Is the polynomial $f(x) = x^4 + tx^3 + (t^2 + 1)x^2 + (t^3 + t)x + (t^4 + t^2)$ irreducible over $k(t)$?

Question

Let $k$ be an algebraically closed field of characteristic 2 and let $k(t)$ be rational function field of one variable. Consider the polynomial $f(x) = x^4 + tx^3 + (t^2 + 1)x^2 + (t^3 + t)x + (t^4 + t^2) \in k(t)[x]$. Is $f(x)$ irreducible over $k(t)$?

Answer 1

First, by Gauss' lemma, $f(x)$ is irreducible over $k(t)$ iff it's irreducible over $k[t]$, so we may do all our calculating in the ring $k[t]$.

The polynomial has no linear factors $(x-a)$, as if it did then $a|t^4+t^2=t^2(t+1)^2$ by the rational root theorem, and each of the 9 possibilities may be plugged in to the polynomial to get a non-zero result.

This leaves the scenario that the polynomial factors as a product of quadratics:

$$f(x)=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+ bd$$

This gives us four equations: $$\begin{eqnarray} a+c&=& t \\ bd&=&t^4+t^2\\ ad+bc &=&t^3+t\\ ac+b+d&=&t^2+1\end{eqnarray}$$

By the mention of Gauss' lemma in the first sentence, we may list the 9 possible values for $b,d$ assuming they lie in $k[t]$ and then check each case.

By symmetry, we have only 5 cases to consider: $b=1,t,t+1,t^2,t(t+1)$ (these give $d=t^4+t^2,t^3+t,t^3+t^2,t^2+1,t(t+1)$ respectively).

We can eliminate most of these situations by considering the third equation, $ad+bc=t^3+t$.

Case 1 gives $a(t^4+t^2)+c=t^3+t$, which may be dismissed by degree considerations- if $a$ is of degree $n$ in $t$, then $c$ must be of degree $4+n$ in $t$, but also is at most degree $n+1$ in $t$ by the equation $a+c=t$.

Case 2 gives $a(t^3+t)+ct=t^3+t$, or $at^3+(a+c)t=at^3+t^2\neq t^3+t$, as $a+c=t$. So this case is impossible.

Case 3 gives $a(t^3+t^2)+c(t+1)=t^3+t$. On the other hand, the equation $ac+b+d=t^2+1$ will help here- plugging in, we have that $ac=t^3+t$, so $a=t^p(t+1)^q$ where $p\in\{0,1,2\}$ and $q\in\{0,1\}$. For each choice, it is clear that $a+c\neq t$, so this case is impossible.

Case 4 gives $a(t^2+1)+ct^2=t^3+t$, or $(a+c)t^2+a=t^3+a=t^3+t$ so $a=t$, $c=0$. Plugging this in to the remaining equation $ac+b+d=t^2+1$, we have $t\cdot0+t^2+t^2+1=1\neq t^2+1$. So this is impossible.

Case 5 may be immediately discarded, as it reduces to $(a+c)(t^2+t)=t(t^2+t)=t^3+t^2\neq t^3+t$.

Therefore the given polynomial is irreducible.