{"id":1821,"date":"2022-06-22T17:45:03","date_gmt":"2022-06-22T20:45:03","guid":{"rendered":"http:\/\/localhost\/?page_id=1821"},"modified":"2022-06-22T21:59:20","modified_gmt":"2022-06-23T00:59:20","slug":"nullstellensatz","status":"publish","type":"page","link":"http:\/\/localhost\/index.php\/ensino\/algebra-comutativa\/nullstellensatz\/","title":{"rendered":"Nullstellensatz"},"content":{"rendered":"
Seja \\(A\\subseteq B\\)<\/span> uma extens\u00e3o integral de dom\u00ednios. Ent\u00e3o \\(A\\)<\/span> \u00e9 corpo se e somente se \\(B\\)<\/span> \u00e9 corpo.<\/div>\n
Assuma que \\(A\\)<\/span> \u00e9 corpo e seja \\(x\\in B\\)<\/span>. Ent\u00e3o \\(x\\)<\/span> \u00e9 integral e existem \\(\\alpha_1,\\ldots,\\alpha_n\\in A\\)<\/span> tais que \\[x^n+\\alpha_1x^{n-1}+\\cdots+\\alpha_{n-1}x+\\alpha_n=0.\\]<\/span> Como \\(B\\)<\/span> \u00e9 dom\u00ednio \\(\\alpha_n\\neq 0\\)<\/span>; ou seja \\[x(x^{n-1}+\\alpha_1x^{n-2}+\\cdots+\\alpha_{n-1})=-\\alpha_n.\\]<\/span> Como \\(A\\)<\/span> \u00e9 corpo, \\(-\\alpha_n\\)<\/span> \u00e9 invert\u00edvel e assim \\(x\\)<\/span> tamb\u00e9m \u00e9 invert\u00edvel.<\/p>\n

Agora assuma que \\(B\\)<\/span> \u00e9 corpo e seja \\(x\\in A\\)<\/span>. O elemento \\(x\\)<\/span> \u00e9 invert\u00edvel em \\(B\\)<\/span>, ent\u00e3o existe \\(1\/x\\in B\\)<\/span> que \u00e9 integral sobre \\(A\\)<\/span>. Logo existem \\(\\alpha_1,\\ldots,\\alpha_n\\in A\\)<\/span> tais que \\[\\frac 1{x^n}+\\alpha_1\\frac 1{x^{n-1}}+\\cdots+\\alpha_{n-1}\\frac 1x+\\alpha_n=0.\\]<\/span> Multiplicando por \\(x^{n-1}\\)<\/span>, obtemos que \\[\\frac 1x+\\alpha_1+\\alpha_2x+\\cdots+\\alpha_{n-1}x^{n-2}+\\alpha_nx^{n-1}=0;\\]<\/span> ou seja \\(1\/x\\in A\\)<\/span>.<\/div>\n

(Nullstellensatz de Zariski) Seja \\(\\mathbb K\\)<\/span> um corpo que \u00e9 tamb\u00e9m uma \\(\\mathbb F\\)<\/span>-alg\u00e9bra finitamente gerada. Ent\u00e3o \\(\\dim_\\mathbb F\\mathbb K\\)<\/span> \u00e9 finita e \\(\\mathbb K\\)<\/span> \u00e9 uma extens\u00e3o alg\u00e9brica de \\(\\mathbb F\\)<\/span>.<\/div>\n
Pelo Teorema de Normaliza\u00e7\u00e3o, existem \\(t_1,\\ldots,t_m\\in \\mathbb K\\)<\/span> algebricamente independentes tais que \\(\\mathbb K\\)<\/span> \u00e9 finitamente gerada sobre \\(\\mathbb F[t_1,\\ldots,t_m]\\)<\/span>. Como \\(\\mathbb K\\)<\/span> \u00e9 corpo, \\(\\mathbb F[t_1,\\ldots,t_m]\\)<\/span> \u00e9 tamb\u00e9m corpo e \\(m=0\\)<\/span>. Logo \\(\\mathbb K\\)<\/span> m\u00f3dulo finito sobre $\\mathbb F$ e \\(\\dim_\\mathbb F\\mathbb K\\)<\/span> \u00e9 finita.<\/div>\n
