Existence of a Polynomial Divisor for Roots in a Field Extension

[Jin314159]

Given a field E which contains a field F. Let f and g be polynomials with coefficients from F. And let a be an element of E such that a is a root of both f and g. Prove that there exists a polynomial h with coefficients in F that divides both f and g.

 

[Hurkyl]

So, what you're looking for is a common factor of f and g, right?

 

[AKG]

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 [itex]a \in E,\ a \notin F,$so$a \neq 0[/itex], 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.