvo isomorfismo.
 \end_layout
 
 \end_inset
 
-
+.
 \end_layout
 
 \begin_layout Standard
@@ -670,7 +617,11 @@ anillo cero
 \series bold
 trivial
 \series default
- al único con un solo elemento, o en el que 
+, 
+\begin_inset Formula $0$
+\end_inset
+
+, al único con un solo elemento, o en el que 
 \begin_inset Formula $1=0$
 \end_inset
 
@@ -928,10 +879,18 @@ nilpotente
 
 .
  Llamamos 
+\series bold
+nilradical
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
 \begin_inset Formula $\text{Nil}(A)$
 \end_inset
 
- al conjunto de elementos de 
+, al conjunto de elementos de 
 \begin_inset Formula $A$
 \end_inset
 
@@ -1028,6 +987,13 @@ cuerpo
 \end_layout
 
 \begin_layout Standard
+\begin_inset Newpage pagebreak
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
 Para 
 \begin_inset Formula $n\geq2$
 \end_inset
@@ -1944,6 +1910,79 @@ ideal
 \end_inset
 
 .
+ En concreto, definiendo la relación de equivalencia 
+\series bold
+módulo
+\series default
+ 
+\begin_inset Formula $I$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+ como 
+\begin_inset Formula $a\equiv b\iff a-b\in I$
+\end_inset
+
+, el conjunto cociente 
+\begin_inset Formula $A\slash I\coloneqq A\slash\equiv$
+\end_inset
+
+ es un anillo con la suma 
+\begin_inset Formula $\overline{a}+\overline{b}\coloneqq\overline{a+b}$
+\end_inset
+
+, el producto 
+\begin_inset Formula $\overline{a}\,\overline{b}\coloneqq\overline{ab}$
+\end_inset
+
+, 
+\begin_inset Formula $0=\overline{0}$
+\end_inset
+
+, 
+\begin_inset Formula $1=\overline{1}$
+\end_inset
+
+, 
+\begin_inset Formula $-\overline{a}=\overline{-a}$
+\end_inset
+
+ y, si 
+\begin_inset Formula $a\in A^{*}$
+\end_inset
+
+, 
+\begin_inset Formula $\overline{a}\in(A/I)^{*}$
+\end_inset
+
+ y 
+\begin_inset Formula $\overline{a}^{-1}=\overline{a^{-1}}$
+\end_inset
+
+, donde 
+\begin_inset Formula $\overline{a}$
+\end_inset
+
+ es la clase de equivalencia de 
+\begin_inset Formula $a$
+\end_inset
+
+, y la 
+\series bold
+proyección canónica
+\series default
+ 
+\begin_inset Formula $p:A\to A/I$
+\end_inset
+
+ es un homomorfismo con núcleo 
+\begin_inset Formula $I$
+\end_inset
+
+.
 \end_layout
 
