Double finite field extension Suppose we are given the field $\mathbb{F}_5$ and
$p(X) = X^2-2 \in \mathbb{F}_5[X]$, an irreducible polynomial over $\mathbb{F}_5$.
Let $\mathbb{K}$ denote the extension of $\mathbb{F}_5$ in which $p(X)$ has a root $\alpha$.  $\mathbb{K}$ is an extension of degree $2$ and of cardinality $5^2=25$.
It is easily verified that $\alpha$ is not a square in $\mathbb{K}$, so $q(Y) = Y^2 - \alpha \in \mathbb{K}[Y]$ is an irreducible polynomial over  $\mathbb{K}$. We can therefore form an extension of $\mathbb{K}$ in which $q(Y)$ has a root $\beta$, i.e. an element such that $\beta^2=\alpha$. I will call this extension $\mathbb{L}$. $\mathbb{L}$ is an extension of degree $2$ over $\mathbb{K}$ and of cardinality $5^4 = 625$.
I am trying to find the minimal polynomial of $\beta$ over $\mathbb{F}_5$ which gives rise to the same extension $\mathbb{L}$, i.e. the polyomial $s(Z) \in \mathbb{F}_5[Z]$ such that $s(\beta)=0$.
$s(Z)$ should be of degree $4$ and $\mathbb{L}$, being the splitting field of $s(Z)$, will contain all its roots which are given by $\beta, \beta^5, \beta^{25}, \beta^{125}$.  So, one way to find $s(Z)$ is to compute and simpify the product $s(Z) = (Z-\beta)(Z-\beta^5)(Z-\beta^{25})(Z-\beta^{125})$. I got $s(Z)=Z^4+3$.
My question: is there another, more intelligent, method to find $s(Z)$ that would work even when the involved fields are of greater cardinality?
 A: You're adding in $\sqrt{2}$ and then $\sqrt{\sqrt{2}}$. So intuitively it's obvious that really you're just adding a root of $x^4-2$.
This is the same polynomial as you got.
More formally speaking, you could verify the isomorphism of rings:
$\mathbb{F}_5[x,y]/\langle x^2 - 2, y^2 -x \rangle \cong \mathbb{F_5}[y]/\langle y^4-2 \rangle$
This amounts to checking that the equality of the ideals
$\langle x^2 - 2, y^2 -x \rangle = \langle y^4-2, y^2 - x\rangle$
The only non-trivial step there is that $x^2-2 \in \langle y^4-2, y^2 - x\rangle$.
A: In general, suppose you have a field extension $K(\beta) / K$, and $\alpha \in K(\beta)$, and suppose that $f(x)$ is the minimal polynomial of $\beta$ over $K(\alpha)$.
Recall that there is a norm function $K(\alpha) \to K$. There are a few different ways to express this e.g. if $L/K$ is the Galois closure of $K(\alpha) / K$, then the norm of $\xi$ is the product of all of the conjugates of $\xi$. i.e. if $m$ is the minimal polynomial of $\xi$ then
$$ N_{K(\alpha) / K}(\xi) = \prod_{\substack{\rho \in L \\ m(\rho) = 0}} \rho $$
There are other approaches involving resultants or determinants too; which one we use isn't especially important....
But the key idea is that we can extend the norm function to polynomials and rational functions: e.g. $K(x, \alpha) / K(x)$ is a field extension, and there is a norm function sending $K(x, \alpha) \to K(x)$. This extends the norm on $K(\alpha) / K$.
And so, we can set $g(x) = N_{K(x,\alpha)/K(x)}(f(x))$. Then $g(x)$ is a polynomial over $K$, $\beta$ is one of its roots, and
$$\deg g(x) = (\deg f(x)) \cdot [K(\alpha) : K] = [K(\beta) : K(\alpha)] \cdot [K(\alpha) : K] = [K(\beta) : K]$$
Consequently, $g(x)$ must be the minimal polynomial of $\beta$ over $K$,
e.g. in your example, the norm of $Y^2 - \alpha$ is
$$ (Y^2 - \alpha)(Y^2 - \bar{\alpha}) = Y^4 - (\alpha + \bar{\alpha})Y^2 + \alpha \bar{\alpha} = Y^4 - 2$$
where $\bar{\alpha} = -\alpha$ is the other square root of $2$. The fact $\alpha + \bar{\alpha} = 0$ can be obtained by looking at the minimal polynomial of $\alpha$: in particular, it is the negation of the coefficient on $x$ (which is zero). The same with $\alpha \bar{\alpha} = -2$.
A: $\{ 1, \alpha, \beta, \alpha \beta \}$ is a $\mathbf{F}_5$-basis for $\mathbf{F}_5(\alpha, \beta)$, and so there must be a linear combination of $1, \beta, \beta^2, \beta^3, \beta^4$, which gives us the minimal polynomial. We can compute these:
$$ \beta^0 = 1$$
$$ \beta^1 = \beta$$
$$ \beta^2 = \alpha$$
$$ \beta^3 = \alpha \beta $$
$$ \beta^4 = 2 $$
and we can do linear algebra to solve. Of course, in this case we can do it by inspection, and see that
$$ \beta^4 - 2 = 0$$
