# Product of $(1-\zeta^k)$ where $\zeta$ is a $p$th root of unity

Let $\zeta$ be a $p$th root of unity over $\mathbb Q$. I am trying to show that

$\prod_{k=1} ^{p-1} (1- \zeta^k) = p$.

So far, all I've been able to do is show that the product is rational, since its fixed by the Galois group of $\mathbb Q(\zeta)$. Any suggestions for this problem?

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You can get displayed equations by putting them in double dollar signs -- they look nicer and are easier to read. –  joriki Apr 20 '12 at 1:05
There are no roots of unity in $\mathbb Q$ other than $\pm1$. –  joriki Apr 20 '12 at 1:06
It's good style to mark edits as edits when they cause someone else's comment to appear wrong. –  joriki Apr 20 '12 at 1:19

We know that

$$x^p-1 = \prod_{k=0}^{p-1} \big(x-\zeta^k\big).$$

We also know that

$$x^p-1 ~~=~~ (x-1)\prod_{k=1}^{p-1} \big(x-\zeta^k\big) ~~=~~ (x-1)\big(1+x+\cdots+x^{p-1}\big)$$

by the geometric sum formula. Divide the right equality by $x-1$.

We end up with the polynomial expansion for $\prod_1^{p-1} (x-\zeta^k)$, so just plug in $x=1$.

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Alternatively, this is what's called the "norm map" $N_{\mathbb{Q}(\zeta)/\mathbb{Q}}(1-\zeta)$. Namely, what you are considering is
$$\prod_{\sigma\in\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})}\sigma(1-\zeta)$$
The cool thing it turns out is that $N_{\mathbb{Q}(\zeta)/\mathbb{Q}}(x)$ is just the determinant of the linear transformation determined by mulltiplication by $x$. In particular, we know that $1,\cdots,\zeta^{p-1}$ is a basis for $\mathbb{Q}(\zeta)/\mathbb{Q}$. Note then that if multiply by $1-\zeta$ we get the matrix
$$\begin{pmatrix}1 & 0 & 0 & \cdots & 1\\ -1 & 1 & 0 & \cdots & 1\\ 0 & -1 & 1 & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & -1 & 2\end{pmatrix}$$
with dimension $(p-1)\times(p-1)$--which has determinant $p$ as desired.