{"id":257,"date":"2024-04-07T12:00:00","date_gmt":"2024-04-07T10:00:00","guid":{"rendered":"https:\/\/www.maths-sup.com\/?p=257"},"modified":"2024-04-07T22:03:52","modified_gmt":"2024-04-07T20:03:52","slug":"demonstration-du-theoreme-de-wedderburn","status":"publish","type":"post","link":"https:\/\/www.maths-sup.com\/index.php\/2024\/04\/07\/demonstration-du-theoreme-de-wedderburn\/","title":{"rendered":"D\u00e9monstration du th\u00e9or\u00e8me de Wedderburn"},"content":{"rendered":"\n

Pr\u00e9liminaires<\/h2>\n\n\n\n

Soit K<\/span> un corps fini.<\/p>\n\n\n\n

On note p>0<\/span> sa caract\u00e9ristique<\/a>.
Comme K<\/span> est un corps, p<\/span> n’a pas de diviseur strict, et donc c’est un nombre premier.<\/p>\n\n\n\n

On note \\mathbb{F}_p<\/span> le corps \\mathbb{Z}\/p\\mathbb{Z}<\/span>.
Alors K<\/span> est un \\mathbb{F}_p<\/span>-espace vectoriel de dimension finie.
On note n<\/span> la dimension de K<\/span> sur \\mathbb{F}_p<\/span>.
Le cardinal de K<\/span> vaut alors p^n<\/span>.
(On a d\u00e9montr\u00e9 au passage que tout corps fini est de cardinal une puissance d’un nombre premier)<\/p>\n\n\n\n

Pour tout entier n>0<\/span>,
on note \\Phi_n(X) \\in \\mathbb{Z}[X]<\/span>,
le n<\/span>-i\u00e8me polyn\u00f4me cyclotomique<\/a>.<\/p>\n\n\n\n

On rappelle que pour tout entier n>0<\/span>, on a<\/p>\n\n\n\n

X^n - 1 = \\prod_{d|n} \\Phi_d(X)<\/span>,<\/p>\n\n\n\n

o\u00f9 d<\/span> parcourt les diviseurs de n<\/span>.<\/p>\n\n\n\n

Th\u00e9or\u00e8me de Wedderburn<\/h2>\n\n\n\n

Th\u00e9or\u00e8me – Wedderburn:<\/strong>
Tout corps fini est commutatif.<\/p>\n\n\n\n

Notons qu’il existe des corps non-commutatifs (par exemple le corps des quaternions<\/a>).<\/p>\n\n\n\n

Preuve du th\u00e9or\u00e8me de Wedderburn<\/h2>\n\n\n\n

Etape 1: L’action par conjugaison<\/h3>\n\n\n\n

On note Z \\subset K<\/span> le centre de<\/strong> K<\/span>.
C’est un corps commutatif<\/strong> et on note q<\/span> son cardinal<\/strong> (c’est une puissance d’un nombre premier, mais ce n’est pas important pour la preuve).<\/p>\n\n\n\n

On cherche \u00e0 d\u00e9montrer que<\/strong> Z = K<\/span><\/p>\n\n\n\n

On note n<\/span> la dimension de K<\/span> sur Z<\/span>.
Le cardinal de K<\/span> vaut alors q^n<\/span>.<\/p>\n\n\n\n

On note K^\\times<\/span> le groupe des inversibles de K<\/span>,
et pareil pour Z^\\times<\/span>.<\/p>\n\n\n\n

On fait agir<\/strong> K^\\times<\/span> par conjugaison sur lui-m\u00eame<\/strong>.
Les points fixes par cette action sont exactement les points de Z^\\times<\/span>.<\/p>\n\n\n\n

Appliquons l’\u00e9quation des classes<\/strong>:
soit x_1, \\dots, x_r \\in K<\/span> une famille de repr\u00e9sentants des orbites non-r\u00e9duit \u00e0 un singleton.<\/p>\n\n\n\n

Alors l’\u00e9quation des classes<\/strong> s’\u00e9crit:<\/p>\n\n\n\n

\\displaystyle{ |K^\\times| = |Z^\\times| + \\sum_{i=1 \\dots r} \\frac{|K^\\times|}{|\\mathrm{Stab}(x_1)|} }<\/span>.<\/p>\n\n\n\n

Pour tout i=1 \\dots r<\/span>,
on note K_i \\subset K<\/span> le commutant de x_i<\/span>:<\/p>\n\n\n\n

K_i := \\{ g \\in K^\\times ~|~ g \\cdot x = x \\cdot g \\}<\/span>.<\/p>\n\n\n\n

On remarque que c’est un corps<\/strong> contenant Z<\/span>.
On note n_i<\/span> la dimention de K_i<\/span> sur Z<\/span>.
On a alors<\/p>\n\n\n\n

|\\mathrm{Stab}(x_i)| = |(K_i)^\\times| = q^{n_i} -1<\/span>.<\/p>\n\n\n\n

L’\u00e9quation des classes se r\u00e9\u00e9crit alors:<\/p>\n\n\n\n

\\displaystyle q^n-1 = (q-1) + \\sum_{i=1 \\dots r} \\frac{q^n - 1}{q^{n_i} - 1}<\/span>.<\/p>\n\n\n\n

