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+#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
+Extensiones normales
+\end_layout
+
+\begin_layout Standard
+Una extensión 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es 
+\series bold
+normal
+\series default
+, o 
+\series bold
+
+\begin_inset Formula $L$
+\end_inset
+
+ es normal sobre 
+\begin_inset Formula $K$
+\end_inset
+
+
+\series default
+, si es algebraica y todo irreducible de 
+\begin_inset Formula $K[X]$
+\end_inset
+
+ con una raíz en 
+\begin_inset Formula $L$
+\end_inset
+
+ tiene todas sus raíces en 
+\begin_inset Formula $L$
+\end_inset
+
+.
+ Basta considerar los irreducibles mónicos de grado al menos 3, pues los
+ de grado 1 tienen su raíz en 
+\begin_inset Formula $K$
+\end_inset
+
+ y, si 
+\begin_inset Formula $f=X^{2}+aX+b$
+\end_inset
+
+ tiene raíces 
+\begin_inset Formula $\alpha_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $\alpha_{2}$
+\end_inset
+
+, 
+\begin_inset Formula $\alpha_{1}+\alpha_{2}=-a$
+\end_inset
+
+ y, si 
+\begin_inset Formula $\alpha_{1}\in L$
+\end_inset
+
+, 
+\begin_inset Formula $\alpha_{2}=-\alpha_{1}-a\in L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ejemplos:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $K\subseteq\overline{K}$
+\end_inset
+
+ es normal.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $[L:K]=2$
+\end_inset
+
+, 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es normal.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Los irreducibles en 
+\begin_inset Formula $K$
+\end_inset
+
+ con una raíz 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+ tiene grado 
+\begin_inset Formula $\text{gr}\text{Irr}(\alpha,K)\leq2$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Si 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es algebraica, todo 
+\begin_inset Formula $K$
+\end_inset
+
+-encaje 
+\begin_inset Formula $\sigma:L\to L$
+\end_inset
+
+ es un 
+\begin_inset Formula $K$
+\end_inset
+
+-automorfismo.
+ 
+\series bold
+Demostración:
+\series default
+ Para 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+, sean 
+\begin_inset Formula $f:=\text{Irr}(\alpha,K)$
+\end_inset
+
+ y 
+\begin_inset Formula $R:=\{\alpha_{1}:=\alpha,\dots,\alpha_{m}\}$
+\end_inset
+
+ el conjunto de las raíces de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $L$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f=\text{Irr}(\alpha_{i},K)$
+\end_inset
+
+ para cada 
+\begin_inset Formula $i$
+\end_inset
+
+ y, como 
+\begin_inset Formula $\sigma$
+\end_inset
+
+ lleva raíces a raíces, 
+\begin_inset Formula $\sigma(R)\subseteq R$
+\end_inset
+
+, pero 
+\begin_inset Formula $\sigma$
+\end_inset
+
+ es inyectiva, luego 
+\begin_inset Formula $\sigma|_{R}:R\to R$
+\end_inset
+
+ es biyectiva y 
+\begin_inset Formula $\alpha\in R=\sigma(R)\subseteq\sigma(L)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, una extensión algebraica 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es normal si y sólo si 
+\begin_inset Formula $L$
+\end_inset
+
+ es el cuerpo de descomposición sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ de un cierto 
+\begin_inset Formula ${\cal P}\subseteq K[X]\setminus0$
+\end_inset
+
+, si y sólo si para cada clausura algebraica 
+\begin_inset Formula $\overline{L}$
+\end_inset
+
+ de 
+\begin_inset Formula $L$
+\end_inset
+
+ y todo 
+\begin_inset Formula $K$
+\end_inset
+
+-encaje 
+\begin_inset Formula $\sigma:L\to\overline{L}$
+\end_inset
+
+ es 
+\begin_inset Formula $\sigma(L)=L$
+\end_inset
+
+, si y sólo si existe una clausura algebraica 
+\begin_inset Formula $\overline{L}$
+\end_inset
+
+ de 
+\begin_inset Formula $L$
+\end_inset
+
+ para la que todo 
+\begin_inset Formula $K$
+\end_inset
+
+-encaje 
+\begin_inset Formula $\sigma:L\to\overline{L}$
