Explicit examples of infinitely many irreducible polynomials in k[x]

My question is the following.

Is it possible to give examples of infinitely many irreducible polynomials in a polynomial ring $k[x]$ with $k$ a field?

I'm interested in this because I'm doing an exercise in which I'm asked to show that there are infinitely many maximal ideals in $k[x]$ by listing them explicitly. Now, I know that if $k$ is infinite then the ideals $I_a := (x - a)$ for $a \in k$ are maximal and this would do. But if the field is finite I really don't know what to do.

I haven't been able to find anything online or in books. I also know that we can prove that there are infinitely many irreducible monic polynomials in $k[x]$ by using Euclid's argument for proving that there are infinitely many prime numbers. But this does not give me explicit maximal ideals, just the existence of infinitely many of them.

Thank you very much for any help.

The best I can come up with for now is the following. If $\alpha$ is an element in some extension of the field $F_p, p$ prime, then its conjugates are well-known to be $\alpha^p$, $\alpha^{p^2}$, $\alpha^{p^3}$, $\ldots$. So, if $\alpha$ happens to be a root of unity of a prime order $\ell$, and $p$ happens to be a generator of the multiplicative group $\mathbf{Z}_\ell^*$, then the minimal polynomial of $\alpha$ is $$ (x-\alpha)(x-\alpha^2)\cdots(x-\alpha^{\ell-1})=\phi_\ell(x)=\sum_{i=0}^{\ell-1}x^i, $$ so $\phi_\ell(x)$ is then irreducible in $F_p[x]$. If Artin's conjecture is true, then this happens for infinitely many rational primes $\ell$. This gives a little bit narrower set of polynomials that is likely to contain an infinite number of irreducible ones. A variant of Artin's conjecture may then also settle the case $k=GF(p^n)$.

No good: depends on the truth of an unknown conjecture, and not very explicit. Feel free to downvote.

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Addendum: (Exercise 3.96 from Finite Fields, Lidl & Niederreiter) An explicit infinite family of irreducible polynomials in $F_2[x]$ is the following. Let $k$ be a positive integer. The polynomial $$ p_k(x)=x^{2\cdot3^k}+x^{3^k}+1\in F_2[x] $$ is irreducible.

My solution to this exercise. The proof of this fact is very similar to the previous construction. The ingredients are the observation that $$ p_k(x)=\frac{x^{3^{k+1}}+1}{x^{3^k}+1} $$ is the cyclotomic polynomial $\phi_{3^{k+1}}(x)$, and the fact that $2$ is a generator of the group of units $U_n$ of the residue class rings $\mathbf{Z}/3^n\mathbf{Z}$. The latter fact is equivalent to showing that the order of $2$ in $U$ is exactly $\phi(3^n)=2\cdot3^{n-1}$. This follows, if we can show that $\nu_3(2^{2\cdot3^{n-1}}-1)=n$, because it is known that a generator $x$ of $U_\ell$ is also a generator of $U_{\ell+1}$ unless it happens that $x^{|U_\ell|}=1$ in $U_{\ell+1}$ (see e.g. Jacobson, Basic Algebra I). But $(x^3-1)=(x-1)(x^2+x+1)$ implies that $$ 2^{2\cdot 3^{n+1}}-1=(2^{2\cdot 3^n}-1)(2^{4\cdot 3^n}+2^{2\cdot 3^n}+1). $$ In the latter factor all three terms are congruent to $1\pmod 9$, so that factor is divisible by $3$ but not divisible by $9$. This is the inductive step.

If $\alpha$ is a primitive root of order $3^{k+1}$, then $\alpha$ is a root of $p_k(x)$. Therefore so are its conjugates $\alpha^{2^j}, j\in\mathbf{N}.$ But we saw that $U_{k+1}=<2>$, so the conjugates are exactly the powers $\alpha^t$, $(t,3)=1$. There are $2\cdot3^k=\deg p_k(x)$ of these, so $p_k(x)$ is the minimal polynomial of $\alpha$ and hence irreducible. Q.E.D.

Define $\phi(y) = 1/(y^p-y-1)$ and define $\phi^n(y) = \phi(\phi(\cdots\phi(y)\cdots))$. I claim that the denominator of $\phi^n(y)$ is irreducible of degree $p^n$ over $\mathbb{F}_p$. This is going to be a long one, but I had a lot of fun finding it so I hope some of you will read it.

The rational function $\phi$ defines a from map $\overline{\mathbb{F}_p} \cup \{ \infty \}$ to itself. This map is $p \to 1$, except that the preimage of $0$ is the single element $\infty$. The image of $0$ is the fixed point $-1$. So $0 \not \in (\phi^n)^{-1}(\infty)$ for any $n$, and thus $(\phi^{n+1})^{-1}(\infty)$ always has $p$ times as many elements as $(\phi^n)^{-1}(\infty)$. So $(\phi^n)^{-1}(\infty)$ has size $p^n$ and we see that the denominator of $\phi$ has degree $p^n$. We must now show that it is irreducible. Let $y_n$ be a root of this denominator; we need to show that $y_n$ is in $GF(p^{p^n})$ and not in $GF(p^{p^{n-1}})$.

