Find the minimal polynomial of $z=u+i \in \mathbb{C}$ in $\mathbb{Q}$ where $u=\sqrt{7}$ There is something that I don't understand in this problem's solution. It finds $f=T^4-12T^2+64 \in \mathbb{Q}[T]$ where $f(z) = 0$.
At this point, I thought I could just use Eisenstein's criterion with $d=2$ to claim that $f$ is irreducible and hence the minimal polynomial.
However, in the solution it claims $f$ is the asked polynomial  using $[\mathbb{Q}(z):\mathbb{Q}]=[\mathbb{Q}(u,i):\mathbb{Q}]=[\mathbb{Q}(u)(i):\mathbb{Q(u)}][\mathbb{Q}(u):\mathbb{Q}]=2.2=4$.
Is my approach correct, if not, why?
Thank you!
Edit: As indicated in the comments, I totally missed that I can't use Eisenstein because $d^2$ divides $a_0$.
 A: If $z=u+i$, then $i=z-u$, so
$$
-1=z^2-2zu+u^2
$$
or
$$
z^2+8=2zu
$$
and, finally,
$$
z^4+16z^2+64=28z^2
$$
which tells the polynomial
$$
T^4-12T^2+64
$$
has $z$ among its roots.
Now, $\mathbb{Q}(z)=\mathbb{Q}(u,i)$, because from the first relation you can derive
$$
u=\frac{z^2+8}{2z}\in\mathbb{Q}(z)
$$
so also $i\in\mathbb{Q}(z)$. The other inclusion is trivial.
Now $[\mathbb{Q}(u,i):\mathbb{Q}]=[\mathbb{Q}(u,i):\mathbb{Q}(u)][\mathbb{Q}(u):\mathbb{Q}]$ and both dimension on the right hand side are equal to $2$.
If you are able to prove that the polynomial is irreducible, by Eisenstein's criterion or other methods, there's nothing wrong in concluding that it's the minimum polynomial. But in some cases this is more difficult than counting dimensions.
In this case Eisenstein's criterion is not applicable. But you can get through doing
\begin{align}
T^4-12T^2+64&=T^4+16T^2+64-28T^2\\
&=(T^2+8)^2-28T^2\\
&=(T^2-2\sqrt{7}\,T+8)(T^2+2\sqrt{7}\,T+8)
\end{align}
which is the factorization into irreducible factors in $\mathbb{R}[T]$. By uniqueness of decompositions, the polynomial is irreducible in $\mathbb{Q}[T]$.
