reserve R for commutative Ring;
reserve A for non degenerated commutative Ring;
reserve I,J,q for Ideal of A;
reserve p for prime Ideal of A;
reserve M,M1,M2 for Ideal of A/q;

theorem Th38:
    for q be proper Ideal of A holds
    sqrt q is maximal implies A/q is local
    proof
      let q be proper Ideal of A;
      set m = sqrt q;
      set E = {I where I is Ideal of (A/q):
        I is quasi-maximal & I <> [#](A/q)};
      assume
A1:   m is maximal;
A2:   (canHom q).:m = nilrad(A/q) by Th37;
      for m1,m2 be object st m1 in m-Spectrum (A/q) & m2 in m-Spectrum (A/q)
      holds m1 = m2
      proof
        let m1,m2 be object;
        assume m1 in m-Spectrum (A/q) & m2 in m-Spectrum (A/q); then
A4:     m1 in E & m2 in E by TOPZARI1:def 7; then
        consider M1 be Ideal of A/q such that
A5:     M1 = m1 and
A6:     M1 is quasi-maximal & M1 <> [#](A/q);
        consider M2 be Ideal of A/q such that
A7:     M2 = m2 and
A8:     M2 is quasi-maximal & M2 <> [#](A/q) by A4;
A9:     M1 is proper by A6;
A10:    M2 is proper by A8;
        set S = {I where I is Ideal of A/q: I is quasi-prime & I <> [#](A/q)};
H:      Spectrum(A/q) = the set of all I where I is prime Ideal of (A/q)
          by TOPZARI1:def 6; then
        M1 in Spectrum (A/q) by A6,A9; then
        meet Spectrum (A/q) c= M1 by SETFAM_1:3; then
A11:    nilrad (A/q) c= M1 by TOPZARI1:19;
        set M = nilrad (A/q);
        M is maximal by A1,TOPZARI1:20,Th30,A2; then
A12:    M is proper quasi-maximal; then
A13:    M1 = M or M1 is non proper by A11;
        M2 in Spectrum (A/q) by H,A8,A10; then
        meet Spectrum (A/q) c= M2 by SETFAM_1:3; then
        nilrad (A/q) c= M2 by TOPZARI1:19; then
A15:    M2 = M or M2 is non proper by A12;
        m1 = m2
        proof
          assume
A16:      m1 <> m2;
          per cases by A13,A15;
            suppose
              M1 = M & M2 = M;
              hence contradiction by A16,A5,A7;
            end;
            suppose
              M1 = M & M2 = [#](A/q);
              hence contradiction by A8;
            end;
            suppose
              M1 = [#](A/q) & M2=M;
              hence contradiction by A6;
            end;
            suppose
              M1 = [#](A/q) & M2= [#](A/q);
              hence contradiction by A16,A5,A7;
            end;
          end;
          hence thesis;
        end;
        hence thesis by TOPZARI1:12;
      end;
