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zz8=j@h=(;wSN62PO@y{th!W@`5;oSk;*!KS2jNW&!AJ6A>b$y^S`r@u0Jd|9dd-?g_D1?|QMsuiE) z3Y=rAZmFu6HY+)Kxqc5rM7|!^FTYOa)n_)Inqwy$+3Y9IIsIuwiJo{wR^QgA_0t%+ z&uly(10rG=MZCVeTmNgfj^gv{*Yjsnapy<7t@>Kj++Ruxl73W->b~j$6=xLfTQCz z!=VQ|e)sKu-_iP-vFJh7!t?((AFa~!tml-nyRN0`_$zqrFI9_aORqMt9=tYO+om$fV)@k#F86*otrqOub9u?t6)twWR$iUq@~&>#)g3OI`?5?GFP?N- ztEO7D(`R*$%l`L9LVq9bd45gw{CRVm{d=D7y!(1*@%_*4mp2TShlAXU_C(1w?L{jZfX79A`MpJG)dHz6VK3aXWM3?Qsdjqd{?E96^M!kuby0Ci aVo^y&QED2Ov9XzvF^2`0s;aBM8y5g_6WgNz diff --git a/TFG_Sixto_José_Martínez_Gan.tex b/TFG_Sixto_José_Martínez_Gan.tex index 1c65303..a9ae767 100644 --- a/TFG_Sixto_José_Martínez_Gan.tex +++ b/TFG_Sixto_José_Martínez_Gan.tex @@ -185,6 +185,7 @@ urlcolor=black]{hyperref} \newcommand{\cuerpofracciones}[1]{\ensuremath{\mathrm{Q}\!\left( #1 \right)}} +\newcommand{\Groebner}{Gr\"obner} \begin{document} \frontmatter % Parte primera del texto @@ -740,7 +741,7 @@ urlcolor=black]{hyperref} Este pequeño resultado nos da una forma bastante rápida de determinar si un monomio pertenece o no a un \submodulo{\Kx} monomial de $\Kx^r$. Démonos cuenta de que esto es justo lo que hacemos cuando intentamos reducir el término líder de $f$ en el algoritmo de la división de \ref{Algoritmo de la División Compuesta}. Con todo esto, ya damos la respuesta final a la pregunta que abre el capítulo. - \begin{teorema} + \begin{teorema} \label{Teorema Buen Funcionamiento División Groebner} Sea $N$ un \submodulo{\Kx} de $\Kx^r$ y sea $G = \left\lbrace g_1, g_2, \dots, g_k \right\rbrace$ una base de Gröbner de $N$. Un elemento $f$ de $\Kx^r$ pertenece a $N$ si y solo si el resto de $f$ dividido por $G$ según el algoritmo \ref{Algoritmo de la División Compuesta} es $0$. \end{teorema} \begin{proof} @@ -778,13 +779,8 @@ urlcolor=black]{hyperref} \end{algorithmic} Al final del algoritmo, $\left\lbrace d_1, \dots, d_t \right\rbrace$ es una base de Gröbner de $N$. \end{definicion} - El funcionamiento de este algoritmo se sustenta en dos pilares. El primero es que, como cada vez que añadimos un nuevo elemento este es un resto, tenemos que los \submodulo{\Kx} \engendrado{\LM{d_1}, \dots, \LM{d_t}} forman una cadena ascendente la cual, por ser $\Kx^r$ noetheriano, debe finalizar en algún momento. Por esto mismo, el algoritmo termina. El segundo es el criterio que garantiza que nuestro conjunto generador sea una base de Gröbner. - \begin{teorema}[criterio de Buchberger] \label{Criterio de Buchberger} - Sea $G$ un subconjunto de $\Kx^r$ equipado con un orden monomial de módulos $>$. $G$ es una base de Gröbner de \engendrado{G} si y solo si para toda pareja de elementos $f$ y $g$ en $G$ se tiene que el resto de $\spoly{f}{g} = 0$ entre $G$ es $0$. - \end{teorema} - \begin{proof} - contenidos... - \end{proof} + El funcionamiento de este algoritmo se sustenta en dos pilares. El primero es que, como cada vez que añadimos un nuevo elemento este es un resto, tenemos que los \submodulo{\Kx} \engendrado{\LM{d_1}, \dots, \LM{d_t}} forman una cadena ascendente la cual, por ser $\Kx^r$ noetheriano, debe finalizar en algún momento. Por esto mismo, el algoritmo termina. El segundo es el teorema \ref{Criterio de Buchberger}, que nos garantiza que cuando este algoritmo pare, el conjunto que tendremos será una base de \Groebner. + \section{Sicigias} @@ -793,7 +789,14 @@ urlcolor=black]{hyperref} \begin{definicion} Sean $A$ un