TITLE: Minimal polynomials of $\sqrt[4]{2}i$ over $\mathbb{Q}$ and $\mathbb{R}$
QUESTION [5 upvotes]: Can someone tell me if this is right:
I would like to find the minimal polynomial of 
(i) $\sqrt[4]{2}i$ over $\mathbb{Q}$
(ii) $\sqrt[4]{2}i$ over $\mathbb{R}$
(i):
$\sqrt[4]{2}i$ is a root of $f(x) = x^4-2$. This is already monic, so to show that this is a minimal polynomial I only need to show that it is irreducible. Edit: To do that, I can use the Eisenstein: $p=2$ does not divide $a_1 = 1$ and $p^2$ does not divide $a_0 = -2$, therefore it is irreducible over $\mathbb{Q}$.
(ii):
This time, $\sqrt[4]{2}i$ is a root of $f(x) = x^2+\sqrt{2}$. For polynomials of degree two it's enough to check if they have a root. The only roots this one has are complex therefore it is irreducible over $\mathbb{R}$ and therefore the minimal polynomial.
So my more general question is: is the way to find a minimal polynomial of an element $a$ in general to find a polynomial $f$ such that $f(a) = 0$ and then to norm $f$ and then to show that $f$ is irreducible?
Many thanks for your help!

REPLY [4 votes]: Part (i) as given is incorrect: lack of roots does not imply irreducibility when the polynomial is of degree greater than 3, because the polynomial could split into irreducible polynomials of degrees greater than 1.
You have shown that $x^4-2$ has no linear factors modulo $3$, but you cannot conclude from this that it is irreducible modulo $3$: it could have two irreducible quadratic factors. And indeed, as chandok writes,
\begin{align*}
(x^2+x+2)(x^2+2x+2) &= x^4 + 2x^3 + 2x^2 + x^3 + 2x^2 + 2x + 2x^2 + 4x + 4\\
&= x^4 + 3x^3 + 6x^2 + 6x + 4 = x^4 + 1 = x^4 - 2,
\end{align*}
so the polynomial is not irreducible over $\mathbb{F}_3$. 
Instead, I would suggest using Eisenstein's Criterion to show $x^4-2$ is irreducible over $\mathbb{Q}$, as it is pretty easy to apply there.
Part (ii) is correct.
As to your final question: there is a top-down approach and a bottoms-up approach. In the top down approach, if you can find any polynomial $f(x)$ that has $a$ as a root, then you know that the minimal polynomial will divide $f(x)$; in fact, it will be an irreducible factor of $f$. So you can try to find which irreducible factor of $f$ has $a$ as a root. This may be $f$ itself, or some factor. For example, to find the minimal polynomial of a $7$th root of unity, you could use the fact that it satisfies $x^7-1$. Then factor $x^7-1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$, and you know that $a$ is a root of either $x-1$ or the second factor. Then you would look at the second factor (since presumably your $a$ is not $1$). And so on.
The bottoms-up approach is described by Jim Belk in his answer, where you "build up" the polynomial by considering powers of $a$ (or alternatively, images under the action of some Galois group).