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Over $F_{q}$ , $x^{q^{n}} - x$ is the product of all monic irreducibles whose degrees divide $n$

Statement

Let $F_{q}$ be a finite field and let $n \geq 1$ . In $F_{q} [t]$ ,

$t^{q^{n}} - t = \prod_{\begin{array}{c} P monic irreducible \deg P ∣ n \end{array}} P (t),$

where each monic irreducible occurs once.

Facts & Assumptions

Given: A finite field $F_{q}$ and a positive integer $n$ .

[L1]

The order of a finite field is a prime power; write $q = p^{r}$ with $r \geq 1$ (Every finite field has order $p^{n}$ for a unique prime characteristic $p$ and positive integer $n$ ).

[L2]

For every prime $p$ and positive integer $s$ , a field of order $p^{s}$ exists (For every prime $p$ and $n \geq 1$ , a field with $p^{n}$ elements exists).

[L3]

A field of order $Q$ is the full root set and a splitting field of $t^{Q} - t$ (A field with $q$ elements is the splitting field of $x^{q} - x$ over its prime subfield).

[L4]

The subfields of a field of order $p^{s}$ have orders $p^{u}$ with $u ∣ s$ (The subfields of $F_{p^{n}}$ are the unique fields $F_{p^{d}}$ for positive divisors $d$ of $n$ ).

[L5]

If an algebraic element has minimal polynomial of degree $d$ , its simple extension has degree $d$ and the corresponding power basis (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[L6]

Every nonzero nonunit polynomial over a field factors into irreducibles (Every nonzero nonunit polynomial over a field factors into irreducible polynomials).

[L7]

A root is repeated exactly when the formal derivative also vanishes there (A root is repeated exactly when it is also a root of the formal derivative).

[L8]

A polynomial of degree at least one over a field has a root in some field extension (Kronecker's one-root step: adjoining a root removes a linear factor and lowers the remaining degree).

[L9]

For an algebraic element $α$ there is a unique monic irreducible $m_{α}$ with $m_{α} (α) = 0$ , and for every $f$ one has $f (α) = 0$ exactly when $m_{α} ∣ f$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

Proof

technique · direct

1.1

By [L1], write $q = p^{r}$ . Use [L2] to choose a field $E$ of order $p^{r n} = q^{n}$ . By [L3], $E$ is the full root set and a splitting field of $t^{q^{n}} - t$ .

givenL1L2L3choose

1.2

Let $P$ be monic irreducible of degree $d \geq 1$ . By [L8] it has a root $α$ in some extension of $F_{q}$ ; since $P$ is monic irreducible and annihilates $α$ , the uniqueness in [L9] makes $P$ the minimal polynomial of $α$ . By [L5], $F_{q} (α)$ has degree $d$ over $F_{q}$ and hence has $q^{d} = p^{r d}$ elements.

givenL1L5L8L9

2.1

The derivative of $t^{q^{n}} - t$ is $- 1$ , which vanishes nowhere. Every irreducible factor of $t^{q^{n}} - t$ has a root in the splitting field $E$ of step 1.1, and a repeated factor would make that root repeated; so [L7] shows that no irreducible factor repeats.

step 1.1givenL7algebra

2.2

If $d ∣ n$ , write $n = e d$ . Applied to the field $F_{q} (α)$ , [L3] gives $α^{q^{d}} = α$ ; iterating this identity $e$ times gives $α^{q^{n}} = α$ . So $t^{q^{n}} - t$ vanishes at $α$ , and since $P$ is its minimal polynomial by step 1.2, [L9] gives $P ∣ t^{q^{n}} - t$ .

step 1.2L3L9algebra

2.3

Conversely, if $P$ divides $t^{q^{n}} - t$ , choose its root $α$ in the splitting field $E$ from step 1.1. Then $F_{q} (α)$ is a subfield of $E$ with order $p^{r d}$ , so [L4] gives $r d ∣ r n$ , and cancellation yields $d ∣ n$ .

step 1.1step 1.2L4choose

3.1

Factor the polynomial by [L6]. Steps 2.2 and 2.3 identify exactly the monic irreducible factors, and step 2.1 gives multiplicity one. Since both sides are monic, their unit factors agree, proving the formula.

step 2.2step 2.3step 2.1L6∎

Depends on

Used by

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Sources

K. Conrad, Finite Fields, Appendix A (standard reference, not scraped)

Over FqFq​, xqn−xxqn−x is the product of all monic irreducibles whose degrees divide nn