anillo, $M$ un \modulo{A} y $f_1, f_2, \dots, f_k$ elementos de $M$. Una sicigia de los elementos $f_1, f_2, \dots, f_k$ es una \nolbreaks{$k$-tupla} $\left( g_1, g_2, \dots, g_k \right)$ de elementos de $A$ de forma que $f_1g_1 + f_2g_2 + \dots + f_kg_k = 0$. \end{definicion} - Nótese que el conjunto de sicigias de $f_1, f_2, \dots, f_k$ forma un \submodulo{A} de $A^k$. + Nótese que el conjunto de sicigias de $f_1, f_2, \dots, f_k$ forma un \submodulo{A} de $A^k$ puesto que es el kernel del homomorfismo de \modulo{A}s + \begin{equation} \label{Definición Funcón Sicigias} + \begin{aligned} + \sigma : A^k &\longrightarrow M \\ + a_1\euno + a_2\edos + \dots + a_k\vectorbasecanonica{k} &\longmapsto a_1f_1 + a_2f_2 + \dots + a_kf_k + \end{aligned} + \end{equation} + \begin{definicion} Sean $A$ un anillo, $M$ un \modulo{A} y $f_1, f_2, \dots, f_k$ elementos de $M$. Denotaremos por $\syz{f_1, f_2, \dots, f_k}$ al \submodulo{A} de $A^k$ formado por todas las sicigias de $f_1, f_2, \dots, f_k$. \end{definicion} @@ -806,6 +809,15 @@ urlcolor=black]{hyperref} &= \definicionconjunto{\left( a_1, a_2, \dots, a_n \right) \in \Kx^n}{\left( a_1, a_2, \dots, a_n, p \right) \in \syz{\partial_1f, \partial_2f, \dots, \partial_nf, f}, p \in \Kx}. \end{align*} + Teniendo ya e + \begin{teorema}[criterio de Buchberger] \label{Criterio de Buchberger} + Sea $G$ un subconjunto de $\Kx^r$, este último equipado con un orden monomial de módulos $>$. $G$ es una base de Gröbner de \engendrado{G} si y solo si para toda pareja de elementos $f$ y $g$ en $G$ se tiene que el resto de $\spoly{f}{g} = 0$ entre $G$ es $0$. + \end{teorema} + \begin{proof} + Si $G$ es una base de \Groebner, entonces, como $\spoly{f}{g} \in \engendrado{G}$, por el teorema \ref{Teorema Buen Funcionamiento División Groebner} tenemos que el resto de su división entre los elementos de $G$ será $0$. + + Como $\Kx^n$ es noetheriano, podemos asumir desde el principio que $G = \conjunto{g_1, g_2, \dots, g_k}$ + \end{proof} Para poder calcular las sicigias necesitaremos dos resultados. El primero nos permite restringir una base de Gröbner a solamente unas coordenadas. \begin{lema} contenidos... @@ -988,22 +1000,70 @@ urlcolor=black]{hyperref} \todo[inline]{Aquí vendría la parte en la que demostramos que $\det \Phi = f$.} Ahora, por un lado, definamos $N := \engendrado{\delta_1, \delta_2, \dots, \delta_n}$ como un \submodulo{\Kx} de \derivaciones{f} donde los $\delta_i$ forman una base y consideremos la inclusión $N \xrightarrow{\iota} \derivaciones{f}$. Volvamos a recordar la aplicación $\Phi$ de \ref{Secuencia Exacta Definición Módulo Derivaciones}. Para aclararnos algo más, escribimos el siguiente esquema de los homomorfismos ---¡ojo!, esto no es una secuencia exacta---: - \begin{equation*} + \begin{equation}\label{Secuencia Demostración Saito} N = \engendrado{\delta_1, \delta_2, \dots, \delta_n} \xrightarrow{\iota} \derivaciones{f} \xrightarrow{\Phi} \derivaciones{\Kx}. - \end{equation*} + \end{equation} Como los $\delta_i \in \derivaciones{\Kx} = \engendrado{\partial_1, \partial_2, \dots, \partial_n}$, tenemos que existe alguna matriz $A \in \mathrm{Mat}\!