Existence of irreducible polynomials over finite field

Question

Let $F$ be a finite field. How do we prove that for each $n \in \mathbb{N}$ there is an irreducible polynomial of degree $n$?

One can assume that $F = \mathbb{F}_{p^m}$ where $p$ is prime. If $n \ge |F|$ then I can construct an irreducible polynomial, namely
$ p(x) = 1 + \prod_{j=1}^{|F|} ( x - a_j )$
where $a_j$ are all the field elements. It is clear that $p(x)$ has no roots in $F$.

This trick doesn't work for $n < |F|$. A counter-example: Let $F = \mathbb{F}_3$ and $p(x) = 1 + (x-1)(x-2)$, then $p(1) = 1$, $p(2) = 1$, $p(0) = 1 + (-1)(-2) = 1 + 2 = 3 = 0 \pmod 3$.

I know there is a way to count them using the Möbius function $\mu(n)$ but I want a proof without it that just shows existence.

Answer 1

The multiplicative group of nonzero elements of any finite field is cyclic; so if $K=\mathbb{F}_{p^n}$, letting $\alpha$ be a generator of the multiplicative group of $K$, we have that $K=\mathbb{F}_p(\alpha)$. In particular, the minimal polynomial of $\alpha$ over $\mathbb{F}_p$, which is irreducible, must have the same degree as $[\mathbb{F}_p(\alpha):\mathbb{F}_p] = [K:F] = n$, so there must exist an irreducible polynomial over $\mathbb{F}_p$ of degree $n$.

Answer 2

I just closed another question as a duplicate of this one but I noticed that the discussion here is incomplete; Arturo's answer

1. assumes the existence of finite fields and
2. uses the theorem that finite subgroups of the multiplicative group of a field are cyclic (let's call this "the cyclic theorem")

but as azimut's 2013 comment remarks, the existence of the finite field $\mathbb{F}_{p^n}$ is more or less equivalent to the existence of an irreducible polynomial over $\mathbb{F}_p$ of degree $n$, so this is arguably circular. The cyclic theorem is also not completely trivial. Here are four non-circular approaches. All of them generalize without modification to the case where $\mathbb{F}_p$ is replaced by a finite field $\mathbb{F}_q$ (and we don't have to know the classification of finite fields in advance), but to avoid the appearance of circularity I will stick to the notation $\mathbb{F}_p$.

Proof 1: Möbius inversion

As far as I know this is the "standard argument," or at least it's the one I learned first. We prove that there exists an irreducible polynomial over $\mathbb{F}_p$ of degree $n$ by counting them using Möbius inversion; we get that the number of (monic) such polynomials is

$$M_n(p) = \frac{1}{n} \sum_{d \mid n} \mu(d) p^{\frac{n}{d}}$$

and then one has to prove an inequality to show that this is always positive. For example, because $\mu(d) \in \{ -1, 0, 1 \}$ we have

$$M_n(p) \ge \frac{p^n - p^{n-1} - \dots - 1}{n}$$

which reduces the problem to showing that

$$p^n > p^{n-1} + \dots + 1 = \frac{p^n - 1}{p - 1}$$

which is clear. This argument does not presuppose the existence of finite fields and avoids having to prove the cyclic theorem.

Proof 2: Splitting fields

This follows azimut's suggestion. In the Möbius inversion argument we begin by using properties of the Frobenius map $x \mapsto x^p$ to show that $x^{p^n} - x \in \mathbb{F}_p[x]$ is the product of all monic irreducible polynomials of degree dividing $n$. This actually already implies that its splitting field must be $\mathbb{F}_{p^n}$, so we can instead proceed by analyzing this splitting field.

Let $L$ be the (minimal) splitting field of $x^{p^n} - x$ over $\mathbb{F}_p$. By taking the formal derivative, $x^{p^n} - x$ is separable, so it has $p^n$ distinct roots. These roots are closed under addition and multiplication (since they are the fixed points of the $n^{th}$ power of the Frobenius map), which means they already, by themselves, form a splitting field inside $L$, hence must be all of $L$; so $|L| = p^n$, meaning $L$ is a finite field of order $p^n$.

