
theorem constpol:
for R being Ring,
    S being RingExtension of R
for a being Element of R, b being Element of S st a = b holds a|R = b|S
proof
let F be Ring, E be RingExtension of F;
let a be Element of F, b be Element of E;
assume AS: a = b;
set ap = a|F, bp = b|E;
X: F is Subring of E by FIELD_4:def 1;
now let n be Element of NAT;
  per cases;
  suppose A: n = 0;
    hence ap.n = b by AS,Th28 .= bp.n by A,Th28;
    end;
  suppose A: n <> 0;
    hence ap.n = 0.F by Th28 .= 0.E by X,C0SP1:def 3
              .= bp.n by A,Th28;
    end;
  end;
hence thesis;
end;