\left( n \times n, \Kx \right)$ de forma que - \begin{equation*} + \begin{equation} \label{Relación Matricial Derivadas} \vectorhorizontalcuatro{\delta_1}{\delta_2}{\dots}{\delta_n} = \vectorhorizontalcuatro{\partial_1}{\partial_2}{\dots}{\partial_n} A - \end{equation*} + \end{equation} Por lo tanto, podemos pensar en que la aplicación $\Phi \circ \iota$ envía a cada $\delta_i$ a $\vectorhorizontalcuatro{\partial_1}{\partial_2}{\dots}{\partial_n} A_i$ donde $A_i$ es la \nolbreaks{$i$-ésima} columna de $A$. Es decir, que las columnas de $A$ se entienden como las coordenadas de los $\delta_i$ en función de la base \conjunto{\partial_1, \partial_2, \dots, \partial_n}. Pensemos ahora en la aplicación \begin{align*} N &\xrightarrow{\bigwedge^n \left( \Phi \circ \iota \right)} \derivaciones{\Kx} \\ \delta_1 \wedge \delta_2 \wedge \dots \wedge \delta_n &\xmapsto{\phantom{\bigwedge^n \left( \Phi \circ \iota \right)}} \partial_1 \wedge \partial_2 \wedge \dots \wedge \partial_n \det A \end{align*} Puede verse, gracias al mismo razonamiento de \ref{Aparición Determinante Función Exterior}, como $\det \left( \Phi \circ \iota \right) = \det\Phi \det\iota = f \det\iota$. Vemos pues que si $N = \derivaciones{f}$, es decir, si \derivaciones{f} es libre, entonces $N = \derivaciones{f}$ y por lo tanto se cumple que $\det\iota$ es una unidad de \Kx. Para ver exactamente esto último, recordemos la regla de Cramer: como $\iota$ es un isomorfismo, necesariamente tiene inversa y eso quiere decir que $\det\iota$ es inversible. Por otro lado, observemos que $\det\left( \Phi \circ \iota \right) = \det A$. Con esto concluimos que si \derivaciones{f} es libre, entonces $\det\left( \delta_1, \delta_2, \dots, \delta_n \right) = \det A = \det \Phi \det \iota = f \det\iota$ donde $\det\iota$ es una unidad de \Kx, como se pedía en el enunciado del teorema. + \begin{wrapfigure}[7]{r}{0.3\textwidth} + \begin{equation*} + \begin{tikzcd} + \left( \dual{M} \right)_\mathrm{Q} \times M_\mathrm{Q} \arrow[r, "\mathrm{eval}"] & \cuerpofracciones{\Kx} & \\ + \phantom{(}\dual{M}\phantom{)_\mathrm{Q}} \times M\hspace{5pt} \arrow[u, shift left=6mm] \arrow[u, shift right=6mm] \arrow[r, "\mathrm{eval}"] & \Kx \arrow[u] & \\ + 0 \hspace{17pt} \phantom{. \times } 0\arrow[u, shift left=6mm] \arrow[u, shift right=6mm] & 0 \arrow[u] & + \end{tikzcd} + \end{equation*} + \end{wrapfigure} - Ahora asumamos que existen $\delta_1, \delta_2, \dots, \delta_n$ de forma que $\det\left( \delta_1, \delta_2, \dots, \delta_n \right) = fu$ con $u$ una unidad de \Kx. - \todo[inline]{Terminar la otra implicación de Saito.} - + Antes de embarcarnos en la otra implicación, necesitaremos un resultado auxiliar. + \begin{lema} + Si $M$ es un \modulo{\Kx} finitamente generado, entonces \dual{M} es reflexivo. + \end{lema} + \begin{proof} + Para cualquier \modulo{A} $B$, denotaremos $B_\mathrm{Q} := B \otimes \cuerpofracciones{A}$. Consideremos el siguiente diagrama de homomorfismos de \modulo{\Kx}: + + + Las inyecciones que aquí mostramos son las naturales: en el caso de las de la izquierda, mandan los elementos $m$ bien de $M$ o \dual{M} a los correspondientes $m \otimes 1$; y en el de la derecha, es directamente la inclusión. Estas son inyecciones si y solo si $M$ es libre de torsión. on el mapa $\mathrm{eval}$ cometemos un abuso de notación y lo reutilizamos en ambos casos aunque no sería igual. Esta aplicación envía un par $\left( f, m \right)$ a $f\!\left( m \right)$ en ambos casos, es decir, evalúa la aplicación en el punto que se le dé. + \todo[inline]{Por terminar.