{"id":1811,"date":"2022-06-22T11:25:43","date_gmt":"2022-06-22T14:25:43","guid":{"rendered":"http:\/\/localhost\/?page_id=1811"},"modified":"2022-06-22T14:25:42","modified_gmt":"2022-06-22T17:25:42","slug":"lying-over-going-down","status":"publish","type":"page","link":"http:\/\/localhost\/index.php\/ensino\/algebra-comutativa\/lying-over-going-down\/","title":{"rendered":"Lying over, going down"},"content":{"rendered":"

Seja \\(S\\)<\/span> uma \\(R\\)<\/span>-\u00e1lgebra, com morfismo \\(\\varphi:R\\to S\\)<\/span>, e seja \\(\\mathfrak p\\in\\mbox{Spec}(S)\\)<\/span>. Ent\u00e3o \\(\\varphi^{-1}(\\mathfrak p)\\)<\/span> pertence ao \\(\\mbox{Spec}(R)\\)<\/span>. Dizemos que o ideal \\(\\mathfrak p\\)<\/span> est\u00e1 acima de<\/em><\/span> \\(\\varphi^{-1}(\\mathfrak p)\\)<\/span>. Quando \\(R\\subseteq S\\)<\/span> e \\(\\varphi\\)<\/span> \u00e9 a inclus\u00e3o, temos que \\(\\varphi^{-1}(\\mathfrak p)=R\\cap \\mathfrak p\\)<\/span>.<\/p>\n

Seja \\(R\\subseteq S\\)<\/span> uma extens\u00e3o de an\u00e9is onde \\(R\\)<\/span> \u00e9 um anel local com ideal maximal \\(\\mathfrak p\\)<\/span>. Seja \\(\\mathfrak q\\)<\/span> um ideal primo de \\(S\\)<\/span> tal que \\(\\mathfrak p\\subseteq \\mathfrak q\\cap R\\)<\/span>. Ent\u00e3o \\(\\mathfrak q\\)<\/span> est\u00e1 acima de \\(\\mathfrak p\\)<\/span>.\n<\/div>\n
Segue que \\(\\mathfrak p= \\mathfrak q\\cap R\\)<\/span> pois \\(\\mathfrak q\\cap R\\neq R\\)<\/span> e \\(\\mathfrak p\\)<\/span> \u00e9 o \u00fanico maximal em \\(R\\)<\/span>.\n<\/div>\n
Seja \\(R\\subseteq S\\)<\/span> uma extens\u00e3o integral de an\u00e9is, \\(\\mathfrak p\\in\\mbox{Spec}(R)\\)<\/span> e \\(\\mathfrak q_1\\subseteq \\mathfrak q_2\\in\\mbox{Spec}(S)\\)<\/span>, e \\(\\mathfrak a\\)<\/span> um ideal arbitr\u00e1rio de \\(S\\)<\/span>.<\/p>\n
    \n
  1. \n

    Se \\(\\mathfrak q\\)<\/span> est\u00e1 acima de \\(\\mathfrak p\\)<\/span> ent\u00e3o \\(\\mathfrak q\\)<\/span> \u00e9 maximal se e somente se \\(\\mathfrak p\\)<\/span> \u00e9 maximal.<\/p>\n<\/li>\n

  2. \n

    Assuma que \\(\\mathfrak q_1\\)<\/span> e \\(\\mathfrak q_2\\)<\/span> est\u00e3o acima de \\(\\mathfrak p\\)<\/span>. Ent\u00e3o \\(\\mathfrak q_1=\\mathfrak q_2\\)<\/span>.<\/p>\n<\/li>\n

  3. \n

    Existe \\(\\mathfrak r\\)<\/span> que est\u00e1 acima de \\(\\mathfrak p\\)<\/span>.<\/p>\n<\/li>\n

  4. \n

    Assuma que \\(\\mathfrak a\\cap R\\subseteq \\mathfrak p\\)<\/span>. Ent\u00e3o existe \\(\\mathfrak r\\)<\/span> em (3) tal que \\(\\mathfrak a\\subseteq \\mathfrak r\\)<\/span>.<\/p>\n<\/li>\n<\/ol>\n<\/div>\n