Seja \\(\\varphi:R\\to S\\)<\/span> um morfismo de \\(\\mathbb F\\)<\/span>-\u00e1lgebras finitamente geradas. Se \\(\\mathfrak m\\)<\/span> \u00e9 maximal em \\(S\\)<\/span>, ent\u00e3o \\(\\varphi^{-1}(\\mathfrak m)\\)<\/span> \u00e9 maximal em \\(R\\)<\/span><\/div>\n
Seja \\(\\mathbb K=S\/\\mathfrak m\\)<\/span> e \\(\\mathfrak n=\\varphi^{-1}(\\mathfrak m)\\)<\/span>. Pelo Nullstellensatz de Zariski, \\(\\mathbb K\\)<\/span> \u00e9 uma extens\u00e3o finita de \\(\\mathbb F\\)<\/span>. Al\u00e9m disso, \\(\\varphi\\)<\/span> induz um morfismo injetivo \\(R\/\\mathfrak n\\to S\/\\mathfrak m\\)<\/span> definido por \\(r+\\mathfrak n\\mapsto \\varphi(r)+\\mathfrak m\\)<\/span>. Logo \\(R\/\\mathfrak n\\subseteq S\/\\mathfrak m\\)<\/span>. Como \\(\\dim_\\mathbb FS\/\\mathfrak m\\)<\/span> \u00e9 finita, \\(\\dim_{\\mathbb F}R\/\\mathfrak n\\)<\/span> \u00e9 finita e o primeiro lema implica que \\(R\/\\mathfrak n\\)<\/span> \u00e9 um corpo. Logo \\(\\mathfrak n\\)<\/span> \u00e9 maximal em \\(R\\)<\/span>.<\/div>\n
Seja \\(\\mathbb F[x_1,\\ldots,x_n]\\)<\/span> uma \u00e1lgebra finitamente gerada e sejam \\(\\alpha_1,\\ldots,\\alpha_n\\in \\mathbb F\\)<\/span>. Mostre que \\((x_1-\\alpha_1,\\ldots,x_n-\\alpha_n)\\)<\/span> \u00e9 um ideal maximal.<\/div>\n
Seja \\(\\mathbb F\\)<\/span> algebricamente fechado e \\(\\mathfrak m\\)<\/span> um ideal maximal de uma \\(\\mathbb F\\)<\/span>-\u00e1lgebra \\(\\mathbb F[y_1,\\ldots,y_n]\\)<\/span> finitamente gerada. Ent\u00e3o existem \\(\\alpha_1,\\ldots,\\alpha_n\\in\\mathbb F\\)<\/span> tais que \\[\\mathfrak m=(x_1-\\alpha_1,\\ldots,x_n-\\alpha_n).\\]<\/span><\/div>\n
Como o ideal \\(\\mathfrak m\\)<\/span> \u00e9 maximal, \\(\\mathbb K=\\mathbb F[x_1,\\ldots,x_n]\/\\mathfrak m\\)<\/span> \u00e9 um corpo e \u00e9 tamb\u00e9m uma \u00e1lgebra finitamente gerada sobre \\(\\mathbb F\\)<\/span>. Pelo Nullstellensatz de Zariski, \\(\\mathbb K\\)<\/span> \u00e9 extens\u00e3o finita de \\(\\mathbb F\\)<\/span> e \\(\\mathbb K=\\mathbb F\\)<\/span> pelo fato que \\(\\mathbb F\\)<\/span> \u00e9 algebricamente fechado. Ou seja, existe um morfismo \\(\\psi: \\mathbb F[x_1,\\ldots,x_n]\\to\\mathbb F\\)<\/span> com n\u00facleo \\(\\mathfrak m\\)<\/span>. Pondo \\(\\alpha_i=\\psi(x_i)\\)<\/span>, \\(x_i-\\alpha_i\\in \\mathfrak m\\)<\/span>; ou seja \\((x_1-\\alpha_1,\\ldots,x_n-\\alpha_n)\\subseteq \\mathfrak m\\)<\/span>. O ideal \\((x_1-\\alpha_1,\\ldots,x_n-\\alpha_n)\\)<\/span> \u00e9 maximal pelo exerc\u00edcio anterior que implica igualdade.<\/div>\n