+\end_inset
+
+ cumple 
+\begin_inset Formula $\sigma(L)=L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Sean 
+\begin_inset Formula ${\cal P}:=\{f_{\alpha}:=\text{Irr}(\alpha,K)\}_{\alpha\in L}\subseteq K[X]\setminus0$
+\end_inset
+
+ y 
+\begin_inset Formula $S$
+\end_inset
+
+ el conjunto de todas las raíces de los polinomios de 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+ en una clausura 
+\begin_inset Formula $\overline{L}$
+\end_inset
+
+ de 
+\begin_inset Formula $L$
+\end_inset
+
+, cada 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+ está en 
+\begin_inset Formula $S$
+\end_inset
+
+ por ser raíz de 
+\begin_inset Formula $f_{\alpha}$
+\end_inset
+
+ y cada 
+\begin_inset Formula $\beta\in S$
+\end_inset
+
+ está en 
+\begin_inset Formula $L$
+\end_inset
+
+ ya que es raíz de un irreducible 
+\begin_inset Formula $f_{\alpha}\in K[X]$
+\end_inset
+
+ que ya tiene una raíz 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ en 
+\begin_inset Formula $L$
+\end_inset
+
+ y por tanto las tiene todas, luego 
+\begin_inset Formula $L=S$
+\end_inset
+
+ es el cuerpo de descomposición de 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+ sobre 
+\begin_inset Formula $K$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\implies3]$
+\end_inset
+
+ Para 
+\begin_inset Formula $f\in{\cal P}$
+\end_inset
+
+ y 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+ raíz de 
+\begin_inset Formula $f$
+\end_inset
+
+, como los 
+\begin_inset Formula $K$
+\end_inset
+
+-encajes llevan raíces a raíces, 
+\begin_inset Formula $\sigma(\alpha)$
+\end_inset
+
+ es raíz de 
+\begin_inset Formula $f$
+\end_inset
+
+ y está en 
+\begin_inset Formula $L$
+\end_inset
+
+, y como 
+\begin_inset Formula $L$
+\end_inset
+
+ está generado por las raíces de los 
+\begin_inset Formula $f\in{\cal P}$
+\end_inset
+
+, 
+\begin_inset Formula $\sigma(L)\subseteq L$
+\end_inset
+
+, luego por la proposición anterior 
+\begin_inset Formula $\sigma$
+\end_inset
+
+ es un 
+\begin_inset Formula $K$
+\end_inset
+
+-automorfismo y 
+\begin_inset Formula $\sigma(L)=L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $3\implies4]$
+\end_inset
+
+ Por la existencia de clausura algebraica.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $4\implies1]$
+\end_inset
+
+ Sean 
+\begin_inset Formula $f\in K[X]$
+\end_inset
+
+ irreducible con una raíz 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+ y 
+\begin_inset Formula $\beta\in\overline{L}$
+\end_inset
+
+ otra raíz de 
+\begin_inset Formula $f$
+\end_inset
+
+, existe un 
+\begin_inset Formula $K$
+\end_inset
+
+-isomorfismo 
+\begin_inset Formula $\sigma':K(\alpha)\to K(\beta)$
+\end_inset
+
+ con 
+\begin_inset Formula $\sigma'(\alpha)=\beta$
+\end_inset
+
+, pero como 
+\begin_inset Formula $K(\beta)\subseteq L$
+\end_inset
+
+, podemos ver 
+\begin_inset Formula $\sigma':K(\alpha)\to\overline{L}$
+\end_inset
+
+ y, como 
+\begin_inset Formula $K(\alpha)\subseteq L$
+\end_inset
+
+ es algebraica por serlo 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+, 
+\begin_inset Formula $\sigma'$
+\end_inset
+
+ se extiende a un 
+\begin_inset Formula $K$
+\end_inset
+
+-homomorfismo 
+\begin_inset Formula $\sigma:L\to\overline{L}$
+\end_inset
+
+, pero por hipótesis 
+\begin_inset Formula $\sigma(L)=L$
+\end_inset
+
+, luego 
+\begin_inset Formula $\beta=\sigma'(\alpha)=\sigma(\alpha)\in L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una extensión finita 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es normal si y sólo si 
+\begin_inset Formula $L$
+\end_inset
+
+ es el cuerpo de descomposición sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ de un polinomio de 
+\begin_inset Formula $K[X]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $L=K(\alpha_{1},\dots,\alpha_{n})$
