Minimal polynomial over a field I have some troubles with this exercise:

Let $a$ and $b$ be the following elements of $\mathbb{Q}(T)$:
  $a:=T^3+T$ and $b:=T^2 -2$
  Compute the minimalpolynomial of $b$ over $\mathbb{Q}(a)$, and use the answer to find a nonzero polynomial $f\in \mathbb{Q}[X_1,X_2]$ such that $f(a,b)=0$

Now the first question would be isn't $\mathbb{Q}(T^3+T)=\mathbb{Q}(T)$ true, because I can work with $T$ as a basis and therefore I can build $a$ and $b$? Or do the elements in $\mathbb{Q}(T^3+T)$ look like this: $1,(T^3+T),(T^3+T)^2...$?
Now over $\mathbb{Q}[X]$ this would be $X-T^2+2$, but I'm a little confused about $\mathbb{Q}(T^3+T)$. Also I guess $f$ would be $f:X_1-T^3+T +X_2-T^2+2$ only if $a,b\in\mathbb{Q}[X_1,X_2]$ but I don't know if they are elements of it...
Thank you for your help.
 A: As explained in the comments we have every reason to look for a cubic minimal polynomial. So we begin by calculating
$$
b^2=T^4-4T^2+4,\qquad b^3=T^6-6T^4+12T^2-8.
$$
We want to write some linear combination of powers of $b$ in terms of powers of $a$. As the highest power of $T$ occurring above was $T^6$ it is natural to suspect that we only need $a$ and $a^2=T^6+2T^4+T^2$. This suggests that we look for a linear combination of $b^3$ and $b^2$ that matches with the degree $6$ and $4$ terms of $a^2$. Such a combination is
$$
b^3+8b^2=T^6+2T^4-4T^2+24=a^2-5T^2+24.
$$
I'm sure that you can find constants $c_1,c_2\in\Bbb{Q}$ such that $c_1b+c_2=5T^2-24$. This gives you the solution 
$$b^3+8b^2+c_1b+(c_2-a^2)=0.$$

The resulting polynomial is surely irreducible. The minimal polynomial cannot be a quadratic as $[\Bbb{Q}(T):\Bbb{Q}(a)]=3$ and cannot contain a quadratic intermediate field. The minimal polynomial cannot be linear for then $b\in\Bbb{Q}(a)$. As $\Bbb{Q}(a,b)=\Bbb{Q(T)}$ this would imply that $T\in \Bbb{Q}(a)$ violating transcendence of $T$ over $\Bbb{Q}$.
