Problem:
Consider the field extension $\mathbb{F}_{5^4}$ over $\mathbb{F}_5$.
(a) Determine the number of elements $a$ satisfying $\mathbb{F}_{5^4} = \mathbb{F}_5(a)$.
(b) Determine the number of irreducible polynomials of degree $4$ in $\mathbb{F}_5[x]$.
My attempt for part (a) was to consider the polynomial $f(x) = x^{5^4}- x$, because I know it splits in $\mathbb{F}_{5^4}$ and the extension is Galois. So if $a$ is a root of $f(x)$ and $a$ is not in $\mathbb{F}_5$, then $a$ is in $\mathbb{F}_{5^4}$. But I don't see a way to access these elements in a way that allows them to be counted.
For part (b) I know that the number of irreducible polynomials of degree $n$ over $\mathbb{F}_p$ is given by the general formula $$\psi(n) = \frac{1}{n} \sum_{d | n} \mu(d) p^{n/d}.$$ So \begin{align} \psi(4) &= \frac{1}{4}[\mu(1)5^4 + \mu(2)5^2 + \mu(4)5^1 ] \\ &= \frac{1}{4}[ 5^4 - 5^2] \\ &= 150 \end{align} However, using the formula wasn't very enlightening. Is there a nice way to use part (a) to make a counting argument?
(a) $\mathbb{F}_5(a)=\mathbb{F}_{5^4}$ iff $\mathbb{F}_5(a)$ is not a strictly smaller subfield of $\mathbb{F}_{5^4}$. Because $4$ has only two proper divisors ($2$ and $1$), there are only two such subfields: $\mathbb{F}_{5^2}$ and $\mathbb{F}_{5^1} \subseteq \mathbb{F}_{5^2}$. Therefore, $\mathbb{F}_5(a)=\mathbb{F}_{5^4}$ iff $a \in \mathbb{F}_{5^4} \setminus \mathbb{F}_{5^2}$.
(b) $\mathbb{F}_5(a)=\mathbb{F}_{5^4}$ iff $a$ is a root of a polynomial of degree $4$ irreducible over $\mathbb F_5$. There are $600=625-25$ possibilities for $a$. They can be grouped in groups of $4$, the roots of the same irreducible polynomial of degree $4$. Therefore, there are $600/4=150$ such polynomials.