Solving equation for a field extension

Question

Let $F(x)=x^{3}+2 x-2,$ let $\alpha \in \mathbb{C}$ be a root of $F,$ and let $K=\mathbb{Q}(\alpha)$

Find $a, b, c \in \mathbb{Q}$ such that $\alpha^{4}=a \alpha^{2}+b \alpha+c$

We have that [$K:\mathbb{Q}]=3$.

I know that such a thing is possible, but do not know in what manner to proceed. Please leave a hint only as this is a homework assignment.

Answer 1

$\alpha$ is a root of $F$, so $\alpha^3 + 2\alpha - 2 = ?$ And you should be able to solve that for $\alpha^4$ (by multiplying by one thing, then moving all non-$\alpha^4$ terms to the right-hand side).

Answer 2

Since $\alpha$ is a root of

$f(x) = x^3 + 2x - 2, \tag 1$

we have

$\alpha^3 + 2\alpha - 2 = 0, \tag 2$

or

$\alpha^3 = -2\alpha + 2; \tag 3$

thus,

$\alpha^4 = -2\alpha^2 + 2\alpha; \tag 4$

therefore,

$a = -2, \tag 5$

$b = 2, \tag 6$

and

$c = 0. \tag 7$

In order to show uniqueness, suppose there are

$q, r, s \in \Bbb Q \tag 8$

with

$\alpha^4 = q\alpha^2 + r\alpha + s; \tag 9$

we subract (4) from (9):

$0 = (q + 2)\alpha^2 + (r - 2)\alpha + s; \tag{10}$

thus $\alpha$ must satisfy the polynomial

$g(x) = (q + 2)x^2 + (r - 2)x + s \in \Bbb Q[x]; \tag{11}$

we observe that $f(x)$ is irreducible over $\Bbb Q$ via the Eisenstein criterion with prime $p = 2$; therefore it is minimal for $\alpha$ over $\Bbb Q$; but this contradicts

$g(\alpha) = 0 \tag{12}$

unless

$g(x) = 0; \tag{13}$

therefore (13) binds and thus

$q = -2 = a, \tag{14}$

$r = 2 = b, \tag{15}$

and

$s = c = 0, \tag{16}$

and we conclude the expression (4) for $\alpha^4$ is unique.