From 9f88e37397a9dcfd32e69a7214317d6ee2330bba Mon Sep 17 00:00:00 2001
From: Juan Marín Noguera <juan.marinn@um.es>
Date: Mon, 19 Apr 2021 20:26:10 +0200
Subject: EALG tema 2

---
 ealg/n2.lyx | 1404 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++---
 1 file changed, 1342 insertions(+), 62 deletions(-)

(limited to 'ealg')

diff --git a/ealg/n2.lyx b/ealg/n2.lyx
index dcd3a77..b2edddd 100644
--- a/ealg/n2.lyx
+++ b/ealg/n2.lyx
@@ -734,7 +734,12 @@ Así, si
 \end_inset
 
  tiene grado finito y primo, no hay ningún cuerpo intermedio entre ellos.
- En efecto, sea 
+ 
+\begin_inset Note Comment
+status open
+
+\begin_layout Plain Layout
+En efecto, sea 
 \begin_inset Formula $E$
 \end_inset
 
@@ -770,6 +775,11 @@ Así, si
 .
 \end_layout
 
+\end_inset
+
+
+\end_layout
+
 \begin_layout Section
 Extensiones generadas y admisibles
 \end_layout
@@ -947,10 +957,14 @@ Llamamos
 \end_inset
 
 , que es la unión dirigida 
-\begin_inset Formula $\bigcup_{\{\alpha_{1},\dots,\alpha_{k}\}\subseteq S}K(\alpha_{1},\dots,\alpha_{k})$
+\begin_inset Formula 
+\[
+\bigcup_{\{\alpha_{1},\dots,\alpha_{k}\}\subseteq S}K(\alpha_{1},\dots,\alpha_{k}),
+\]
+
 \end_inset
 
-, donde 
+ donde 
 \begin_inset Formula $K(\alpha_{1},\dots,\alpha_{k})$
 \end_inset
 
@@ -1068,53 +1082,6 @@ K\subseteq E_{1}\cap E_{2}\subseteq E_{1},E_{2}\subseteq E_{1}E_{2}\subseteq L.
 \end_inset
 
 
-\end_layout
-
-\begin_layout Standard
-\begin_inset Note Comment
-status open
-
-\begin_layout Plain Layout
-Sean 
-\begin_inset Formula $p,q\in\mathbb{N}$
-\end_inset
-
- primos, 
-\begin_inset Formula $\mathbb{Q}\subseteq\mathbb{Q}[\sqrt{p}]$
-\end_inset
-
- y 
-\begin_inset Formula $\mathbb{Q}\subseteq\mathbb{Q}[\sqrt{q}]$
-\end_inset
-
- son admisibles y
-\begin_inset Formula 
-\[
-\mathbb{Q}[\sqrt{p}]\mathbb{Q}[\sqrt{q}]=\{a+b\sqrt{p}+c\sqrt{q}+d\sqrt{pq}\}_{a,b,c,d\in\mathbb{Q}}=\mathbb{Q}(\sqrt{p}+\sqrt{q}).
-\]
-
-\end_inset
-
-
-\series bold
-Demostración:
-\series default
- 
-\begin_inset Note Note
-status open
-
-\begin_layout Plain Layout
-TODO Hay que hacer mierdas de irreducibles para que salga bien, a.pdf::29.
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\end_inset
-
-
 \end_layout
 
 \begin_layout Section
@@ -1234,10 +1201,14 @@ Si
 \end_inset
 
 -encaje, entonces 
-\begin_inset Formula $[L:K]\mid[L':K]$
+\begin_inset Formula 
+\[
+[L:K]\mid[L':K],
+\]
+
 \end_inset
 
-, con igualdad si y solo si 
+ con igualdad si y solo si 
 \begin_inset Formula $\sigma$
 \end_inset
 
