# Splitting field of $f=t^{4}+2\in \mathbb{Z}_{3}[t]$

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 ?

• I think you can use Hensel lemma – Bonbon Feb 3 at 15:36
• Can you simplify $\mathbb{F}_{3}(\sqrt{2})$? Hint: it is a finite field. – Adam Higgins Feb 3 at 15:37
• I just looked up the Hensel lemma and I think it's quite a bit too advanced for my current grasp of algebra – Christian Singer Feb 3 at 15:39
• I know that all splitting fields of $f$ are isomorphic but just for now I don't see a quick way to simplify $\mathbb{F}_{3}(\sqrt{2})$ – Christian Singer Feb 3 at 15:41
• @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 $\mathbb{F}_{3^{2}} = \mathbb{F}_{9}$. – Adam Higgins Feb 3 at 16:13

## 1 Answer

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 $$s^2=2.$$

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