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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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  • $\begingroup$ Presumably you mean $F_{p^2}[x]/(x^2+1)$? $\endgroup$
    – Zev Chonoles
    Commented Mar 25, 2011 at 19:17
  • $\begingroup$ Yes. I meant that! $\endgroup$
    – Niti
    Commented Mar 25, 2011 at 19:19
  • $\begingroup$ 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. $\endgroup$
    – Zev Chonoles
    Commented Mar 25, 2011 at 19:22
  • $\begingroup$ 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. $\endgroup$
    – Zev Chonoles
    Commented Mar 25, 2011 at 19:23
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    $\begingroup$ 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. $\endgroup$
    – Zev Chonoles
    Commented Mar 25, 2011 at 19:27

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OK, this is about imitating the formula for a complex cube root of unity. Write p as 12k - 1. The real issue is only why 3 to the power 3k should act as square root of 3 in this field. Square it and apply Fermat's little theorem to see why. (There is a missing factor 2 in the formula you gave.)

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  • $\begingroup$ Hi Charles, where exactly do you think the 2 is missing? $\endgroup$
    – Niti
    Commented Mar 26, 2011 at 10:52
  • $\begingroup$ Please write down the formula for a complex cube root of unity. $\endgroup$
    – Charles Matthews
    Commented Mar 26, 2011 at 11:14
  • $\begingroup$ lets say z^3=1. Then z in complex (Re{z} + i* Im{z}): Re{z} = (p-1)/2, Im{z} = 3^((p+1)/4) mod(p) $\endgroup$
    – Niti
    Commented Mar 26, 2011 at 12:49
  • $\begingroup$ Sigh. Like in en.wikipedia.org/wiki/Root_of_unity#Examples. The formula you are proposing mimics that formula, but it is all over the factor 2. Anyway I guess I gave the wrong reason about the square root of 3, since quadratic reciprocity should enter. But like everyone said, MO is the wrong forum; and your statement had a couple of errors also. $\endgroup$
    – Charles Matthews
    Commented Mar 26, 2011 at 13:34
  • $\begingroup$ Thank you very much! I got it now. You are my hero. Anyway I am surprised why all of you get so uppset. I made some errors but... I am just trying to learn. $\endgroup$
    – Niti
    Commented Mar 26, 2011 at 16:48

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