@@ -1876,9 +1847,9 @@ Un
 \end_inset
 
 .
-\end_layout
+\begin_inset Note Comment
+status open
 
-\begin_deeper
 \begin_layout Enumerate
 \begin_inset Argument item:1
 status open
@@ -1951,7 +1922,11 @@ Si
 
 \end_layout
 
-\end_deeper
+\end_inset
+
+
+\end_layout
+
 \begin_layout Enumerate
 Si 
 \begin_inset Formula $\alpha\in K$
@@ -1997,7 +1972,7 @@ Para
 \begin_deeper
 \begin_layout Standard
 Son raíces de 
-\begin_inset Formula $X^{2}-p\in\mathbb{Q}[X]$
+\begin_inset Formula $X^{2}-m\in\mathbb{Q}[X]$
 \end_inset
 
 .
@@ -2395,10 +2370,14 @@ Si
 \end_inset
 
  con 
-\begin_inset Formula $\sum_{i=0}^{n-1}a_{i}\alpha^{i}=0$
+\begin_inset Formula 
+\[
+\sum_{i=0}^{n-1}a_{i}\alpha^{i}=0,
+\]
+
 \end_inset
 
-, pero entonces 
+ pero entonces 
 \begin_inset Formula $g:=\sum_{i=0}^{n-1}a_{i}X^{i}\in K[X]\setminus0$
 \end_inset
 
@@ -2649,6 +2628,138 @@ status open
 
 \end_layout
 
+\begin_layout Standard
+Sean 
+\begin_inset Formula $p,q\in\mathbb{N}$
+\end_inset
+
+ primos, 
+\begin_inset Formula $\mathbb{Q}\subseteq\mathbb{Q}[\sqrt{p}]$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{Q}\subseteq\mathbb{Q}[\sqrt{q}]$
+\end_inset
+
+ son admisibles y
+\begin_inset Formula 
+\[
+\mathbb{Q}[\sqrt{p}]\mathbb{Q}[\sqrt{q}]=\{a+b\sqrt{p}+c\sqrt{q}+d\sqrt{pq}\}_{a,b,c,d\in\mathbb{Q}}=\mathbb{Q}(\sqrt{p}+\sqrt{q}).
+\]
+
+\end_inset
+
+
+\series bold
+Demostración:
+\series default
+ Para 
+\begin_inset Formula $p=q$
+\end_inset
+
+ esto es obvio, por lo que supondremos 
+\begin_inset Formula $p\neq q$
+\end_inset
+
+.
+ Sean 
+\begin_inset Formula $F_{p}:=\mathbb{Q}[\sqrt{p}]$
+\end_inset
+
+ y 
+\begin_inset Formula $F_{q}:=\mathbb{Q}[\sqrt{q}]$
+\end_inset
+
+, 
+\begin_inset Formula $F_{p}$
+\end_inset
+
+ y 
+\begin_inset Formula $F_{q}$
+\end_inset
+
+ son admisibles por ser ambos subcuerpos de 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Claramente 
