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


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 ?

  • $\begingroup$ I think you can use Hensel lemma $\endgroup$ – Bonbon Feb 3 at 15:36
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

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 $s^2=2.$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.