Primitive pth root of unity of a prime p

**tttcomrader** · April 12th 2008, 01:32 PM

Let $K= \mathbb {Q}[ \omega ]$ , where $\omega$ is a primitive pth root of unity for an odd prime p, and write N for $N_{K}$ .

a) Let f be the minimum polynomial of $\omega$ . Show that $f'( \omega ) = \frac {p}{ \omega ( \omega - 1)}$

proof so far:

Now, $x^p - 1 = (x-1)f(x)$
Then $px^{p-1} = (x-1)f'(x) + f(x)$

Plug in $\omega$ , then $f( \omega ) =1$ , then we have:

$p \omega ^{p-1} = ( \omega - 1) f'( \omega) + 1$
$\frac {p \omega ^{p-1} - 1} { \omega - 1 } = f' ( \omega )$

Am I doing this right so far?

b) Show that $disc( \omega ) = -p^{p-2}$ , in addition, $disc ( \omega ) = p^ {p-2}$ iff $p \equiv 1 \ (mod \ 4)$

I just can't really get a hand on disc, any hint?

Thank you!

**ThePerfectHacker** · April 12th 2008, 05:28 PM

Originally Posted by tttcomrader

Let $K= \mathbb {Q}[ \omega ]$ , where $\omega$ is a primitive pth root of unity for an odd prime p, and write N for $N_{K}$ .

a) Let f be the minimum polynomial of $\omega$ . Show that [tex] f'( \omega ) = \frac {p}{ \omega ( \omega - 1)}

proof so far:

Now, $x^p - 1 = (x-1)f(x)$
Then $px^{p-1} = (x-1)f'(x) + f(x)$

Plug in $\omega$ , then $f( \omega ) =1$ , then we have:

$p \omega ^{p-1} = ( \omega - 1) f'( \omega) + 1$
$\frac {p \omega ^{p-1} - 1} { \omega - 1 } = f' ( \omega )$

Am I doing this right so far?

I would do it a little differently. The mimimal polynomial is $1+x+...+x^{p-1}$ . It happens to be irreducible so that is why it is minimal (you never said why you poynomial was minimal). And then differenciate term-by-term.

b) Show that $disc( \omega ) = -p^{p-2}$ , in addition, $disc ( \omega ) = p^ {p-2}$ iff $p \equiv 1 \ (mod \ 4)$

I just can't really get a hand on disc, any hint?

I never seen "disk" before.

**tttcomrader** · April 23rd 2008, 06:38 AM

Disc is the discriminant, so is the determinant of $\{ \omega ^{1}, \omega ^{2}, . . . , \omega ^{p-1} \}$

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