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Let $f(X)$ be the minimal polynomial of something like $\zeta + \frac{1}{\zeta}$, where $\zeta$ is a primitive $p$-th root of unity for some prime $p > 2$.

I'd like to show that $f(X) \equiv (X-2)^{(p-1)/2}\pmod p$, that is, the only zero of $f$ is $2$.

It is not difficult to see that $f(X)$, over $\mathbb{C}$, has exactly the zeros $\eta + \frac{1}{\eta}$ where $\eta$ runs through the primitive $p$-th roots of unity. Over $\mathbb{F}_p$, the only $p$-th roots of unity are $1$, so "heuristically" every $\eta + \frac{1}{\eta} \equiv 1+1 = 2$ mod $p$, and $2$ is the only root of $f$ in $\overline{\mathbb{F}_p}$. Can this argument be made rigorous? Thanks!

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There is a unique homomorphism $\mathbf{Z}[\zeta] \to \mathbf{F}_p$, because there is a unique $p$-th root of unity in $\mathbf{F}_p$.

All of the roots of $f(x)$ are in $\mathbf{Z}[\zeta]$, so $f(x)$ splits into linear factors over $\mathbf{Z}[\zeta]$. If we let $\bar{f}(x)$ be the reduction fo $f$ modulo $p$, then applying the homomorphism $\mathbf{Z}[\zeta] \to \mathbf{F}_p$ tells us that $\bar{f}(x)$ also splits into linear factors over $\mathbf{F}_p$, and furthermore its roots are precisely the images of the roots of $f(x)$. In particular, every root is $2$ as you describe.

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Let $g(X,Y) = (X-Y-Y^{p-1})(X-Y^2-Y^{p-2})\cdots (X-Y^{\frac{p-1}{2}}-Y^{\frac{p+1}{2}}) \in \mathbb{Q}[X,Y]$, so that $g(X,\zeta)=f(X) \in \mathbb{Q}[X]\subset\mathbb{Q}(\zeta)[X]$.

We can identify $\mathbb{Q}(\zeta)(X)$ with $\mathbb{Q}(X)[Y]/\Phi_p (Y)$, where $\Phi_p (Y) = (Y^p-1)/(Y-1)$ is the p-th cyclotomic polynomial. Since the image of $g$ in the quotient is $f(X)$, by the division algorithm there is some $h(X,Y)\in\mathbb{Q}(X)[Y]$ with $g(X,Y)=f(X)+h(X,Y)\Phi_p (Y)$. By Gauss' Lemma, we can assume $h(X,Y)\in \mathbb{Z}[X,Y]$.

Since $\Phi_p (1) \equiv 0\pmod{p}$, setting $Y=1$ gives us $f(X)\equiv g(X,1) \equiv (X-1-1)(X-1-1)\cdot (X-1-1) \equiv (X-2)^{\frac{p-1}{2}} \pmod{p}$.

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Since the ring of integers in $K=\mathbb{Q}(\zeta+\frac{1}{\zeta})$ is $\mathbb{Z}[\zeta+\frac{1}{\zeta}]$, we can write prime decomposition of $p$ in $K$ in terms of the factorization of the minimal polynomial $f(X)$ modulo $p$. i.e. If $$ f(X)=f_1(X)^{e_1}\cdots f_r(X)^{e_r}$$ modulo $p$, then the prime $p$ in $\mathbb{Q}$ is decomposed in $K$ as $$ Q_1^{e_1}\cdots Q_r^{e_r}$$ where $$Q_i= (p, f_i(\zeta+\frac{1}{\zeta})). $$

Since $p$ is totally ramified in $\mathbb{Q}(\zeta)$, it follows that $p$ is also totally ramified in $K$.

Therefore, the factorization of $f$ modulo $p$ should be $(X-\alpha)^{\frac{p-1}{2}}$ where $\alpha\in \mathbb{F}_p$.

Now determination of $\alpha$ follows from classical identity: $$ (2-2\cos(2\pi/p))(2-2\cos(4\pi/p))\cdots (2-2\cos((p-1)2\pi/p))=p^2$$

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