diff options
| -rw-r--r-- | ac/n.lyx | 14 | ||||
| -rw-r--r-- | ac/n2.lyx | 699 |
2 files changed, 713 insertions, 0 deletions
@@ -171,5 +171,19 @@ filename "n1.lyx" \end_layout +\begin_layout Chapter +Anillos noetherianos +\end_layout + +\begin_layout Standard +\begin_inset CommandInset include +LatexCommand input +filename "n2.lyx" + +\end_inset + + +\end_layout + \end_body \end_document diff --git a/ac/n2.lyx b/ac/n2.lyx new file mode 100644 index 0000000..a4645f1 --- /dev/null +++ b/ac/n2.lyx @@ -0,0 +1,699 @@ +#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 +Retículos +\end_layout + +\begin_layout Standard +Un +\series bold +conjunto ordenado +\series default + +\begin_inset Formula $(A,\leq)$ +\end_inset + + cumple la +\series bold +condición de cadena ascendente +\series default + ( +\series bold +ACC +\series default +, +\emph on +\lang english +Ascending Chain Condition +\emph default +\lang spanish +) si para +\begin_inset Formula $\{a_{n}\}_{n}\subseteq A$ +\end_inset + + con cada +\begin_inset Formula $a_{n}\leq a_{n+1}$ +\end_inset + + existe +\begin_inset Formula $n_{0}\in\mathbb{N}$ +\end_inset + + tal que para +\begin_inset Formula $n\geq n_{0}$ +\end_inset + + es +\begin_inset Formula $a_{n}=a_{n_{0}}$ +\end_inset + +, si y sólo si todo +\begin_inset Formula $S\subseteq A$ +\end_inset + + no vacío tiene un elemento maximal. +\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 + +Probamos el contrarrecíproco. + Si +\begin_inset Formula $S\subseteq A$ +\end_inset + + no tiene elementos maximales, sea +\begin_inset Formula $s_{1}\in S$ +\end_inset + + arbitrario, como +\begin_inset Formula $s_{1}$ +\end_inset + + no es maximal, existe +\begin_inset Formula $s_{2}\in S$ +\end_inset + + son +\begin_inset Formula $s_{1}<s_{2}$ +\end_inset + +, y por inducción se puede construir una secuencia +\begin_inset Formula $\{s_{n}\}_{n}\subseteq S\subseteq A$ +\end_inset + + con cada +\begin_inset Formula $s_{n}<s_{n+1}$ +\end_inset + + y +\begin_inset Formula $A$ +\end_inset + + no cumple la ACC. +\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 + +Dada +\begin_inset Formula $\{a_{n}\}_{n}\subseteq A$ +\end_inset + + con cada +\begin_inset Formula $a_{n}\leq a_{n+1}$ +\end_inset + +, como +\begin_inset Formula $\{a_{n}\}_{n}\neq\emptyset$ +\end_inset + +, tiene un maximal +\begin_inset Formula $a_{n_{0}}$ +\end_inset + +, y para +\begin_inset Formula $n\geq n_{0}$ +\end_inset + +, +\begin_inset Formula $a_{n}\geq a_{n_{0}}$ +\end_inset + + y por tanto +\begin_inset Formula $a_{n}=a_{n_{0}}$ +\end_inset + +. + +\end_layout + +\begin_layout Standard +\begin_inset Formula $(A,\leq)$ +\end_inset + + cumple la +\series bold +condición de cadena descendente +\series default + ( +\series bold +DCC +\series default +, +\emph on +\lang english +Descending Chain Condition +\emph default +\lang spanish +) si +\begin_inset Formula $(A,\geq)$ +\end_inset + + cumple la ACC, si y sólo si todo +\begin_inset Formula $S\subseteq A$ +\end_inset + + no vacío tiene un elemento minimal. +\end_layout + +\begin_layout Standard +Un +\series bold +retículo +\series default + es un conjunto parcialmente ordenado en que todo subconjunto de dos elementos + tiene supremo e ínfimo, y