+\begin_inset Formula $S:=\{a+b\sqrt{p}+c\sqrt{q}+d\sqrt{pq}\}_{a,b,c,d\in\mathbb{Q}}\subseteq F_{p}F_{q}$
+\end_inset
+
+.
+ Sea ahora 
+\begin_inset Formula $\alpha:=\sqrt{p}+\sqrt{q}\in S$
+\end_inset
+
+, 
+\begin_inset Formula 
+\begin{multline*}
+\alpha^{2}=p+q+2\sqrt{pq}\implies\alpha^{2}-(p+q)=2\sqrt{pq}\implies\\
+\implies\alpha^{4}-2(p+q)\alpha^{2}+2(p+q)^{2}=4pq\implies\alpha^{4}-2(p+q)\alpha^{2}+2(p-q)^{2}=0,
+\end{multline*}
+
+\end_inset
+
+luego 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ es raíz del polinomio 
+\begin_inset Formula $X^{4}-2(p+q)X^{2}+2(p-q)^{2}\in\mathbb{Q}[X]$
+\end_inset
+
+ y por tanto es algebraico, con lo que 
+\begin_inset Formula $\mathbb{Q}[\alpha]=\mathbb{Q}(\alpha)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $S$
+\end_inset
+
+ es un anillo, pues es cerrado para restas y productos y contiene al 1,
+ y como además 
+\begin_inset Formula $\mathbb{Q}\cup\{\alpha\}\subseteq S$
+\end_inset
+
+, 
+\begin_inset Formula $\mathbb{Q}[\alpha]\subseteq S$
+\end_inset
+
+.
+ Finalmente, como 
+\begin_inset Formula 
+\[
+\frac{1}{\alpha}=\frac{1}{\sqrt{p}+\sqrt{q}}\frac{\sqrt{p}-\sqrt{q}}{\sqrt{p}-\sqrt{q}}=\frac{\sqrt{p}-\sqrt{q}}{p-q}\in\mathbb{Q}(\alpha),
+\]
+
+\end_inset
+
+
+\begin_inset Formula $\sqrt{p}-\sqrt{q}\in\mathbb{Q}(\alpha)$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\sqrt{p},\sqrt{q}\in\mathbb{Q}(\alpha)$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $F_{p},F_{q}\subseteq\mathbb{Q}(\alpha)$
+\end_inset
+
+ y 
+\begin_inset Formula $F_{p}F_{q}\subseteq\mathbb{Q}(\alpha)$
+\end_inset
+
+.
+ Con esto 
+\begin_inset Formula $S\subseteq F_{p}F_{q}\subseteq\mathbb{Q}(\alpha)=\mathbb{Q}[\alpha]\subseteq S$
+\end_inset
+
+, luego estos conjuntos son iguales.
+\end_layout
+
 \begin_layout Standard
 Sean 
 \begin_inset Formula $K$
@@ -3156,17 +3267,1186 @@ Un
 
