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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?

Rotdat
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  • What is $\mathbb{Q}_3$? – morrowmh Dec 19 '20 at 02:52
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    I don't understand why you say that you can't look over $\mathbb{F}_3$. Yes, it's not enough to just check that $f$ has no roots since it's of degree $4$, but you can also explicitly write down all degree $2$ irreducible polynomials over $\mathbb{F}_3$ and show that they are not factors of $f$. If $f$ has no roots and no degree $2$ factors, then it is irreducible. – Viktor Vaughn Dec 19 '20 at 02:57
  • @RichardD.James: Oh, you're right, I did not think about that! Still, I am not sure on how to show that. I could try to assume that $f$ decomposes into two quadratic factors and show a contradiction with their coefficients. But the system of equation I worked out in this case does not seem to be that trivial. – Rotdat Dec 19 '20 at 03:12
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    @MichaelMorrow: This is the field of the $3$-adic numbers. – Rotdat Dec 19 '20 at 03:12
  • @Rotdat You don't have to solve any systems of equations. There are only $3$ monic degree $2$ irreducible polynomials over $\mathbb{F}_3$. You can explicitly write these down. If you divide $f$ by each of these using polynomial long division and show you get a nonzero remainder, you've just shown that $f$ has no degree $2$ factor over $\mathbb{F}_3$. – Viktor Vaughn Dec 19 '20 at 03:52
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    I find it surprising if you could not adapt any of the answers to your question https://math.stackexchange.com/q/3920993/96384 to this one, where the polynomial in some ways is easier than there. – Torsten Schoeneberg Dec 19 '20 at 05:35
  • Because it's cyclotomic with the wrong order, maybe? The fifth order cyclotomic polynomial gives primitive fifth roots of unity, but 3-adics contain only square roots and $5|2$ is false. – Oscar Lanzi Dec 31 '20 at 13:24

1 Answers1

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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.

Lubin
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