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Let $p > 2$ be a prime, and let $\zeta = e^{2\pi i/p}.$ Then we know that the minimal polynomial of $\zeta$ is $$f(x) = 1 + x + x^2 + \cdots + x^{p-1} = \prod_{i=1}^{p-1}(x - \zeta^i).$$ Now $\mathrm{Norm}(1-\zeta) = \prod_{i=1}^{p-1}(1 - \zeta^i) = f(1) = p,$ where the norm is the product of the conjugates $\sigma_i(\zeta),$ with the $\sigma_i$ being the $p-1$ embeddings of $\mathbb{Q}(\zeta)$ into $\mathbb{C}$. Furthermore, $\mathrm{Norm}(1-\zeta) = \mathrm{Norm}(\zeta - 1)$ as $p-1$ is even.

Now after this, my notes say that $$f(x+1) = \mathrm{Norm}(x + 1 - \zeta) = \mathrm{Norm}(x - (\zeta -1)) = \text{minimal polynomial of }\zeta - 1.$$ However, I do not really understand how the last equality comes from. Is it not possible that the minimal polynomial of $\zeta - 1$ divides $f(x+1)?$ I'm assuming it uses the previous statement that the norms of $1-\zeta$ and $\zeta - 1$ are the same, but I cannot see how? Can anybody help please?

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    $\begingroup$ $\zeta$ is a primitive $p^{th}$ root of $1$, so $\zeta-1$ can't have degree less than $p-1$. It's minimum polynomial will have the same degree as $f$ $\endgroup$ – sharding4 May 31 '17 at 22:32
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    $\begingroup$ By definition of the minimal polynomial, If $f(x)$ is the minimal polynomial of $\alpha$ then $f(x+1)$ is the minimal polynomial of $\alpha-1$. $\endgroup$ – reuns Jun 1 '17 at 2:51

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