 \end_deeper
 \begin_layout Enumerate
-\begin_inset Note Note
-status open
+Para toda raíz 
+\begin_inset Formula $\beta$
+\end_inset
 
-\begin_layout Plain Layout
-a1::46, C2.35
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+, 
+\begin_inset Formula $f$
+\end_inset
+
+ tiene 
+\begin_inset Formula $m$
+\end_inset
+
+ raíces en 
+\begin_inset Formula $K(\beta)$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Al ser 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ y 
+\begin_inset Formula $\beta$
+\end_inset
+
+ raíces de un mismo irreducible, 
+\begin_inset Formula $\text{Gal}(K(\alpha)/K)\cong\text{Gal}(K(\beta)/K)$
+\end_inset
+
+, luego 
+\begin_inset Formula $|\text{Gal}(K(\beta)/K)|=m$
+\end_inset
+
+, y el resultado se obtiene del punto anterior.
 \end_layout
 
+\end_deeper
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ no tiene raíces múltiples (por ejemplo, si 
+\begin_inset Formula $\text{car}K=0$
 \end_inset
 
+), entonces 
+\begin_inset Formula $m\mid n$
+\end_inset
 
+.
 \end_layout
 
+\begin_deeper
+\begin_layout Standard
+Las 
+\begin_inset Formula $n$
+\end_inset
+
+ raíces se reparten en extensiones 
+\begin_inset Formula $K(\alpha)$
+\end_inset
+
+ de 
+\begin_inset Formula $m$
+\end_inset
+
+ elementos, y si una raíz 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ estuviera en una extensión 
+\begin_inset Formula $K(\beta)$
+\end_inset
+
+, siendo 
+\begin_inset Formula $\beta$
+\end_inset
+
+ otra raíz de 
+\begin_inset Formula $f$
+\end_inset
+
+, entonces 
+\begin_inset Formula $K(\beta)\subseteq K(\alpha)$
+\end_inset
+
+ con 
+\begin_inset Formula $[K(\beta):K]=[K(\alpha):K]$
+\end_inset
+
+, luego 
+\begin_inset Formula $K(\alpha)=K(\beta)$
+\end_inset
+
+ y ninguna raíz está en dos extensiones distintas.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Algunos grupos de Galois
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\text{Gal}(\mathbb{R}/\mathbb{Q})=1.
+\]
+
+\end_inset
+
+
+\begin_inset Note Comment
+status open
+
+\begin_layout Plain Layout
+
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $\sigma:\mathbb{R}\to\mathbb{R}$
+\end_inset
+
+ un 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+-automorfismo, y queremos ver que entonces 
+\begin_inset Formula $\sigma=1_{\mathbb{R}}$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $r,s\in\mathbb{R}$
+\end_inset
+
+, si 
+\begin_inset Formula $r<s$
+\end_inset
+
+, existe 
+\begin_inset Formula $a\in\mathbb{R}^{*}$
+\end_inset
+
+ con 
+\begin_inset Formula $s-r=a^{2}$
+\end_inset
+
+, y como 
+\begin_inset Formula $a\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $\sigma(a)\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $\sigma(s)-\sigma(r)=\sigma(a^{2})=\sigma(a)^{2}>0$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $\sigma(r)<\sigma(s)$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\sigma$
+\end_inset
+
+ conserva el orden, luego para 
+\begin_inset Formula $t\in\mathbb{R}$
+\end_inset
+
+, si 
+\begin_inset Formula $\sigma(t)<t$
+\end_inset
+
+, sea 
+\begin_inset Formula $q\in\mathbb{Q}$
+\end_inset
+
+ con 
+\begin_inset Formula $\sigma(t)<q<t$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\sigma(t)<q=\sigma(q)<\sigma(t)\#$
+\end_inset
+
+, y si 
+\begin_inset Formula $\sigma(t)>t$
+\end_inset
+
+, sea 
+\begin_inset Formula $q\in\mathbb{Q}$
+\end_inset
+
+ con 
+\begin_inset Formula $t<q<\sigma(t)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\sigma(t)<\sigma(q)=q<\sigma(t)\#$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $\sigma(t)=t$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+sremember{GyA}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+orden
