Factor $X^7 − 1$ into irreducibles in $\mathbb{Z}_{127}[X]$

Question

Problem: Factor $X^7 − 1$ into irreducibles in $\mathbb{Z}_{127} [X]$

I'm having a lot of trouble understanding how to approach this, and this course topic in general. Any help (or even better, a worked solution) would be great.

I recognise that 127 is a prime number. I'm assuming the use of $2^7 $($=128 \equiv 1 \pmod{127})$ is applicable somewhere.

Edit: thank you for all your comments! And typesetting suggestions

Answer 1

You are looking for the seventh roots of unity. In the algebraic closure of $\mathbb F_{127}$ they form a cyclic group of order $7$. Because $7$ is prime every element different from the neutral element (which is $1$) will generate the group. Since you've found already another root, namely $2$, you just need to compute the powers of $2$ and get everything. So you have $$X^7-1 = (X-1)(X-2)(X-4)(X-8)(X-16)(X-32)(X-64).$$