Seja \\(\\mathbb F\\)<\/span> um corpo algebricamente fechado e seja \\(\\mathfrak a\\subset \\mathbb F[x_1,\\ldots,x_n]\\)<\/span> um ideal pr\u00f3prio na \u00e1lgebra dos polin\u00f4mios. Ent\u00e3o \\[V(\\mathfrak a)=\\{(\\alpha_1,\\ldots,\\alpha_n)\\in\\mathbb F^n\\mid f(\\alpha_1,\\ldots,\\alpha_n)=0\\mbox{ para todo }f\\in \\mathfrak a\\}\\neq \\emptyset.\\]<\/span><\/div>\n
Seja \\(\\mathfrak m=(x_1-\\alpha_1,\\ldots,x_n-\\alpha_n)\\)<\/span> um ideal maximal que cont\u00e9m \\(\\mathfrak a\\)<\/span>. Ent\u00e3o \\[(\\alpha_1,\\ldots,\\alpha_n)\\in V(\\mathfrak m)\\subseteq V(\\mathfrak a).\\]<\/span><\/div>\n
Seja \\(R\\)<\/span> um anel e considere \\(R[x_1,\\ldots,x_n]\\)<\/span>. Assuma que \\(f\\in R[x_1]\\)<\/span>. Ent\u00e3o \\[R[x_1,\\ldots,x_n]\/(fR[x_1,\\ldots,x_n])\\cong (R[x_1]\/(fR[x_1])[x_2,\\ldots,x_n].\\]<\/span><\/div>\n
Seja \\(R=\\mathbb F[x_1,\\ldots,x_n]\\)<\/span> um anel de polin\u00f4mios e \\(\\mathfrak m\\subseteq R\\)<\/span> um ideal maximal. Ent\u00e3o \\(\\mathfrak m\\)<\/span> \u00e9 gerado por \\(n\\)<\/span> elementos.<\/div>\n
Indu\u00e7\u00e3o por \\(n\\)<\/span>. Quando \\(n=0\\)<\/span>, \u00e9 trivial pois o \u00fanico ideal maximal de \\(\\mathbb F\\)<\/span> \u00e9 o zero. Assuma que o resultado est\u00e1 verdadeiro para \\(\\mathbb F[x_1,\\ldots,x_{n-1}]\\)<\/span> e seja \\(\\mathfrak m\\)<\/span> um ideal maximal de \\(R=\\mathbb F[x_1,\\ldots,x_{n}]\\)<\/span>. Seja \\(\\mathbb K=R\/\\mathfrak m\\)<\/span>. Pelo Nullstellensatz de Zariski, \\(\\mathbb K\\)<\/span> \u00e9 um corpo e \u00e9 uma extens\u00e3o finita de \\(\\mathbb F\\)<\/span>. Seja \\(R_0=\\mathbb F[x_1]\\)<\/span> e \\(\\mathfrak n=\\mathfrak m\\cap R_0\\)<\/span>. Ent\u00e3o \\(\\mathfrak n=(f_1)_{R_0}\\)<\/span> e seja \\(\\mathbb L=R_0\/\\mathfrak n\\)<\/span>. Pelo corol\u00e1rio acima, \\(\\mathfrak n\\)<\/span> \u00e9 maximal. Ora \\(R\/(\\mathfrak nR)\\cong \\mathbb L[x_2,\\ldots,x_n]\\)<\/span> e \\(\\mathfrak m\/(\\mathfrak nR)\\)<\/span> \u00e9 maximal em \\(R\/(\\mathfrak nR)\\cong \\mathbb L[x_2,\\ldots,x_n]\\)<\/span>. Pela hip\u00f3tese da indu\u00e7\u00e3o, \\(\\mathfrak m\/(\\mathfrak nR)\\)<\/span> \u00e9 gerado por \\(f_2,\\ldots,f_n\\)<\/span> e \\(\\mathfrak m\\)<\/span> \u00e9 gerado por \\(f_1,f_2,\\ldots,f_n\\)<\/span>.<\/div>\n
Assuma que \\(R=\\mathbb F[x_1,\\ldots,x_n]\\)<\/span> \u00e9 uma \u00e1lgebra finitamente gerada e \\(\\mathfrak a\\neq R\\)<\/span> \u00e9 um ideal. Ent\u00e3o \\[\\sqrt{(\\mathfrak a)}=\\bigcap_{\\mathfrak a\\subseteq \\mathfrak m}\\mathfrak m\\]<\/span> onde a interse\u00e7\u00e3o est\u00e1 tomada para todos os ideais maximais de \\(R\\)<\/span> que cont\u00eam \\(\\mathfrak a\\)<\/span>.<\/div>\n
