5
$\begingroup$

I have the polynomial $f(T)=T^2+T+1$; then, for which primes $p$ does $f(T)$ have roots in $\Bbb F_p$?

I tried this way: since the three roots of $f(T)$ are generated from the cubic root of $1$, we need it to be contained in the field $\Bbb F_p$; namely that $m^3\equiv 1$ $($mod $p)$ for some $m\in \Bbb F_p$. For example in $\Bbb F_7$ we have that $2^3\equiv 1 $ $($mod $7)$ and in fact in this field $2$ is a root of the polynomial $f$; however I don't know how to describe in general in which fields $f(T)$ is reducible and in which is not.

Thank you :)

$\endgroup$
  • $\begingroup$ note: $4$ is also a root in $\mathbb F_7$ $\endgroup$ – J. W. Tanner Jun 24 at 23:08
  • $\begingroup$ The cube root is a slight distraction because $1$ is always a cube root of $1$. $\endgroup$ – user10354138 Jun 24 at 23:09
  • $\begingroup$ $f$ has only at most two roots in any field… $\endgroup$ – Bernard Jun 24 at 23:13
4
$\begingroup$

Hint: Use $4f(T)=(2T+1)^2+3$, and quadratic reciprocity.

$\endgroup$
4
$\begingroup$

You're just about there. As you've observed, $f(T)$ has roots in $\Bbb Z / p \Bbb Z$ if and only if $1$ has non-trivial cube roots in that field. And since the multiplicative group has order $\lvert (\Bbb Z / p \Bbb Z)^* \rvert = p-1$ and it's cyclic, that occurs exactly when $p \equiv 1 \pmod{6}$.

$\endgroup$
  • $\begingroup$ Thank you very much $\endgroup$ – Dorian Jun 25 at 8:11

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.