+\series default
+ de 
+\begin_inset Formula $G$
+\end_inset
+
+ al cardinal del conjunto.
+ Algunos grupos:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un anillo, [...] 
+\begin_inset Formula $(A^{*},\cdot)$
+\end_inset
+
+ es su 
+\series bold
+grupo de unidades
+\series default
+ [...]
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+3.
+\end_layout
+
+\end_inset
+
+Dada una familia 
+\begin_inset Formula $(G_{i})_{i\in I}$
+\end_inset
+
+ de grupos, 
+\begin_inset Formula $\prod_{i\in I}G_{i}$
+\end_inset
+
+ [...] con el producto componente a componente.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+4.
+\end_layout
+
+\end_inset
+
+Llamamos 
+\series bold
+grupo cíclico
+\series default
+ de orden 
+\begin_inset Formula $n\in\mathbb{N}^{*}$
+\end_inset
+
+ a 
+\begin_inset Formula $C_{n}:=\{1,a,a^{2},\dots,a^{n-1}\}$
+\end_inset
+
+ con [...] 
+\begin_inset Formula $a^{i}a^{j}:=a^{[i+j]_{n}}$
+\end_inset
+
+ [...].
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+eremember
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+sremember{GyA}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+5.
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $n\in\mathbb{N}^{*}$
+\end_inset
+
+, llamamos 
+\series bold
+grupo diédrico
+\series default
+ de orden 
+\begin_inset Formula $2n$
+\end_inset
+
+ a 
+\begin_inset Formula 
+\[
+D_{n}:=\{1,a,a^{2},\dots,a^{n-1},b,ab,a^{2}b,\dots,a^{n-1}b\}
+\]
+
+\end_inset
+
+con la operación 
+\begin_inset Formula $(a^{i_{1}}b^{j_{1}})(a^{i_{2}}b^{j_{2}}):=a^{[i_{1}+(-1)^{j_{1}}i_{2}]_{n}}b^{[j_{1}+j_{2}]_{2}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+[...] 
+\series bold
+Teorema de Lagrange:
+\series default
+ Si 
+\begin_inset Formula $G$
+\end_inset
+
+ es un grupo finito y 
+\begin_inset Formula $H\leq G$
+\end_inset
+
+, 
+\begin_inset Formula $|G|=|H|[G:H]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+[...] Permutaciones entre conjuntos finitos, 
+\begin_inset Formula $S_{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+[, con la composición].
+ [...] Llamamos 
+\series bold
+grupo alternado
+\series default
+ [...] a 
+\begin_inset Formula $A_{n}:=\ker\text{sgn}$
+\end_inset
+
+, el subgrupo de 
+\begin_inset Formula $S_{n}$
+\end_inset
+
+ de las permutaciones pares.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+eremember
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+grupo de Klein
+\series default
+ es 
+\begin_inset Formula $C_{2}\times C_{2}$
+\end_inset
+
+.
+ Dado un cuerpo 
+\begin_inset Formula $K$
+\end_inset
+
+, todo subgrupo finito de 
+\begin_inset Formula $(K^{*},\cdot)$
+\end_inset
+
+ es cíclico.
+ En particular, si 
+\begin_inset Formula $p$
+\end_inset
+
+ es primo, 
+\begin_inset Formula $\mathbb{Z}_{p}^{*}$
+\end_inset
+
+ es cíclico.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $p\in\mathbb{Z}^{+}$
+\end_inset
+
+ primo y 
+\begin_inset Formula $\xi:=e^{2\pi i/p}$
+\end_inset
+
+, 
+\begin_inset Formula $\text{Gal}(\mathbb{Q}(\xi)/\mathbb{Q})\cong C_{p-1}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $\text{Irr}(\xi,\mathbb{Q})=X^{p-1}+\dots+X^{2}+X+1$
+\end_inset
+
+ tiene 
+\begin_inset Formula $p-1$
+\end_inset
+
+ raíces, los 
+\begin_inset Formula $\xi^{k}$
+\end_inset
+
+ para 
+\begin_inset Formula $k\in\{1,\dots,p-1\}$
+\end_inset
+
+, que están en 
+\begin_inset Formula $\mathbb{Q}(\xi)$
+\end_inset
+
+, luego 
+\begin_inset Formula $\text{Gal}(\mathbb{Q}(\xi)/\mathbb{Q})$
+\end_inset
+
+ tiene 
+\begin_inset Formula $p-1$
+\end_inset
+
+ elementos 