 \begin_layout Itemize
@@ -1972,17 +2011,3317 @@ Sean
 \end_inset
 
 .
-\begin_inset Note Note
+ Entonces 
+\begin_inset Formula $f(ax)=f(a)f(x)=f(a)0=0$
+\end_inset
+
+ y 
+\begin_inset Formula $f(x+y)=f(x)+f(y)=0+0=0$
+\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 $a\equiv a',b\equiv b'\in A$
+\end_inset
+
+, sean 
+\begin_inset Formula $x\coloneqq a-a',y\coloneqq b-b'\in I$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a+b=a'+x+b'+y=a'+b'+(x+y)$
+\end_inset
+
+ con 
+\begin_inset Formula $x+y\in I$
+\end_inset
+
+, luego 
+\begin_inset Formula $a+b\equiv a'+b'$
+\end_inset
+
+ y la suma está bien definida.
+ Además 
+\begin_inset Formula $ab=(a'+x)(b'+y)=a'b'+a'y+b'x+xy$
+\end_inset
+
+ con 
+\begin_inset Formula $a'y+b'x+xy\in I$
+\end_inset
+
+, luego 
+\begin_inset Formula $ab\equiv a'+b'$
+\end_inset
+
+ y el producto está bien definido.
+ Entonces es fácil ver que 
+\begin_inset Formula $A/I$
+\end_inset
+
+ es un anillo con los neutros y simétricos indicados.
+ Además, 
+\begin_inset Formula $p(1)=[1]$
+\end_inset
+
+, 
+\begin_inset Formula $p(a+b)=[a+b]=[a]+[b]=p(a)+p(b)$
+\end_inset
+
+ y del mismo modo 
+\begin_inset Formula $p(ab)=p(a)p(b)$
+\end_inset
+
+, y 
+\begin_inset Formula $p(x)=[x]=0\iff x-0=x\in I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\begin_inset Formula ${\cal I}(A)$
+\end_inset
+
+ al conjunto de ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Todo anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ tiene al menos el 
+\series bold
+ideal trivial
+\series default
+ 
+\begin_inset Formula $0\coloneqq\{0\}$
+\end_inset
+
+ y el 
+\series bold
+ideal impropio
+\series default
+ 
+\begin_inset Formula $A$
+\end_inset
+
+, que es el único que contiene una unidad.
+ En efecto, si 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ y existe 
+\begin_inset Formula $u\in I\cap A^{*}$
+\end_inset
+
+, para 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, 
+\begin_inset Formula $a=(au^{-1})u\in I$
+\end_inset
+
+.
+ 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ es 
+\series bold
+propio
+\series default
+, 
+\begin_inset Formula $I\triangleleft A$
+\end_inset
+
+, si no es impropio.
+\end_layout
+
+\begin_layout Standard
+La intersección de toda familia de ideales de 
+\begin_inset Formula $A$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Dados un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ y un subconjunto 
+\begin_inset Formula $S\subseteq A$
+\end_inset
+
+, llamamos 
+\series bold
+ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+ generado por 
+\begin_inset Formula $G$
+\end_inset
+
+
+\series default
+ a
+\begin_inset Formula 
+\[
+(S)\coloneqq\bigcap\{I\trianglelefteq A:G\subseteq I\}=\{a_{1}s_{1}+\dots+a_{n}s_{n}\}_{n\in\mathbb{N},a\in A^{n},s\in S^{n}},
+\]
+
+\end_inset
+
+y decimos que 
+\begin_inset Formula $S$
+\end_inset
+
+ es un 
+\series bold
+conjunto generador
+\series default
+ de 
+\begin_inset Formula $(S)$
+\end_inset
+
+.
+ En efecto, 
+\begin_inset Formula $\bigcap\{I\trianglelefteq A:S\subseteq I\}$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contiene a 
+\begin_inset Formula $S$
+\end_inset
+
+ y es el menor de ellos, pero todo ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contenga a 
+\begin_inset Formula $S$
+\end_inset
+
+ debe contener a las combinaciones 
+\begin_inset Formula $A$
+\end_inset
+
+-lineales finitas de elementos de 
+\begin_inset Formula $S$
+\end_inset
+
+, y el conjunto de estas combinaciones es claramente un ideal, luego ambos
+ conjuntos son iguales.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ es 
+\series bold
+finitamente generado
+\series default
+ (FG) si existe 
+\begin_inset Formula $S\subseteq I$
+\end_inset
+
+ finito tal que 
+\begin_inset Formula $I=(S)$
+\end_inset
+
+, en cuyo caso, si 
+\begin_inset Formula $S=\{b_{1},\dots,b_{n}\}$
+\end_inset
+
+, escribimos 
+\begin_inset Formula $I\eqqcolon(b_{1},\dots,b_{n})$
+\end_inset
+
+.
+ Un 
+\series bold
+ideal principal
+\series default
+ de un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ es uno de la forma 
+\begin_inset Formula $Ab\coloneqq(b)$
+\end_inset
+
+ para algún 
+\begin_inset Formula $b\in A$
+\end_inset
+
+.
+ Por ejemplo, 
+\begin_inset Formula $0=(0)$
+\end_inset
+
+ y 
+\begin_inset Formula $A=(1)$
+\end_inset
+
+.
+ Dados 
+\begin_inset Formula $b\in A$
+\end_inset
+
+ e 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+, 
+\begin_inset Formula $(b)\subseteq I$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $b\in I$
+\end_inset
+
+, y en particular para 
+\begin_inset Formula $b'\in A$
+\end_inset
+
+, 
+\begin_inset Formula $(b)\subseteq(b')$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $b'\mid b$
+\end_inset
+
+ y en un dominio 
+\begin_inset Formula $(b)=(b')$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $b$
+\end_inset
+
+ y 
+\begin_inset Formula $b'$
+\end_inset
+
+ son asociados.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $A$
+\end_inset
+
+ es un cuerpo si y sólo si sus únicos ideales son 0 y 
+\begin_inset Formula $A$
+\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 $I\trianglelefteq A$
+\end_inset
+
+, si existe 
+\begin_inset Formula $e\in I\setminus\{0\}$
+\end_inset
+
+, 
+\begin_inset Formula $1=e^{-1}e\in I$
+\end_inset
+
+ e 
+\begin_inset Formula $I=A$
+\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
+
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ no fuera un cuerpo, sea 
+\begin_inset Formula $e\in A\setminus0$
+\end_inset
+
+ no invertible, 
+\begin_inset Formula $1\notin(e)$
+\end_inset
+
+, pues no existe 
+\begin_inset Formula $f\in A$
+\end_inset
+
+ tal que 
+\begin_inset Formula $ef=1$
+\end_inset
+
+, luego 
+\begin_inset Formula $0\subsetneq(e)\subsetneq A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+dominio de ideales principales
+\series default
+ (DIP) es uno en que todos los ideales son principales, como 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{K}[x]$
+\end_inset
+
+ para todo cuerpo 
+\begin_inset Formula $\mathbb{K}$
+\end_inset
+
+.
+ Todo DIP es un DFU.
+\end_layout
+
+\begin_layout Standard
+No todos los ideales son finitamente generados.
+ En efecto, dado un anillo no trivial 
+\begin_inset Formula $A$
+\end_inset
+
+, en 
+\begin_inset Formula $A^{\mathbb{N}}$
+\end_inset
+
+ con las operaciones componente a componente, 
+\begin_inset Formula $A^{(\mathbb{N})}$
+\end_inset
+
+ formado por los elementos de 
+\begin_inset Formula $A^{\mathbb{N}}$
+\end_inset
+
+ con una cantidad finita de entradas no nulas es un ideal de 
+\begin_inset Formula $A^{\mathbb{N}}$
+\end_inset
+
+, pero no es finitamente generado porque si tomamos una cantidad finita
+ de elementos del ideal, hay un índice a partir del cual todos tienen solo
+ ceros y no generan elementos de 
+\begin_inset Formula $A^{(\mathbb{N})}$
+\end_inset
+
+ con un 1 después de esta posición.
+\end_layout
+
+\begin_layout Standard
+Dados subconjuntos 
+\begin_inset Formula $S_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $S_{2}$
+\end_inset
+
+ de un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+, llamamos 
+\begin_inset Formula $S_{1}+S_{2}\coloneqq\{x+y\}_{x\in S_{1},y\in S_{2}}$
+\end_inset
+
+ y 
+\begin_inset Formula $S_{1}\cdot S_{2}\coloneqq\{xy\}_{x\in S_{1},y\in S_{2}}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $S_{1}\eqqcolon\{a\}$
+\end_inset
+
+, llamamos 
+\begin_inset Formula $a+S_{2}\coloneqq\{a\}+S_{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $a\cdot S_{2}\coloneqq\{a\}\cdot S_{2}$
+\end_inset
+
+.
+ Por ejemplo, para 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, 
+\begin_inset Formula $a+I$
+\end_inset
+
+ es la clase de equivalencia de 
+\begin_inset Formula $A$
+\end_inset
+
+ en 
+\begin_inset Formula $A/I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+ideal suma
+\series default
+ de 
+\begin_inset Formula $I,J\trianglelefteq A$
+\end_inset
+
+ es el ideal 
+\begin_inset Formula $I+J$
+\end_inset
+
+
+\begin_inset Note Comment
+status open
+
+\begin_layout Plain Layout
+, pues si 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, 
+\begin_inset Formula $x,x'\in I$
+\end_inset
+
+ e 
+\begin_inset Formula $y,y'\in J$
+\end_inset
+
+, 
+\begin_inset Formula $(x+y)+(x'+y')=(x+x')+(y+y')\in I+J$
+\end_inset
+
+ y 
+\begin_inset Formula $a(x+y)=ax+ay\in I+J$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $S_{1},S_{2}\subseteq A$
+\end_inset
+
+, 
+\begin_inset Formula $(S_{1})+(S_{2})=(S_{1}\cup S_{2})$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $I+J=(I\cup J)$
+\end_inset
+
+.
+\begin_inset Note Comment
+status open
+
+\begin_layout Itemize
+\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 $a$
+\end_inset
+
+ está en todo 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ con 
+\begin_inset Formula $S_{1}\subseteq I$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ en todo 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ con 
+\begin_inset Formula $S_{2}\subseteq I$
+\end_inset
+
+, en particular 
+\begin_inset Formula $a$
+\end_inset
+
+ está en todo 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ con 
+\begin_inset Formula $S_{1}\cup S_{2}\subseteq I$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ también.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
 status open
 