    (1) Considere o mapa \\(\\psi:R\/\\mathfrak p\\to S\/\\mathfrak q\\)<\/span> definida por \\(r+\\mathfrak p\\mapsto r+\\mathfrak q\\)<\/span>. O mapa \\(\\psi\\)<\/span> est\u00e1 bem definido e \\[\\ker\\psi=\\{r+\\mathfrak p\\in R\/\\mathfrak p\\mid r+\\mathfrak q=0\\}.\\]<\/span> Por outro lado, se \\(r+\\mathfrak q=0\\)<\/span>, ent\u00e3o \\(r\\in \\mathfrak q\\cap R=\\mathfrak p\\)<\/span> e \\(r+\\mathfrak p=0\\)<\/span>. Logo \\(\\psi\\)<\/span> \u00e9 injetiva. Portanto \\(R\/\\mathfrak p\\subseteq S\/\\mathfrak q\\)<\/span> \u00e9 uma extens\u00e3o de an\u00e9is. Se \\(s\\in S\\)<\/span> ent\u00e3o \\[s^n+\\beta_1s^{n-1}+\\cdots+\\beta_n=0\\]<\/span> com \\(\\beta_i\\in R\\)<\/span> e \\[\\begin{aligned}
    \n &(s+\\mathfrak q)^n+(\\beta_1+\\mathfrak p)(s+\\mathfrak q)^{n-1}+\\cdots+(\\beta_n+\\mathfrak p)\\\\&=
    \n (s+\\mathfrak q)^n+(\\beta_1+\\mathfrak q)(s+\\mathfrak q)^{n-1}+\\cdots+(\\beta_n+\\mathfrak q)\\\\&=
    \n (s^n+\\beta_1s^{n-1}+\\cdots+\\beta_n)+\\mathfrak q=0.
    \n \\end{aligned}\\]<\/span> Portanto, \\(R\/\\mathfrak p\\subseteq S\/\\mathfrak q\\)<\/span> \u00e9 integral. Por um lema anterior, um \u00e9 corpo se e somente se outro \u00e9.<\/p>\n

    (2) Assuma primeiro que \\((R,\\mathfrak p)\\)<\/span> \u00e9 um anel local. Ent\u00e3o \\(\\mathfrak q_1\\)<\/span> e \\(\\mathfrak q_2\\)<\/span> s\u00e3o maximais por item (1) e segue que \\(\\mathfrak q_1=\\mathfrak q_2\\)<\/span>.<\/p>\n

    No caso geral, seja \\(U=R\\setminus\\mathfrak p\\)<\/span> e considere as localiza\u00e7\u00f5es \\(R_{(\\mathfrak p)}=U^{-1}R\\)<\/span> e \\(U^{-1}S\\)<\/span>. O mapa de mergulho \\(U^{-1}R\\to U^{-1}S\\)<\/span> \u00e9 injetivo, pois se \\(r\/u\\in U^{-1}R\\)<\/span> tal que \\(r\/u=0\/1\\in U^{-1}S\\)<\/span> ent\u00e3o \\(rw=0\\)<\/span> com algum \\(w\\in R\\setminus \\mathfrak p\\)<\/span> e \\(r\/u=0\/1\\in U^{-1}R\\)<\/span>. Ent\u00e3o \\(U^{-1}R\\subseteq U^{-1}S\\)<\/span> \u00e9 uma extens\u00e3o de an\u00e9is. Afirmamos que, a extens\u00e3o \\(U^{-1}R\\subseteq U^{-1}S\\)<\/span> \u00e9 integral. De fato, se \\(s\/u\\in U^{-1}S\\)<\/span>, ent\u00e3o \\(U^{-1}R[s\/u]=U^{-1}R[s]\\)<\/span>. Mas \\(s\\)<\/span> \u00e9 integral sobre \\(R\\)<\/span> e satisfaz uma equa\u00e7\u00e3o integral \\[s^n+\\beta_1s^{n+1}+\\cdots+\\beta_n=0\\]<\/span> com coeficientes \\(\\beta_i\\in R\\)<\/span>. A mesma equa\u00e7\u00e3o ser\u00e1 uma equa\u00e7\u00e3o integral para \\(s\/1\\)<\/span> com coeficientes em \\(U^{-1}R\\)<\/span>. Portanto \\(U^{-1}R[s\/u]=
    \n U^{-1}[s]\\)<\/span> \u00e9 finitamente gerado como \\(U^{-1}R\\)<\/span>-m\u00f3dulo e \\(s\/u\\)<\/span> \u00e9 integral sobre \\(U^{-1}R\\)<\/span>.<\/p>\n

