Let f(x), g(x), h(x), epsilon F[x] such that f(x) does not equal 0. Prove that: If f(x) | g(x) and f(x) | [g(x)^2 +h(x)], then f(x). Hint: g(x)^2 = g(x) * g(x) Solution given f(x) | g(x) =>g(x)=k.f(x) =>g(x)2=k2.f(x)2 =>g(x)2=l.f(x) eq1 where l=k2.f(x) and f(x) | [g(x)^2 +h(x)] =>[g(x)^2 +h(x)]=m.f(x) eq2 eq2-eq1 =>h(x)=(l-m)f(x)=p.f(x) where p=l-m there fore f(x)|h(x) then f(x). Hint: g(x)^2 = g(x) * g(x).

1. Let R be an integral domain and F a field such that R is a subring of F, and every element of F is
equal to the quotient of two elements of R. Show that F is isomorphic to Q(R) (the field of
quotients of R).
Solution
If x belongs to F, then x is of the form p/q where p and q are elements of R.
Thus let p/q be of the form simplest or p and q have no common factor between them.
Then each x can be expressed uniquely as p/q form.
Thus there exists a bijective mapping from F to Q(R)
Hence F is isomorphic to Q(R)