 \begin_layout Plain Layout
-TODO pg.
- 14, seguir por 11–13, luego por 15.
+\begin_inset Formula $\supseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $a=a_{1}s_{1}+\dots+a_{n}s_{n}\in(S_{1}\cup S_{2})$
+\end_inset
+
+ con los 
+\begin_inset Formula $a_{i}\in A$
+\end_inset
+
+ y los 
+\begin_inset Formula $s_{i}\in S_{1}\cup S_{2}$
+\end_inset
+
+, podemos suponer que 
+\begin_inset Formula $s_{1},\dots,s_{k}\in G_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $s_{k+1},\dots,s_{n}\in G_{2}$
+\end_inset
+
+ para cierto 
+\begin_inset Formula $k$
+\end_inset
+
+, luego 
+\begin_inset Formula $a$
+\end_inset
+
+ es suma de un elemento de 
+\begin_inset Formula $(S_{1})$
+\end_inset
+
+ y uno de 
+\begin_inset Formula $(S_{2})$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+ideal producto
+\series default
+ de 
+\begin_inset Formula $I,J\trianglelefteq A$
+\end_inset
+
+ es 
+\begin_inset Formula 
+\[
+IJ\coloneqq(I\cdot J)=\{x_{1}y_{1}+\dots+x_{n}y_{n}\}_{n\in\mathbb{N},x\in I^{n},y\in J^{n}}\subseteq I\cap J.
+\]
+
+\end_inset
+
+
+\begin_inset Note Comment
+status open
+
+\begin_layout Description
+\begin_inset Formula $1\subseteq2]$
+\end_inset
+
+ Los elementos de 
+\begin_inset Formula $I\cdot J$
+\end_inset
+
+ son de la forma 
+\begin_inset Formula $a_{1}x_{1}y_{1}+\dots+a_{n}x_{n}y_{n}$
+\end_inset
+
+ con los 
+\begin_inset Formula $a_{i}\in A$
+\end_inset
+
+, los 
+\begin_inset Formula $x_{i}\in I$
+\end_inset
+
+ y los 
+\begin_inset Formula $y_{i}\in J$
+\end_inset
+
+, pero cada 
+\begin_inset Formula $a_{i}x_{i}\in I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\subseteq1]$
+\end_inset
+
+ Es una suma finita de elementos de 
+\begin_inset Formula $I\cdot J$
+\end_inset
+
+, luego está en 
+\begin_inset Formula $(I\cdot J)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\subseteq3]$
+\end_inset
+
+ Cada 
+\begin_inset Formula $x_{i}y_{i}\in I,J$
+\end_inset
+
+, luego cada 
+\begin_inset Formula $x_{i},y_{i}\in I\cap J$
+\end_inset
+
+ y la suma está en 
+\begin_inset Formula $I\cap J$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $S_{1},S_{2}\subseteq A$
+\end_inset
+
+, 
+\begin_inset Formula $(S_{1})(S_{2})=(S_{1}\cdot S_{2})$
+\end_inset
+
+, y en particular el producto de ideal principales es un ideal principal.
+\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
+
+Usando combinaciones lineales, los elementos de 
+\begin_inset Formula $(S_{1})(S_{2})$
+\end_inset
+
+ son sumas de elementos de la forma 
+\begin_inset Formula $a_{i}x_{i}b_{j}y_{j}$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{i},b_{j}\in A$
+\end_inset
+
+, 
+\begin_inset Formula $x_{i}\in S_{1}$
+\end_inset
+
+ e 
+\begin_inset Formula $y_{j}\in S_{2}$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $x_{i}y_{j}\in S_{1}\cdot S_{2}$
+\end_inset
+
+ y, como 
+\begin_inset Formula $(S_{1}\cdot S_{2})$
+\end_inset
+
+ es un ideal, 
+\begin_inset Formula $(a_{i}b_{j})(x_{i}y_{j})$
+\end_inset
+
+ está en 
+\begin_inset Formula $(S_{1}\cdot S_{2})$
+\end_inset
+
+ y la suma de estos elementos también.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\supseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Los elementos de 
+\begin_inset Formula $(S_{1}\cdot S_{2})$
+\end_inset
+
+ tienen forma 
+\begin_inset Formula $s=a_{1}x_{1}y_{1}+\dots+a_{n}x_{n}y_{n}$
+\end_inset
+
+ con los 
+\begin_inset Formula $a_{i}\in A$
+\end_inset
+
+, los 
+\begin_inset Formula $x_{i}\in S_{1}$
+\end_inset
+
+ y los 
+\begin_inset Formula $y_{i}\in S_{2}$
+\end_inset
+
+, pero 
+\begin_inset Formula $a_{i}x_{i}\in(S_{1})$
+\end_inset
+
+ e 
+\begin_inset Formula $y_{i}\in(S_{2})$
+\end_inset
+
+, luego cada 
+\begin_inset Formula $a_{i}x_{i}y_{i}\in(S_{1})(S_{2})=((S_{1})\cdot(S_{2}))$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $s\in(S_{1})(S_{2})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $I,J\trianglelefteq A$
+\end_inset
+
+, en general 
+\begin_inset Formula $I\cdot J$
+\end_inset
+
+ no es un ideal.
+ En efecto, sean 
+\begin_inset Formula $A\coloneqq\mathbb{Z}[x,y]=\mathbb{Z}[x][y]$
+\end_inset
+
+ e 
+\begin_inset Formula $I\coloneqq(x,y)\trianglelefteq A$
+\end_inset
+
+, entonces 
+\begin_inset Formula $x^{2},y^{2},xy\in I\cdot I$
+\end_inset
+
+, y si 
+\begin_inset Formula $I\cdot I$
+\end_inset
+
+ fuera un ideal sería 
+\begin_inset Formula $p\coloneqq x^{2}+xy+y^{2}\in I\cdot I$
+\end_inset
+
+ y por tanto habría 
+\begin_inset Formula $q=a_{0}x+b_{0}y+\dots,r=a_{1}x+b_{1}y+\dots\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $p=qr$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $a_{0}a_{1},b_{0}b_{1},a_{0}b_{1}+b_{0}a_{1}=1$
+\end_inset
+
+, pero como los coeficientes son enteros las dos primeras ecuaciones implican
+ 
+\begin_inset Formula $a_{0}=a_{1},b_{0}=b_{1}\in\{\pm1\}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $a_{0}b_{1}+b_{0}a_{1}\in\{\pm2\}\#$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Para 
+\begin_inset Formula $I,J\trianglelefteq A$
+\end_inset
+
+, en general 
+\begin_inset Formula $IJ\neq I\cap J$
+\end_inset
+
+, pues por ejemplo en 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+ es 
+\begin_inset Formula $(2)\cap(4)=(4)$
+\end_inset
+
+ pero 
+\begin_inset Formula $(2)(4)=(8)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Isomorfía
+\end_layout
+
+\begin_layout Standard
+Dado un homomorfismo de anillos 
+\begin_inset Formula $f:A\to B$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $J\trianglelefteq B$
+\end_inset
+
+, la 
+\series bold
+contracción
+\series default
+ de 
+\begin_inset Formula $J$
+\end_inset
+
+ es 
+\begin_inset Formula $f^{-1}(J)\trianglelefteq A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sea 
+\begin_inset Formula $\pi:B\to B/J$
+\end_inset
+
+ la proyección canónica, 
+\begin_inset Formula $J=\pi^{-1}(0)$
+\end_inset
+
+, luego 
+\begin_inset Formula $f^{-1}(J)=(\pi\circ f)^{-1}(0)$
+\end_inset
+
+ es un ideal.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+, 
+\begin_inset Formula $f(I)\trianglelefteq\text{Im}f$
+\end_inset
+
+, y llamamos 
+\series bold
+extensión
+\series default
+ de 
+\begin_inset Formula $I$
+\end_inset
+
+ relativa a 
+\begin_inset Formula $f$
+\end_inset
+
+ a 
+\begin_inset Formula $(f(I))$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $f(a)\in\text{Im}f$
+\end_inset
+
+ para cierto 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $f(x),f(y)\in f(I)$
+\end_inset
+
+ con 
+\begin_inset Formula $x,y\in I$
+\end_inset
+
+, 
+\begin_inset Formula $f(x)+f(y)=f(x+y)\in f(I)$
+\end_inset
+
+ y 
+\begin_inset Formula $f(a)f(x)=f(ax)\in f(I)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+En general 
+\begin_inset Formula $f(I)$
+\end_inset
+
+ no es ideal de 
+\begin_inset Formula $B$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+La inclusión 
+\begin_inset Formula $\iota:\mathbb{Z}\to\mathbb{Q}$
+\end_inset
+
+ es un homomorfismo de anillos y 
+\begin_inset Formula $\mathbb{Z}\trianglelefteq A$
+\end_inset
+
+, pero 
