Suppose F, E are fields and F is a subring of E. Suppose g is an element of E and is algebraic over F, that is p(g)=0 for some nonzero polynomial p(x) in F[x]. Then there must be a nonzero polynomial m(x) smallest degree among those nonzero polynomials in F[x] with g as a root. Prove m(x) cannot be factored as a(x)b(x) for polynomials a(x), b(x) of smaller degree in F[x].
The suppose q(x) in F[x] with q(g)=0. Show q(x) is an element of m(x)F[x]
(so ker(Eg:F[x]-->E) = m(x)F[x])
Then show that the subring F[g] of E is isomorphic to the quotient ring
1) Prove m(x) cannot be factored as a(x)b(x) for polynomials a(x), b(x) of smaller degree in F[x].
Suppose, it can. Then, m(x)=a(x)b(x), and m(g)=a(g)b(g)=0. Since a(g) and b(g) are elements of the field E, and so E has no zero divisors, we have a(g)=0 or b(g)=0. But if a(g)=0, then since the degree of a(x) is less than the degree of m(x), we have a contradiction with the fact that m(x) has the ...
Ring theory is assessed in the solution.