Determine number of divisors of $x^n -1$ of degree $k$. Why are they such a number?

Question

Let us consider the codes $C[n,k]$ where $n=31$ and $k=5$ and base field $\mathbb{F}_2$. I know there are ${31}\choose{k}$ divisors of $x^ {31}-1$ of degree $k$. But I can't understand why. Does anyone have an idea?

Answer 1

The field extension $_{32} / _2$ can be built as the smallest splitting field of $x^{32}-x$ (it is the standard proof of existence for a field of order $p^m$). So $x^{31}-1 = (x-α_1)...(x-α_{31})$ with $_{32} - \{0\} = \{α_1,...,α_{31}\}$, and any divisor of $x^{31}-1$ of degree $k$ can be obtained by selecting $k$ factors from $(x-α_1)...(x-α_{31})$ (order of selection irrelevant).