} + \end{proof} + Ahora asumamos que existen $\delta_1, \delta_2, \dots, \delta_n$ de forma que $\det\left( \delta_1, \delta_2, \dots, \delta_n \right) = uf$ con $u$ una unidad de \Kx{} y volvamos al esquema de \ref{Secuencia Demostración Saito}. A partir de este, sabemos que tenemos + \begin{equation*} + \begin{array}{rcccl} + \bigwedge^n N & \longrightarrow & \bigwedge^n \derivaciones{f} & \subseteq & \bigwedge^n \derivaciones{\Kx} \\ + \engendrado{\delta_1 \wedge \dots \wedge \delta_n} & \longrightarrow & \engendrado{f \partial_1 \wedge \dots \wedge \partial_n} & \subseteq & \engendrado{\partial_1 \wedge \dots \wedge \partial_n} + \end{array} + \end{equation*} + Ahora bien, recordemos que + \begin{equation*} + \delta_1 \wedge \delta_2 \wedge \dots \wedge \delta_n = \det\left( \delta_1, \delta_2, \dots, \delta_n \right) \partial_1 \wedge \partial_2 \wedge \dots \wedge \partial_n = uf \partial_1 \wedge \partial_2 \wedge \dots \wedge \partial_n. + \end{equation*} + Al ser $u$ simplemente una unidad, hemos llegado a que $\bigwedge^n N = \bigwedge^n \derivaciones{f}$. Por último, lo que buscamos es quitar ese producto exterior. Ya sabemos que $ N \subseteq \derivaciones{f}$, por lo que solo nos hace falta ver la otra implicación. Sea $\delta \in \derivaciones{f}$. Por ser este un subconjunto de \derivaciones{\Kx}, lo podemos escribir como + \begin{equation*} + \delta = p_1\partial_1 + p_2\partial_2 + \dots + p_n\partial_n. + \end{equation*} Fijémonos ahora en la relación \ref{Relación Matricial Derivadas}. Recordando la regla de Cramer, sabemos que + \begin{equation*} + A^{-1} = \frac{1}{\det A} \mathrm{adj}\!\left( A \right) = \frac{1}{uf}\mathrm{adj}\!\left( A \right) = \frac{1}{f}B + \end{equation*} donde $\mathrm{adj}\!\left( A \right)$ es la matriz adjunta de $A$. $A^{-1}$ está en $\mathrm{Mat}\!\left( n \times n, \cuerpofracciones{\Kx} \right)$ pero los adjuntos de la matriz $A$ están en \Kx, por lo que podemos expresar $A^{-1}$ como vemos al final de la igualdad con $B \in \mathrm{Mat}\!\left( n \times n, \Kx\right)$. Por lo tanto, como + \begin{equation*} + \vectorhorizontalcuatro{\partial_1}{\partial_2}{\dots}{\partial_n} = \vectorhorizontalcuatro{\delta_1}{\delta_2}{\dots}{\delta_n}A^{-1} = \frac{1}{f}\vectorhorizontalcuatro{\delta_1}{\delta_2}{\dots}{\delta_n}B, + \end{equation*} + reescribimos $\delta$ como + \begin{align*} + \delta &= \frac{b_1}{f}\delta_1 + \frac{b_2}{f}\delta_2 + \dots + \frac{b_n}{f}\delta_n, \quad \text{donde $b_i \in \Kx$, o también} \\ + &= \overline{b}_1\delta_1 + \overline{b}_2\delta_2 + \dots + \overline{b}_n\delta_n, \quad \text{donde $\overline{b}_i \in \cuerpofracciones{\Kx}$.} + \end{align*} + Para poder garantizar que estos $\overline{b}_i$ están en \Kx, acordémonos de que $\delta \in \derivaciones{f}$ y $\bigwedge^n \derivaciones{f} = \bigwedge^n N$, por lo que si en $\delta_1 \wedge \delta_2 \wedge \dots \wedge \delta_n$ sustituimos $\delta_i$ por $\delta$ nos queda + \begin{equation*} + \delta_1 \wedge \delta_2 \wedge \dots \wedge \delta_{i-1} \wedge \delta \wedge \delta_{i+1} \wedge \dots \wedge \delta_{n-1} \wedge \delta_n \in \bigwedge^n N. + \end{equation*} + Percatémonos de que esta expresión, una vez eliminamos todos los productos exteriores nulos queda como $\overline{b}_i \delta_1 \wedge \delta_2 \wedge \dots \wedge \delta_n \in \bigwedge^n N$, lo cual es solo posible si $\overline{b}_i \in \Kx$. Con esto concluimos que $\delta \in N$ y, por tanto, que $N = \derivaciones{f}$. \appendix \chapter{Un ejemplo completo de cálculo}