diff options
Diffstat (limited to 'ac/n2.lyx')
| -rw-r--r-- | ac/n2.lyx | 512 |
1 files changed, 505 insertions, 7 deletions
@@ -1095,6 +1095,69 @@ Demostración: \end_layout \begin_layout Standard +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +begin{exinfo} +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard +Si +\begin_inset Formula $A$ +\end_inset + + es noetheriano: +\end_layout + +\begin_layout Enumerate +Todo ideal suyo contiene una potencia de su radical. +\end_layout + +\begin_layout Enumerate +si +\begin_inset Formula $b\in A$ +\end_inset + + es cancelable y no unidad, +\begin_inset Formula $\bigcap_{n\in\mathbb{N}}(b^{n})$ +\end_inset + + puede ser no trivial, pero no contiene elementos cancelables. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A$ +\end_inset + + tiene una cantidad finita de primos minimales. +\end_layout + +\begin_layout Standard +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +end{exinfo} +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard \series bold Teorema de la base de Hilbert: @@ -1258,16 +1321,20 @@ Así, si \begin_layout Standard Dados -\begin_inset Formula $I,J\trianglelefteq A$ +\begin_inset Formula $I\trianglelefteq A$ +\end_inset + + y +\begin_inset Formula $S\subseteq A$ \end_inset , llamamos -\begin_inset Formula $(I:J)=\{a\in A:aJ\subseteq I\}$ +\begin_inset Formula $(I:S)=\{a\in A:aS\subseteq I\}$ \end_inset . -\begin_inset Formula $I\subseteq(I:J)$ +\begin_inset Formula $I\subseteq(I:S)$ \end_inset , pues para @@ -1275,13 +1342,182 @@ Dados \end_inset , -\begin_inset Formula $xJ\subseteq xA\subseteq I$ +\begin_inset Formula $xS\subseteq xA\subseteq I$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +begin{exinfo} +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard +Dados +\begin_inset Formula $I,J\trianglelefteq A$ +\end_inset + +, +\begin_inset Formula $X,Y\subseteq A$ +\end_inset + +, +\begin_inset Formula $\{K_{\lambda}\}_{\lambda\in\Lambda}\subseteq{\cal L}(A)$ +\end_inset + + y +\begin_inset Formula $\{Z_{\lambda}\}_{\lambda\in\Lambda}\subseteq{\cal P}(A)$ +\end_inset + +: +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $(I:X)$ +\end_inset + + es el mayor +\begin_inset Formula $L\trianglelefteq A$ +\end_inset + + con +\begin_inset Formula $LX\subseteq I$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $I\subseteq J\implies(I:X)\subseteq(J:X)$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $X\subseteq Y\implies(I:Y)\subseteq(I:X)$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $(I:X)=(I:(X))$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $(I:A)=I$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $(I:0)=A$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $(A:X)=A$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $((I:X):Y)=(I:X\cdot Y)$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(I:\bigcup_{\lambda}Z_{\lambda}\right)=\bigcap_{\lambda}(I:Z_{\lambda})$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(I:\sum_{\lambda}K_{\lambda}\right)=\bigcap_{\lambda}(I:K_{\lambda})$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(\bigcap_{\lambda}K_{\lambda}:J\right)=\bigcap_{\lambda}(K_{\lambda}:J)$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Llamamos +\series bold +anulador +\series default + de +\begin_inset Formula $X$ +\end_inset + + en +\begin_inset Formula $A$ +\end_inset + + a +\begin_inset Formula $\text{ann}_{A}(X)\coloneqq(0:X)=\{a\in A:aX=0\}$ +\end_inset + +, t entonces +\begin_inset Formula $\text{ann}(X)=\text{ann}((X))$ +\end_inset + +, +\begin_inset Formula $\text{ann}\left(\bigcup_{\lambda}Z_{\lambda}\right)=\bigcap_{\lambda}\text{ann}(Z_{\lambda})$ +\end_inset + + y +\begin_inset Formula $\text{ann}\left(\sum_{\lambda}K_{\lambda}\right)=\bigcap_{\lambda}\text{ann}(Z_{\lambda})$ \end_inset . \end_layout \begin_layout Standard +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +end{exinfo} +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard \series bold Teorema de Cohen: @@ -2165,12 +2401,275 @@ Dado \end_inset . +\end_layout + +\end_deeper +\begin_layout Enumerate +\begin_inset Formula $\text{Jac}(A)=\text{Nil}(A)$ +\end_inset + + es nilpotente. +\end_layout + +\begin_deeper +\begin_layout Standard +\begin_inset Formula $J\coloneqq\text{Jac}(A)=\bigcap\text{MaxSpec}(A)=\bigcap\text{Spec}(A)=\text{Nil}(A)$ +\end_inset + +. + Como +\begin_inset Formula $A$ +\end_inset + + es artiniano, la cadena +\begin_inset Formula $J\supseteq J^{2}\supseteq J^{3}\supseteq\dots$ +\end_inset + + se estabiliza en un cierto +\begin_inset Formula $I=J^{n}$ +\end_inset + +, y queremos ver que +\begin_inset Formula $I=0$ +\end_inset + +. + Si no lo fuera, +\begin_inset Formula $\Omega\coloneqq\{K\trianglelefteq A:KI\neq0\}\neq\emptyset$ +\end_inset + +, pues +\begin_inset Formula $A\in\Omega$ +\end_inset + +, con lo que tiene un minimal +\begin_inset Formula $K$ +\end_inset + +. + Como +\begin_inset Formula $KI\neq0$ +\end_inset + +, existe +\begin_inset Formula $x\in K$ +\end_inset + + con +\begin_inset Formula $xI=(x)I\neq0$ +\end_inset + +, luego +\begin_inset Formula $(x)\in\Omega$ +\end_inset + + y, como +\begin_inset Formula $(x)\subseteq K$ +\end_inset + +, +\begin_inset Formula $K=(x)$ +\end_inset + +. + Ahora bien, +\begin_inset Formula $I^{2}=J^{2n}=J^{n}=I$ +\end_inset + +, luego +\begin_inset Formula $0\neq xI=xI^{2}=(xI)I$ +\end_inset + +, con lo que +\begin_inset Formula $xI\in\Omega$ +\end_inset + + y está contenido en +\begin_inset Formula $(x)$ +\end_inset + + y por tanto +\begin_inset Formula $xI=(x)$ +\end_inset + +. + En particular +\begin_inset Formula $x\in xI$ +\end_inset + +, luego existe +\begin_inset Formula $y\in I$ +\end_inset + + con +\begin_inset Formula $x=xy$ +\end_inset + +, y por inducción +\begin_inset Formula $x=xy^{n}$ +\end_inset + + para todo +\begin_inset Formula $n\in\mathbb{N}$ +\end_inset + +, pues si +\begin_inset Formula $x=xy^{n-1}$ +\end_inset + + entonces +\begin_inset Formula $x=(xy)y^{n-1}=xy^{n}$ +\end_inset + +. + Ahora bien, +\begin_inset Formula $y\in I\subseteq J=\text{Nil}(A)$ +\end_inset + +, luego existe +\begin_inset Formula $n$ +\end_inset + + con +\begin_inset Formula $y^{n}=0$ +\end_inset + + y por tanto +\begin_inset Formula $x=xy^{n}=0$ +\end_inset + +, pero +\begin_inset Formula $xI\neq0\#$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Enumerate +0 es producto finito de ideales maximales. +\end_layout + +\begin_deeper +\begin_layout Standard +\begin_inset Formula $\text{MaxSpec}(A)$ +\end_inset + + es finito, digamos +\begin_inset Formula $\text{MaxSpec}(A)=\{M_{1},\dots,M_{r}\}$ +\end_inset + +, y entonces +\begin_inset Formula $\text{Jac}(A)=M_{1}\cap\dots\cap M_{r}=M_{1}\cdots M_{r}$ +\end_inset + + por ser los +\begin_inset Formula $M_{i}$ +\end_inset + + comaximales dos a dos, pero existe +\begin_inset Formula $n\in\mathbb{N}$ +\end_inset + + con +\begin_inset Formula $\text{Jac}(A)^{n}=0$ +\end_inset + +, luego +\begin_inset Formula $0=M_{1}^{n}\cdots M_{r}^{n}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Standard +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +begin{exinfo} +\end_layout + +\end_inset + +Dado un anillo artiniano +\begin_inset Formula $A$ +\end_inset + +, sean +\begin_inset Formula $\text{Spec}(A)=\{M_{1},\dots,M_{k}\}$ +\end_inset + + y +\begin_inset Formula $n\in\mathbb{N}$ +\end_inset + + con +\begin_inset Formula $\text{Jac}(A)^{n}=0$ +\end_inset + +, +\begin_inset Formula $A\cong\frac{A}{M_{1}^{n}}\times\dots\times\frac{A}{M_{k}^{n}}$ +\end_inset + +, con cada +\begin_inset Formula $\frac{A}{M_{i}^{k}}$ +\end_inset + + local y artiniano. +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +end{exinfo} +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard + +\series bold +Teorema de Akizuki: +\series default + Un anillo +\begin_inset Formula $A$ +\end_inset + + es artiniano si y sólo si es noetheriano y +\begin_inset Formula $\dim A=0$ +\end_inset + +. \begin_inset Note Note status open \begin_layout Plain Layout -TODO ejercicios 1.8 en adelante en tema 1, y luego la última página del tema - 2. +No escribo la demostración hasta tenerla completa. + Por ahora tenemos: +\end_layout + +\begin_layout Itemize +Una demostración de Saorín, que al no usar conceptos que todavía no hemos + visto es bastante enrevesada y no llega a probar que artiniano implica + noetheriano. +\end_layout + +\begin_layout Itemize +Una que hay al final del tema 4 en la página 61 (67/122) de los apuntes. +\end_layout + +\begin_layout Plain Layout +Seguramente me quede con la segunda. \end_layout \end_inset @@ -2178,6 +2677,5 @@ TODO ejercicios 1.8 en adelante en tema 1, y luego la última página del tema \end_layout -\end_deeper \end_body \end_document |