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It can be brute force proven that every irreducible degree 3 polynomial in $\mathbb{Z}_3[x]$ is also irreducible in $\mathbb{F}_9$. I haven't been able to find a good proof of why this is true (other than going through case by case for each irreducible polynomial), or any generalizations of it. In my class we are using $\mathbb{F}_9=\{a+b\gamma \mid \gamma^2=2, a,b\in \mathbb{Z}_3\}$. I am looking for a method of proving the statement without brute forcing it. I've tried showing that if $(x-c)$ is a factor with $c \notin \mathbb{Z}_3$ then the polynomial we get by expanding $(x-c)(x^2+px+q)$ is not in $\mathbb{Z}_3[x]$, but I have not been able to work this out since the coefficients $p,q$ might be in $\mathbb{F}_9$. The problem becomes much easier if $p,q \in \mathbb{Z}_3$, so any proper justification that they might necessarily have to be would also be helpful.

For example $(x-(1+\gamma))(x^2+(1+\gamma)x+(1+2\gamma))=x^3+x+1$ but in this case it is reducible. How might we show that for an irreduible, there is no factorization in $\mathbb{F}_9$?

Conor_Meise
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    A zero of an irreducible polynomial of degree 3 gives rise to a field extension of degree 3. That does not fit inside a field extension of degree 2. – lhf Mar 13 '25 at 18:52
  • @lhf Could you add on to this maybe with a resource or any other information? I have not encountered this before and want to see the reasoning why. – Conor_Meise Mar 13 '25 at 18:58
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    I answered below. Notation however: You really should replace in your post $\mathbb{Z}_3$ with $\mathbb{F}_3$, or be clear that you are working with $3$-adic integers. The way it is now will probably cause confusion to other readers. I am assuming based on your tags that you do indeed mean $\mathbb{Z}_3$ to be $\mathbb{Z}/3\mathbb{Z}$ i.e., the field on $3$ elements? – Mike Mar 13 '25 at 19:14
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    The general result is that an irreducible polynomial of degree $m$ in $\Bbb{F}q(x)$ remains irreducible in $\Bbb{F}{q^n}(x)$ whenever $\gcd(m,n)=1$. I'm fairly sure we have covered this on the site somewhere. I don't have the time to search for it right now, sorry. – Jyrki Lahtonen Mar 13 '25 at 19:47
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    I deleted my answer as this probably is a duplicate question, where the more general question as per @JyrkiLahtonen's post was likely answered. – Mike Mar 13 '25 at 20:41
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    See here for the result @JyrkiLahtonen mentions. The question is about extensions of $\mathbb Q$ instead of $\mathbb F_p$, but the proof is exactly the same. – anankElpis Mar 13 '25 at 20:52
  • This could serve as a duplicate target, but surely this oldie needs to be mentioned as the mother thread. – Jyrki Lahtonen Mar 29 '25 at 19:34

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