diff options
Diffstat (limited to 'ac')
| -rw-r--r-- | ac/n.lyx | 33 | ||||
| -rw-r--r-- | ac/n2.lyx | 1504 |
2 files changed, 1523 insertions, 14 deletions
@@ -7,10 +7,6 @@ \textclass book \begin_preamble \input{../defs} -\usepackage[x11names, svgnames, rgb]{xcolor} -%\usepackage[utf8]{inputenc} -\usepackage{tikz} -\usetikzlibrary{snakes,arrows,shapes} \end_preamble \use_default_options true \begin_modules @@ -157,6 +153,35 @@ Alberto del Valle Robles. Clases de Manuel Saorín Castaño. \end_layout +\begin_layout Standard +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +begin{exinfo} +\end_layout + +\end_inset + +Los párrafos marcados como este proceden de ejercicios del libro. +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +end{exinfo} +\end_layout + +\end_inset + + +\end_layout + \begin_layout Chapter Anillos conmutativos \end_layout @@ -519,6 +519,10 @@ Análogamente, es cocompacto. \end_layout +\begin_layout Section +Retículos de ideales +\end_layout + \begin_layout Standard Dado un anillo \begin_inset Formula $A$ @@ -632,10 +636,6 @@ Sean . \end_layout -\begin_layout Section -Anillos noetherianos y artinianos -\end_layout - \begin_layout Standard Un anillo \begin_inset Formula $A$ @@ -649,7 +649,8 @@ noetheriano \begin_inset Formula ${\cal L}(A)$ \end_inset - cumple la ACC y + cumple la ACC, si y sólo si todos sus ideales son finitamente generados, + y es \series bold artiniano \series default @@ -664,8 +665,8 @@ Si un anillo es noetheriano o artiniano, también lo es cualquier anillo \begin_deeper \begin_layout Standard -El teorema de la correspondencia establece una biyección que conserva la - inclusión entre los ideales de +El teorema de la correspondencia establece una biyección entre los ideales + de \begin_inset Formula $A/I$ \end_inset @@ -677,17 +678,1499 @@ El teorema de la correspondencia establece una biyección que conserva la \begin_inset Formula $I$ \end_inset -, conservando la ACC o DCC. + que conserva la inclusión y por tanto las condiciones ACC y DCC. +\end_layout + +\end_deeper +\begin_layout Enumerate +Los anillos con una cantidad finita de ideales son noetherianos y artinianos, + y en particular lo son los cuerpos. +\end_layout + +\begin_layout Enumerate +Los DIPs son noetherianos. +\end_layout + +\begin_deeper +\begin_layout Standard +Todos sus ideales son finitamente generados. +\end_layout + +\end_deeper +\begin_layout Enumerate +Un anillo con un elemento cancelable y no invertible no es artiniano. +\end_layout + +\begin_deeper +\begin_layout Standard +Sean +\begin_inset Formula $A$ +\end_inset + + el anillo y +\begin_inset Formula $x\in A$ +\end_inset + + el elemento, +\begin_inset Formula $(x)\supsetneq(x^{2})\supsetneq\dots\supsetneq(x^{k})\supsetneq\dots$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Enumerate +Los dominios que no son cuerpos no son artinianos, y en particular los DIPs + son noetherianos pero no artinianos. +\end_layout + +\begin_deeper +\begin_layout Standard +Por ser dominios todos los elementos no nulos son cancelables, y por no + ser cuerpo hay un elemento no nulo no invertible. +\end_layout + +\end_deeper +\begin_layout Enumerate +Los anillos de polinomios no son artinianos. +\end_layout + +\begin_deeper +\begin_layout Standard +\begin_inset Formula $X$ +\end_inset + + es cancelable y no invertible. +\end_layout + +\end_deeper +\begin_layout Enumerate +Dado un anillo +\begin_inset Formula $A$ +\end_inset + + no trivial, +\begin_inset Formula $A^{\mathbb{N}}$ +\end_inset + + y +\begin_inset Formula $A[X_{1},X_{2},\dots]$ +\end_inset + + no son noetherianos ni artinianos. +\end_layout + +\begin_deeper +\begin_layout Standard +Para +\begin_inset Formula $A^{\mathbb{N}}$ +\end_inset + +, los +\begin_inset Formula $I_{n}\coloneqq\{a:\forall k>n,a_{k}=0\}$ +\end_inset + + cumplen +\begin_inset Formula $I_{1}\subsetneq I_{2}\subsetneq\dots$ +\end_inset + + y los +\begin_inset Formula $J_{n}\coloneqq\{a:\forall k<n,a_{k}=0\}$ +\end_inset + + cumplen +\begin_inset Formula $J_{1}\supsetneq J_{2}\supsetneq\dots$ +\end_inset + +. + Para +\begin_inset Formula $A[X_{1},X_{2},\dots]$ +\end_inset + + esto ocurre con los +\begin_inset Formula $I_{n}\coloneqq(X_{1},\dots,X_{n})$ +\end_inset + + y los +\begin_inset Formula $J_{n}\coloneqq(X_{n},X_{n+1},\dots)$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Enumerate +Si un anillo es un +\begin_inset Formula $K$ +\end_inset + +-espacio vectorial de dimensión finita en que todos los ideales son subespacios + vectoriales entonces es noetheriano y artiniano. +\end_layout + +\begin_deeper +\begin_layout Standard +Si la dimensión es +\begin_inset Formula $d\in\mathbb{N}$ +\end_inset + +, una cadena estricta de ideales tiene a lo sumo +\begin_inset Formula $d+1$ +\end_inset + + elementos. +\end_layout + +\end_deeper +\begin_layout Enumerate +Para todo cuerpo +\begin_inset Formula $K$ +\end_inset + +, +\begin_inset Formula $A=\frac{K[X]}{(X^{n})}$ +\end_inset + + es noetheriano y artiniano. +\end_layout + +\begin_deeper +\begin_layout Standard +Es un +\begin_inset Formula $K$ +\end_inset + +-espacio vectorial con base +\begin_inset Formula $\overline{1},\overline{X},\dots,\overline{X}^{n-1}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Enumerate +Que un anillo sea noetheriano o artiniano no implica que sus subanillos + lo sean. +\end_layout + +\begin_deeper +\begin_layout Standard +\begin_inset Formula $K[X_{1},X_{2},\dots]$ +\end_inset + + no es noetheriano ni artiniano pero es subanillo de su cuerpo de fracciones. +\end_layout + +\end_deeper +\begin_layout Section +Anillos noetherianos +\end_layout + +\begin_layout Standard +Si +\begin_inset Formula $A$ +\end_inset + + es noetheriano, todo +\begin_inset Formula $X\subseteq A$ +\end_inset + + admite un +\begin_inset Formula $X_{0}\subseteq X$ +\end_inset + + finito con +\begin_inset Formula $(X_{0})=(X)$ +\end_inset + +, pues +\begin_inset Formula $(X)=(b_{1},\dots,b_{m})$ +\end_inset + + con cada +\begin_inset Formula $b_{i}=\sum_{j=1}^{k_{j}}a_{ij}x_{ij}$ +\end_inset + + para ciertos +\begin_inset Formula $a_{ij}\in A$ +\end_inset + + y +\begin_inset Formula $x_{ij}\in X$ +\end_inset + + y por tanto +\begin_inset Formula $(X)=(x_{11},\dots,x_{1j_{1}},\dots,x_{m1},\dots,x_{mk_{m}})$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Todo dominio noetheriano es un dominio de factorización. + +\series bold +Demostración: +\series default + Supongamos que +\begin_inset Formula $D$ +\end_inset + + es noetheriano pero no es un DF, de modo que el conjunto +\begin_inset Formula $S$ +\end_inset + + de elementos no nulos de +\begin_inset Formula $D$ +\end_inset + + que no se factorizan, es decir, que no están en +\begin_inset Formula $\{up_{1}\cdots p_{n}\}_{u\in D^{*}}^{p_{1},\dots,p_{n}\in D\text{ irreducibles}}$ +\end_inset + +, no es vacío. + Entonces +\begin_inset Formula $\Omega\coloneqq\{(a)\}_{a\in S}$ +\end_inset + + tiene un maximal +\begin_inset Formula $(a)$ +\end_inset + + con +\begin_inset Formula $a\in S$ +\end_inset + +, y como +\begin_inset Formula $a$ +\end_inset + + no es nulo, invertible ni irreducible, existen +\begin_inset Formula $b,c\in D$ +\end_inset + + no asociados de +\begin_inset Formula $a$ +\end_inset + + con +\begin_inset Formula $a=bc$ +\end_inset + +. + Entonces +\begin_inset Formula $(b),(c)\supsetneq(a)$ +\end_inset + + y por tanto no están en +\begin_inset Formula $\Omega$ +\end_inset + +, luego +\begin_inset Formula $b,c\notin S$ +\end_inset + +, y claramente +\begin_inset Formula $b,c\neq0$ +\end_inset + +, de modo que +\begin_inset Formula $b$ +\end_inset + + y +\begin_inset Formula $c$ +\end_inset + + se factorizan y por tanto también lo hace +\begin_inset Formula $a\#$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Si +\begin_inset Formula $A$ +\end_inset + + es noetheriano, todo ideal contiene un producto finito de ideales primos, + y en particular 0 es un producto finito de ideales primos. + +\series bold +Demostración: +\series default + De no ser así, el conjunto +\begin_inset Formula $\Omega$ +\end_inset + + de ideales que no contienen un producto finito de primos es no vacío y + tiene un maximal +\begin_inset Formula $I\in\Omega$ +\end_inset + +. + Como +\begin_inset Formula $I$ +\end_inset + + no es primo, existen +\begin_inset Formula $a,b\notin I$ +\end_inset + + con +\begin_inset Formula $ab\in I$ +\end_inset + +, luego +\begin_inset Formula $I+(a),I+(b)\supsetneq I$ +\end_inset + + e +\begin_inset Formula $I+(a),I+(b)\notin\Omega$ +\end_inset + +, con lo que existen +\begin_inset Formula $P_{i}\trianglelefteq_{\text{p}}A$ +\end_inset + + y +\begin_inset Formula $Q_{j}\trianglelefteq_{\text{p}}A$ +\end_inset + + con +\begin_inset Formula $P'\coloneqq\prod_{i}P_{i}\subseteq I+(a)$ +\end_inset + + y +\begin_inset Formula $Q'\coloneqq\prod_{i}Q_{i}\subseteq I+(b)$ +\end_inset + +. + Ahora bien, los elementos de +\begin_inset Formula $P'Q'$ +\end_inset + + son suma de elementos +\begin_inset Formula $pq$ +\end_inset + + con +\begin_inset Formula $p\in P'\subseteq I+(a)$ +\end_inset + + y +\begin_inset Formula $q\in Q'\subseteq I+(b)$ +\end_inset + +, de modo que existen +\begin_inset Formula $x,y\in I$ +\end_inset + + y +\begin_inset Formula $s,t\in A$ +\end_inset + + con +\begin_inset Formula $p=x+sa$ +\end_inset + + y +\begin_inset Formula $b=y+tb$ +\end_inset + +, y entonces +\begin_inset Formula $pq=(x+sa)(y+tb)=xy+(tb)x+(sa)y+(st)(ab)\in I$ +\end_inset + +, ya que +\begin_inset Formula $ab\in I$ +\end_inset + +, y por tanto +\begin_inset Formula $P'Q'\subseteq I\#$ +\end_inset + +. +\end_layout + +\begin_layout Standard + +\series bold +Teorema de la base de Hilbert: +\series default + Un anillo +\begin_inset Formula $A$ +\end_inset + + es noetheriano si y sólo si lo es +\begin_inset Formula $A[X]$ +\end_inset + +, si y sólo si lo es cualquiera de los +\begin_inset Formula $A[X_{1},\dots,X_{n}]$ +\end_inset + +. +\end_layout + +\begin_layout Description +\begin_inset Formula $1\implies2]$ +\end_inset + + Probamos el contrarrecíproco. + Sea +\begin_inset Formula $I\trianglelefteq A[X]$ +\end_inset + + no finitamente generado, por inducción y usando el buen orden de +\begin_inset Formula $\mathbb{N}$ +\end_inset + + definimos una secuencia +\begin_inset Formula $\{f_{i}\}_{i=1}^{\infty}\subseteq