Number of elements $a\in\mathbb{F}_{5^4}$ such that $\mathbb{F}_{5^4}=\mathbb{F}_5(a)$

Question

Determine the number of elements $a\in\mathbb{F}_{5^4}$ such that $\mathbb{F}_{5^4}=\mathbb{F}_5(a)$, and find the number of irreducible polynomials of degree $4$ in $\mathbb{F}_5[x]$.

My thoughts: Once we know the number of elements $a$, I think the number of irreducible polynomials of degree $4$ will be the same. This is because adjoining a root of an irreducible gets you an extension with the same degree as the irreducible, and $\mathbb{F}_{5^4}$ over $\mathbb{F}_5$ is of degree $4$. But how can we find the number of such $a$? I am thinking along the lines of a primitive element, like a generator for the multiplicative group $\mathbb{F}_{5^4}^\times$. But I am not sure if this is what we want. Any hints?

Answer 1

Any subfield of $\mathbb{F}_{5^4}$ must have size $x=5^d$ with $5^4=x^k$ for natural numbers $d,k$. Thus $dk=4$ and the only proper subfields are $\mathbb{F}_{5}, \mathbb{F}_{5^2}$, with one copy of each as there are precisely $24$ roots of the polynomial $x^{24}-1$ in $\mathbb{F}_{5^4}$.

Thus the number of values of $a$ which do not lie in a subfield is precisely $5^4-25=600$.

Each irreducible polynomial over $\mathbb{F}_{5}$ will have 4 distinct roots from this set of 600, so we may group the 600 elements into 150 sets of 4, where the elements of each set have the same minimal polynomial.

Thus there are 150 monic irreducible polynomials. There are 4 possible leading coefficients, so in total 600 irreducible polynomials.

Note this is the same as the number of values of $a$, but only by coincidence as the degree was 4 and 5-1=4 too.

Answer 2

Correction (I made the same mistake). We were conflating the notions of primitive element for the multiplicative group and elements $a$ such that $\Bbb F_{5^4}\cong\Bbb F_5(a)$

There are $5^4-5^2=600$ elements of the latter type, because that's how many elements of $\Bbb F_{5^4}$ are not contained in $\Bbb F_{5^2}$.

Answer 3

One way to think of this is that $\mathbb{F}_{5}(\alpha)$ is defined to be the smallest subfield of $\overline{F}_{5}$ that contains both $\mathbb{F}_{5}$ and $\alpha$. So you can see that you need $\alpha \in \mathbb{F}_{5^{4}}$ but not in any proper subfield. You can easily count the $\alpha$ that satisfy this requirement.