+\begin_inset Formula $\{\sigma_{k}\}_{k=1}^{p-1}$
+\end_inset
+
+ donde 
+\begin_inset Formula $\sigma_{k}(\xi):=\xi^{k}$
+\end_inset
+
+.
+ Además, la biyección 
+\begin_inset Formula $\mathbb{Z}_{p}^{*}=\{1,2,\dots,p-1\}\to\text{Gal}(\mathbb{Q}(\xi)/\mathbb{Q})$
+\end_inset
+
+ dada por 
+\begin_inset Formula $k\mapsto\sigma_{k}$
+\end_inset
+
+ es un isomorfismo, pues 
+\begin_inset Formula $1\mapsto1_{\mathbb{Q}(\xi)}$
+\end_inset
+
+ y, para 
+\begin_inset Formula $j,k\in\mathbb{Z}_{p}^{*}$
+\end_inset
+
+, 
+\begin_inset Formula $(\sigma_{j}\sigma_{k})(\xi)=\sigma_{j}(\xi^{k})=\sigma_{j}(\xi)^{k}=\xi^{jk}=\sigma_{jk}(\xi)$
+\end_inset
+
+, luego 
+\begin_inset Formula $\text{Gal}(\mathbb{Q}(\xi)/\mathbb{Q})\cong\mathbb{Z}_{p}^{*}\cong C_{p-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Extensiones finitamente generadas
+\end_layout
+
+\begin_layout Standard
+Una extensión 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es 
+\series bold
+finitamente generada
+\series default
+ si existen 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in L$
+\end_inset
+
+ con 
+\begin_inset Formula $L=K(\alpha_{1},\dots,\alpha_{n})$
+\end_inset
+
+.
+ 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ es finita si y solo si es finitamente generada y algebraica, si y solo
+ si existen 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in L$
+\end_inset
+
+ algebraicos sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ tales que 
+\begin_inset Formula $L=K(\alpha_{1},\dots,\alpha_{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Toda extensión finita es algebraica, y dada una base 
+\begin_inset Formula $(\alpha_{1},\dots,\alpha_{n})$
+\end_inset
+
+ de 
+\begin_inset Formula $L$
+\end_inset
+
+ como 
+\begin_inset Formula $K$
+\end_inset
+
+-espacio vectorial, 
+\begin_inset Formula $L=K(\alpha_{1},\dots,\alpha_{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\implies3]$
+\end_inset
+
+ Obvio.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $3\implies1]$
+\end_inset
+
+ Para 
+\begin_inset Formula $n=1$
+\end_inset
+
+, 
+\begin_inset Formula $[L:K]=\text{gr}\text{Irr}(\alpha_{1},K)<+\infty$
+\end_inset
+
+.
+ Para 
+\begin_inset Formula $n>1$
+\end_inset
+
+, supuesto esto probado para 
+\begin_inset Formula $1,\dots,n-1$
+\end_inset
+
+, 
+\begin_inset Formula $K\subseteq K(\alpha_{1})$
+\end_inset
+
+ es finita y, como 
+\begin_inset Formula $\alpha_{2},\dots,\alpha_{n}$
+\end_inset
+
+ son algebraicos sobre 
+\begin_inset Formula $K(\alpha_{1})$
+\end_inset
+
+, 
+\begin_inset Formula $K(\alpha_{1})\subseteq K(\alpha_{1})(\alpha_{2},\dots,\alpha_{n})=K(\alpha_{1},\dots,\alpha_{n})=L$
+\end_inset
+
+ es finita, y 
+\begin_inset Formula $[L:K]=[L:K(\alpha_{1})][K(\alpha_{1}):K]$
+\end_inset
+
+ es finito.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ una extensión y 
+\begin_inset Formula $S\subseteq L$
+\end_inset
+
+ un subconjunto cuyos elementos son algebraicos sobre 
+\begin_inset Formula $K$
+\end_inset
+
+, entonces 
+\begin_inset Formula $K\subseteq K(S)$
+\end_inset
+
+ es una extensión algebraica, pues para 
+\begin_inset Formula $\alpha\in K(S)$
+\end_inset
+
+, 
+\begin_inset Formula $\alpha\in K(\alpha_{1},\dots,\alpha_{k})$
+\end_inset
+
+ para ciertos 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{k}\in S$
+\end_inset
+
+, que son algebraicos, luego por lo anterior 
+\begin_inset Formula $K\subseteq K(\alpha_{1},\dots,\alpha_{k})$
+\end_inset
+
+ es algebraica y por tanto 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ es algebraico.
+\end_layout
+
+\begin_layout Standard
+La 
+\series bold
+clausura algebraica