    Como \\(U^{-1}R=R_{(\\mathfrak p)}\\)<\/span> \u00e9 local com \u00fanico ideal maximal \\(\\mathfrak pU^{-1}R\\)<\/span> e como \\(\\mathfrak pU^{-1}R\\subseteq \\mathfrak q_i U^{-1}S\\cap U^{-1}R\\)<\/span> para \\(i\\in\\{1,2\\}\\)<\/span>, e o lema anterior implica que \\(\\mathfrak q_i U^{-1}S\\)<\/span> est\u00e1 acima de \\(\\mathfrak pU^{-1}R\\)<\/span>. Ora, o par\u00e1grafo anterior implica que \\(\\mathfrak q_1U^{-1}S=\\mathfrak q_2U^{-1}S\\)<\/span>. Como \\(\\mathfrak q_i\\cap U=\\emptyset\\)<\/span>, e a correspond\u00eancia entre os primos em \\(\\mbox{Spec}(S)\\)<\/span> que interceptam trivialmente com \\(U\\)<\/span> e os primos de \\(U^{-1}S\\)<\/span> implica que \\(\\mathfrak q_1=\\mathfrak q_2\\)<\/span>.<\/p>\n

    (3) Assuma primeiro que \\((R,\\mathfrak p)\\)<\/span> \u00e9 um anel local. O anel \\(S\\)<\/span> tem um ideal maximal \\(\\mathfrak q\\)<\/span> e \\(\\mathfrak q\\cap R\\)<\/span> \u00e9 maximal por afirma\u00e7\u00e3o\u00a0(1). Logo \\(\\mathfrak q\\cap R\\)<\/span> deve ser igual a \\(\\mathfrak p\\)<\/span>.<\/p>\n

    No caso geral, considere as localiza\u00e7\u00f5es \\(U^{-1}R\\)<\/span> e \\(U^{-1}S\\)<\/span> como na demonstra\u00e7\u00e3o da afirma\u00e7\u00e3o\u00a0(2). Assuma que \\(\\mathfrak r'\\)<\/span> \u00e9 um ideal maximal de \\(U^{-1}S\\)<\/span>. Ent\u00e3o \\(\\mathfrak r'\\cap U^{-1}R\\)<\/span> maximal em \\(U^{-1}R\\)<\/span> e portanto \\(\\mathfrak r'\\cap U^{-1}R=\\mathfrak pU^{-1}R\\)<\/span>. Definimos \\(\\mathfrak r=\\varphi_S^{-1}(\\mathfrak r')\\)<\/span>. J\u00e1 monstramos da demonstra\u00e7\u00e3o da afirma\u00e7\u00e3o\u00a0(2) que o mapa can\u00f4nico \\(U^{-1}R\\to U^{-1}S\\)<\/span> \u00e9 uma inclus\u00e3o e assim temos o seguinte diagrama comutativo:
    \n\\[
    \n\\begin{CD}
    \n R @>>> S \\\\
    \n\t\t@V{\\varphi_R}VV @VV{\\varphi_S}V\\\\
    \n U^{-1}R @>>> U^{-1}S
    \n \\end{CD}\\]<\/span> Calculando a pr\u00e9-imagem em \\(R\\)<\/span> de \\(\\mathfrak r'\\subseteq U^{-1}S\\)<\/span> pela composi\u00e7\u00e3o dos mapas das duas maneiras no diagrama, obtemos que \\[R\\cap \\mathfrak r=R\\cap \\varphi_S^{-1}(\\mathfrak r')=\\varphi_R^{-1}(\\mathfrak r'\\cap U^{-1}R)=\\varphi_{R}^{-1}(\\mathfrak pU^{-1}R)=\\mathfrak p\\]<\/span><\/p>\n