+\end_inset
+
+, sean 
+\begin_inset Formula $f_{\alpha}:=\text{Irr}(\alpha,K)$
+\end_inset
+
+ para 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+ y 
+\begin_inset Formula $S$
+\end_inset
+
+ el conjunto de las raíces de 
+\begin_inset Formula $f_{\alpha_{1}},\dots,f_{\alpha_{n}}$
+\end_inset
+
+, cada 
+\begin_inset Formula $\beta\in S$
+\end_inset
+
+ está en 
+\begin_inset Formula $L$
+\end_inset
+
+ por ser raíz de un 
+\begin_inset Formula $f_{\alpha_{i}}\in K[X]$
+\end_inset
+
+ teniendo 
+\begin_inset Formula $f_{\alpha_{i}}$
+\end_inset
+
+ una raíz en 
+\begin_inset Formula $L$
+\end_inset
+
+ y siendo 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ normal, luego 
+\begin_inset Formula $S\subseteq L$
+\end_inset
+
+ y 
+\begin_inset Formula $K(S)\subseteq L$
+\end_inset
+
+, pero 
+\begin_inset Formula $\{\alpha_{1},\dots,\alpha_{n}\}\subseteq S$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $L=K(\alpha_{1},\dots,\alpha_{n})\subseteq K(S)$
+\end_inset
+
+, luego 
+\begin_inset Formula $L=K(S)$
+\end_inset
+
+ es el cuerpo de descomposición sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ de 
+\begin_inset Formula $\{f_{\alpha_{1}},\dots,f_{\alpha_{n}}\}$
+\end_inset
+
+ y por tanto de 
+\begin_inset Formula $f_{\alpha_{1}}\cdots f_{\alpha_{n}}$
+\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
+
+Por el teorema.
+\end_layout
+
+\begin_layout Standard
+Dada una torre 
+\begin_inset Formula $K\subseteq E\subseteq L$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es normal, 
+\begin_inset Formula $E\subseteq L$
+\end_inset
+
+ también.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si 
+\begin_inset Formula $L$
+\end_inset
+
+ es cuerpo de descomposición de un 
+\begin_inset Formula ${\cal P}\subseteq K[X]\setminus0$
+\end_inset
+
+ sobre 
+\begin_inset Formula $K$
+\end_inset
+
+, también lo es sobre 
+\begin_inset Formula $E$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Que 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ sea normal no implica que 
+\begin_inset Formula $K\subseteq E$
+\end_inset
+
+ lo sea.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt[3]{2},e^{2\pi i/3})$
+\end_inset
+
+ es normal por ser el cuerpo de descomposición de 
+\begin_inset Formula $X^{3}-2$
+\end_inset
+
+, pero 
+\begin_inset Formula $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt[3]{2})$
+\end_inset
+
+ no lo es ya que solo una raíz del irreducible 
+\begin_inset Formula $X^{3}-2$
+\end_inset
+
+ está en 
+\begin_inset Formula $\mathbb{Q}(\sqrt[3]{2})$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Que 
+\begin_inset Formula $K\subseteq E$
+\end_inset
+
+ y 
+\begin_inset Formula $E\subseteq L$
+\end_inset
+
+ sean normales no implica que 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ lo sea.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{2})$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt[4]{2})$
+\end_inset
+
+ son normales por tener grado 2, pero 
+\begin_inset Formula $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt[4]{2})$
+\end_inset
+
+ no lo es ya que dos de las raíces del irreducible 
+\begin_inset Formula $X^{4}-2$
+\end_inset
+
+ no están en 
+\begin_inset Formula $\mathbb{Q}(\sqrt[4]{2})$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ y 
+\begin_inset Formula $K\subseteq M$
+\end_inset
+
+ extensiones admisibles, si 
+\begin_inset Formula $L$
+\end_inset
+
+ es el cuerpo de descomposición sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ de un 
+\begin_inset Formula ${\cal P}\subseteq K[X]\setminus0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $LM$
+\end_inset
+
+ es el cuerpo de descomposición sobre 
+\begin_inset Formula $M$
+\end_inset
+
+ de 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+.
