I've been slowly grinding my way through a bunch of finite field/cyclotomic polynomial questions in my assignment, and I'm really stuck here.
The question asks: Prove that if $q$ is a prime modulo such that p is a primitive root modulo q, then $x^{q-1}+...+x+1$ is irreducible over $\mathbb{F}_p$ (Slightly confusing wording?)
From previous questions I've shown $x^n-1$ splits over E, an extension of $\mathbb{F}_p$ iff $n$ divides $p^d-1$, and that the degree of the splitting field of $x^n-1$ over $\mathbb{F}_p$ is the order of $p$ in $(\mathbb{Z}/n\mathbb{Z})^*$ (the multiplicative group).
Also, this related question and its answer makes no sense to me
Thanks
Edit: I totally misunderstood what "primitive root modulo q" meant. I now understand $p^{q-1}\equiv 1 \mod q$ and so $p$ has order $q-1$ in $(\mathbb{Z}/q\mathbb{Z})^*$. Still not sure where to go from here though.
Try this:
The hypothesis is that $p^{q-1}$ is the smallest power of $p$ that is congruent to $1$ modulo $q$.
Now, what are the orders of the (cyclic) groups $\Bbb F_{p^m}^\times$? They are, of course, $p^m-1$, and a field $\Bbb F_{p^m}$ contains a primitive $q$-th root of unity if and only if $q|(p^m-1)$, i.e. if and only if $p^m\equiv1\pmod q$. Thus our hypothesis says that $\Bbb F_{p^{q-1}}$ is the first extension of $\Bbb F_p$ that contains a $q$-th root of unity $\zeta_q$. In other words, $\zeta_q$ generates an extension of degree $q-1$, so $X^{q-1}+\cdots+X+1$ is irreducible over $\Bbb F_p$