TITLE: Explicit examples of infinitely many irreducible polynomials in k[x]
QUESTION [9 upvotes]: 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.

REPLY [3 votes]: Another, easier, way. Let $\ell$ be a prime dividing $p-1$ and, if $\ell=2$, assume that $p \equiv 1 \bmod 4$. Such an $\ell$ exists for every $p \geq 5$. Let $Q \in \mathbb{F}_p$ not be an $\ell$-th power. I claim that $x^{\ell^n}-Q$ is irreducible for every $n$.
Lemma Let $\ell$ be prime, let $k$ be a field which contains the $\ell$-th roots of unity, and let $a \in k$. Then $x^{\ell} -a$ is either irreducible or splits into linear factors.
Proof Let $b$ be an $\ell$-th root of $a$ in the algebraic closure of $k$. So the roots of $x^{\ell}-a$ are $b$, $b \zeta$, $b \zeta^2$, ..., $b \zeta^{\ell-1}$. Suppose that $g(x)$ is a nontrivial factor of $x^{\ell}-a$, with $g(x) = x^d + \cdots + (-1)^d g_0$. Then $g_0 = \pm \zeta^j b^d$ for some $j$ and we deduce that $b^d \in k$. Since $\ell$ is prime and $0 < d < \ell$, we can find $f$ and $g$ such that $df-g \ell =1$. So $(b^d)^f a^{-g} = b$ is in $k$. We have showed that, if $x^{\ell}-a$ has a nontrivial factor then $a$ has an $\ell$-th root in $k$ and, since $k$ contains the $\ell$-th roots of unity, this shows that $x^{\ell}-a$ splits in $k$. $\square$.
Let 
$$x_1^{\ell} = Q \quad x_2^{\ell}=x_1 \quad x_3^{\ell} = x_2 \quad \cdots.$$
Let $K_i = \mathbb{F}_p(x_1, x_2, \ldots, x_i)$. We show by induction on $i$ that $[K_i:K_{i-1}] = \ell$. So $K_n = GF(p^{\ell^n})$ and $x_n$ is in $GF(p^{\ell^n})$ and not $GF(p^{\ell^{n-1}})$. So the minimal polynomial of $x_n$ has degree $\ell^n$ and must be $x^{\ell^n}-Q$, showing that $x^{\ell^n}-Q$ is irreducible as desired.
So, we want to show that $[K_i:K_{i-1}]=\ell$. By the lemma, $[K_i:K_{i-1}]$ is either $1$ or $\ell$. To rule out the latter case, we must show that $x_i$ is not an $\ell$-th power in $K_i$. Suppose, for the sake of contradiction that $b^{\ell} = x_i$. Then $N_{K_i/\mathbb{F}_p}(b)^{\ell} = N(x_i)$. Since, inductively, the minimal polynomial of $x_i$ is $z^{\ell^i} - Q$ and (also inductively) $K_i = GF(p^{\ell^i})$, we get $N(x_i) = (-1)^{\ell-1} Q$. 
By hypothesis, $Q$ is not an $\ell$-th power in $\mathbb{F}_p$ and (using that $p \equiv 1 \bmod 4$ for $\ell=2$) neither is $(-1)^{\ell-1} Q$. So we have a contradiction. QED