+\series default
+ de una extensión 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ o de 
+\begin_inset Formula $K$
+\end_inset
+
+ en 
+\begin_inset Formula $L$
+\end_inset
+
+ es
+\begin_inset Formula 
+\[
+\overline{K}_{L}:=\{\alpha\in L:\alpha\text{ es algebraico sobre }K\}.
+\]
+
+\end_inset
+
+Es un cuerpo, pues para 
+\begin_inset Formula $\alpha,\beta\in\overline{K}_{L}$
+\end_inset
+
+, 
+\begin_inset Formula $K(\alpha,\beta)$
+\end_inset
+
+ es una extensión algebraica de 
+\begin_inset Formula $K$
+\end_inset
+
+ que contiene a 1, 
+\begin_inset Formula $\alpha-\beta$
+\end_inset
+
+, 
+\begin_inset Formula $\alpha\beta$
+\end_inset
+
+ y, si 
+\begin_inset Formula $\beta\neq0$
+\end_inset
+
+, a 
+\begin_inset Formula $\alpha\beta^{-1}$
+\end_inset
+
+, y que al ser algebraica está contenida en 
+\begin_inset Formula $\overline{K}_{L}$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $\overline{K}_{L}$
+\end_inset
+
+ es el mayor cuerpo intermedio de 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ algebraico sobre 
+\begin_inset Formula $K$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+cuerpo de números algebraicos
+\series default
+ es un cuerpo 
+\begin_inset Formula $L$
+\end_inset
+
+ entre 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+ y 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\mathbb{Q}\subseteq L$
+\end_inset
+
+ es finita.
+ Llamamos 
+\series bold
+cuerpo de los números algebraicos
+\series default
+ a 
+\begin_inset Formula ${\cal A}:=\overline{\mathbb{Q}}_{\mathbb{C}}$
+\end_inset
+
+, y 
+\series bold
+números algebraicos
+\series default
+ a los elementos de 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+, de modo que para todo cuerpo de números algebraicos 
+\begin_inset Formula $L$
+\end_inset
+
+, 
+\begin_inset Formula $L\subseteq{\cal A}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\mathbb{Q}\subseteq{\cal A}$
+\end_inset
+
+ es algebraica pero no finita, pues para un primo 
+\begin_inset Formula $p$
+\end_inset
+
+ arbitrariamente grande, 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+ contiene a 
+\begin_inset Formula $\mathbb{Q}(\xi_{p})$
+\end_inset
+
+ para 
+\begin_inset Formula $\xi_{p}:=e^{2\pi i/p}$
+\end_inset
+
+, pero 
+\begin_inset Formula $[\mathbb{Q}(\xi_{p}):\mathbb{Q}]=p-1$
+\end_inset
+
+, luego 
+\begin_inset Formula $[{\cal A}:\mathbb{Q}]\geq p-1$
+\end_inset
+
+ para todo primo 
+\begin_inset Formula $p$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $[{\cal A}:\mathbb{Q}]=\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Propiedades de extensiones
+\end_layout
+
+\begin_layout Standard
+Una 
+\series bold
+torre de extensiones
+\series default
+ es una secuencia de extensiones de cuerpos de la forma 
+\begin_inset Formula $(K_{i-1}\subseteq K_{i})_{i=1}^{n}$
+\end_inset
+
+, escrita como 
+\begin_inset Formula $K_{0}\subseteq K_{1}\subseteq\dots\subseteq K_{n}$
+\end_inset
+
+, y cada extensión de la secuencia es una 
+\series bold
+subextensión
+\series default
+ de 
+\begin_inset Formula $K_{0}\subseteq K_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una propiedad de extensiones es 
+\series bold
+multiplicativa en torres
+\series default
+ si para cada torre de extensiones 
+\begin_inset Formula $K\subseteq L\subseteq M$
+\end_inset
+
+, 
+\begin_inset Formula $K\subseteq M$
+\end_inset
+
+ cumple la propiedad si y solo si la cumplen 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ y 
+\begin_inset Formula $L\subseteq M$
+\end_inset
+
+.
+ Son multiplicativas en torres:
+\end_layout
+
+\begin_layout Enumerate
