Why is $x^4+x^3+x^2+x+1$ irreducible over $\mathbb{Q}_3$? I know that checking $f = x^4+x^3+x^2+x+1 \in \mathbb{Q}_3[x]$ over its residue field $\mathbb{F}_3$ is not enough since it has degree $4$. I also know that $f$ is supposed to be a polynomial where a primitive $5$-th root of unity vanishes but I would like to show the irreducibility more directly. Stuff like the Eisenstein criterion cannot be applied to.
Could you please help me with this problem?
 A: Here’s how you can do it, and I think it’s the best way.
First, because $\Bbb Z_3$, the $3$-adic integers, are a Unique Factorization Domain, irreducibility over $\Bbb Q_3$ is the same as irreducibility over $\Bbb Z_3$. (I’ve left off some details, but it’s the same as the story of $\Bbb Q$ versus $\Bbb Z$.)
Second, what you say about irreducibility over $\Bbb F_3$ not being enough to conclude irreducibility over $\Bbb Z_3$, that’s wrong. A factorization over $\Bbb Z_3$ will induce a factorization over $\Bbb F_3$.
Now I show why $x^4+x^3+x^2+x+1$, which you know is $(x^5-1)/(x-1)$, is $\Bbb F_3$-irreducible. Its roots are fifth roots of unity unequal to $1$. And it’s irreducible if and only if adjoining a fifth root of unity requires a degree-four extension of the base field $\Bbb F_3$. How do you decide this? See what the first field $\Bbb F_{3^m}$ is that contains fifth roots of unity.
How do you do that? Look at $(\Bbb F_{3^m})^\times$, the (cyclic) group of nonzero elements, $3^m-1$ in number, and see whether it has an element of period five, i.e. see whether $5|(3^m-1)$, i.e. see whether $3^m\equiv1\pmod5$. Oh of course, that’s $81=3^4$. (You could also have asked for the period of $3$ in $(\Bbb F_5)^\times$.) No matter what, there’s your proof.
