Basis of $\mathbb{Q}(t)$ over $\mathbb{Q}(t^3-2)$ While solving some problems in galois theory. I got to know that $\mathbb{Q}(t)$ over $\mathbb{Q}(t^3-2)$ is a finite extension.
But I am not sure how do I get a basis explicitly.
Can anyone help me with this. I know the minimal polynomial but still I would like to get a basis explicitly.
Further I have following question.
Can I write $\mathbb{Q}(t) \equiv \mathbb{Q}(t^3-2)/(f(t)$ for some polynomial $f(t)$?
I can see that
$\mathbb{Q}(t) =\mathbb{Q}(t^3-2)(t) \equiv \frac{\mathbb{Q}(t^3-2)[x]}{(x^3-2-(t^3-2))}$
so can I say that $\{1,t,t^2\}$ is a basis but I would like to see how any element is being spanned by these explicitly. I know linear independence is difficult to prove because that basically means that the polynomial is irreducible.
 A: First note that any polynomial in $ P\in\mathbb{Q}[t]$ can be written uniquely in the form $P=A+Bt+Ct^2$ where $A,B,C\in\mathbb{Q}[t^3-2]$. Then the equation of the form $PQ=1$ in $\mathbb{Q}(t)=\mathbb{Q}(t^3-2)[x]/(x^3-2-(t^3-2))$ can be solved in the form $Q=A'+B't+C't^2$, now $A',B',C'\in\mathbb{Q}(t^3-2)$. The reason is that by using the relation $x^3-2=t^3-2$ we can reduce the left hand side of the equation to degree $\le2$, and we can compare the coefficients, which amounts to solving 3 linear equations in $\mathbb{Q}(t^3-2)$. The solution must exist, sice if not, the ideal $  (P)\in\mathbb{Q}(t^3-2)[x]/(x^3-2-(t^3-2))$ do not contain 1, whic occurs only when $P=0$. And we have obtained every element $\frac{P}{Q}\in \mathbb{Q}(t)$ in the desired form.
Let us work on the exapmle $$ \frac{P}{Q}=\frac{t^4-2t}{t^4+1}.$$
Let us denote $T=t^3-2$. First, we want to represent $P$ and $Q^{-1}$ as polynomials of degree $\le 2$ in $\mathbb{Q}(T)$. We can write
$$\begin{cases} P=Tt \\ 
Q=(T+2)t+1 \end{cases}.$$
Next, using the relation $t^3=T+2$, $$(at^2+bt+c)Q=(a+b(T+2))t^2+(b+c(T+2))t+a(T+2)^2+c=1$$
and we can compare the coefficients. We obtain $$Q^{-1}=\frac{(T+2)^2}{(T+2)^4+1}t^2-\frac{T+2}{(T+2)^4+1}t+\frac{1}{(T+2)^4+1}.$$
Multiplying $P$ and $Q^{-1}$, and again using $t^3=T+2$,$$\frac{P}{Q}=-\frac{T(T+2)}{(T+2)^4+1}t^2+\frac{T}{(T+2)^4+1}t+\frac{T(T+2)^3}{(T+2)^4+1}.$$