Quocientando por \\(\\mathfrak a\\)<\/span>, podemos assumir que \\(\\mathfrak a=0\\)<\/span>. Como \\(\\sqrt{0}\\)<\/span> \u00e9 o nilradical que est\u00e1 contido no radical de Jacobson, temos que \\(\\sqrt{0}\\subseteq \\bigcap \\mathfrak m\\)<\/span>. Ora assuma que \\(f\\not\\in \\sqrt{0}\\)<\/span>. Ent\u00e3o \\(f\\)<\/span> n\u00e3o \u00e9 nilpotente, e a localiza\u00e7\u00e3o \\(R_f\\neq 0\\)<\/span>. Seja \\(\\mathfrak n\\)<\/span> um ideal maximal de \\(R_f\\)<\/span>. Como \\(R_f\\cong R[X]\/(1-fX)\\)<\/span>, temos que \\(R_f\\)<\/span> \u00e9 finitamente gerada como \\(\\mathbb F\\)<\/span>-\u00e1lgebra e assim o corpo \\(R_f\/\\mathfrak n\\)<\/span> \u00e9 uma extens\u00e3o finita de \\(\\mathbb F\\)<\/span>. Seja \\(\\mathfrak m=\\mathfrak n\\cap R\\)<\/span>. Pelo corol\u00e1rio em cima, \\(\\mathfrak m\\)<\/span> \u00e9 maximal e \\(R\/\\mathfrak m\\)<\/span> \u00e9 um corpo. Mas \\(f\\)<\/span> \u00e9 invert\u00edvel em \\(R_f\\)<\/span> e assim \\(f\/1\\not\\in \\mathfrak n\\)<\/span> e \\(f\\not\\in\\mathfrak m\\)<\/span>. Logo \\(f\\not\\in\\bigcap\\mathfrak m\\)<\/span>.<\/div>\n","protected":false},"excerpt":{"rendered":"

Seja \\(A\\subseteq B\\) uma extens\u00e3o integral de dom\u00ednios. Ent\u00e3o \\(A\\) \u00e9 corpo se e somente se \\(B\\) \u00e9 corpo. Assuma que \\(A\\) \u00e9 corpo e seja \\(x\\in B\\). Ent\u00e3o \\(x\\) \u00e9 integral e existem \\(\\alpha_1,\\ldots,\\alpha_n\\in A\\) tais que \\[x^n+\\alpha_1x^{n-1}+\\cdots+\\alpha_{n-1}x+\\alpha_n=0.\\] Como \\(B\\) \u00e9 dom\u00ednio \\(\\alpha_n\\neq 0\\); ou seja \\[x(x^{n-1}+\\alpha_1x^{n-2}+\\cdots+\\alpha_{n-1})=-\\alpha_n.\\] Como \\(A\\) \u00e9 corpo, \\(-\\alpha_n\\) \u00e9 invert\u00edvel … Continue reading Nullstellensatz<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1808,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":[],"_links":{"self":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/1821"}],"collection":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/comments?post=1821"}],"version-history":[{"count":5,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/1821\/revisions"}],"predecessor-version":[{"id":1841,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/1821\/revisions\/1841"}],"up":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/1808"}],"wp:attachment":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/media?parent=1821"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}