+ En efecto, sean 
+\begin_inset Formula $S$
+\end_inset
+
+ el conjunto de raíces de los polinomios de 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+ en 
+\begin_inset Formula $L$
+\end_inset
+
+, como 
+\begin_inset Formula $L=K(S)$
+\end_inset
+
+, 
+\begin_inset Formula $LM=ML=MK(S)=M(S)$
+\end_inset
+
+.
+ Ser normal es estable por levantamientos, pues lo es ser algebraica y ser
+ cuerpo de descomposición de un 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ y 
+\begin_inset Formula $K\subseteq M$
+\end_inset
+
+ son normales y admisibles, 
+\begin_inset Formula $K\subseteq LM$
+\end_inset
+
+ es normal.
+ En efecto, 
+\begin_inset Formula $L$
+\end_inset
+
+ es el cuerpo de descomposición sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ de un 
+\begin_inset Formula ${\cal P}\subseteq K[X]\setminus0$
+\end_inset
+
+ y 
+\begin_inset Formula $M$
+\end_inset
+
+ el de un 
+\begin_inset Formula ${\cal Q}\subseteq K[X]\setminus0$
+\end_inset
+
+, luego si 
+\begin_inset Formula $S$
+\end_inset
+
+ el conjunto de las raíces de polinomios de 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+ y 
+\begin_inset Formula $T$
+\end_inset
+
+ es el de las raíces de polinomios de 
+\begin_inset Formula $Q$
+\end_inset
+
+, 
+\begin_inset Formula $LM=K(S)K(T)=K(S\cup T)$
+\end_inset
+
+ es el cuerpo de descomposición sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ de 
+\begin_inset Formula ${\cal P}\cup{\cal Q}$
+\end_inset
+
+ y por tanto es normal.
+\end_layout
+
+\begin_layout Section
+Clausura normal
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $\{L_{i}\}_{i\in I}$
+\end_inset
+
+ es una familia de extensiones admisibles de 
+\begin_inset Formula $K$
+\end_inset
+
+ y cada 
+\begin_inset Formula $K\subseteq L_{i}$
+\end_inset
+
+ es normal, también lo es 
+\begin_inset Formula $K\subseteq\bigcap_{i\in I}L_{i}$
+\end_inset
+
+.
+ En efecto, si 
+\begin_inset Formula $f\in K[X]$
+\end_inset
+
+ es irreducible con una raíz en 
+\begin_inset Formula $\bigcap_{i\in I}L_{i}$
+\end_inset
+
+, entonces tiene todas sus raíces en todos los 
+\begin_inset Formula $L_{i}$
+\end_inset
+
+ y por tanto en 
+\begin_inset Formula $\bigcap_{i\in I}L_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ algebraica y 
+\begin_inset Formula $\overline{L}$
+\end_inset
+
+ una clausura normal de 
+\begin_inset Formula $L$
+\end_inset
+
+, la 
+\series bold
+clausura normal
+\series default
+ de 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ en 
+\begin_inset Formula $\overline{L}$
+\end_inset
+
+ es la menor extensión normal de 
+\begin_inset Formula $K$
+\end_inset
+
+ entre 
+\begin_inset Formula $L$
+\end_inset
+
+ y 
+\begin_inset Formula $\overline{L}$
+\end_inset
+
+, y viene dada por
+\begin_inset Formula 
+\[
+N:=\bigcap\{E\text{ intermedio en }L\subseteq\overline{L}:K\subseteq E\text{ normal}\}.
