Finding minimal polynomial when the straight equation must be reduced. I'm asked to calculate $Irr(\sqrt[4]{2},F_3)$ this is in principle easy, I managed to find that the polinomial $X^4 - 2$ factorizes as the product of two irreducible $(x^2+x+2)(x^2+2x+2)$. Hence, since $\sqrt[4]{2}$ is a root of the first polinomial it has to be a root of some of these factors. 
However, how can I decide from what factor is $\sqrt[4]{2}$ a root? Is there a simple reasoning for this? 
 A: Your notation is unfamiliar; I think you're asking for the minimial polynomial of $\sqrt[4]{2}$ over the field $\mathbb{F}_3$ of three elements?
One thing you're overlooking is that the notation $\sqrt[4]{2}$ is ambiguous: there are four distinct elements of $\mathbb{F}_9$ that could claim to be given by such notation. Two of them are roots of $x^2 + x + 2$, and two of them are roots of $x^2 + 2x + 2$.
You have to have some way of deciding which of these four fourth roots of $2$ you're interested in before you can determine which of the two quadratics is the minimal polynomial.
For example, if you have a specific construction of $\mathbb{F}_9$ you can do arithmetic in, and are using $\sqrt[4]{2}$ to refer to a specific element of that field, you can just plug it into both quadratics to see which one gives zero.
A: Since we are working over $\mathbf F_3$, it would perhaps be simpler to replace the polynomial $X^4 - 2$ by $f=X^4 + 1=(X^2+X-1)(X^2-X-1)$ , whose roots in a given algebraic closure $\bar {\mathbf F}_{3}$ of $\mathbf F_3$ are the distinct $8$-th roots of $1$, wich form a cyclic group  of order $8$, generated by an arbitrary primitive $8$-th root of $1$, say $\omega$ (or $\sqrt[4] 2$ in your notation). Note that all such primitive roots are of the form $\omega^k$, with $k$ coprime to $8$, so their number is $4$.
Because the only quadratic extension of $\mathbf F_3$ in $\bar {\mathbf F}_{3}$ is $\mathbf F_{3^2}$,  the splitting field of the polynomial $f$ is no other than $\mathbf F_{3^2}$. The same uniqueness property also shows that both the irreducible polynomials $f_1=(X^2+X-1)$ and $f_2=(X^2-X-1)$ have the same splitting field $\mathbf F_{3^2}$. Choose for instance $\omega$ to be a root of $f_1$; the product of the roots of $f_1$ being $-1$, the other root is $\omega^3$. This implies that the roots of $f_2$ are $\omega^5$ and $\omega^7$.