Now we can apply the cyclic theorem as in Arturo's answer. The standard argument for the cyclic theorem, in turn, is conceptually similar to the Möbius inversion argument, so we still end up having to do similar work, and we also have to know something about splitting fields, so this proof requires more field theory than proof 1.

Proof 3: cyclotomic polynomials

The strategy here is to show directly that there is an element of some finite extension of $\mathbb{F}_p$ with multiplicative order $p^n - 1$, which must therefore be a cyclic generator of the multiplicative group of the finite field $\mathbb{F}_{p^n}$, but without assuming either the existence of this finite field or using the cyclic theorem. We need the following basic facts about the cyclotomic polynomials:

• The polynomials $\Phi_n(x) = \prod_{\gcd(k, n) = 1} (x - e^{\frac{2 \pi i k}{n}})$ whose roots over $\mathbb{C}$ are the primitive $n^{th}$ roots of unity have integer coefficients, and
• $x^n - 1 = \prod_{d \mid n} \Phi_d(x)$.

The second statement follows from the definition of the cyclotomic polynomials (or can be used to define them) while the first follows from either a little Galois theory (this is potentially circular, though) or by induction on the second statement.

The significance of the fact that $\Phi_n(x)$ has integer coefficients is that we can reduce it $\bmod p$; in fact we can consider its roots over any field. The roots of $x^n - 1$ over any field are (by definition) the $n^{th}$ roots of unity, and:

Lemma: Over any field $F$ of characteristic not dividing $n$, the roots of $\Phi_n(x)$ are exactly the primitive $n^{th}$ roots of unity; that is, they consist of $\alpha \in F$ such that $\alpha^n = 1$ but $\alpha^d \neq 1$ for $d \nmid n$.

Proof. Taking the formal derivative gives that $x^n - 1$ is separable over $F$, so its roots are distinct. Hence if $\alpha^n = 1$ then $\alpha$ is a root of exactly one cyclotomic polynomial $\Phi_d(x), d \mid n$, which gives $\alpha^d = 1$. So $\alpha$ is primitive iff $d = n$. $\Box$

Over $\mathbb{F}_p$ we now consider the cyclotomic polynomial $\Phi_{p^n-1}(x)$. Let $f(x)$ be any irreducible factor of it, so that we can construct the finite extension $\mathbb{F}_p[\alpha]/f(\alpha)$ containing a root $\alpha$ of it. By the lemma, $\alpha$ is a primitive $p^n-1$-th root of unity, so in particular satisfies

$$\alpha^{p^n} = \alpha.$$

Because it's a primitive $p^n-1$-th root of unity it does not have order $p^d-1$ for any $d \mid n$, which means its orbit under the action of Frobenius has size exactly $n$, which in turn means $\deg f = n$. So $f$ is an irreducible polynomial of degree $n$, and the finite extension $\mathbb{F}_p[\alpha]/f(\alpha)$ is a finite field of order $\mathbb{F}_{p^n}$ whose multiplicative group is cyclic generated by $\alpha$.

I like this argument because it feels quite minimal; we don't need the cyclic theorem and in fact prove it for finite fields, and we don't even need the concept of a splitting field.

Proof 4: Cauchy's theorem and $GL_n(\mathbb{F}_p)$

This is a nice argument I learned about recently here on math.SE, which uses cyclotomic polynomials in a different way. Edit: Unfortunately, it has some awkward exceptional cases! Our starting point is the simple counting argument that produces the order of the group $GL_n(\mathbb{F}_p)$, namely

$$|GL_n(\mathbb{F}_p)| = (p^n - 1)(p^n - p) \dots (p^n - p^{n-1}).$$

This means the order is divisible by $p^n - 1$. Suppose we could find a prime $\ell$ dividing $p^n - 1$ but not dividing $p^k - 1$ for any $k < n$; this is called a primitive prime divisor of $p^n - 1$. First I'll explain how this lets us finish. By Cauchy's theorem, $GL_n(\mathbb{F}_p)$ has an element $X$ of order $\ell$. We'll show that the characteristic polynomial $f(x) = \det(xI - X)$ of $X$ is irreducible.