    (4) Aplique (3) para o extens\u00e3o \\(R\/(\\mathfrak a\\cap R)\\subseteq S\/\\mathfrak a\\)<\/span>.<\/div>\n

    \n<\/span> Seja \\(R\\subseteq S\\)<\/span> uma extens\u00e3o de an\u00e9is e \\(f\\in R[x]\\)<\/span> um polin\u00f4mio m\u00f4nico. Assuma que \\(f=gh\\)<\/span> onde \\(g,h\\in S[x]\\)<\/span> e \\(g\\)<\/span> \u00e9 m\u00f4nico.<\/p>\n
      \n
    1. \n

      Existe uma extens\u00e3o \\(T\\)<\/span> de \\(R\\)<\/span> tal que \\(f(x)=\\prod (x-x_i)\\)<\/span> em \\(T[x]\\)<\/span>. Al\u00e9m disso, \\(T\\)<\/span> \u00e9 \\(R\\)<\/span>-m\u00f3dulo livre de posto \\(d!\\)<\/span> onde \\(d=\\deg f\\)<\/span>.<\/p>\n<\/li>\n

    2. \n

      \\(h\\)<\/span> \u00e9 m\u00f4nico e os coeficientes de \\(g\\)<\/span> e \\(h\\)<\/span> s\u00e3o integrais sobre \\(R\\)<\/span>.<\/p>\n<\/li>\n<\/ol>\n<\/div>\n

      (1) Seja \\(R_1=R[x]\/(f)\\)<\/span>. Ent\u00e3o \\(R_1\\)<\/span> \u00e9 gerado por \\(1,\\bar x,\\bar x^2,\\ldots,\\bar x^{d-1}\\)<\/span>. Assuma que \\[\\alpha_0+\\alpha_1\\bar x+\\cdots+\\alpha_{d-1}\\bar x^{d-1}\\]<\/span> \u00e9 uma depend\u00eancia linear n\u00e3o trivial com \\(\\alpha_i\\in R\\)<\/span>. Ent\u00e3o \\(\\bar x\\)<\/span> \u00e9 ra\u00edz do polin\u00f4mio \\[q(x)=\\alpha_0+\\cdots+\\alpha_{d-1}x^{d-1}\\]<\/span> e \\(q(x)\\mid f(x)\\)<\/span> mas isso \u00e9 imposs\u00edvel porque \\(\\deg q(x)<\\deg f(x)\\)<\/span> e \\(f(x)\\)<\/span> \u00e9 m\u00f4nico. Assim \\(R_1\\)<\/span> \u00e9 \\(R\\)<\/span>-m\u00f3dulo livre de posto \\(d\\)<\/span> e \\(R\\subseteq R_1\\)<\/span>. Ademais, \\(f(\\bar x)=0\\)<\/span> e \\(f(x)=f_1(x)(x-\\bar x)\\)<\/span> com \\(f_1\\in R_1[x]\\)<\/span>. Note que \\(f_1\\)<\/span> \u00e9 m\u00f4nico e \\(\\deg{f_1}=d-1\\)<\/span>. Repetimos o processo para \\(f_1\\)<\/span> e no final obtemos \\(T\\)<\/span>, as ra\u00edzes \\(x_1=\\bar x,\\ldots,x_d\\in T\\)<\/span>.<\/p>\n

      (2) O coeficiente l\u00edder de \\(f\\)<\/span> \u00e9 o produto dos coeficientes l\u00edderes de \\(f\\)<\/span> e \\(g\\)<\/span>. Como \\(f\\)<\/span> \u00e9 m\u00f4nico, \\(g\\)<\/span> \u00e9 m\u00f4nico, obtemos que \\(h\\)<\/span> \u00e9 m\u00f4nico. Aplicando (1), obtemos uma extens\u00e3o \\(Q_1\\)<\/span> de \\(S\\)<\/span> tal que \\(g=\\prod (x-y_i)\\)<\/span> e uma extens\u00e3o \\(Q_2\\)<\/span> de \\(S\\)<\/span> tal que \\(h=\\prod(x-z_i)\\)<\/span>. Os elementos \\(y_i\\)<\/span> e \\(z_i\\)<\/span> s\u00e3o integrais sobre \\(R\\)<\/span>, pois s\u00e3o ra\u00edzes de \\(f\\)<\/span>. Mas os coeficientes de \\(g\\)<\/span> e \\(h\\)<\/span> s\u00e3o polin\u00f4mios em \\(y_i\\)<\/span> e \\(z_i\\)<\/span> e eles tamb\u00e9m s\u00e3o integrais sobre \\(R\\)<\/span>.<\/div>\n