+\begin_inset Formula $\iota(\mathbb{Z})=\mathbb{Z}$
+\end_inset
+
+ no es ideal de 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+
+\series bold
+Teorema de la correspondencia:
+\end_layout
+
+\begin_layout Enumerate
+Dado un homomorfismo de anillos 
+\begin_inset Formula $f:A\to B$
+\end_inset
+
+, la extensión es una biyección 
+\begin_inset Formula 
+\[
+\{I\trianglelefteq A:\ker f\subseteq I\}\to\{J\trianglelefteq\text{Im}f\},
+\]
+
+\end_inset
+
+ y su inversa es la contracción.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Primero vemos que las imágenes están donde deben.
+ Si 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+, sabemos que 
+\begin_inset Formula $f(I)\trianglelefteq\text{Im}f$
+\end_inset
+
+, y si 
+\begin_inset Formula $J\trianglelefteq\text{Im}f$
+\end_inset
+
+, sabemos que 
+\begin_inset Formula $f^{-1}(J)\trianglelefteq A$
+\end_inset
+
+ y, como 
+\begin_inset Formula $0\in J$
+\end_inset
+
+, 
+\begin_inset Formula $f^{-1}(0)=\ker f\subseteq f^{-1}(J)$
+\end_inset
+
+.
+ Ahora vemos que la extensión y la contracción son inversas una de la otra.
+ Por teoría de conjuntos, para todo 
+\begin_inset Formula $J\subseteq\text{Im}f$
+\end_inset
+
+, 
+\begin_inset Formula $f(f^{-1}(J))=J$
+\end_inset
+
+, y para todo 
+\begin_inset Formula $I\subseteq A$
+\end_inset
+
+, 
+\begin_inset Formula $I\subseteq f^{-1}(f(I))$
+\end_inset
+
+, por lo que solo hay que ver que 
+\begin_inset Formula $f^{-1}(f(I))\subseteq I$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $x\in f^{-1}(f(I))$
+\end_inset
+
+, 
+\begin_inset Formula $f(x)\in f(I)$
+\end_inset
+
+, luego existe 
+\begin_inset Formula $y\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $f(x)=f(y)$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $f(x-y)=0$
+\end_inset
+
+, 
+\begin_inset Formula $x-y\in\ker f\subseteq I$
+\end_inset
+
+ y 
+\begin_inset Formula $x=y+(x-y)\in I$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $I$
+\end_inset
+
+ es un ideal de 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $p:A\to A/I$
+\end_inset
+
+ es la proyección canónica, 
+\begin_inset Formula $\rho:\{J\trianglelefteq A:I\subseteq J\}\to\{K\trianglelefteq A/I\}$
+\end_inset
+
+ dada por 
+\begin_inset Formula $\rho(J)\coloneqq J/I\coloneqq p(J)=\{x+I\}_{x\in J}$
+\end_inset
+
+ es una biyección que conserva la inclusión de los elementos.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Basta aplicar el punto anterior a 
+\begin_inset Formula $p$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $I,J\trianglelefteq A$
+\end_inset
+
+ y 
+\begin_inset Formula $p:A\to A/I$
+\end_inset
+
+ la proyección canónica, 
+\begin_inset Formula $p(J)=(J+I)/I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Hay tantos ideales de 
+\begin_inset Formula $\mathbb{Z}_{n}$
+\end_inset
+
+ como divisores positivos de 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ En efecto, como 
+\begin_inset Formula $\mathbb{Z}_{n}=\mathbb{Z}/(n)$
+\end_inset
+
+, por el teorema anterior hay una biyección entre 
+\begin_inset Formula ${\cal I}(\mathbb{Z}_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $\{I\trianglelefteq\mathbb{Z}:(n)\subseteq I\}$
+\end_inset
+
+, pero 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+ es un DIP, luego estos elementos se corresponden precisamente con los 
+\begin_inset Formula $m\in\mathbb{Z}$
+\end_inset
+
+ con 
+\begin_inset Formula $(n)\subseteq(m)$
+\end_inset
+
+, que son aquellos con 
+\begin_inset Formula $m\mid n$
+\end_inset
+
+, y tomamos solo los 
+\begin_inset Formula $m$
+\end_inset
+
+ positivos ya que los negativos son sus asociados.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema del factor:
+\series default
+ Sean 
+\begin_inset Formula $f:A\to B$
+\end_inset
+
+ un homomorfismo de anillos, 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ y 
+\begin_inset Formula $p:A\to A/I$
+\end_inset
+
+ la proyección canónica, existe un homomorfismo 
+\begin_inset Formula $\overline{f}:A/I\to B$
+\end_inset
+
+ con 
+\begin_inset Formula $\overline{f}\circ p=f$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $I\subseteq\ker f$
+\end_inset
+
+, en cuyo caso 
+\begin_inset Formula $\overline{f}$
+\end_inset
+
+ es único y 
+\begin_inset Formula $\ker\overline{f}=\ker f/I$
+\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
+
+Para 
+\begin_inset Formula $x\in I$
+\end_inset
+
+, 
+\begin_inset Formula $f(x)=\overline{f}(p(x))=h(0+I)=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $I\subseteq\ker f$
+\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
+
+Sea 
+\begin_inset Formula $\overline{f}:A/I\to B$
+\end_inset
+
+ con 
+\begin_inset Formula $\overline{f}\circ p=f$
+\end_inset
+
+, para 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, 
+\begin_inset Formula $\overline{f}(a+I)=f(a)$
+\end_inset
+
+, lo que prueba la unicidad.
+ Para la existencia, si 
+\begin_inset Formula $a+I=b+I$
+\end_inset
+
+, 
+\begin_inset Formula $f(b)=f(a+b-a)=f(a)+f(b-a)=f(a)$
+\end_inset
+
+ porque 
+\begin_inset Formula $b-a\in I\subseteq\ker f$
+\end_inset
+
+.
+ Finalmente 
+\begin_inset Formula $x+I\in\ker\overline{f}\iff\overline{f}(x+I)=f(x)=0\iff x\in\ker f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teoremas de isomorfía:
+\end_layout
+
+\begin_layout Enumerate
+Para un isomorfismo de anillos 
+\begin_inset Formula $f:A\to B$
+\end_inset
+
+, 
+\begin_inset Formula $A/\ker f\cong\text{Im}f$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $B'\coloneqq\text{Im}f$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $B$
+\end_inset
+
+, luego 
+\begin_inset Formula $f:A\to B'$
+\end_inset
+
+ es un homomorfismo suprayectivo y, por el teorema del factor, existe 
+\begin_inset Formula $\overline{f}:A/\ker f\to B'$
+\end_inset
+
+ con 
+\begin_inset Formula $\overline{f}\circ p=f$
+\end_inset
+
+ y 
+\begin_inset Formula $\ker\overline{f}=\ker f/\ker f=0$
+\end_inset
+
+, que es pues inyectivo y es suprayectivo por serlo 
+\begin_inset Formula $f$
+\end_inset
+
+, de modo que es un isomorfismo.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ y 
+\begin_inset Formula $S$
+\end_inset
+
+ un subanillo de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $I+S$
+\end_inset
+
+ es un subanillo de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $I\cap S\trianglelefteq S$
+\end_inset
+
+, 
+\begin_inset Formula $I\trianglelefteq I+S$
+\end_inset
+
+ y 
+\begin_inset Formula $S/(I\cap S)\cong(I+S)/I$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Es fácil comprobar que 
+\begin_inset Formula $I+S$
+\end_inset
+
+ es subanillo de 
+\begin_inset Formula $A$
+\end_inset
+
+ y claramente 
+\begin_inset Formula $I\cap S\trianglelefteq S$
+\end_inset
+
+ e 
+\begin_inset Formula $I\trianglelefteq I+S$
+\end_inset
+
+.
+ Sea ahora el homomorfismo 
+\begin_inset Formula $f:S\to(I+S)/I$
+\end_inset
+
+ dado por 
+\begin_inset Formula $f(a)=a+I$
+\end_inset
+
+, 
+\begin_inset Formula $f(a)=0\iff a\in I$
+\end_inset
+
+, luego 
