0
$\begingroup$

I'm currently reading a proof that the minimal polynomial of $\alpha$, a primitive pth root of unity, over $\mathbb{Q}$ is $$1+t+\ldots +t^{p-1},$$ yet the author simply states "Clearly $\alpha$ is a zero." in the proof, and I can't seem to prove that bit myself.

$\endgroup$
  • $\begingroup$ Oh, now I see that if I multiply the polynomial by $(t-1)$ I get $t^p-1$, which zeroes are clearly $1, \alpha , \ldots , \alpha ^{p-1}$. $\endgroup$ – Leo Apr 29 at 0:27
2
$\begingroup$

Note $(1+t+\cdots + t^{p-1} ) (1-t)= t^p -1$, and that $\alpha$ is a root of $t^p -1$, since it is a pth root of unity, but not of $1-t$, since it is primitive. Then it must be a root of $1+\cdots+t^{p-1}$, as the author says.

| cite | improve this answer | |
$\endgroup$

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.