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In order to determine the splitting field of $t^{4}+2\in \mathbb{Z}_{3}[t]$, I first "guessed" the roots $1$ and $2$ then, by polynomial division, obtained

$t^{4}+2=(t^{2}+1)(t+2)(t+1)$

Since for $t^{2}+1$ to be $0$ I'd need $\sqrt{2}$, I came to the conclusion that $\mathbb{Z}_{3}(\sqrt{2})$ is the spitting field of $f$ over $\mathbb{Z}_{3}$

Is my idea correct ?

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  • $\begingroup$ I think you can use Hensel lemma $\endgroup$ – Bonbon Feb 3 at 15:36
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    $\begingroup$ Can you simplify $\mathbb{F}_{3}(\sqrt{2})$? Hint: it is a finite field. $\endgroup$ – Adam Higgins Feb 3 at 15:37
  • $\begingroup$ I just looked up the Hensel lemma and I think it's quite a bit too advanced for my current grasp of algebra $\endgroup$ – Christian Singer Feb 3 at 15:39
  • $\begingroup$ I know that all splitting fields of $f$ are isomorphic but just for now I don't see a quick way to simplify $\mathbb{F}_{3}(\sqrt{2})$ $\endgroup$ – Christian Singer Feb 3 at 15:41
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    $\begingroup$ @ChristianSinger The splitting field of $t^{4} + 2$ over $\mathbb{F}_{3}$ is equivalent to the splitting field $K$ of the irreducible polynomial $t^{2} + 1$ over $\mathbb{F}_{3}$. It is clear that $K$ is a degree two extension of $\mathbb{F}_{3}$. The unique such extension is $\mathbb{F}_{3^{2}} = \mathbb{F}_{9}$. $\endgroup$ – Adam Higgins Feb 3 at 16:13
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Your approach is correct, but I don't think that the expression $\sqrt2$ is used in this context. I would say that the splitting field of $t^4+2$ is $\mathbb{Z}_3[t]/(t^2+1)$ or that it is$$\{a+bs\,|\,a,b\in\mathbb{Z}_3\},$$where $s^2=2.$

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  • $\begingroup$ Yeah, I see your point. $\endgroup$ – Christian Singer Feb 3 at 15:37
  • $\begingroup$ I hope that it is clear that I am not criticizing your reasoning, but just the choice of notation. $\endgroup$ – José Carlos Santos Feb 3 at 15:40
  • $\begingroup$ Yeah sure! I appreciate your response :-) $\endgroup$ – Christian Singer Feb 3 at 15:42

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