To get you started:
Case 1: a is in F, then h(t) = (t - a) divides both f and g, and it clearly takes coefficients from F.
Case 2: a is not in F. Now f(t) and g(t) have a unique factorization in the form:
f(t) = c(\phi _1 (t))^{n_1}(\phi _2 (t))^{n_2} \dots (\phi _k (t))^{n_k}
Where c is some element of F, \forall \ i,\ \phi _i is a monic irreducible polynomial, and n_i \in \mathbb{N}. We also know that for some i, let it arbitrarily be 1, \phi _1 (t) = (t - a). For simplicity sake, let c = 1 (and something should be similarly true for g(t)). So:
f(t) = (t - a)^{n_1}(\phi _2 (t))^{n_2} \dots (\phi _k (t))^{n_k}
g(t) = (t - a)^{m_1}(\psi _2 (t))^{m_2} \dots (\psi _l (t))^{m_l}
You need to show that if f and g are to have only coefficients in F, then for some i and j such that 1 \leq i \leq k and 1 \leq j \leq l, \phi _i = \psi _j. This will reduce to showing that any polynomial with (t - a) as a factor will have to have some factor with coefficients in F for the whole thing to have coefficients in F. We can write:
f(t) = (t - a)(t^N + a_{N - 1}t^{N - 1} + \dots + a_0)
= t^{N + 1} + (a_{N - 1} - a)t^N + (a_{N - 2} - a_{N - 1}a)t^{N - 1} + \dots + (a_0 - a_1a)t + a_0a
Note that N must be at least 1, otherwise if N were 0, you'd have f(t) being (t - a) times a non-zero constant, C, and Ct - Ca = f(t), but if Ca is in F, then C is not in F (recall that a was assumed (the whole point of the second case) to not be in F, so it's not zero, so it's inverse exists and is in E but not in F (otherwise if 1/a was in F, then since F is a field, 1/(1/a) = a would be in F, contradiction), so the coefficient of "t" is C which is not in F, so N really must be at least one).
At this point, I would try doing some induction on N to find out what form the polynomial (t^N + a_{N - 1}t^{N - 1} + \dots + a_0) must take in order for (a_{N - 1} - a),\ (a_{N - 2} - a_{N - 1}a),\ \dots ,\ (a_0 - a_1a),\ a_0a are in F. Then show that all polynomials f and g such that (t^N + a_{N - 1}t^{N - 1} + \dots + a_0) takes the required form must be factorizable by some polynomial h over F.
Start with n = 1. f(t) = (t - a)(t + a_0) = t^2 + (a_0 - a)t - a_0a. Now a \in E,\ a \notin F,soa \neq 0, but a_0a \in F, so a_0 = k/a for some k /in E. If k = 0, then the coefficient of "t" is not in F, so k \neq 0. Now, we have:
a_0 - a = k/a - a \in F
You want to convince yourself that the k is unique. Assume for some other K \in F,\ K \neq k where clearly, K \neq 0, we have that K/a - a \in F. Then:
K/a - a - (k/a - a) \in F
K/a - k/a \in F
(1/a)(K - k) \in F
K - k = La for some L \in F. Now if L \neq 0, then K = k + La but k \in F, La \notin F, so k + La \notin F while K \in F, contradiction, so L = 0, and hence K = k, so k is unique. Now, if you can show that every polynomial over F that has (t - a) as a factor must also have (t + k/a) as a factor, then you know that every such polynomial will have (t - a)(t + k/a) = t² + (k/a - a)t - k as it's factor, and h(t) = t² + (k/a - a)t - k is a polynomial over F.
I hope this is on the right track. I would hope you know some theorems that you can use to do this problem, I'm working with no special knowledge of the subject.
