Algebraic over $\Bbb Q$ Hi I was wondering if anyone could help me with the following problem I'm trying to solve it is below.

Let $\alpha=\sqrt[4]{20}$, then put $F=\{a_0+a_1\alpha+a_2\alpha^2+a_3\alpha^3:a_0,a_1,a_2,a_3\in \Bbb Q\}$.
The questions are:
(i) Prove that $F=\Bbb Q(\alpha)$ and
(ii) evaluate $(1+\alpha^3)^{-1}$ as an element of $F$ and verify your answer.

My attempt is as follows
(i) $\alpha=\sqrt[4]{20}$ $=>\alpha^4-20=0$
Now let $m(x)=x^4-20$ then $m(x)\in\Bbb Q[x]$ and $m(x)$ is monic
Now $m(\alpha)=\alpha^4-20=0$ thus $m(x)$ is the minimum polynomial of $\alpha$ over $\Bbb Q$ $<=> m(x)$ is irreducible in $\Bbb Q[x]$
Now $m(x)\in\Bbb Z[x]$ and is irreducible in $\Bbb Z[x]$ by E.I.T using prime $5$, and since $m(x)$ is irred in $\Bbb Z[x]$ this implies $m(x)$ is irreducible in $\Bbb Q[x]$
So $x^4-20$ is the min poly of $\alpha$ over $\Bbb Q$ since $\alpha$ is algebraic over $\Bbb Q$
thus $[\Bbb Q(\alpha):\Bbb Q]=deg(m(x))=4$.
Thus $1,\alpha,\alpha^2,\alpha^3$ is a basis for $\Bbb Q(\alpha)$ over $\Bbb Q$
Thus $\Bbb Q(\alpha)=a_0+a_1(\alpha)+a_2(\alpha^2)+a_3(\alpha^3):a_0,a_1,a_2,a_3\in \Bbb Q=F$ as required
This is the end of my solution for part (i) I can't see anywhere I'm going wrong with this part but I put it on here in case I have made an error somewhere?
part (ii)
Evaluate $(1+\alpha^3)^{-1}$
Put $g(x)=1+x^3)$ then I need to show that the $gcd(m(x),g(x))=1$ as I need to find an $R(x),S(x):1=R(x)(1+x^3)+S(x)(x^4-20)$
this is as far as I have got, I dont know where to go from here so any help would be highly appreciated
 A: You don't need to do any polynomial division. In part (i), you have proved $\Bbb{Q}(\alpha)$ has $1, \alpha, \alpha^2$ and $\alpha^3$ as a basis over $\Bbb{Q}$. In part (ii) you use this: to invert $1 + \alpha^3$, you have to solve:
$$
\begin{align*}
1 &= (1 + \alpha^3)(a_0 + a_1\alpha + a_2\alpha^2 + a_3\alpha^3)\\
  &= (a_0 + 20a_1) + (a_1 + 20a_2)\alpha + (a_2 + 20a_3)\alpha^2 + (a_3 + a_0)\alpha^3
\end{align*}
$$
where the second equation follows by multiplying out and using $\alpha^4 = 20$.
By part (i), the coefficients of $\alpha^3$, $\alpha^2$ and $\alpha$ must be zero and the constant term must be $1$, so you must have:
$$
\begin{array}{rllr}
a_3 &= -a_0 &\\
a_2 &= -20a_3 &= 20a_0\\
a_1 &= -20a_2 &= -400a_0\\
a_0 &= 1 -20a_1 &= 1 + 8000a_0
\end{array}
$$
Now you can easily solve the last equation for $a_0$ and read off the other $a_i$, to get
$$
(1 + \alpha^3)^{-1} = \frac{1}{7999}(-1 + 400\alpha -20\alpha^2 + \alpha^3).
$$
A: Computing $(1+\alpha^3)^{-1}$ looks a little hairy using the extended Euclidean algorithm, but you can pull a little trick: first compute $(\alpha+20)^{-1}$ since that will terminate in a single long division, and then note that $\alpha(\alpha+20)^{-1}=(1+\alpha^3)^{-1}$
So, to be explicit:
$\alpha^4 - 20 = (\alpha^3 - 20 \alpha^2 + 400 \alpha - 8000) (\alpha + 20) + 159980$,
and you already have 
$(\alpha + 20)^{-1}\equiv \frac{\alpha^3 - 20 \alpha^2 + 400 \alpha - 8000}{-159980}$
Then multiplying by $\alpha$ you get
$$
(1+\alpha^3)^{-1}\equiv\frac{\alpha^4 - 20 \alpha^3 + 400 \alpha^2 - 8000\alpha}{-159980}\\ \equiv\frac{20- 20 \alpha^3 + 400 \alpha^2 - 8000\alpha}{-159980}\\ \equiv\frac{1-  \alpha^3 + 20 \alpha^2 - 400\alpha}{-7999}
$$
