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For every finite field $F_{q}$ and every $n \geq 1$ , a monic irreducible polynomial of degree $n$ exists

Statement

For every finite field $F_{q}$ and every integer $n \geq 1$ , there exists a monic irreducible polynomial in $F_{q} [t]$ of degree $n$ .

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 $q = p^{r}$ (Every finite field has order $p^{n}$ for a unique prime characteristic $p$ and positive integer $n$ ).

[L2]

A field of order $p^{r n} = q^{n}$ exists (For every prime $p$ and $n \geq 1$ , a field with $p^{n}$ elements exists).

[L3]

The group $E^{\times}$ is cyclic (The multiplicative group $F_{q}^{\times}$ of a finite field is cyclic).

[L4]

Since $r ∣ r n$ , a field of order $p^{r n}$ has a unique subfield of order $p^{r} = q$ (The subfields of $F_{p^{n}}$ are the unique fields $F_{p^{d}}$ for positive divisors $d$ of $n$ ).

[L5]

Finite fields of the same order are isomorphic (Finite fields of the same order are isomorphic).

[L6]

An algebraic element's monic irreducible minimal polynomial has degree equal to the degree of its simple extension (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[L7]

An element is algebraic over a base field when some nonzero polynomial over that field vanishes at it (Algebraic and transcendental elements and algebraic extensions).

Proof

technique · contradiction

1.1

By [L1], write $q = p^{r}$ . Choose by [L2] a field $E$ of order $p^{r n} = q^{n}$ , and by [L3] a generator $a$ of the cyclic group $E^{\times}$ , whose order is $q^{n} - 1$ .

givenL1L2L3choose

2.1

By [L4] and [L5], identify the unique order- $q$ subfield of $E$ with the given $F_{q}$ . Since $E$ is finite, the powers $a^{0}, a^{1}, \dots, a^{∣ E ∣}$ cannot be pairwise distinct, so $a^{i} = a^{j}$ for some $i < j$ and $a$ is a root of the nonzero polynomial $t^{j} - t^{i} \in F_{q} [t]$ ; by [L7], $a$ is algebraic over $F_{q}$ . Let $m_{a}$ be its minimal polynomial over that subfield and put $d = \deg m_{a}$ . Then $F_{q} (a)$ has $q^{d}$ elements by [L6] and is a subfield of $E$ , so $d \leq n$ .

step 1.1L4L5L6L7algebra

3.1

Suppose, for contradiction, that $d < n$ . Then $F_{q} (a)$ is a proper subfield whose multiplicative group has only $q^{d} - 1 < q^{n} - 1$ elements and cannot contain an element of order $q^{n} - 1$ . This contradicts the choice of $a$ .

step 1.1step 2.1assume-contraalgebra

4.1

Therefore $d = n$ , and $m_{a}$ is the required monic irreducible polynomial.

step 3.1L6discharge-contradiction∎

Depends on

Used by

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Sources

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

For every finite field FqFq​ and every n≥1n≥1, a monic irreducible polynomial of degree nn exists