+\]
+
+\end_inset
+
+Como 
+\series bold
+teorema
+\series default
+:
+\end_layout
+
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $S\subseteq L$
+\end_inset
+
+ con 
+\begin_inset Formula $L=K(S)$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal P}:=\{\text{Irr}(\alpha,K)\}_{\alpha\in S}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $N$
+\end_inset
+
+ es el único cuerpo de descomposición de 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+ sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ en 
+\begin_inset Formula $\overline{L}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sea 
+\begin_inset Formula $R$
+\end_inset
+
+ el conjunto de raíces en 
+\begin_inset Formula $\overline{L}$
+\end_inset
+
+ de los polinomios de 
+\begin_inset Formula ${\cal P}$
+\end_inset
+
+, queremos ver que 
+\begin_inset Formula $N=K(R)$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $S\subseteq R\subseteq L$
+\end_inset
+
+, 
+\begin_inset Formula $L=K(S)\subseteq K(R)\subseteq\overline{L}$
+\end_inset
+
+ y 
+\begin_inset Formula $K\subseteq K(R)$
+\end_inset
+
+ es normal, se tiene 
+\begin_inset Formula $N\subseteq K(R)$
+\end_inset
+
+, y como 
+\begin_inset Formula $K\subseteq N$
+\end_inset
+
+ es normal y cada 
+\begin_inset Formula $\alpha\in L\subseteq N$
+\end_inset
+
+, todas las raíces de los 
+\begin_inset Formula $\text{Irr}(\alpha,K)$
+\end_inset
+
+ están en 
+\begin_inset Formula $N$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $R\subseteq N$
+\end_inset
+
+ y 
+\begin_inset Formula $K(R)\subseteq N$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es finita si y sólo si lo es 
+\begin_inset Formula $K\subseteq N$
+\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
+
+Sea 
+\begin_inset Formula $L=:K(\alpha_{1},\dots,\alpha_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $S:=\{\alpha_{1},\dots,\alpha_{n}\}$
+\end_inset
+
+, el conjunto 
+\begin_inset Formula $R$
+\end_inset
+
+ de raíces de los polinomios 
+\begin_inset Formula $\text{Irr}(\alpha,K)$
+\end_inset
+
+ con 
+\begin_inset Formula $\alpha\in S$
+\end_inset
+
+ es finito y, por el punto anterior, 
+\begin_inset Formula $N=K(R)$
+\end_inset
+
+, luego 
+\begin_inset Formula $K\subseteq N$
+\end_inset
+
+ es algebraica y finitamente generada y por tanto finita.
+\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 $K\subseteq L\subseteq N$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Dos clausuras normales de 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ en distintas clausuras algebraicas de 
+\begin_inset Formula $L$
+\end_inset
+
+ son 
+\begin_inset Formula $L$
+\end_inset
+
+-isomorfas.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula $L=K(L)$
+\end_inset
+
+, 
+\begin_inset Formula $N$
+\end_inset
+
+ es un cuerpo de descomposición de 
+\begin_inset Formula $\{\text{Irr}(\alpha,K)\}_{\alpha\in L}$
+\end_inset
+
+ sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ y por tanto sobre 
+\begin_inset Formula $L$
+\end_inset
+
+ y todos los cuerpos tales son 
+\begin_inset Formula $L$
+\end_inset
+
+-isomorfos.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es finita, existen 
+\begin_inset Formula $E_{1},\dots,E_{r}\subseteq\overline{L}$
+\end_inset
+
+ 
+\begin_inset Formula $K$
+\end_inset
+
+-isomorfos a 
+\begin_inset Formula $L$
+\end_inset
+
+ con 
+\begin_inset Formula $N=E_{1}\cdots E_{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $L=:K(\alpha_{1},\dots,\alpha_{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $R=\{\beta_{1},\dots,\beta_{r}\}$
+\end_inset
+
+ el conjunto de raíces de los 
+\begin_inset Formula $f_{i}:=\text{Irr}(\alpha_{i},K)$
+\end_inset
+
+ en 
+\begin_inset Formula $\overline{L}$
+\end_inset
+
+, cada 
+\begin_inset Formula $\beta_{j}$
+\end_inset
+
+ es conjugado con un 
+\begin_inset Formula $\alpha_{i}$
+\end_inset
+
+, luego existe un 
+\begin_inset Formula $K$
+\end_inset
+
+-isomorfismo 
+\begin_inset Formula $\sigma_{j}:K(\alpha_{i})\to K(\beta_{j})$
+\end_inset
+
+ con 
+\begin_inset Formula $\sigma_{j}(\alpha_{i})=\beta_{j}$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $N$
