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Suppose that we have the finite field $\mathbb{Z}/p\mathbb{Z}$, and we are considering adjoining the roots of degree $k$ monic polynomials to the field (whose coefficients are taken from this field). So we have a total of $p^k$ possible polynomials that could factor to some degree.

Is there a formula for how many of these polynomials completely factor into linear factors?

I found this answer on Stack Exchange concerning irreducible polynomials.

Matt Groff
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    This is just the number of ways to choose $\alpha_1,\dots,\alpha_k\in\mathbb Z/p\mathbb Z$ which are not necessarily distinct, and where you identify reorderings. – Kenta S Dec 31 '21 at 22:10
  • @KentaS: If you can prove this I will accept this answer. – Matt Groff Dec 31 '21 at 22:12

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There is a bijection: $$\{\text{monic polynomials }f \text{ of degree $k$ which split completely}\}\to\{\text{multisets of size $k$, consisting of elements of }\mathbb Z/p\},$$ sending $f$ to the set of solutions to $f(x)=0$, which all lie in $\mathbb Z/p$ due to our assumption that $f$ splits into linear factors. In the other direction, take a multiset $S$ to the polynomial $\prod_{\alpha\in S}(x-\alpha)$. [Here, a multiset is a set with multiplicity.]

The number of such multisets is ${k+p-1\choose p-1}$—imagine having $k$ objects in a line, and putting $p-1$ dividers into them to make $p$ possibly empty groups. We can instead think about $k+p$ objects, and putting $p-1$ dividers, where we require the groups to be nonempty. There are $k+p-1$ places to put the dividers, so the total number is ${k+p-1\choose p-1}$.

Alternatively, consider the coefficient of $x^k$ in the generating function $(1+x+x^2+\dots)^p=\frac1{(1-x)^p}$. The formula for this is proved here.

Kenta S
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