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?

Answer

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 $x^2-2 \in \langle y^4-2, y^2 - x\rangle$.