+\begin_inset Formula $\ker f=I\cap S$
+\end_inset
+
+, e 
+\begin_inset Formula $\text{Im}f=(I+S)/I$
+\end_inset
+
+, pues un 
+\begin_inset Formula $(x+s)+I\in(I+S)/I$
+\end_inset
+
+ arbitrario, con 
+\begin_inset Formula $x\in I$
+\end_inset
+
+ y 
+\begin_inset Formula $s\in S$
+\end_inset
+
+, es la imagen por 
+\begin_inset Formula $f$
+\end_inset
+
+ de 
+\begin_inset Formula $s$
+\end_inset
+
+.
+ Entonces, por el primer teorema de isomorfía, 
+\begin_inset Formula $S/(I\cap S)\cong(I+S)/I$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $I,J\trianglelefteq A$
+\end_inset
+
+ con 
+\begin_inset Formula $I\subseteq J$
+\end_inset
+
+, 
+\begin_inset Formula $(A/I)/(J/I)\cong A/J$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sea 
+\begin_inset Formula $p:A\to A/J$
+\end_inset
+
+ la proyección canónica, como 
+\begin_inset Formula $I\subseteq J=\ker f$
+\end_inset
+
+, el teorema del factor nos da un homomorfismo 
+\begin_inset Formula $\overline{p}:A/I\to A/J$
+\end_inset
+
+ con 
+\begin_inset Formula $\ker\overline{p}=\ker p/I=J/I$
+\end_inset
+
+, que es suprayectivo por serlo 
+\begin_inset Formula $p$
+\end_inset
+
+, y el resultado se obtiene del primer teorema de isomorfía.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Formula $I,J\trianglelefteq A$
+\end_inset
+
+ son 
+\series bold
+comaximales
+\series default
+ si 
+\begin_inset Formula $I+J=A$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $\exists a\in I,b\in J:a+b=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema chino de los restos:
+\series default
+ Dados anillos 
+\begin_inset Formula $A,B_{1},\dots,B_{n}$
+\end_inset
+
+ y homomorfismos 
+\begin_inset Formula $g_{i}:A\to B_{i}$
+\end_inset
+
+ con 
+\begin_inset Formula $K_{i}\coloneqq\ker g_{i}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\phi:A\to B_{1}\times\dots\times B_{n}$
+\end_inset
+
+ dado por 
+\begin_inset Formula $\phi(x)=(g_{1}(x),\dots,g_{n}(x))$
+\end_inset
+
+ es un homomorfismo con núcleo 
+\begin_inset Formula $K_{1}\cap\dots\cap K_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si los 
+\begin_inset Formula $g_{i}$
+\end_inset
+
+ son suprayectivos y los 
+\begin_inset Formula $K_{i}$
+\end_inset
+
+ son comaximales dos a dos, 
+\begin_inset Formula $\phi$
+\end_inset
+
+ es suprayectivo, 
+\begin_inset Formula $\ker\phi=K_{1}\cdots K_{n}$
+\end_inset
+
+ y 
+\begin_inset Formula $A/(K_{1}\cdots K_{n})\cong B_{1}\times\dots\times B_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para 
+\begin_inset Formula $n\in\{0,1\}$
+\end_inset
+
+ es claro, por lo que suponemos 
+\begin_inset Formula $n\geq2$
+\end_inset
+
+.
+ Primero vemos que 
+\begin_inset Formula $K_{1}$
+\end_inset
+
+ es comaximal con 
+\begin_inset Formula $K_{2}\cdots K_{n}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $n=2$
+\end_inset
+
+ esto es claro.
+ Si 
+\begin_inset Formula $n=3$
+\end_inset
+
+, existen 
+\begin_inset Formula $a,a'\in K_{1}$
+\end_inset
+
+, 
+\begin_inset Formula $b\in K_{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $b'\in K_{3}$
+\end_inset
+
+ con 
+\begin_inset Formula $1=a+b=a'+b'$
+\end_inset
+
+, luego 
+\begin_inset Formula $1=aa'+ab'+ba'+bb'$
+\end_inset
+
+ con 
+\begin_inset Formula $bb'\in K_{2}K_{3}$
+\end_inset
+
+ y el resto de sumandos en 
+\begin_inset Formula $K_{1}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $n>3$
+\end_inset
+
+ basta hacer inducción.
+ Al ser 
+\begin_inset Formula $K_{2}\cap\dots\cap K_{n}$
+\end_inset
+
+ más grande que 
+\begin_inset Formula $K_{2}\cdots K_{n}$
+\end_inset
+
+, también es comaximal con 
+\begin_inset Formula $K_{1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sean ahora 
+\begin_inset Formula $a\in K_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $b\in K_{2}\cap\dots\cap K_{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $a+b=1$
+\end_inset
+
+, como 
+\begin_inset Formula $g_{1}(a)=0$
+\end_inset
+
+, 
+\begin_inset Formula $g_{1}(b)=g_{1}(a)+g_{1}(b)=g_{1}(a+b)=1$
+\end_inset
+
+, y para 
+\begin_inset Formula $j\in\{2,\dots,n\}$
+\end_inset
+
+, 
+\begin_inset Formula $g_{j}(b)=0$
+\end_inset
+
+.
+ Sea ahora 
+\begin_inset Formula $x\in B_{1}$
+\end_inset
+
+ arbitrario, por suprayectividad existe 
+\begin_inset Formula $u\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $g_{1}(u)=x$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $g_{1}(ub)=g_{1}(u)g_{1}(b)=x$
+\end_inset
+
+ y, para 
+\begin_inset Formula $j\in\{2,\dots,n\}$
+\end_inset
+
+, 
+\begin_inset Formula $g_{j}(ub)=g_{j}(u)g_{j}(b)=0$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $\phi(u)=(x,0,\dots,0)$
+\end_inset
+
+.
+ Por simetría, para cada 
+\begin_inset Formula $i$
+\end_inset
+
+ y 
+\begin_inset Formula $x\in B_{i}$
+\end_inset
+
+ existe 
+\begin_inset Formula $u$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\phi(u)=(0,\dots,0,x,0,\dots,0)$
+\end_inset
+
+ con 
+\begin_inset Formula $x$
+\end_inset
+
+ en la 
+\begin_inset Formula $i$
+\end_inset
+
+-ésima posición, y como todo elemento de 
+\begin_inset Formula $B_{1}\times\dots\times B_{n}$
+\end_inset
+
+ es suma de elementos de esta forma, 
+\begin_inset Formula $\phi$
+\end_inset
+
+ es suprayectiva.
+\end_layout
+
+\begin_layout Standard
+Veamos ahora que 
+\begin_inset Formula $\ker\phi=K_{1}\cap\dots\cap K_{n}=K_{1}\cdots K_{n}$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $n=2$
+\end_inset
+
+, sea 
+\begin_inset Formula $x\in K_{1}\cap K_{2}$
+\end_inset
+
+, existen 
+\begin_inset Formula $a\in K_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $b\in K_{2}$
+\end_inset
+
+ con 
+\begin_inset Formula $a+b=1$
+\end_inset
+
+, luego 
+\begin_inset Formula $x=1x=ax+bx$
+\end_inset
+
+, pero como 
+\begin_inset Formula $a\in K_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $x\in K_{2}$
+\end_inset
+
+, 
+\begin_inset Formula $ax\in K_{1}K_{2}$
+\end_inset
+
+, y análogamente 
+\begin_inset Formula $xb\in K_{1}K_{2}$
+\end_inset
+
+, luego 
+\begin_inset Formula $x\in K_{1}K_{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $K_{1}\cap K_{2}\subseteq K_{1}K_{2}$
+\end_inset
+
+, y la otra inclusión la sabemos.
+ Para 
+\begin_inset Formula $n>2$
+\end_inset
+
+ basta hacer inducción.
+ La última afirmación se debe al primer teorema de isomorfía.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Ideales notables
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $I\triangleleft A$
+\end_inset
+
+ es 
+\series bold
+maximal
+\series default
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $I\trianglelefteq_{\text{m}}A$
+\end_inset
+
+, si es maximal en el conjunto de ideales propios de 
+\begin_inset Formula $A$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $A/I$
+\end_inset
+
+ es un cuerpo.
+ 
+\series bold
+Demostración:
+\series default
+ Como 
+\begin_inset Formula $J\mapsto J/I$
+\end_inset
+
+ conserva la inclusión, 
+\begin_inset Formula $I$
+\end_inset
+
+ es maximal si y sólo si lo es 
+\begin_inset Formula $I/I=0$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $A/I$