Set $\phi(y_n) = y_{n-1}$, $\phi(y_{n-1}) = y_{n-2}$, etcetera. We rewrite these relations as $$y_1^p-y_1 = 1 \quad y_2^p-y_2 = \frac{y_1+1}{y_1} \quad y_3^p-y_3 = \frac{y_2+1}{y_2} \quad y_4^p-y_4 = \frac{y_3+1}{y_3} \cdots$$ Define $K_i=\mathbb{F}_p(y_1, y_2, \ldots, y_i)$. We will establish, below, the following claim:

Key Claim The polynomial $z^p-z = \frac{y_i+1}{y_i}$ is irreducible over $K_i$.

Thus $[K_{i+1}:K_{i}]=p$ for every $i$. So $y_n \in K_n = GF(p^{p^n})$ and $y_n \not\in K_{n-1} = GF(p^{p^{n-1}})$. This shows that the minimal polynomial of $y_n$ over $\mathbb{F}_p$ has degree $p^{p^n}$, so the denominator of $\phi$ must be irreducible.

Our goal now is to establish the claim. We first need some lemmas about polynomials of the form $x^p-x=a$. This is the subject of Artin-Schrier theory (see also here).

Let $k$ be a field of characteristic $p$ and let $a \in k$.

Lemma 1 Let $u$ be a root of $x^p-x=a$. Then the other roots are $u+1$, $u+2$, ..., $u+p-1$.

Proof Straightforward. $\square$

Lemma 2 If $a$ is of the form $b^p-b$, for $b \in k$, then $x^p-x-a = \prod_{i=0}^{p-1} (x-b-i)$. If $a$ is not of the form $b^p-b$, for $b \in k$, then $x^p-x-a$ is irreducible.

Proof The first statement is obvious.

We prove the contrapositive of the second statement. Suppose that $f$ has a nontrivial factor $g(x)$; let $g(x) = x^e - g_1 x^{e-1} + \cdots$. Let $u$ be a root of $f(x)$ and let the roots of $g$ be $u+i_1$, $u+i_2$, ..., $u+i_e$. Then $g_1 = \sum_{j=1}^e (u+i_j) =e u + \sum i_j$ is in $k$, so $u = (g_1 - \sum i_j)/e$ is in $k$. $\square$

Lemma 3 Let $x^p-x-a$ be irreducible and let $u$ be a root of $x^p-x=a$. Then $$Tr_{k(u)/k} \frac{1}{u} = \frac{-1}{a}.$$

Proof Let $z=1/u$. Then $z^p + (1/a) z^{p-1} - (1/a)=0$. This polynomial is irreducible, so it is the minimal polynomial of $z$, so $Tr(z)$ is the coefficient of $z^{p-1}$. $\square$

Lemma 4 Let $x^p-x-a$ be irreducible and let $u$ be a root of $x^p-x=a$. Then $$Tr_{k(u)/k} \frac{eu+f}{gu+h} = \frac{eh-fg}{g^2 a}.$$ for $e$, $f$, $g$, $h \in \mathbb{F}_p$, and $g \neq 0$.

Proof $$Tr \left( \frac{eu+f}{gu+h} \right) = Tr \left( \frac{e}{g} \right) + \frac{fg-eh}{g^2} Tr\left( \frac{1}{u+h/g} \right).$$ Since $e/g \in k$, and $\deg k(u)/k=p$, we have $Tr(e/g) = 0$. Since $h/g \in k$, the element $u+h/g$ is another root of $x^p-x=a$, so $Tr(1/(u+h/g)) = Tr(1/u) = -1/a$ by Lemma 3. $\square$

Lemma 5 Let $k$ be a finite field. If $Tr_{k/\mathbb{F}_p}(a) \neq 0$, then $a$ is not of the form $b^p-b$.

Proof Again, we prove the contrapositive. If $a=b^p-b$, then $$Tr_{k/\mathbb{F}_p}(a) = Tr_{k/\mathbb{F}_p}(b^p)-Tr_{k/\mathbb{F}_p}(b)= Tr_{k/\mathbb{F}_p}(b)^p-Tr_{k/\mathbb{F}_p}(b)= 0. \quad \square$$

We now move to the main claim. We want to show that $z^p-z=\frac{y_i+1}{y_i}$ is irreducible. So, by Lemmas 2 and 5, we must show that $Tr_{K_{i}/\mathbb{F}_p}\left( \frac{y_{i}+1}{y_i} \right) \neq 0$. We compute this trace as the composite of traces $K_{i} \to K_{i-1} \to \cdots \to K_2 \to K_1 \to \mathbb{F}_p$. By our inductive hypotheses, each of these individual field extensions is a degree $p$ Artin-Schrier extension, so Lemma 4 is relevant. Using Lemma 4 repeatedly: $$Tr_{K_{i}/K_{i-1}} \left( \frac{y_{i}+1}{y_{i}} \right) = \frac{- y_{i-1}}{y_{i-1}+1}$$ $$Tr_{K_{i-1}/K_{i-2}} \left(\frac{- y_{i-1}}{y_{i-1}+1}\right) = \frac{y_{i-2}}{y_{i-2}+1}$$ $$Tr_{K_{i-2}/K_{i-3}} \left(\frac{ y_{i-2}}{y_{i-2}+1}\right) = \frac{- y_{i-3}}{y_{i-3}+1}$$ etcetera. At the end of the day, we get $Tr_{K_{i}/\mathbb{F}_p}\left( \frac{y_{i}+1}{y_i} \right) = \pm 1 \neq 0$ as desired. By Lemmas 2 and 5, this proves the claim.