+\end_inset
+
+ es el cuerpo de descomposición de 
+\begin_inset Formula $\{f_{1},\dots,f_{n}\}$
+\end_inset
+
+ sobre 
+\begin_inset Formula $K$
+\end_inset
+
+, lo es sobre 
+\begin_inset Formula $K(\alpha_{i})$
+\end_inset
+
+ y sobre 
+\begin_inset Formula $K(\beta_{j})$
+\end_inset
+
+, luego 
+\begin_inset Formula $\sigma_{j}$
+\end_inset
+
+ se extiende a un 
+\begin_inset Formula $K$
+\end_inset
+
+-automorfismo 
+\begin_inset Formula $\overline{\sigma}_{j}:N\to N$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $E_{j}:=\overline{\sigma}_{j}(L)$
+\end_inset
+
+ es un subcuerpo de 
+\begin_inset Formula $N$
+\end_inset
+
+ 
+\begin_inset Formula $K$
+\end_inset
+
+-isomorfo a 
+\begin_inset Formula $L$
+\end_inset
+
+ con 
+\begin_inset Formula $\beta_{j}=\sigma_{j}(\alpha_{i})=\overline{\sigma}_{j}(\alpha_{i})\in\overline{\sigma}_{j}(L)=E_{j}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $N=K(\beta_{1},\dots,\beta_{r})=K(\beta_{1})\cdots K(\beta_{r})\subseteq E_{1}\cdots E_{r}\subseteq N$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Extensiones separables
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $K$
+\end_inset
+
+ un cuerpo, 
+\begin_inset Formula $f\in K[X]$
+\end_inset
+
+ es 
+\series bold
+separable
+\series default
+ si no tiene raíces múltiples en un cuerpo de descomposición de 
+\begin_inset Formula $f$
+\end_inset
+
+ sobre 
+\begin_inset Formula $K$
+\end_inset
+
+.
+ Dada una extensión 
+\begin_inset Formula $L$
+\end_inset
+
+ de 
+\begin_inset Formula $K$
+\end_inset
+
+, 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+ es 
+\series bold
+separable
+\series default
+ sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ si es algebraico sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ e 
+\begin_inset Formula $\text{Irr}(\alpha,K)$
+\end_inset
+
+ es separable.
+ Entonces 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es 
+\series bold
+separable
+\series default
+ si cada 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+ es separable sobre 
+\begin_inset Formula $K$
+\end_inset
+
+.
+ 
+\begin_inset Formula $K$
+\end_inset
+
+ es 
+\series bold
+perfecto
+\series default
+ si todo irreducible en 
+\begin_inset Formula $K[X]$
+\end_inset
+
+ es separable, si y sólo si toda extensión algebraica de 
+\begin_inset Formula $K$
+\end_inset
+
+ es separable sobre 
+\begin_inset Formula $K$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dado un cuerpo 
+\begin_inset Formula $K$
+\end_inset
+
+ y un 
+\begin_inset Formula $f\in K[X]$
+\end_inset
+
+ irreducible:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $\text{car}K=0$
+\end_inset
+
+, 
+\begin_inset Formula $f$
+\end_inset
+
+ es separable.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sea 
+\begin_inset Formula $L$
+\end_inset
+
+ el cuerpo de descomposición, como 
+\begin_inset Formula $\text{car}K=0$
+\end_inset
+
+ y 
+\begin_inset Formula $f$
+\end_inset
+
+ es irreducible en 
+\begin_inset Formula $K[X]$
+\end_inset
+
+, 
+\begin_inset Formula $f$
+\end_inset
+
+ no tiene raíces múltiples en 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $p:=\text{car}K\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $f\in K[X]$
+\end_inset
+
+ es irreducible, 
+\begin_inset Formula $f$
+\end_inset
+
+ es no separable si y sólo si 
+\begin_inset Formula $f\in K[X^{p}]$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Para 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+, como 
+\begin_inset Formula $f$
+\end_inset
+
+ es irreducible en 
+\begin_inset Formula $K[X]$
+\end_inset
+
+ con alguna raíz en 
+\begin_inset Formula $L$
+\end_inset
+
+, 
+\begin_inset Formula $f$
+\end_inset
+
+ tiene raíces múltiples en 
+\begin_inset Formula $L$
+\end_inset
+
+ si y solo si 
+\begin_inset Formula $f\in K[X^{p}]$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $K$
+\end_inset
+
+ es de característica 0, finito o algebraicamente cerrado, 
+\begin_inset Formula $K$
+\end_inset
+
+ es perfecto.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si 
+\begin_inset Formula $\text{car}K=0$
+\end_inset
+
+ ya lo hemos visto.