      Seja \\(R\\)<\/span> um dom\u00ednio normal, \\(\\mathbb K=\\mbox{Frac}(R)\\)<\/span> e \\(\\mathbb K\\subseteq \\mathbb L\\)<\/span> uma extens\u00e3o de corpos. Seja \\(y\\in\\mathbb L\\)<\/span> integral sobre \\(R\\)<\/span> e \\(f\\in\\mathbb K[x]\\)<\/span> o polin\u00f4mio minimal (m\u00f4nico) de \\(y\\)<\/span>. Ent\u00e3o \\(f\\in R[x]\\)<\/span> e \\(f(y)=0\\)<\/span> \u00e9 depend\u00eancia integral.<\/div>\n
      Como \\(y\\in \\mathbb L\\)<\/span> \u00e9 integral, existe \\(g\\in R[x]\\)<\/span> m\u00f4nico tal que \\(g(y)=0\\)<\/span>. Pela defini\u00e7\u00e3o do polin\u00f4mio minimal, \\(f(x)\\mid g(x)\\)<\/span> considerados como polin\u00f4mios em \\(\\mathbb K[x]\\)<\/span> e escreva \\(g=fh\\)<\/span> com \\(h\\in \\mathbb K[x]\\)<\/span>. Ent\u00e3o os coeficientes de \\(f\\)<\/span> s\u00e3o integrais sobre \\(R\\)<\/span>. Como \\(R\\)<\/span> \u00e9 normal, \\(f\\in R[x]\\)<\/span>.<\/div>\n
      [ex:prim]<\/span> Seja \\(S\\)<\/span> uma \\(R\\)<\/span> \u00e1lgebra com o mapa \\(\\varphi: R\\to S\\)<\/span>. Seja \\(\\mathfrak p\\in\\mbox{Spec}(R)\\)<\/span> tal que \\(\\varphi^{-1}(\\mathfrak pS)=\\mathfrak p\\)<\/span>. Ent\u00e3o existe um ideal \\(\\mathfrak q\\in\\mbox{Spec}(S)\\)<\/span> tal que \\(\\varphi^{-1}(\\mathfrak q)=\\mathfrak p\\)<\/span>.<\/div>\n
      Seja \\(R\\subseteq S\\)<\/span> uma extens\u00e3o integral de dom\u00ednios com \\(R\\)<\/span> normal, e \\(\\mathfrak p\\subset \\mathfrak q\\)<\/span> primos em \\(R\\)<\/span> e \\(\\mathfrak q'\\)<\/span> um primo em \\(S\\)<\/span> que est\u00e1 acima de \\(\\mathfrak q\\)<\/span>. Ent\u00e3o existe um primo \\(\\mathfrak p'\\)<\/span> em \\(S\\)<\/span> que est\u00e1 acima de \\(\\mathfrak p\\)<\/span> e \\(\\mathfrak p'\\subset \\mathfrak q'\\)<\/span>.<\/div>\n
      Considere a localiza\u00e7\u00e3o \\(S_{(\\mathfrak q')}\\)<\/span>. Como \\(R\\)<\/span> e \\(S\\)<\/span> s\u00e3o dom\u00ednios, temos que \\(R\\subseteq S\\subseteq S_{(\\mathfrak q')}\\)<\/span>. Primeiro afirmamos que \\(\\mathfrak pS_{(\\mathfrak q)}\\cap R=\\mathfrak p\\)<\/span>. A inclus\u00e3o \\(\\mathfrak p\\subseteq \\mathfrak pS_{(\\mathfrak q)}\\cap R\\)<\/span> vale trivialmente, ent\u00e3o precisa-se provar que \\(\\mathfrak pS_{(\\mathfrak q)}\\cap R\\subseteq \\mathfrak p\\)<\/span>.<\/p>\n