I$ +\end_inset + + donde cada +\begin_inset Formula $f_{i}$ +\end_inset + + es un elemento de +\begin_inset Formula $I\setminus(f_{1},\dots,f_{i-1})$ +\end_inset + + de grado mínimo. + Llamando +\begin_inset Formula $n_{i}$ +\end_inset + + al grado de +\begin_inset Formula $f_{i}$ +\end_inset + + y +\begin_inset Formula $b_{i}$ +\end_inset + + a su coeficiente principal, +\begin_inset Formula $(b_{1})\subseteq(b_{1},b_{2})\subseteq(b_{1},b_{2},b_{3})\subseteq\dots$ +\end_inset + + es una cadena ascendente de ideales de +\begin_inset Formula $A$ +\end_inset + +, y queda ver que los contenidos son estrictos. + En efecto, si fuera +\begin_inset Formula $b_{k}\in(b_{1},\dots,b_{k-1})$ +\end_inset + +, digamos +\begin_inset Formula $b_{k}=\sum_{i=1}^{k-1}a_{i}b_{i}$ +\end_inset + + para ciertos +\begin_inset Formula $a_{i}\in A$ +\end_inset + +, como +\begin_inset Formula $n_{1}\leq n_{2}\leq\dots$ +\end_inset + +, podemos tomar +\begin_inset Formula $f_{k}-\sum_{i=1}^{k-1}a_{i}f_{i}X^{n_{k}-n_{i}}$ +\end_inset + +, que está en +\begin_inset Formula $I\setminus(f_{1},\dots,f_{k-1})$ +\end_inset + + y tiene grado menor que +\begin_inset Formula $n_{k}\#$ +\end_inset + +. +\end_layout + +\begin_layout Description +\begin_inset Formula $2\implies1]$ +\end_inset + + +\begin_inset Formula $A\cong A[X]/(X)$ +\end_inset + + es noetheriano. +\end_layout + +\begin_layout Description +\begin_inset Formula $1\iff3]$ +\end_inset + + Por inducción en +\begin_inset Formula $[1\iff2]$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Así, si +\begin_inset Formula $K$ +\end_inset + + es un cuerpo, los anillos de la forma +\begin_inset Formula $K[X_{1},\dots,X_{n}]/I$ +\end_inset + + son noetherianos. + Si +\begin_inset Formula $A$ +\end_inset + + es un subanillo noetheriano de +\begin_inset Formula $B$ +\end_inset + + y +\begin_inset Formula $b_{1},\dots,b_{n}\in B$ +\end_inset + +, entonces +\begin_inset Formula $A[b_{1},\dots,b_{n}]$ +\end_inset + + es noetheriano, pues la evaluación +\begin_inset Formula $\epsilon_{b}:A[X_{1},\dots,X_{n}]\to A[b_{1},\dots,b_{n}]$ +\end_inset + + es un homomorfismo y por tanto +\begin_inset Formula $A[b_{1},\dots,b_{n}]\cong A[X_{1},\dots,X_{n}]/\ker\epsilon_{b}$ +\end_inset + +. + En particular los +\begin_inset Formula $\mathbb{Z}[\sqrt{m}]$ +\end_inset + + son noetherianos, aunque muchos no son DFUs. +\end_layout + +\begin_layout Standard +Dados +\begin_inset Formula $I,J\trianglelefteq A$ +\end_inset + +, llamamos +\begin_inset Formula $(I:J)=\{a\in A:aJ\subseteq I\}$ +\end_inset + +. + +\begin_inset Formula $I\subseteq(I:J)$ +\end_inset + +, pues para +\begin_inset Formula $x\in I$ +\end_inset + +, +\begin_inset Formula $xJ\subseteq xA\subseteq I$ +\end_inset + +. +\end_layout + +\begin_layout Standard + +\series bold +Teorema de Cohen: +\series default + Un anillo es noetheriano si y sólo si todos sus ideales primos son finitamente + generados. +\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 + +Obvio. +\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 + +Probamos el contrarrecíproco. + Sean +\begin_inset Formula $A$ +\end_inset + + no noetheriano y +\begin_inset Formula $\Omega$ +\end_inset + + el conjunto de ideales de +\begin_inset Formula $A$ +\end_inset + + no finitamente generados, la unión de una cadena de elementos de +\begin_inset Formula $\Omega$ +\end_inset + + está en +\begin_inset Formula $\Omega$ +\end_inset + +. + En efecto, dada una cadena +\begin_inset Formula $\{I_{\lambda}\}_{\lambda\in\Lambda}\subseteq\Omega$ +\end_inset + +, si fuera +\begin_inset Formula $\bigcup_{\lambda}I_{\lambda}\eqqcolon(a_{1},\dots,a_{n})$ +\end_inset + +, por ser +\begin_inset Formula $\{I_{\lambda}\}_{\lambda}$ +\end_inset + + una cadena existe un +\begin_inset Formula $\mu\in\Lambda$ +\end_inset + + con +\begin_inset Formula $a_{1},\dots,a_{n}\in I_{\mu}$ +\end_inset + +, luego +\begin_inset Formula $(a_{1},\dots,a_{n})\subseteq I_{\mu}\subseteq(a_{1},\dots,a_{n})$ +\end_inset + + e +\begin_inset Formula $I_{\mu}$ +\end_inset + + es finitamente generado. +\begin_inset Formula $\#$ +\end_inset + + Esto nos permite aplicar el lema de Zorn y obtener un elemento maximal + +\begin_inset Formula $P$ +\end_inset + + de +\begin_inset Formula $\Omega$ +\end_inset + +, y queda ver que +\begin_inset Formula $P$ +\end_inset + + es primo. + Supongamos que no lo fuera. + +\begin_inset Formula $P\neq A$ +\end_inset + +, pues +\begin_inset Formula $A=(1)$ +\end_inset + + es finitamente generado, luego existen +\begin_inset Formula $a,b\notin P$ +\end_inset + + con +\begin_inset Formula $ab\in P$ +\end_inset + +. + Entonces +\begin_inset Formula $P\subsetneq P+(a)$ +\end_inset + +, luego +\begin_inset Formula $P+(a)\notin\Omega$ +\end_inset + + y existen +\begin_inset Formula $p_{1},\dots,p_{n}\in P$ +\end_inset + + y +\begin_inset Formula $r_{1},\dots,r_{n}\in A$ +\end_inset + + con +\begin_inset Formula $P+(a)=(p_{1}+r_{1}a,\dots,p_{n}+r_{n}a)=(p_{1},\dots,p_{n},a)$ +\end_inset + +. + Para la segunda igualdad: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Argument item:1 +status open + +\begin_layout Plain Layout +\begin_inset Formula $\subseteq]$ +\end_inset + + +\end_layout + +\end_inset + +Cada +\begin_inset Formula $p_{i}+r_{i}a\in(p_{i},a)\subseteq(p_{1},\dots,p_{n},a)$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Argument item:1 +status open + +\begin_layout Plain Layout +\begin_inset Formula $\supseteq]$ +\end_inset + + +\end_layout + +\end_inset + +Claramente +\begin_inset Formula $a\in P+(a)$ +\end_inset + +, y entonces cada +\begin_inset Formula $p_{i}=(p_{i}+r_{i}a)-r_{i}a\in P+(a)$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Formula $(P:(a))=\{c\in A:c(a)=(ca)\subseteq P\}=\{c\in A:ac\in P\}$ +\end_inset + +, y entonces +\begin_inset Formula $P\subsetneq(P:(a))$ +\end_inset + + ya que +\begin_inset Formula $ab\in P$ +\end_inset + + pero +\begin_inset Formula $b\notin P$ +\end_inset + +. + Por tanto +\begin_inset Formula $(P:(a))=(q_{1},\dots,q_{m})$ +\end_inset + + para ciertos +\begin_inset Formula $q_{1},\dots,q_{m}$ +\end_inset + + con cada +\begin_inset Formula $q_{j}a\in P$ +\end_inset + +. + Entonces +\begin_inset Formula $P=(p_{1},\dots,p_{n},q_{1}a,\dots,q_{m}a)$ +\end_inset + +: +\end_layout + +\begin_layout Itemize +\begin_inset Argument item:1 +status open + +\begin_layout Plain Layout +\begin_inset Formula $\subseteq]$ +\end_inset + + +\end_layout + +\end_inset + +Para +\begin_inset Formula $x\in P$ +\end_inset + +, +\begin_inset Formula $x\in P+(a)=(p_{1},\dots,p_{n},a)$ +\end_inset + +, luego +\begin_inset Formula $x=s_{1}p_{1}+\dots+s_{n}p_{n}+ra$ +\end_inset + + para ciertos +\begin_inset Formula $s_{j},r\in A$ +\end_inset + +, y como +\begin_inset Formula $x\in P$ +\end_inset + + y los +\begin_inset Formula $p_{i}\in P$ +\end_inset + +, +\begin_inset Formula $ra\in P$ +\end_inset + + y por tanto +\begin_inset Formula $ra\in(P:(a))=(q_{1},\dots,q_{m})$ +\end_inset + +, con lo que +\begin_inset Formula $r=t_{1}q_{1}+\dots+t_{m}q_{m}$ +\end_inset + + para ciertos +\begin_inset Formula $t_{j}\in A$ +\end_inset + + y por tanto +\begin_inset Formula $x=s_{1}p_{1}+\dots+s_{n}p_{n}+t_{1}q_{1}a+\dots+t_{m}q_{m}a\in(p_{1},\dots,p_{n},q_{1}a,\dots,q_{m}a)$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Argument item:1 +status open + +\begin_layout Plain Layout +\begin_inset Formula $\supseteq]$ +\end_inset + + +\end_layout + +\end_inset + +Cada +\begin_inset Formula $p_{i}$ +\end_inset + + y cada +\begin_inset Formula $q_{j}a$ +\end_inset + + está en +\begin_inset Formula $P$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Con esto +\begin_inset Formula $P$ +\end_inset + + es finitamente generado. +\begin_inset Formula $\#$ +\end_inset + + \end_layout \end_deeper +\begin_layout Standard +Como +\series bold +teorema +\series default +, un anillo +\begin_inset Formula $A$ +\end_inset + + es noetheriano si y sólo si lo es +\begin_inset Formula $A\llbracket X\rrbracket$ +\end_inset + +, si y sólo si lo es cualquiera de los +\begin_inset Formula $A\llbracket X_{1},\dots,X_{n}\rrbracket$ +\end_inset + +. +\end_layout + +\begin_layout Description +\begin_inset Formula $1\implies2]$ +\end_inset + + Sea +\begin_inset Formula $P\trianglelefteq_{\text{p}}A\llbracket X\rrbracket$ +\end_inset + + y queremos ver que +\begin_inset Formula $P$ +\end_inset + + es finitamente generado. + Como la evaluación en 0 +\begin_inset Formula $\epsilon:A\llbracket X\rrbracket\to A$ +\end_inset + + es suprayectiva, +\begin_inset Formula $\varepsilon(P)$ +\end_inset + + es un ideal de +\begin_inset Formula $A$ +\end_inset + + y por tanto es finitamente generado, digamos +\begin_inset Formula $\varepsilon(P)=(b_{1},\dots,b_{n})$ +\end_inset + + con cada +\begin_inset Formula $b_{i}=\varepsilon(f_{i})$ +\end_inset + + para cierto +\begin_inset Formula $f_{i}\in P$ +\end_inset + +. + Si +\begin_inset Formula $X\in P$ +\end_inset + + entonces +\begin_inset Formula $P=(b_{1},\dots,b_{n},X)$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Argument item:1 +status open + +\begin_layout Plain Layout +\begin_inset Formula $\subseteq]$ +\end_inset + + +\end_layout + +\end_inset + +Para +\begin_inset Formula $f\in P$ +\end_inset + +, +\begin_inset Formula $f=gX+b$ +\end_inset + + para ciertos +\begin_inset Formula $g\in A\llbracket X\rrbracket$ +\end_inset + + y +\begin_inset Formula $b\in\varepsilon(P)=(b_{1},\dots,b_{n})$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Argument item:1 +status open + +\begin_layout Plain Layout +\begin_inset Formula $\supseteq]$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Formula $X\in P$ +\end_inset + + y cada +\begin_inset Formula $b_{i}=f_{i}-g_{i}X$ +\end_inset + + para cierto +\begin_inset Formula $g_{i}\in A\llbracket X\rrbracket$ +\end_inset + +, luego +\begin_inset Formula $b_{i}\in P$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Si +\begin_inset Formula $X\notin P$ +\end_inset + + entonces +\begin_inset Formula $P=(f_{1},\dots,f_{n})$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Argument item:1 +status open + +\begin_layout Plain Layout +\begin_inset Formula $\subseteq]$ +\end_inset + + +\end_layout + +\end_inset + +Sea +\begin_inset Formula $g\in P$ +\end_inset + +, queremos ver que existen +\begin_inset Formula $(a_{ij})_{1\leq i\leq n}^{j\in\mathbb{N}}\in A\llbracket X\rrbracket$ +\end_inset + + tales que, para +\begin_inset Formula $j\in\mathbb{N}$ +\end_inset + +, existe +\begin_inset Formula $g'\in P$ +\end_inset + + con +\begin_inset Formula $f=a_{1j}f_{1}+\dots+a_{nj}f_{n}+g'X^{j}$ +\end_inset + +, y +\begin_inset Formula $a_{ij}$ +\end_inset + + y +\begin_inset Formula $a_{kj}$ +\end_inset + + tienen los mismos coeficientes hasta el de grado +\begin_inset Formula $\min\{i,k\}-1$ +\end_inset + +. + Para +\begin_inset Formula $j=0$ +\end_inset + +, tomamos los +\begin_inset Formula $a_{i0}=0$ +\end_inset + + y +\begin_inset Formula $g'=g$ +\end_inset + +. + Para +\begin_inset Formula $j>0$ +\end_inset + +, probado esto para +\begin_inset Formula $j-1$ +\end_inset + +, +\begin_inset Formula $g=a_{1,k-1}f_{1}+\dots+a_{n,k-1}f_{n}+g'X^{j-1}$ +\end_inset + + con +\begin_inset Formula $g'\in P$ +\end_inset + +, pero como +\begin_inset Formula $\epsilon(g')\in\epsilon(P)$ +\end_inset + +, existen +\begin_inset Formula $x_{i}$ +\end_inset + + con +\begin_inset Formula $\epsilon(g')\eqqcolon\sum_{i=1}^{n}x_{i}b_{i}$ +\end_inset + +, con lo que +\begin_inset Formula $h_{0}\coloneqq g'-\sum_{i=1}^{n}x_{i}f_{i}$ +\end_inset + + está en +\begin_inset Formula $P$ +\end_inset + + y tiene término independiente 0, luego +\begin_inset Formula $h_{0}=hX$ +\end_inset + + con +\begin_inset Formula $h\in P$ +\end_inset + + ya que +\begin_inset Formula $X\notin P$ +\end_inset + + y +\begin_inset Formula $P$ +\end_inset + + es primo, y como +\begin_inset Formula $g'=hX+\sum_{i=1}^{n}x_{i}f_{i}$ +\end_inset + +, +\begin_inset Formula $g=(a_{1,j-1}+x_{1}X^{j-1})f_{1}+\dots+(a_{n,j-1}+x_{n}X^{j-1})f_{n}+hX^{j}$ +\end_inset + +, y hacemos los +\begin_inset Formula $a_{ij}\coloneqq a_{i,j-1}+x_{i}X^{j-1}$ +\end_inset + +. + Con esto, para +\begin_inset Formula $i\in\{1,\dots,n\}$ +\end_inset + + definimos +\begin_inset Formula $c_{i}\in A\llbracket X\rrbracket$ +\end_inset + + de modo que +\begin_inset Formula $c_{ik}\coloneqq a_{i,k+1,k}$ +\end_inset + +, y entonces +\begin_inset Formula $g=\sum_{i=1}^{n}c_{i}f_{i}$ +\end_inset + +. + En efecto, para el coeficiente de grado +\begin_inset Formula $j$ +\end_inset + +, +\begin_inset Formula +\begin{multline*} +\left(\sum_{i=1}^{n}c_{i}f_{i}\right)_{j}=\sum_{i=1}^{n}\sum_{k=1}^{j}c_{ik}f_{i,j-k}=\sum_{i=1}^{n}\sum_{k=1}^{j}a_{i,k+1,k}f_{i,j-k}=\\ +=\sum_{i=1}^{n}\sum_{k=1}^{j}a_{i,j+1,k}f_{i,j-k}=\sum_{i=1}^{n}(a_{i,j+1}f_{i})_{j}=g_{j}. +\end{multline*} + +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Argument item:1 +status open + +\begin_layout Plain Layout +\begin_inset Formula $\supseteq]$ +\end_inset + + +\end_layout + +\end_inset + +Todo +\begin_inset Formula $f_{i}\in P$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Description +\begin_inset Formula $2\implies1]$ +\end_inset + + +\begin_inset Formula $A\cong A\llbracket X\rrbracket/(X)$ +\end_inset + +, que es noetheriano. +\end_layout + +\begin_layout Description +\begin_inset Formula $1\iff3]$ +\end_inset + + Por inducción en +\begin_inset Formula $[1\iff2]$ +\end_inset + +. +\end_layout + +\begin_layout Section +Anillos artinianos +\end_layout + +\begin_layout Standard +La +\series bold +dimensión de Krull +\series default + de un anillo +\begin_inset Formula $A$ +\end_inset + + es +\begin_inset Formula +\[ +\dim