Etape 2<\/h3>\n\n\n\n

Soit i \\in \\{1, \\dots, r\\} <\/span>.
Montrons que<\/strong> n_i<\/span> divise<\/strong> n<\/span>.<\/p>\n\n\n\n

Une premi\u00e8re m\u00e9thode simple consiste \u00e0 remarquer que K<\/span> est un K_i<\/span>-espace vectoriel de dimension fini,
et donc |K|<\/span> est une puissance de |K_i|<\/span>,
ce qui permet de conclure: n_i | n<\/span>.<\/p>\n\n\n\n

Cependant K_i<\/span> n’est pas un corps commutatif,
il faut donc admettre la notion de dimension sur les espaces-vectoriels sur des corps non-commutatifs.
On propose une preuve alternative plus \u00e9l\u00e9mentaire.<\/p>\n\n\n\n

On sait que \\frac{q^n - 1}{q^{n_i} - 1}<\/span> est entier,
donc (q^{n_i} - 1)<\/span> divise<\/strong> (q^n - 1)<\/span>.<\/p>\n\n\n\n

L’astuce est de consid\u00e9rer la division euclidienne<\/strong> de n<\/span> par n_i<\/span>,
et de montrer que le reste est nul.<\/p>\n\n\n\n

On note n = q \\cdot n_i + r<\/span> la division euclidienne de n<\/span> par n_i<\/span>.
On a alors<\/p>\n\n\n\n

q^n -1 = (q^{n_i \\cdot q} -1) \\times q^r + (q^r -1)<\/span>.<\/p>\n\n\n\n

Comme (q^{n_i} - 1)<\/span> divise (q^n - 1)<\/span> et (q^{n_i \\cdot q} -1)<\/span>,
On en d\u00e9duit qu’il divise aussi (q^r -1)<\/span>.<\/p>\n\n\n\n

Comme (q^r -1) < (q^{n_i} - 1)<\/span>,
on en d\u00e9duit que r=0<\/span>.
Donc n_i<\/span> divise n<\/span>.<\/p>\n\n\n\n

Etape 3: utilisation du polyn\u00f4me cyclotomique<\/h2>\n\n\n\n

On rappelle que \\Phi_n(X)<\/span> divise X^n -1<\/span> et (X^n -1)\/(X^{n_i} -1)<\/span> dans \\mathbb{Z}[X]<\/span>, pour tout i=1 \\dots r<\/span>.<\/p>\n\n\n\n

En utilisant l’\u00e9quation des classe,
on en d\u00e9duit que \\Phi_n(q)<\/span> divise<\/strong> (q - 1)<\/span>.<\/p>\n\n\n\n

Etape 4: Fin de la preuve<\/h3>\n\n\n\n

Il faut d\u00e9montrer que<\/strong> n =1<\/span>.
On raisonne par l’absurde.<\/p>\n\n\n\n

Supposons que n \\geq 2<\/span>.
Montrons que<\/strong> |\\Phi_n(q)| > |q-1|<\/span>,
ce qui contredira \\Phi_n(q) ~|~ (q-1)<\/span>.<\/p>\n\n\n\n

On note \\zeta = \\exp(2i\\pi\/n)<\/span> la racine primitive.
On a:<\/p>\n\n\n\n

\\displaystyle \\Phi_n(q) = \\prod_{\\mathrm{pgcd}(d, n) = 1} (q - \\zeta^d)<\/span>.<\/p>\n\n\n\n

Pour tout d<\/span>, on a<\/p>\n\n\n\n

|q - \\zeta^d| \\geq \\mathrm{Re}(q - \\zeta^d) = q-1<\/span>.<\/p>\n\n\n\n

L’in\u00e9galit\u00e9 est strict car \\zeta^d<\/span> n’est jamais r\u00e9el (puisque n \\geq 2<\/span>).
Donc |\\Phi_n(q)| > |q-1|<\/span>.
CQFD<\/p>\n\n\n\n

Le\u00e7ons concern\u00e9es<\/h2>\n\n\n\n
    \n
  • 101 – Groupe op\u00e9rant sur un ensemble. Exemples et applications.<\/li>\n\n\n\n
  • 112 – Corps \ffinis. Applications.<\/li>\n<\/ul>\n\n\n\n

    Si tu as des remarques ou des questions n’h\u00e9sites \u00e0 les \u00e9crire en commentaire.<\/p>\n\n\n\n


    <\/p>\n\n\n\n


    <\/p>\n\n\n\n


    <\/p>\n","protected":false},"excerpt":{"rendered":"

    Pr\u00e9liminaires Soit un corps fini. On note sa caract\u00e9ristique.Comme est un corps, n’a pas de diviseur strict, et donc c’est un nombre premier. On note le corps .Alors est un -espace vectoriel de dimension finie.On note la dimension de sur .Le cardinal de vaut alors .(On a d\u00e9montr\u00e9 au passage […]<\/p>\n","protected":false},"author":2,"featured_media":175,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[11,28,25,26,27],"tags":[],"class_list":["post-257","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-action-de-groupe","category-algebre","category-corps-fini","category-lecon-101","category-lecon-112"],"yoast_head":"\nD\u00e9monstration du th\u00e9or\u00e8me de Wedderburn - Maths-Sup<\/title>\n<meta name=\"description\" content=\"D\u00e9monstration du th\u00e9or\u00e8me de Wedderburn \u00e9tapes par \u00e9tapes \u2705. 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