      Seja \\(y\\in \\mathfrak pS_{(\\mathfrak q')}\\cap R\\)<\/span> com \\(y\\in R\\setminus \\mathfrak p\\)<\/span>. Assuma que \\(y=x\/s\\)<\/span> onde \\(x\\in \\mathfrak pS\\)<\/span> e \\(s \\in S\\setminus\\mathfrak q'\\)<\/span>. Seja \\(\\mathbb K=\\mbox{Frac}(R)\\)<\/span>.<\/p>\n

      Afirmamos primeiro que o polin\u00f4mio minimal de \\(x\\)<\/span> sobre \\(\\mathbb K\\)<\/span> \u00e9 um polin\u00f4mio m\u00f4nico em \\(R[t]\\)<\/span> tal que os coeficientes n\u00e3o l\u00edderes percentem a \\(\\mathfrak p\\)<\/span>. Ponha \\(x=\\sum_{i\\leq k} y_ix_i\\)<\/span> onde \\(y_i\\in \\mathfrak p\\)<\/span> e \\(x_i\\in S\\)<\/span> e seja \\(T=R[x_1,\\ldots,x_k]\\)<\/span>. Ent\u00e3o \\(x_i\\)<\/span> \u00e9 integral sobre \\(R\\)<\/span> para todo \\(i\\)<\/span> e \\(T\\)<\/span> \u00e9 m\u00f3dulo-finito sobre \\(R\\)<\/span>. Al\u00e9m disso, \\[xt=\\sum_{i\\leq k}y_ix_it\\in \\sum_{i\\leq k} y_iT\\subseteq \\mathfrak pT\\mbox{ para todo }t\\in T\\]<\/span> e segeu que \\(xT\\subseteq \\mathfrak pT\\)<\/span>. Assuma que \\(g(t)\\in R[t]\\)<\/span> \u00e9 o polin\u00f4mio carater\u00edstico de \\(\\mu_x:T\\to T\\)<\/span> definida pela multiplica\u00e7\u00e3o por \\(x\\)<\/span>. Ent\u00e3o \\(g(t)\\in R[x]\\)<\/span> \u00e9 um polin\u00f4mio m\u00f4nico tal que os coeficientes n\u00e3o l\u00edderes de \\(g(t)\\)<\/span> pertencem a \\(\\mathfrak p\\)<\/span> e \\(g(x)=0\\)<\/span>. O polin\u00f4mio minimal \\(f(t)\\)<\/span> de \\(x\\)<\/span> sobre \\(\\mathbb K\\)<\/span> \u00e9 um divisor de \\(g(t)\\)<\/span> e escreva \\(g(t)=f(t)h(t)\\)<\/span> onde \\(f\\)<\/span> e \\(h\\)<\/span> s\u00e3o m\u00f4nicos. Pelo Lema\u00a0[lem:coef_int]<\/a>, \\(f(t),h(t)\\in R[t]\\)<\/span>, pois \\(R\\)<\/span> \u00e9 normal. Al\u00e9m disso, \\(g\\equiv t^n\\pmod{\\mathfrak p}\\)<\/span>, ent\u00e3o \\(f\\equiv t^r\\)<\/span> e \\(h\\equiv t^{n-r}\\pmod \\mathfrak p\\)<\/span> por fatora\u00e7\u00e3o \u00fanica em \\(\\mbox{Frac}(R\/\\mathfrak p)[x]\\)<\/span>. Assim os coeficientes n\u00e3o l\u00edderes de \\(f\\)<\/span> e \\(h\\)<\/span> s\u00e3o em \\(\\mathfrak p\\)<\/span>. Ent\u00e3o a afirma\u00e7\u00e3o est\u00e1 provada. Assuma que o polin\u00f4mio minimal de \\(x\\)<\/span> est\u00e1 na forma \\[\\label{eq:pol_min}
      \n f(t)=t^n+a_1t^{n-1}+\\cdots+a_n\\in R[t]\\]<\/span> onde \\(a_i\\in \\mathfrak p\\)<\/span>.<\/p>\n

