I am trying to figure out the splitting field of $x^3 - x + 1$ over $\mathbb{F}_3$ .

It will be the field $\mathbb{F}_3[x]/\langle x^3 -x +1\rangle$ but I have to show irreducibility too.

All I know is that there is a rule that states that if a polynomial of degree $n$ over a finite field is irreducible if it is a factor of $x^{p^n -1} -1$. But in this case, I have to do long division of $x^{26} -1$. Is there any other way this can be done ?

Can mod-2 test be used here ?

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    $\begingroup$ For a cubic you can check whether it has a root to determine if it is reducible or irreducible. $\endgroup$ – sharding4 Jun 29 '18 at 14:36
  • $\begingroup$ @sharding4 Is it right if I simply say that elements of $\mathbb{F}_3$ , i.e 0,1 and 2 are not roots of the polynomial ? $\endgroup$ – The Doctor Jun 29 '18 at 14:38
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    $\begingroup$ Yes. For a quadratic or cubic you can say that the polynomial is irreducible if it has no root over the field in question. $\endgroup$ – sharding4 Jun 29 '18 at 14:39
  • $\begingroup$ @sharding4 How do I next show that the polynomial indeed has a root in the splitting field I thus obtained ? $\endgroup$ – The Doctor Jun 29 '18 at 14:41
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    $\begingroup$ If you have obtained the splitting field it contains the root by definition ;-) If you think about it a moment, it should become clear that $x+\langle x^3-x+1 \rangle $ is a root. $\endgroup$ – sharding4 Jun 29 '18 at 14:48

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