+\end_inset
+
+ no es trivial y no tiene ideales propios no nulos (si tuviera alguno, contendrí
+a a 0 y 0 no sería maximal), si y sólo si es un cuerpo no trivial, pero
+ sabemos que 
+\begin_inset Formula $A/I$
+\end_inset
+
+ es no trivial porque 
+\begin_inset Formula $I$
+\end_inset
+
+ es propio.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+espectro maximal
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $\text{MaxSpec}(A)$
+\end_inset
+
+, al conjunto de ideales maximales de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+, la biyección 
+\begin_inset Formula $\{J\in{\cal I}(A):I\subseteq J\}\to{\cal I}(A/I)$
+\end_inset
+
+ del teorema de la correspondencia se restringe a una biyección 
+\begin_inset Formula $\{J\in\text{MaxSpec}(A):I\subseteq J\}\to\text{MaxSpec}(A/I)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una 
+\series bold
+cadena
+\series default
+ en un conjunto ordenado es un subconjunto que, con el mismo orden, es totalment
+e ordenado.
+ Un conjunto ordenado es 
+\series bold
+inductivo
+\series default
+ si toda cadena no vacía tiene cota superior.
+ 
+\series bold
+Lema de Zorn:
+\series default
+ Todo conjunto inductivo no vacío tiene un elemento maximal.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $I\triangleleft A$
+\end_inset
+
+, existe un ideal maximal de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contiene a 
+\begin_inset Formula $I$
+\end_inset
+
+, y en particular todo anillo no trivial tiene al menos un elemento maximal.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $\Omega\coloneqq\{J\triangleleft A:I\subseteq J\}$
+\end_inset
+
+, 
+\begin_inset Formula $\Omega\neq\emptyset$
+\end_inset
+
+ porque 
+\begin_inset Formula $I\in\Omega$
+\end_inset
+
+.
+ Considerándolo ordenado por inclusión, si 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ es una cadena no vacía en 
+\begin_inset Formula $\Omega$
+\end_inset
+
+, 
+\begin_inset Formula $\bigcup{\cal C}$
+\end_inset
+
+ es una cota superior, pues si 
+\begin_inset Formula $x,y\in\bigcup{\cal C}$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, sean 
+\begin_inset Formula $I,J\in{\cal C}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $x\in I$
+\end_inset
+
+ e 
+\begin_inset Formula $y\in J$
+\end_inset
+
+, entonces por ejemplo 
+\begin_inset Formula $I\subseteq J$
+\end_inset
+
+, luego 
+\begin_inset Formula $x,y\in J\subseteq\bigcup{\cal C}$
+\end_inset
+
+ y 
+\begin_inset Formula $x+y,ax,ay\in J\subseteq\bigcup{\cal C}$
+\end_inset
+
+ para 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $\bigcup{\cal C}$
+\end_inset
+
+ es un ideal, contiene a 
+\begin_inset Formula $I$
+\end_inset
+
+ y no es propio porque ninguno de los elementos de 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ contiene a 1.
+ Entonces, por el lema de Zorn, 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ tiene un elemento maximal 
+\begin_inset Formula $J$
+\end_inset
+
+, que es un ideal propio que contiene a 
+\begin_inset Formula $I$
+\end_inset
+
+ y es maximal porque, para 
+\begin_inset Formula $J'\triangleleft A$
+\end_inset
+
+ con 
+\begin_inset Formula $J\subseteq J'$
+\end_inset
+
+, 
+\begin_inset Formula $J'\in\Omega$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $J'=J$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $I\triangleleft A$
+\end_inset
+
+ es 
+\series bold
+primo
+\series default
+, 
+\begin_inset Formula $I\trianglelefteq_{\text{p}}A$
+\end_inset
+
+, si 
+\begin_inset Formula $\forall x,y\in A,(xy\in I\implies x\in I\lor y\in I)$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $\forall n\in\mathbb{N}^{*},\forall x_{1},\dots,x_{n}\in A,(x_{1}\cdots x_{n}\in I\implies\exists k:x_{k}\in I)$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $A/I$
+\end_inset
+
+ es un dominio, si y sólo si 
+\begin_inset Formula $\forall n\in\mathbb{N}^{*},\forall I_{1},\dots,I_{n}\trianglelefteq A,(I_{1}\cdots I_{n}\subseteq J\implies\exists k:I_{k}\subseteq J)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\iff2]$
+\end_inset
+
+ Trivial.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\iff3]$
+\end_inset
+
+ Para 
+\begin_inset Formula $x,y\in A$
+\end_inset
+
+, 
+\begin_inset Formula $xy\in I\implies x\in I\lor y\in I$
+\end_inset
+
+ si y sólo si, en 
+\begin_inset Formula $A/I$
+\end_inset
+
+, 
+\begin_inset Formula $\overline{x}\,\overline{y}=\overline{xy}=0\implies\overline{x}=0\lor\overline{y}=0$
+\end_inset
+
+, y que esto se de para todo 
+\begin_inset Formula $\overline{x},\overline{y}\in A/I$
+\end_inset
+
+ equivale a que 
+\begin_inset Formula $A/I$
+\end_inset
+
+ sea un dominio.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\implies4]$
+\end_inset
+
+ Si fuera cada 
+\begin_inset Formula $I_{k}\nsubseteq J$
+\end_inset
+
+, sea 
+\begin_inset Formula $x_{k}\in I_{k}\setminus J$
+\end_inset
+
+ para cada 
+\begin_inset Formula $k$
+\end_inset
+
+, 
+\begin_inset Formula $x_{1}\cdots x_{n}\in I_{1}\cdots I_{n}\subseteq J$
+\end_inset
+
+, pero si 
+\begin_inset Formula $J$
+\end_inset
+
+ es primo existe 
+\begin_inset Formula $k$
+\end_inset
+
+ con 
+\begin_inset Formula $x_{j}\in J\#$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $4\implies1]$
+\end_inset
+
+ Sean 
+\begin_inset Formula $a_{1},a_{2}\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{1}a_{2}\in J$
+\end_inset
+
+, 
+\begin_inset Formula $(a_{1})(a_{2})=(a_{1}a_{2})\subseteq J$
+\end_inset
+
+, luego por hipótesis 
+\begin_inset Formula $(a_{1})\subseteq J$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $a_{1}\in J$
+\end_inset
+
+ o 
+\begin_inset Formula $(a_{2})\subseteq J$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $a_{2}\in J$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+espectro primo
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $\text{Spec}(A)$
+\end_inset
+
+, es el conjunto de todos los ideales primos.
+ 
+\begin_inset Formula $\text{MaxSpec}(A)\subseteq\text{Spec}(A)$
+\end_inset
+
+, pues todo cuerpo es un dominio.
+ Para 
+\begin_inset Formula $I\triangleleft A$
+\end_inset
+
+, la biyección 
+\begin_inset Formula $\{J\in{\cal I}(A):I\subseteq J\}\to{\cal I}(A/I)$
+\end_inset
+
+ se restringe a una biyección 
+\begin_inset Formula $\{J\in\text{Spec}(A):I\subseteq J\}\to\text{Spec}(A/I)$
+\end_inset
+
+.
+ En efecto, para 
+\begin_inset Formula $J\in{\cal I}(A)$
+\end_inset
+
+ con 
+\begin_inset Formula $I\subseteq J$
+\end_inset
+
+, 
+\begin_inset Formula $(A/I)/(J/I)\cong A/J$
+\end_inset
+
+ por el tercer teorema de isomorfía, con lo que 
+\begin_inset Formula $J$
+\end_inset
+
+ es primo si y sólo si 
+\begin_inset Formula $A/J$
+\end_inset
+
+ es un dominio, si y sólo si lo es 
+\begin_inset Formula $(A/I)/(J/I)$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $J/I$
+\end_inset
+
+ es primo en 
+\begin_inset Formula $A/I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+En un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+:
 \end_layout
 
