Number of invertible elements in $\mathbb{F}_q[X]/\langle X^p-1\rangle$ with $p=\operatorname{char} \mathbb{F}_q$

Question

I need to find the number of invertible elements in $\mathbb{F}_q[X]/\langle X^p-1\rangle$ with $p=\operatorname{char} \mathbb{F}_q$, which is equal to the number of invertible $p\times p$ circulant matrices over $\mathbb{F}_q$.

If $n$ and $q$ are coprime then the number of $n\times n$ invertible circulant matrices over $\mathbb{F}_q$ is calculated, see here. But what if $n$ and $q$ are not coprime (in particular, $n=p=\operatorname{char}\mathbb{F}_q$)? In this case no extension field of $\mathbb{F}_p$ contains a primitive $n$-th root of unity.

Answer 1

As $X^p-1=(X-1)^p$ in characteristic $p$, then $\mathbb{F}_q[X]/\langle X^p-1\rangle\cong\mathbb{F}_q[X]/\langle Y^p\rangle$ (set $Y=X-1$). As $Y$ is nilpotent, the invertible elements are $a_0+a_1Y+\cdots +a_{p-1}Y^{p-1}$ with $a_0\ne0$. There are $(q-1)q^{p-1}$ of them.

Answer 2

Hint: We have $X^p-1 = (X-1)^p$ in any field of characteristic $p>0$. The invertible elements of $F_q[X]/\langle X^p-1\rangle$ are the polynomials (residue classes) $g(X)$ which are relatively prime to $X-1$.