I was reading the post Is $\mathbb{F}_3(t,t^{1/3})/\mathbb{F}_3(t)$ a normal extension? Is it separable?
I do not understand, why we can use Eisenstein's criterion to show that $x^3 – t$ is irreducible in $\mathbb{F}_3(t)[x]$. As far as I know, Eisenstein's criterion is specific to $\mathbb{Z}[x]$ and generalises by a lemma from Gauss to $\mathbb{Q}[x]$, but not to finite fields or their extensions in general – am I wrong in saying so?
If we cannot use Eisenstein's criterion, what would be the best way to show that the polynomial is indeed irreducible?
Best Answer
To prove $x^3 - t$ is irreducible, it suffices to show there are no roots thereof.
Recall that elements $z$ of $\mathbb{F}_3(t)$ are of the form $z = \frac{P(t)}{Q(t)}$ where $P, Q \in \mathbb{F}_3[x]$ and $Q \neq 0$. We define $degree(z) = degree(P) - degree(Q)$ (you must verify that this is well defined).
Note that $degree$ is a monoid homomorphism $degree : (\mathbb{F}_3(t), \cdot) \to (\mathbb{Z} \cup \{-\infty\}, +)$.
Now if we had $z$ such that $z^3 = t$, then we would have $3 degree(z) = degree(t) = 1$. But there is no $k \in \mathbb{Z} \cup \{-\infty\}$ such that $3k = 1$.