# nontrivial cube root of unity [closed]

Hi,

I have a finite field Fp with p = 11 mod(12) and I am trying to get the third nontrivial root of unity in Fp^2 = Fp^2[x]/(x^2+1). So, i need x where x^3=1.

Somehow I came into a source saying that it would be: (p-1)/2 + (3^((p+1)/4)) mod(p))*i where i^2=-1. But it seems not to be correct: Any idea?

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## closed as off topic by Franz Lemmermeyer, Zev Chonoles, Mariano Suárez-Alvarez♦, Daniel Litt, Felipe VolochMar 25 '11 at 20:19

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Presumably you mean $F_{p^2}[x]/(x^2+1)$? – Zev Chonoles Mar 25 '11 at 19:17
Yes. I meant that! – Niti Mar 25 '11 at 19:19
You can edit your question - the "edit" button is right underneath the tags. Also, you mean "nontrivial cube root of unity", not "third nontrivial root of unity", because every element of a finite field other than 0 and 1 is a nontrivial root of unity. – Zev Chonoles Mar 25 '11 at 19:22
Also, I'm afraid that your question is not at the right level for MO, but there are many other sites where your question would be better suited. See mathoverflow.net/faq#whatnot. – Zev Chonoles Mar 25 '11 at 19:23
Right. However, the polynomial $x^2+1$ already has its roots in $F_{p^2}$, so $F_{p^2}[x]/(x^2+1)$ cannot be a finite field. – Zev Chonoles Mar 25 '11 at 19:27