es +\series bold +completo +\series default + si todo subconjunto tiene supremo e ínfimo. + +\begin_inset Formula $({\cal L},\leq)$ +\end_inset + + es un retículo completo si y sólo si lo es +\begin_inset Formula $({\cal L},\geq)$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Sea +\begin_inset Formula ${\cal L}$ +\end_inset + + un retículo completo, para +\begin_inset Formula $S\subseteq{\cal L}$ +\end_inset + +, llamamos +\begin_inset Formula $\bigvee S\coloneqq\sup_{{\cal L}}S$ +\end_inset + + y +\begin_inset Formula $\bigwedge S\coloneqq\inf_{{\cal L}}S$ +\end_inset + +. + Un +\begin_inset Formula $x\in{\cal L}$ +\end_inset + + es +\series bold +compacto +\series default + si para +\begin_inset Formula $S\subseteq{\cal L}$ +\end_inset + + no vacío con +\begin_inset Formula $x=\bigvee S$ +\end_inset + + existe +\begin_inset Formula $F\subseteq S$ +\end_inset + + finito con +\begin_inset Formula $x=\bigvee F$ +\end_inset + +, y es +\series bold +cocompacto +\series default + si para +\begin_inset Formula $S\subseteq{\cal L}$ +\end_inset + + no vacío con +\begin_inset Formula $x=\bigwedge S$ +\end_inset + + existe +\begin_inset Formula $F\subseteq S$ +\end_inset + + finito con +\begin_inset Formula $x=\bigwedge F$ +\end_inset + +. + +\begin_inset Formula $\{a_{n}\}_{n}$ +\end_inset + + tiene un maximal +\end_layout + +\begin_layout Standard +Un retículo completo +\begin_inset Formula $({\cal L},\leq)$ +\end_inset + + cumple la ACC si y sólo si todo +\begin_inset Formula $x\in{\cal L}$ +\end_inset + + es compacto. +\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 + +Sean +\begin_inset Formula $x\in{\cal L}$ +\end_inset + + y +\begin_inset Formula $S\subseteq{\cal L}$ +\end_inset + + no vacío con +\begin_inset Formula $x=\bigvee S$ +\end_inset + +, +\begin_inset Formula $\Sigma\coloneqq\{\bigvee F\}_{F\subseteq S\text{ finito no vacío}}$ +\end_inset + + tiene un elemento maximal +\begin_inset Formula $\bigvee F$ +\end_inset + + con +\begin_inset Formula $F\subseteq S$ +\end_inset + + finito, y queremos ver que +\begin_inset Formula $x=\bigvee F$ +\end_inset + +. + Sean +\begin_inset Formula $t\in S$ +\end_inset + + arbitrario y +\begin_inset Formula $F'\coloneqq F'\cup\{t\}$ +\end_inset + +, entonces +\begin_inset Formula $\bigvee F\leq\bigvee F'$ +\end_inset + +, pero como +\begin_inset Formula $\bigvee F'\in\Sigma$ +\end_inset + + y +\begin_inset Formula $\bigvee F$ +\end_inset + + es maximal, +\begin_inset Formula $\bigvee F=\bigvee F'$ +\end_inset + +, de modo que +\begin_inset Formula $t\leq\bigvee F$ +\end_inset + +. + Como +\begin_inset Formula $t\in S$ +\end_inset + + es arbitrario, +\begin_inset Formula $\bigvee S\leq\bigvee F$ +\end_inset + + y por tanto +\begin_inset Formula $x=\bigvee S=\bigvee F$ +\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 + +Sean +\begin_inset Formula $\{s_{n}\}_{n}\subseteq{\cal L}$ +\end_inset + + con cada +\begin_inset Formula $s_{n}\leq s_{n+1}$ +\end_inset + + y +\begin_inset Formula $x\coloneqq\bigvee_{n}s_{n}$ +\end_inset + +, como +\begin_inset Formula $x$ +\end_inset + + es compacto, +\begin_inset Formula $x=\bigvee_{k=1}^{r}s_{n_{k}}$ +\end_inset + + para ciertos +\begin_inset Formula $n_{1}<\dots<n_{r}$ +\end_inset + +, luego +\begin_inset Formula $x=s_{n_{r}}$ +\end_inset + + y, para +\begin_inset Formula $n\geq n_{r}$ +\end_inset + +, +\begin_inset Formula $s_{n_{r}}\leq s_{n}$ +\end_inset + + y por tanto, como +\begin_inset Formula $s_{n_{r}}$ +\end_inset + + es maximal, +\begin_inset Formula $s_{n_{r}}=s_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Análogamente, +\begin_inset Formula $({\cal L},\leq)$ +\end_inset + + cumple la DCC si y sólo si todo +\begin_inset Formula $x\in{\cal L}$ +\end_inset + + es cocompacto. +\end_layout + +\begin_layout Standard +Dado un anillo +\begin_inset Formula $A$ +\end_inset + +, +\begin_inset Formula $({\cal L}(A),\subseteq)$ +\end_inset + + es un retículo completo con supremo +\begin_inset Formula $\bigvee S=\sum S=\{a_{1}+\dots+a_{n}\}_{n\in\mathbb{N},\{a_{1},\dots,a_{n}\}\subseteq\bigcup S}$ +\end_inset + + e ínfimo +\begin_inset Formula $\inf S=\bigcap S$ +\end_inset + +. + +\begin_inset Formula $I\trianglelefteq A$ +\end_inset + + es compacto en +\begin_inset Formula $({\cal L}(A),\subseteq)$ +\end_inset + + si y sólo si es finitamente generado. +\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 + + +\begin_inset Formula $I=\bigvee_{x\in I}(x)$ +\end_inset + +, por lo que existen +\begin_inset Formula $x_{1},\dots,x_{n}\in I$ +\end_inset + + tales que +\begin_inset Formula $I=\bigvee_{i=1}^{n}(x_{i})=(\{x_{i}\}_{i=1}^{n})$ +\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 + +Sean +\begin_inset Formula $I\eqqcolon(x_{1},\dots,x_{n})$ +\end_inset + + y +\begin_inset Formula $S\subseteq{\cal L}(A)$ +\end_inset + + no vacío con +\begin_inset Formula $I=\bigvee S$ +\end_inset + +, para cada +\begin_inset Formula $i$ +\end_inset + +, como +\begin_inset Formula $x_{i}\in I$ +\end_inset + +, existen +\begin_inset Formula $a_{i1}\in J_{i1},\dots,a_{ik_{i}}\in J_{ik_{i}}$ +\end_inset + + con +\begin_inset Formula $x_{i}=a_{i1}+\dots+a_{ik_{i}}$ +\end_inset + +, de modo que todo elemento de +\begin_inset Formula $I$ +\end_inset + + se puede expresar como combinación lineal de los +\begin_inset Formula $a_{ij}$ +\end_inset + + y por tanto +\begin_inset Formula $I=\bigvee_{ij}J_{ij}$ +\end_inset + +. +\end_layout + +\begin_layout Section +Anillos noetherianos y artinianos +\end_layout + +\begin_layout Standard +Un anillo +\begin_inset Formula $A$ +\end_inset + + es +\series bold +noetheriano +\series default + si +\begin_inset Formula ${\cal L}(A)$ +\end_inset + + cumple la ACC y +\series bold +artiniano +\series default + si cumple la DCC. + Ejemplos: +\end_layout + +\begin_layout Enumerate +Si un anillo es noetheriano o artiniano, también lo es cualquier anillo + cociente suyo. +\end_layout + +\begin_deeper +\begin_layout Standard +El teorema de la correspondencia establece una biyección que conserva la + inclusión entre los ideales de +\begin_inset Formula $A/I$ +\end_inset + + y los de +\begin_inset Formula $A$ +\end_inset + + que contienen a +\begin_inset Formula $I$ +\end_inset + +, conservando la ACC o DCC. +\end_layout + +\end_deeper +\begin_layout Enumerate +Un anillo +\begin_inset Note Note +status open + +\begin_layout Plain Layout +TODO pg 28 (22) +\end_layout + +\end_inset + + +\end_layout + +\end_body +\end_document |