The number of primitive elements for $\mathbb{F}_{p^{n}}$ is given by the following formula:$$(p-1)\cdot\sum_{d\mid n}\mu(d)\cdot p^{\tfrac{n}{d}},$$ where 'the primitive element for $\mathbb{F}_{p^{n}}$' means 'an element $\alpha\in\mathbb{F}_{p^{n}}$ such that $\mathbb{F}_{p^{n}}=\mathbb{F}_{p}(\alpha)$', and $\mu$ is the Möbius function.
My question is that 'what is the total number of elements in the following set:$$\left\{\beta\in\mathbb{F}_{p^{n}}\,:\,\mathbb{F}_{p^{n}}=K(\beta)\right\}$$ if $K=\mathbb{F}_{p^{m}}$ with $1\neq m\mid n$?'
Please, give some comments or any references. Thank you!
Actually, there is a general result. The number of primitive elements in $\mathbb{F}_{q^n}$ over $\mathbb{F}_q$ is $$\sum_{d\mid n}\mu(d)q^{n/d},$$ since the number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_q$ is $$\frac{1}{n}\sum_{d\mid n}\mu(d)q^{n/d}.$$ Therefore, the number of primitive elements in $\mathbb{F}_{p^n}$ over $\mathbb{F}_{p^m}$ is $$\sum_{d\mid (n/m)}\mu(d)p^{n/d}.$$