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

Question

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.

Answer 1

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.

Answer 2

If $k$ is infinite, it's easy as you say.

Suppose that $k$ is finite. Then, for any positive integer $n\geq 2$, there exists an irreducible polynomial in $k[t]$ of degree $n$.

Consider the finite field $\mathbf{F}_p$, where $p$ is a prime number.

Edit: (might help a bit.) The irreducible polynomials of degree less or equal to $n$ in $\mathbf{F}_p[x]$ are the irreducible factors of $X^{p^n}-X$. In fact, $$X^{p^n} - X = \prod_{f \textrm{ monic irreducible}, (\deg f) \vert n} f.$$ As a fun sidefact, this implies that the number of irreducible polynomials $x_n$ of degree $n$ satisfies $\sum_{d\vert n } dx_d = p^n$.

Anyway, to get explicit examples you'll have to factorize the above polynomial for each $n$. This is a hard, but possible in theory. For small $p$ and $n$ you can do this. For example, for $p=2$ and $n=2$, you have to factor $X^4-X = X(X^3-1)=X(X-1)(X^2+X+1)$ in $\mathbf{F}_2$.