 reserve R for commutative Ring;
 reserve A,B for non degenerated commutative Ring;
 reserve h for Function of A,B;
 reserve I0,I,I1,I2 for Ideal of A;
 reserve J,J1,J2 for proper Ideal of A;
 reserve p for prime Ideal of A;
 reserve S,S1 for non empty Subset of A;
 reserve E,E1,E2 for Subset of A;
 reserve a,b,f for Element of A;
 reserve n for Nat;
 reserve x,o,o1 for object;
 reserve m for maximal Ideal of A;
 reserve p for prime Ideal of A;
 reserve k for non zero Nat;

theorem Th28:
  sqrt J = meet PrimeIdeals(A,J)
    proof
      for x be object st x in sqrt J holds x in meet PrimeIdeals(A,J)
      proof
        let x be object;
        assume x in sqrt J; then
        x in {a where a is Element of A: ex n be Element of NAT st a|^n in J}
          by IDEAL_1:def 24; then
        consider y be Element of A such that
A2:    y = x and
A3:    ex n being Element of NAT st y|^n in J;
       consider k be Element of NAT such that
A4:    y|^k in J by A3;
       y in meet PrimeIdeals(A,J)
       proof
         for Y be set holds Y in PrimeIdeals(A,J) implies y in Y
         proof
           let Y be set;
           Y in PrimeIdeals(A,J) implies y in Y
           proof
             assume Y in PrimeIdeals(A,J); then
             consider Y1 be prime Ideal of A such that
A6:          Y1 = Y and
A7:          J c= Y1;
             reconsider k as non zero Nat by A4,A7,Lm24;
             y|^k in Y1 by A4,A7;
             hence thesis by A6,Th23;
           end;
           hence thesis;
         end;
         hence thesis by SETFAM_1:def 1;
       end;
       hence thesis by A2;
     end; then
A9:  sqrt J c= meet PrimeIdeals(A,J);
     set N1 = meet PrimeIdeals(A,J);
    for x be object st not x in sqrt J holds not x in N1
    proof
      let x be object;
      assume
A10:  not x in sqrt J;
      per cases;
        suppose x in A; then
          reconsider x1 = x as Element of A;
          consider p be prime Ideal of A such that
A12:      not x1 in p and
A13:      J c= p by A10,Th9;
          p in PrimeIdeals(A,J) by A13;
          hence thesis by A12, SETFAM_1:def 1;
        end;
        suppose
A15:      not x in A;
          not x in N1
          proof
            assume
A16:        x in N1;
            consider m be prime Ideal of A such that
A17:        J c= m by Th11;
            m in PrimeIdeals(A,J) by A17; then
            x in m by A16,SETFAM_1:def 1;
            hence contradiction by A15;
          end;
          hence thesis;
        end;
      end; then
      meet PrimeIdeals(A,J) c= sqrt J;
      hence thesis by A9;
  end;