A\coloneqq\text{Kdim}A\coloneqq\sup\{n\in\mathbb{N}:\exists P_{0},\dots,P_{n}\trianglelefteq_{\text{p}}A:P_{0}\subsetneq\dots\subsetneq P_{n}\}\in\mathbb{N}\cup\{\infty\}, +\] + +\end_inset + +y se tiene +\begin_inset Formula $\text{Spec}A=\text{MaxSpec}A\iff\dim A=0$ +\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 existen +\begin_inset Formula $P,Q\trianglelefteq_{\text{p}}A$ +\end_inset + + con +\begin_inset Formula $P\subsetneq Q$ +\end_inset + +, +\begin_inset Formula $P\in\text{Spec}A\setminus\text{MaxSpec}A$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Argument item:1 +status open + +\begin_layout Plain Layout +\begin_inset Formula $\impliedby]$ +\end_inset + + +\end_layout + +\end_inset + +Si existe +\begin_inset Formula $P\in\text{Spec}A\setminus\text{MaxSpec}A$ +\end_inset + +, sabemos que +\begin_inset Formula $P$ +\end_inset + + está contenido (estrictamente) en un maximal +\begin_inset Formula $Q$ +\end_inset + +, que debe ser primo, luego +\begin_inset Formula $P\subsetneq Q$ +\end_inset + + y +\begin_inset Formula $\dim A\geq1$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Dado un anillo artiniano +\begin_inset Formula $A$ +\end_inset + +: +\end_layout + \begin_layout Enumerate -Un anillo +\begin_inset Formula $\dim A=0$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Standard +Dado +\begin_inset Formula $P\trianglelefteq_{\text{p}}A$ +\end_inset + +, +\begin_inset Formula $A/P$ +\end_inset + + es un dominio por ser +\begin_inset Formula $P$ +\end_inset + + primo y es artiniano por serlo +\begin_inset Formula $A$ +\end_inset + +, pero los dominios no cuerpos no son artinianos, luego +\begin_inset Formula $A/P$ +\end_inset + + es un cuerpo y por tanto +\begin_inset Formula $P$ +\end_inset + + es maximal y +\begin_inset Formula $\text{Spec}A=\text{MaxSpec}A$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Enumerate +\begin_inset Formula $\text{Spec}A=\text{MaxSpec}A$ +\end_inset + + es finito. +\end_layout + +\begin_deeper +\begin_layout Standard +\begin_inset Formula $\Omega\coloneqq\{\bigcap{\cal M}\}_{{\cal M}\subseteq\text{MaxSpec}A}\neq\emptyset$ +\end_inset + +, pues +\begin_inset Formula $\emptyset\neq\text{MaxSpec}A\subseteq\Omega$ +\end_inset + +, con lo que tiene un minimal +\begin_inset Formula $I\coloneqq M_{1}\cap\dots\cap M_{k}\in\Omega$ +\end_inset + + con los +\begin_inset Formula $M_{i}\trianglelefteq_{\text{m}}A$ +\end_inset + +. + Para +\begin_inset Formula $M\trianglelefteq_{\text{m}}A$ +\end_inset + +, +\begin_inset Formula $M\cap I=M\cap M_{1}\cap\dots\cap M_{k}\in\Omega$ +\end_inset + + y por tanto +\begin_inset Formula $M\cap I\subseteq I$ +\end_inset + +, con lo que +\begin_inset Formula $I\subseteq M$ +\end_inset + +, pero +\begin_inset Formula $M_{1}\cdots M_{k}\subseteq M_{1}\cap\dots\cap M_{k}=I\subseteq M$ +\end_inset + + y, como +\begin_inset Formula $M$ +\end_inset + + es primo, algún +\begin_inset Formula $M_{i}\subseteq M$ +\end_inset + +, de modo que +\begin_inset Formula $M_{i}=M$ +\end_inset + + por ser +\begin_inset Formula $M$ +\end_inset + + maximal y +\begin_inset Formula $\text{MaxSpec}(A)=\{M_{1},\dots,M_{k}\}$ +\end_inset + +. \begin_inset Note Note status open \begin_layout Plain Layout -TODO pg 28 (22) +TODO ejercicios 1.8 en adelante en tema 1, y luego la última página del tema + 2. \end_layout \end_inset @@ -695,5 +2178,6 @@ TODO pg 28 (22) \end_layout +\end_deeper \end_body \end_document |