      Afirmamos que a equa\u00e7\u00e3o minimal de \\(s\\)<\/span> sobre \\(\\mathbb K\\)<\/span> \u00e9 \\[m(t)= t^n+b_1t^{n-1}+\\cdots+b_n=0\\mbox{ com }b_i=a_i\/y^i\\in\\mathbb K.\\]<\/span> Substituindo \\(s=x\/y\\)<\/span> em \\(m(t)\\)<\/span> d\u00e1 zero. Por outro lado, se existir uma equa\u00e7\u00e3o com \\(\\ell<n\\)<\/span> na forma \\[s^\\ell+\\beta_1 s^{\\ell-1}+\\cdots+\\beta_\\ell=0,\\]<\/span> ent\u00e3o a mesma implica que \\[x^\\ell+\\beta_1y_1x^{\\ell-1}+\\cdots +\\beta_\\ell y_1^\\ell=0\\]<\/span> que seria uma equa\u00e7\u00e3o integral para \\(x\\)<\/span>. Mas a equa\u00e7\u00e3o minimal de \\(x\\)<\/span> tem grau \\(n\\)<\/span> e assim \\(m(t)\\)<\/span> deve ser a equa\u00e7\u00e3o minimal de \\(s\\)<\/span> sobre \\(\\mathbb K\\)<\/span>.<\/p>\n

      Como \\(y\\not\\in \\mathfrak p\\)<\/span>, \\(b_i\\in\\mathfrak p\\)<\/span> pois \\(a_i=b_iy^i\\in\\mathfrak p\\)<\/span>. Ent\u00e3o \\(s^n\\in\\mathfrak pS\\subseteq \\mathfrak qS\\subseteq\\mathfrak q'\\)<\/span>. Obtemos que \\(s\\in\\mathfrak q'\\)<\/span>, que \u00e9 uma contradi\u00e7\u00e3o. Assim \\(y\\in\\mathfrak p\\)<\/span> e \\(\\mathfrak pS_{(\\mathfrak q')}\\cap R\\subseteq\\mathfrak p\\)<\/span> e temos que \\(\\mathfrak pS_{(\\mathfrak q')}\\cap R=\\mathfrak p\\)<\/span> .<\/p>\n

      Pelo exerc\u00edcio anterior, existe um primo \\(\\mathfrak p''\\subseteq S_{(\\mathfrak q')}\\)<\/span> com \\(\\mathfrak p''\\cap R=\\mathfrak p\\)<\/span>. Al\u00e9m disso, \\(\\mathfrak p''\\subseteq \\mathfrak q'R_{(\\mathfrak q')}\\)<\/span> pois \\(\\mathfrak q'R_{(\\mathfrak q')}\\)<\/span> \u00e9 o \u00fanico maximal. Defina \\(\\mathfrak p'=\\mathfrak p''\\cap S\\)<\/span>. Ent\u00e3o \\(\\mathfrak p'\\cap R=\\mathfrak p''\\cap S\\cap R=\\mathfrak p\\)<\/span> e \\(\\mathfrak p'\\subseteq \\mathfrak q'\\)<\/span>.<\/div>\n","protected":false},"excerpt":{"rendered":"

      Seja \\(S\\) uma \\(R\\)-\u00e1lgebra, com morfismo \\(\\varphi:R\\to S\\), e seja \\(\\mathfrak p\\in\\mbox{Spec}(S)\\). Ent\u00e3o \\(\\varphi^{-1}(\\mathfrak p)\\) pertence ao \\(\\mbox{Spec}(R)\\). Dizemos que o ideal \\(\\mathfrak p\\) est\u00e1 acima de \\(\\varphi^{-1}(\\mathfrak p)\\). Quando \\(R\\subseteq S\\) e \\(\\varphi\\) \u00e9 a inclus\u00e3o, temos que \\(\\varphi^{-1}(\\mathfrak p)=R\\cap \\mathfrak p\\). Seja \\(R\\subseteq S\\) uma extens\u00e3o de an\u00e9is onde \\(R\\) \u00e9 um anel … Continue reading Lying over, going down<\/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\/1811"}],"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=1811"}],"version-history":[{"count":5,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/1811\/revisions"}],"predecessor-version":[{"id":1816,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/1811\/revisions\/1816"}],"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=1811"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}