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $I_{1},\dots,I_{n}\trianglelefteq A$
+\end_inset
+
+ con intersección 
+\begin_inset Formula $J$
 \end_inset
 
+ prima, algún 
+\begin_inset Formula $I_{k}=J$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $J$
+\end_inset
+
+ está contenido en cada 
+\begin_inset Formula $I_{k}$
+\end_inset
+
+ y, como 
+\begin_inset Formula $I_{1}\cdots I_{n}\subseteq I_{1}\cap\dots\cap I_{n}=J$
+\end_inset
+
+, algún 
+\begin_inset Formula $I_{k}\subseteq J$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ y 
+\begin_inset Formula $J_{1},\dots,J_{n}\trianglelefteq_{\text{p}}A$
+\end_inset
+
+, si 
+\begin_inset Formula $I\subseteq J_{1}\cup\dots\cup J_{n}$
+\end_inset
+
+, 
+\begin_inset Formula $I$
+\end_inset
+
+ está contenido en algún 
+\begin_inset Formula $J_{k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para 
+\begin_inset Formula $n=1$
+\end_inset
+
+ es trivial.
+ Si 
+\begin_inset Formula $n>1$
+\end_inset
+
+, supuesto esto probado para 
+\begin_inset Formula $n-1$
+\end_inset
+
+, si fuera 
+\begin_inset Formula $I\nsubseteq J_{k}$
+\end_inset
+
+ para todo 
+\begin_inset Formula $k$
+\end_inset
+
+, para cada 
+\begin_inset Formula $i$
+\end_inset
+
+, 
+\begin_inset Formula $I\subseteq J_{k}$
+\end_inset
+
+ para 
+\begin_inset Formula $k\neq i$
+\end_inset
+
+ y por tanto existe 
+\begin_inset Formula $a_{i}\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{i}\notin\bigcup_{k\neq i}J_{k}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $a_{i}\in J_{i}$
+\end_inset
+
+.
+ Sea entonces 
+\begin_inset Formula $b_{i}\coloneqq\prod_{k\neq i}a_{k}$
+\end_inset
+
+, 
+\begin_inset Formula $b_{i}\in I$
+\end_inset
+
+ y 
+\begin_inset Formula $b_{i}\in\bigcap_{k\neq i}J_{k}$
+\end_inset
+
+, pero 
+\begin_inset Formula $b_{i}\notin J_{i}$
+\end_inset
+
+ porque es el producto de elementos fuera de 
+\begin_inset Formula $J_{i}$
+\end_inset
+
+ y 
+\begin_inset Formula $J_{i}$
+\end_inset
+
+ es primo.
+ Entonces 
+\begin_inset Formula $b\coloneqq\sum_{k=1}^{n}b_{k}\in I=\bigcup_{k=1}^{n}J_{k}$
+\end_inset
+
+ y debe haber un 
+\begin_inset Formula $k$
+\end_inset
+
+ con 
+\begin_inset Formula $b\in J_{k}$
+\end_inset
+
+, pero de ser así, como 
+\begin_inset Formula $b_{i}\in J_{k}$
+\end_inset
+
+ para 
+\begin_inset Formula $i\neq k$
+\end_inset
+
+, sería 
+\begin_inset Formula $b-\sum_{i\neq k}b_{i}=b_{k}\in J_{k}\#$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Dado un conjunto ordenado 
+\begin_inset Formula $(S,\leq)$
+\end_inset
+
+, su 
+\series bold
+opuesto
+\series default
+ es 
+\begin_inset Formula $(S,\leq)^{\text{op}}\coloneqq(S,\geq)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $(S,\leq)$
+\end_inset
+
+ es 
+\series bold
+contra-inductivo
+\series default
+ si su opuesto es inductivo, es decir, si toda subcadena no vacía tiene
+ una cota inferior.
+ 
+\series bold
+Lema de Zorn dual:
+\series default
+ Todo conjunto contra-inductivo tiene al menos un elemento minimal.
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+primo minimal
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+ es un elemento minimal de 
+\begin_inset Formula $\text{Spec}(A)$
+\end_inset
+
+.
+ Llamamos 
+\begin_inset Formula $\text{MinSpec}(A)$
+\end_inset
+
+ al conjunto de primos minimales de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ y 
+\begin_inset Formula $Q\trianglelefteq_{\text{p}}A$
+\end_inset
+
+ contiene a 
+\begin_inset Formula $I$
+\end_inset
+
+, 
+\begin_inset Formula $Q$
+\end_inset
+
+ contiene un 
+\series bold
+primo minimal sobre 
+\begin_inset Formula $I$
+\end_inset
+
+
+\series default
+, un minimal entre los ideales primos que contienen a 
+\begin_inset Formula $I$
+\end_inset
+
+, y en particular todo 
+\begin_inset Formula $Q\trianglelefteq_{\text{p}}A$
+\end_inset
+
+ contiene un primo minimal.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $\Omega\coloneqq\{P\trianglelefteq_{\text{p}}A:I\subseteq P\subseteq Q\}$
+\end_inset
+
+, 
+\begin_inset Formula $\Omega\neq\emptyset$
+\end_inset
+
+ porque 
+\begin_inset Formula $Q\in\Omega$
+\end_inset
+
+.
+ Sea entonces 
+\begin_inset Formula ${\cal C}$
+\end_inset
+
+ una cadena no vacía en 
+\begin_inset Formula $\Omega$
+\end_inset
+
+, 
+\begin_inset Formula $\bigcap{\cal C}\in\Omega$
+\end_inset
+
+, pues es un ideal propio entre 
+\begin_inset Formula $I$
+\end_inset
+
+ y 
+\begin_inset Formula $Q$
+\end_inset
+
+ y, usando el contrarrecíproco de la definición de primo, si 
+\begin_inset Formula $x,y\notin\bigcap{\cal C}$
+\end_inset
+
+, sean 
+\begin_inset Formula $J_{1},J_{2}\in{\cal C}$
+\end_inset
+
+ con 
+\begin_inset Formula $x\notin J_{1}$
+\end_inset
+
+ e 
+\begin_inset Formula $y\notin J_{2}$
+\end_inset
+
+, si por ejemplo 
+\begin_inset Formula $J_{1}\subseteq J_{2}$
+\end_inset
+
+, 
+\begin_inset Formula $x,y\notin J_{1}$
+\end_inset
+
+, luego 
+\begin_inset Formula $xy\notin J_{1}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $xy\notin\bigcap{\cal C}$
+\end_inset
+
+.
+ Entonces, por el lema de Zorn dual, 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ tiene un minimal 
+\begin_inset Formula $J$
+\end_inset
+
+, que es un ideal primo de 
+\begin_inset Formula $A$
+\end_inset
+
+ entre 
+\begin_inset Formula $I$
+\end_inset
+
+ y 
+\begin_inset Formula $Q$
+\end_inset
+
+ y, si 
+\begin_inset Formula $J'$
+\end_inset
+
+ es un ideal primo de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contiene a 
+\begin_inset Formula $I$
+\end_inset
+
+ con 
+\begin_inset Formula $J'\subseteq J$
+\end_inset
+
+, entonces 
+\begin_inset Formula $J'\subseteq Q$
+\end_inset
+
+ y 
+\begin_inset Formula $J'\in\Omega$
+\end_inset
+
+, luego 
+\begin_inset Formula $J'=J$
+\end_inset
+
+ y 
+\begin_inset Formula $J$
+\end_inset
+
+ es minimal sobre 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ es un 
+\series bold
+radical
+\series default
+ de 
+\begin_inset Formula $A$
+\end_inset
+
+, 
+\begin_inset Formula $I\trianglelefteq_{\text{r}}A$
+\end_inset
+
+, si 
+\begin_inset Formula $\forall x\in A,\forall n\in\mathbb{N},(x^{n}\in I\implies x\in I)$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $\forall x\in A,(x^{2}\in I\implies x\in I)$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $A/I$
+\end_inset
+
+ es reducido.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Obvio.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\implies1]$
+\end_inset
+
+ Sean 
+\begin_inset Formula $x\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $x^{n}\in I$
+\end_inset
+
+, si 
+\begin_inset Formula $n=0$
+\end_inset
+
+, 
+\begin_inset Formula $1\in I$
+\end_inset
+
+ y 
+\begin_inset Formula $x\in I$
+\end_inset
+
+, y si 
+\begin_inset Formula $n=1$
+\end_inset
+
+ es obvio.
+ En otro caso, si 
+\begin_inset Formula $n$
+\end_inset
+
+ es par, 
+\begin_inset Formula $x^{n/2}\in I$
+\end_inset
+
+, y si es impar, 
+\begin_inset Formula $x^{n+1}=xx^{n}\in I$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $x^{(n+1)/2}\in I$