Subproof 4a. Pass to a splitting field of $f(x)$; then all of the roots $\alpha$ of $f(x)$ have order either $1$ or $\ell$. If $\alpha$ is any root with order $\ell$, then its orbit under the Frobenius has size $d$ where $d$ is minimal such that $\alpha^{p^d} = \alpha$, or equivalently $\alpha^{p^d-1} = 1$, or equivalently $\ell \mid p^d - 1$. By hypothesis, $d = n$. This means the orbit of $\alpha$ under Frobenius consists of every root of $f(x)$, so $f(x)$ is irreducible as desired.

Subproof 4b. We'll show that the action of $X$ on $\mathbb{F}_p^n$ has no nontrivial invariant subspaces. Since a factorization of the characteristic polynomial produces such subspaces, this implies that the characteristic polynomial is irreducible.

Suppose otherwise; if $X$ has a nonzero invariant subspace of dimension $m < n$, then the cyclic group $C_{\ell}$ it generates embeds into $GL_m(\mathbb{F}_p)$, hence $\ell \mid |GL_m(\mathbb{F}_p)|$. But by hypothesis this is not possible.

To find $\ell$ we can apply Zsigmondy's theorem, which can be proven using cyclotomic polynomials similar to the lemma in proof 3. Slightly specialized to this case, it says:

Proposition: Let $a \ge 2, n \ge 2$. Then $a^n - 1$ has a primitive prime divisor (a prime divisor not dividing $a^k - 1$ for any $k < n$), unless

• $n = 2$ and $a + 1$ is a power of $2$, or
• $n = 6$ and $a = 2$.

So we can find the required prime $\ell$ in all cases except the above. The $n = 2, p = 2^q - 1$ case requires $p$ to be a Mersenne prime (or prime power), but in this case it's easy to prove that monic irreducible quadratic polynomials exist by counting reducible ones (there are $p^2$ monic quadratics and $p + {p \choose 2}$ of them are reducible), without using the full strength of any of the other proofs. The $n = 6, p = 2$ case requires exhibiting a single irreducible polynomial of degree $6$ over $\mathbb{F}_2$ which is a finite calculation and can also be done with a direct counting argument. Sadly I hadn't previously noticed these exceptional cases exist and they somewhat dampen the fun of this argument.

The mistake I had previously made was the following. Notice that if $\ell \mid p^n - 1$ but $\ell \nmid p^k - 1$ for $k < n$ then by the cyclotomic factorization, $\ell$ must divide $\Phi_n(p)$. The lemma in proof 3 implies:

Corollary: The primes $\ell$ which divide $\Phi_n(p)$ but don't divide $n$ are exactly the primes such that the multiplicative order of $p \bmod \ell$ is $n$; symbolically, $\text{ord}_{\ell}(p) = n$.

Since the multiplicative order of $p \bmod \ell$ is unique, this is very close to saying that if $\ell \mid \Phi_n(p)$ then $\ell$ is the desired primitive prime divisor. It's not hard to show that $\Phi_n(p) > 1$, so it has at least one prime divisor. But I had overlooked that we need to rule out the possibility that $\ell \mid n$. Zsigmondy's theorem exactly characterizes when this possibility can and can't be ruled out; in the first case above $n = 2, \ell = 2 \mid \Phi_n(p) = p + 1$ and in the second case $n = 6, \ell = \Phi_n(2) = 3$.

Answer 3

The answer given by Arturo can be improved for any finite field $F$. If $|F|=p^k$ then we will consider the extension $\mathbb{F}_{p^{nk}}$ over $\mathbb{F}_p$. Then we will have $$ \mathbb{F}_p \subseteq F \subseteq \mathbb{F}_{p^{nk}}. $$ And now $[\mathbb{F}_{p^{nk}}:F]=\frac{nk}{k}=n.$ Since any finite extension of a finite field is simple we have $\mathbb{F}_{p^{nk}}=F(\alpha)$ for some $\alpha \in \mathbb{F}_{p^{nk}}$. Here the minimal polynomial of $\alpha$ does the job.