+ Si 
+\begin_inset Formula $K$
+\end_inset
+
+ es finito y 
+\begin_inset Formula $f\in K[X]$
+\end_inset
+
+ es irreducible, existe una raíz 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ en un cierto 
+\begin_inset Formula $K(\alpha)$
+\end_inset
+
+, que es finito y por tanto está formado por las raíces de 
+\begin_inset Formula $X^{|K(\alpha)|}-X$
+\end_inset
+
+, que no tiene raíces múltiples, pero como 
+\begin_inset Formula $f=\text{Irr}(\alpha,K)$
+\end_inset
+
+, 
+\begin_inset Formula $f\mid X^{|K(\alpha)|}\mid X$
+\end_inset
+
+ y 
+\begin_inset Formula $f$
+\end_inset
+
+ no tiene raíces múltiples.
+ Si 
+\begin_inset Formula $K$
+\end_inset
+
+ es algebraicamente cerrado, los únicos irreducibles son de la forma 
+\begin_inset Formula $X-a$
+\end_inset
+
+ y no tienen raíces múltiples.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Además:
+\end_layout
+
+\begin_layout Enumerate
+No todos los polinomios irreducibles son separables.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $K:=\mathbb{Z}_{p}(T)$
+\end_inset
+
+ y 
+\begin_inset Formula $f(X):=X^{p}-T\in K[X]$
+\end_inset
+
+, 
+\begin_inset Formula $f$
+\end_inset
+
+ es irreducible por Eisenstein al ser 
+\begin_inset Formula $T$
+\end_inset
+
+ irreducible en 
+\begin_inset Formula $\mathbb{Z}_{p}[T]$
+\end_inset
+
+, pero si 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ es una raíz de 
+\begin_inset Formula $f$
+\end_inset
+
+ en una extensión de 
+\begin_inset Formula $K$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\alpha^{p}=T$
+\end_inset
+
+ y, en 
+\begin_inset Formula $K(\alpha)[X]$
+\end_inset
+
+, 
+\begin_inset Formula $f(X)=X^{p}-\alpha^{p}=(X-\alpha)^{p}$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ es raíz múltiple.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ y 
+\begin_inset Formula $K\subseteq F$
+\end_inset
+
+ son admisibles y 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+ es separable sobre 
+\begin_inset Formula $K$
+\end_inset
+
+, lo es sobre 
+\begin_inset Formula $F$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $f:=\text{Irr}(\alpha,K)$
+\end_inset
+
+ y 
+\begin_inset Formula $g:=\text{Irr}(\alpha,F)$
+\end_inset
+
+, como 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ es raíz de 
+\begin_inset Formula $f\in F[X]$
+\end_inset
+
+, 
+\begin_inset Formula $g\mid f$
+\end_inset
+
+, y como 
+\begin_inset Formula $f$
+\end_inset
+
+ no tiene raíces múltiples, tampoco las tiene 
+\begin_inset Formula $g$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Dada la torre 
+\begin_inset Formula $K\subseteq F\subseteq L$
+\end_inset
+
+, si 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es separable, también lo son 
+\begin_inset Formula $K\subseteq F$
+\end_inset
+
+ y 
+\begin_inset Formula $F\subseteq L$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Todo 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+ es separable sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ y, por lo anterior, sobre 
+\begin_inset Formula $F$
+\end_inset
+
+, luego 
+\begin_inset Formula $F\subseteq L$
+\end_inset
+
+ es separable.
+ 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\end_body
+\end_document