set K = GF p;
A1: p > 1 by INT_2:def 4;
now :: thesis: for K1 being strict Subfield of GF p holds K1 = GF p
let K1 be strict Subfield of GF p; :: thesis: K1 = GF p
set C = the carrier of (GF p);
set C1 = the carrier of K1;
set n1 = p - 1;
reconsider n1 = p - 1 as Element of NAT by A1, NAT_1:20;
A2: for x being set st x in GF p holds
x in K1
proof
A3: for n being Element of NAT st n in Segm p holds
n in the carrier of K1
proof
defpred S1[ Nat] means p in the carrier of K1;
0 in Segm p by NAT_1:44;
then 0. (GF p) = 0 by SUBSET_1:def 8;
then 0. K1 = 0 by Def1;
then A4: S1[ 0 ] ;
A5: now :: thesis: for n being Element of NAT st 0 <= n & n < n1 & S1[n] holds
S1[n + 1]
let n be Element of NAT ; :: thesis: ( 0 <= n & n < n1 & S1[n] implies S1[n + 1] )
assume A6: ( 0 <= n & n < n1 ) ; :: thesis: ( S1[n] implies S1[n + 1] )
assume A7: S1[n] ; :: thesis: S1[n + 1]
A8: 1. K1 = 1. (GF p) by Def1
.= 1 by A1, INT_3:14 ;
then A9: [1,n] in [: the carrier of K1, the carrier of K1:] by A7, ZFMISC_1:87;
A10: the addF of K1 = the addF of (GF p) || the carrier of K1 by Def1;
A11: 1 + n < n1 + 1 by A6, XREAL_1:6;
n < n1 + 1 by A6, NAT_1:13;
then A12: ( 1 in Segm p & n in Segm p ) by A1, NAT_1:44;
the addF of K1 . (1,n) = (addint p) . (1,n) by A9, A10, FUNCT_1:49
.= (1 + n) mod p by A12, GR_CY_1:def 4
.= 1 + n by A11, NAT_D:63 ;
hence S1[n + 1] by A7, A8, BINOP_1:17; :: thesis: verum
end;
A13: for n being Element of NAT st 0 <= n & n <= n1 holds
S1[n] from INT_1:sch 7(A4, A5);
thus for n being Element of NAT st n in Segm p holds
S1[n] :: thesis: verum
proof
let n be Element of NAT ; :: thesis: ( n in Segm p implies S1[n] )
assume A14: n in Segm p ; :: thesis: S1[n]
( 0 <= n & n < n1 + 1 ) by A14, NAT_1:44;
then ( 0 <= n & n <= n1 ) by NAT_1:13;
hence S1[n] by A13; :: thesis: verum
end;
end;
thus for x being set st x in GF p holds
x in K1 by A3; :: thesis: verum
end;
GF p is strict Subfield of GF p by Th1;
then GF p is strict Subfield of K1 by A2, Th7;
hence K1 = GF p by Th4; :: thesis: verum
end;
then for K1 being Field st K1 is strict Subfield of GF p holds
K1 = GF p ;
hence GF p is prime ; :: thesis: verum