+\end_inset
+
+.
+ En cualquier caso 
+\begin_inset Formula $x^{k}\in I$
+\end_inset
+
+ para cierto 
+\begin_inset Formula $k$
+\end_inset
+
+ con 
+\begin_inset Formula $1\leq k<n$
+\end_inset
+
+, y repitiendo el proceso llegamos a que 
+\begin_inset Formula $x\in I$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\iff3]$
+\end_inset
+
+ 
+\begin_inset Formula $x^{n}\in I\implies x\in I$
+\end_inset
+
+ es equivalente a que, en 
+\begin_inset Formula $A/I$
+\end_inset
+
+, 
+\begin_inset Formula $\overline{x}^{n}=\overline{x^{n}}=0\implies\overline{x}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Newpage pagebreak
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Propiedades:
+\end_layout
+
+\begin_layout Enumerate
+Todo ideal primo es radical.
+\end_layout
+
+\begin_layout Enumerate
+La intersección de una familia de radicales es un radical.
+\end_layout
+
+\begin_layout Enumerate
+Para 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+, la biyección del teorema de la correspondencia se restringe a una entre
+ los radicales de 
+\begin_inset Formula $A$
+\end_inset
+
+ que contienen a 
+\begin_inset Formula $I$
+\end_inset
+
+ y los radicales de 
+\begin_inset Formula $A/I$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sea 
+\begin_inset Formula $J\trianglelefteq A$
+\end_inset
+
+ con 
+\begin_inset Formula $I\subseteq J$
+\end_inset
+
+, por el tercer teorema de isomorfía, 
+\begin_inset Formula $(A/I)/(J/I)\cong A/J$
+\end_inset
+
+, luego 
+\begin_inset Formula $J$
+\end_inset
+
+ es un radical de 
+\begin_inset Formula $A$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $A/J$
+\end_inset
+
+ es reducido, si y sólo si lo es 
+\begin_inset Formula $(A/I)/(J/I)$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $J/I$
+\end_inset
+
+ es un radical de 
+\begin_inset Formula $A/I$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Un 
+\series bold
+subconjunto multiplicativo
+\series default
+ de un anillo 
+\begin_inset Formula $A$
+\end_inset
+
+ es un 
+\begin_inset Formula $S\subseteq A$
+\end_inset
+
+ cerrado para el producto y que contiene al 1.
+
+\series bold
+ Lema de Krull:
+\series default
+ Sean 
+\begin_inset Formula $A$
+\end_inset
+
+ un anillo, 
+\begin_inset Formula $S\subseteq A$
+\end_inset
+
+ un subconjunto multiplicativo e 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ disjunto de 
+\begin_inset Formula $S$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula ${\cal I}_{I,S}\coloneqq\{J\trianglelefteq A:I\subseteq J,J\cap S=\emptyset\}$
+\end_inset
+
+ es un conjunto inductivo no vacío.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+La unión de elementos de una cadena vacía de 
+\begin_inset Formula ${\cal I}_{I,S}$
+\end_inset
+
+ está en 
+\begin_inset Formula ${\cal I}_{I,S}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Todo elemento maximal de 
+\begin_inset Formula ${\cal I}_{I,S}$
+\end_inset
+
+ es un ideal primo de 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $J$
+\end_inset
+
+ maximal de 
+\begin_inset Formula ${\cal I}_{I,S}$
+\end_inset
+
+ y supongamos que existen 
+\begin_inset Formula $x,y\in A$
+\end_inset
+
+ con 
+\begin_inset Formula $x,y\notin J$
+\end_inset
+
+ pero 
+\begin_inset Formula $xy\in J$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $J\subsetneq(x)+J$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $(x)+J\notin{\cal I}_{I,S}$
+\end_inset
+
+ y, como es un ideal que contiene a 
+\begin_inset Formula $I$
+\end_inset
+
+, debe contener un elemento de 
+\begin_inset Formula $S$
+\end_inset
+
+, 
+\begin_inset Formula $a_{1}x+b_{1}\in((x)+J)\cap S$
+\end_inset
+
+, donde 
+\begin_inset Formula $a_{1}\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $b_{1}\in J$
+\end_inset
+
+, y análogamente existe 
+\begin_inset Formula $a_{2}y+b_{2}\in((y)+J)\cap S$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{2}\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $b_{2}\in J$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $s\coloneqq(a_{1}x+b_{1})(a_{2}y+b_{2})=a_{1}a_{2}xy+a_{1}xb_{2}+b_{1}a_{2}y+b_{1}b_{2}\in S$
+\end_inset
+
+, pero como 
+\begin_inset Formula $xy,b_{1},b_{2}\in J$
+\end_inset
+
+, 
+\begin_inset Formula $s\in J\cap S\#$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Para 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+, llamamos 
+\series bold
+radical
+\series default
+ de 
+\begin_inset Formula $I$
+\end_inset
+
+ a
+\begin_inset Formula 
+\[
+\sqrt{I}\coloneqq\{x\in A:\exists n\in\mathbb{N}:x^{n}\in I\}=\bigcap\{J\trianglelefteq_{\text{r}}A:I\subseteq J\}=\bigcap\{J\trianglelefteq_{\text{p}}A:I\subseteq J\},
+\]
+
+\end_inset
+
+y en particular
+\begin_inset Formula 
+\[
+\sqrt{0}=\text{Nil}(A)=\bigcap\{J\trianglelefteq_{\text{r}}A\}=\bigcap\{J\trianglelefteq_{\text{p}}A\}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\subseteq2]$
+\end_inset
+
+ Sean 
+\begin_inset Formula $x\in A$
+\end_inset
+
+, 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $x^{n}\in I$
+\end_inset
+
+ y 
+\begin_inset Formula $J\trianglelefteq_{\text{r}}A$
+\end_inset
+
+ con 
+\begin_inset Formula $I\subseteq J$
+\end_inset
+
+, 
+\begin_inset Formula $x^{n}\in J$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $x\in J$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\subseteq3]$
+\end_inset
+
+ Todo ideal primo es radical.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $3\subseteq1]$
+\end_inset
+
+ Sea 
+\begin_inset Formula $x\in A$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall n\in\mathbb{N},x^{n}\notin I$
+\end_inset
+
+, y queremos ver que existe 
+\begin_inset Formula $P\trianglelefteq_{\text{p}}A$
+\end_inset
+
+ con 
+\begin_inset Formula $I\subseteq P$
+\end_inset
+
+ y 
+\begin_inset Formula $x\notin P$
+\end_inset
+
+.
+ 
+\begin_inset Formula $S\coloneqq\{x^{n}\}_{n\in\mathbb{N}}$
+\end_inset
+
+ es un subconjunto multiplicativo de 
+\begin_inset Formula $A$
+\end_inset
+
+ y por el lema de Krull existe un maximal 
+\begin_inset Formula $P$
+\end_inset
+
+ de 
+\begin_inset Formula ${\cal I}_{I,S}$
+\end_inset
+
+ que es primo, de modo que 
+\begin_inset Formula $P\trianglelefteq_{\text{p}}A$
+\end_inset
+
+, 
+\begin_inset Formula $I\subseteq P$
+\end_inset
+
+ y 
+\begin_inset Formula $x\notin P$
+\end_inset
+
+ porque 
+\begin_inset Formula $x\in S$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $I\subseteq\sqrt{I}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $I$
+\end_inset
+
+ es radical si y sólo si 
+\begin_inset Formula $I=\sqrt{I}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $I\trianglelefteq A$
+\end_inset
+
+ y 
+\begin_inset Formula $J\trianglelefteq_{\text{r}}A$
+\end_inset
+
+ con 
+\begin_inset Formula $I\subseteq J$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\sqrt{I}\subseteq J$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $I\triangleleft A$
+\end_inset
 
+ es un radical si y sólo si es intersección de ideales primos.
 \end_layout
 
 \end_body