+Ser finita.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Se debe a que 
+\begin_inset Formula $[M:K]=[M:L][L:K]$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Ser algebraica.
+\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 $K\subseteq L$
+\end_inset
+
+ es algebraica porque, para 
+\begin_inset Formula $\alpha\in L$
+\end_inset
+
+, 
+\begin_inset Formula $\alpha\in M$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ es algebraico sobre 
+\begin_inset Formula $K$
+\end_inset
+
+, y 
+\begin_inset Formula $L\subseteq M$
+\end_inset
+
+ lo es porque, para 
+\begin_inset Formula $\alpha\in M$
+\end_inset
+
+, 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ es algebraico sobre 
+\begin_inset Formula $K$
+\end_inset
+
+ y por tanto sobre 
+\begin_inset Formula $L$
+\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
+
+Para 
+\begin_inset Formula $\alpha\in M$
+\end_inset
+
+, existe 
+\begin_inset Formula $f:=\sum_{i=0}^{n}a_{i}X^{i}\in L[X]$
+\end_inset
+
+ que tiene a 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ como raíz.
+ Pero cada 
+\begin_inset Formula $a_{i}\in L$
+\end_inset
+
+ es algebraico sobre 
+\begin_inset Formula $K$
+\end_inset
+
+, luego 
+\begin_inset Formula $K\subseteq L'=K(a_{0},\dots,a_{n-1})$
+\end_inset
+
+ es finita, y como 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ es algebraico sobre 
+\begin_inset Formula $L'$
+\end_inset
+
+ por ser raíz de 
+\begin_inset Formula $f\in L'[X]$
+\end_inset
+
+, 
+\begin_inset Formula $L'\subseteq L'(\alpha)$
+\end_inset
+
+ es finita, de modo que 
+\begin_inset Formula $K\subseteq L'(\alpha)$
+\end_inset
+
+ es algebraica y 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ es algebraico sobre 
+\begin_inset Formula $K$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Ser finitamente generada.
+\end_layout
+
+\begin_layout Standard
+Una propiedad relativa a extensiones es 
+\series bold
+estable por levantamientos
+\series default
+ si, dadas dos extensiones admisibles 
+\begin_inset Formula $K\subseteq L$
+\end_inset
+
+ y 
+\begin_inset Formula $K\subseteq M$
+\end_inset
+
+, si 
+\begin_inset Formula $K\subseteq M$
+\end_inset
+
+ cumple la propiedad, 
+\begin_inset Formula $L\subseteq LM$
+\end_inset
+
+ también.
+ Son estables por levantamientos:
+\end_layout
+
+\begin_layout Enumerate
+Ser algebraica.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Si 
+\begin_inset Formula $K\subseteq M$
+\end_inset
+
+ es algebraica, los 
+\begin_inset Formula $\alpha\in M$
+\end_inset
+
+ son algebraicos sobre 
+\begin_inset Formula $L$
+\end_inset
+
+ al serlo sobre 
+\begin_inset Formula $K$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $L\subseteq L(M)=LM$
+\end_inset
+
+ es algebraica.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Ser finitamente generada.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Sean 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in M$
+\end_inset
+
+ tales que 
+\begin_inset Formula $M=K(\alpha_{1},\dots,\alpha_{n})$
+\end_inset
+
+, 
+\begin_inset Formula 
+\[
+LM=LK(\alpha_{1},\dots,\alpha_{n})=L(\alpha_{1},\dots,\alpha_{m}),
+\]
+
+\end_inset
+
+ pues 
+\begin_inset Formula $LK(\alpha_{1},\dots,\alpha_{n})$
+\end_inset
+
+ es el menor cuerpo que contiene a 
+\begin_inset Formula $L\cup K\cup\{\alpha_{1},\dots,\alpha_{n}\}=L\cup\{\alpha_{1},\dots,\alpha_{n}\}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Ser finita.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Equivale a ser algebraica y finitamente generada.
+\end_layout
+
+\end_deeper
 \end_body
 \end_document
-- 
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