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
F[x]/m(x)F[x].

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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 ...

Please assist me with the attached homomorphism questions. Thanks!
Example:
? Let f: G -->H be a group homomorphism with kernel K = Ker(f), show that f is one to one if and only if K = ...

Here's my problem:
If A and B are subsets of a group G, define
AB = {ab | a 2 A, b 2 B}. Now suppose phi: G -> G0 is a homomorphism of groups and N = Ker(phi) is its kernel.
(i) If H is a subgroup of G, show that HN = NH. (Warning: this is an equation of sets; proceed
accordingly; do not assume that G is abelian.)
(i

Note:
S4 means symmetric group of degree 4
A4 means alternating group of degree 4
e is the identity
Is there a group homomorphism $:S4 -> A4, with
kernel $ = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}?

I am stuck with this question.
If phi : F -> R is a nonzero homomorphism from a field F to a ring R, show that phi is one-to-one (Hint: recall that for a ring homomorphism, phi is one-to-one if and only if Ker(phi)={0}. )
Can you help with this?
Many thanks in advance

Let G be a group and let H be a normal subgroup of G. Let m be the index of H in G (that is, the number of cosets of H). Prove that for any a we have am H.
(b) Give an example of group G, a subgroup H of index in, and an element a G such that am is not in H. (Of course, your subgroup H had better not be normal.)
(4) (a) Suppos