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| -rw-r--r-- | ealg/n2.lyx | 1400 |
1 files changed, 1340 insertions, 60 deletions
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 @@ -1070,53 +1084,6 @@ K\subseteq E_{1}\cap E_{2}\subseteq E_{1},E_{2}\subseteq E_{1}E_{2}\subseteq L. \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 Grupos de Galois \end_layout @@ -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 @@ -2651,6 +2630,138 @@ status open \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 $K$ \end_inset @@ -3156,17 +3267,1186 @@ Un \end_deeper \begin_layout Enumerate -\begin_inset Note Note +Para toda raíz +\begin_inset Formula $\beta$ +\end_inset + + 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 -a1::46, C2.35 +\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 |