Relation bewteen $p-1$-roots of unity mod $p$, $p$ prime. My goal is to prove Kummer's Lemma, and I am trying to unserstand the proof of the Kummer criterion for the regular prime (pag 236 of Edwards, Harold M.,Fermat's Last Theorem. A Genetic Introduction to Algebraic Number Theory). Let $\beta$ be a primitive $p-1$-root of unity and $\gamma$ a $p-1$-root of unity mod $p$. So both of them are in particular $p-1$-root of unity mod $p$. 
If $\psi(x)$ is a polynomial with integer coefficient, then $\psi(\beta)\psi(\beta^3)\cdots\psi(\beta^{p-2})\equiv\psi(\gamma)\psi(\gamma^3)\cdots\psi(\gamma^{p-2})\pmod p$. This, explay Edward, because "the integer $\gamma$ satisfies mod $p$ all the relations satisfied by the complex $\beta$, bu i can't understand this passage..
 A: The first thing to mention is that, as a primitive $(p-1)$th root of unity, $\beta \in \Bbb C$ has the property that $\beta^{p-1} = 1$. Similarly, as a primitive root modulo $p$ (which is the definition of $\gamma$ given in the text you refer to), $\gamma$ has the property $\gamma^{p-1} \equiv 1 \bmod p$ (and $\gamma^k \not\equiv 1 \bmod p$ for any $k < p-1$). 
Now, consider $\psi(X) = b_1 X + \dots + b_{p-1} X^{p-1}$ with $pb_i = \gamma\gamma_{i-1} - \gamma_i$ (where $\gamma_{i}$ is an integer such that $\gamma_{i} \equiv \gamma^{i} \bmod p$, which you know exists because $\gamma$ is a primitive root mod $p$). The product of $\psi(\beta), \psi(\beta^3), \dots, \psi(\beta^{p-2})$ is an integer; to see this, note that this product is invariant under the action of $\operatorname{Gal}(\Bbb Q(\beta_{p-1})/\Bbb Q)$, which only fixes elements of $\Bbb Q$ (hence also $\Bbb Z$) and since the coefficients are sums and products of integers, this is true for the considered product. Thus, you have
$$\psi(\beta) \psi(\beta^3) \cdots \psi(\beta^{p-2}) \equiv \psi(\gamma) \psi(\gamma^3) \cdots \psi(\gamma^{p-2}) \bmod p$$
which is what the text is referring to (i.e. these integers coincide modulo the prime $p$ because $\gamma$ satisfies the same relations as $\beta$).

There were a few typos in the answer I gave before! Let me expand on the explanation a bit.
To show that $\psi(\beta)\psi(\beta^{3}) \dots \psi(\beta^{p-2}) \in \Bbb Z$, note that
$$\sigma_i(\psi(\beta)) = \sigma_i(b_1\beta + b_2\beta^2 + \dots + b_{p-1}\beta^{p-1}) = b_1\sigma_i(\beta) + \dots + b_{p-1}\sigma_i(\beta^{p-1})$$
since $\sigma_i$ is a field automorphism of $\Bbb Q(\beta_{p-1})/\Bbb Q$.
To reason why we should take the product over the odd powers of $\beta$, we note that, since $p$ is itself odd, $p-1$ is even, and so any primitive $(p-1)$th root of unity must itself be an odd power of $\beta$, and any automorphism $\sigma_i \in \operatorname{Gal}(\Bbb Q(\beta_{p-1})/\Bbb Q)$ simply permutes the odd powers of $\beta$. In particular, this means that the product
$$\psi(\beta)\psi(\beta^3) \cdots \psi(\beta^{p-2}),$$
whose coefficients are simply sums and products of the $b_j$ (defined in terms of the integers $\gamma\gamma_{j-1} - \gamma_j$), is invariant under the action of $\operatorname{Gal}(\Bbb Q(\beta_{p-1})/\Bbb Q)$.
To reason that $\psi(\beta)\psi(\beta^3)\cdots \psi(\beta^{p-2}) \equiv \psi(\gamma)\psi(\gamma^{3}) \cdots \psi(\gamma^{p-2}) \bmod p$, the explanation that, modulo $p$, $\gamma$ satisfies the same relations as $\beta$ suffices (though one should not forget that $\beta \bmod p$ makes no sense as $\beta \notin \Bbb Z$, this is why I keep writing modulo in bold; modulo $p$, the congruence written above is simply just relabelling $\beta$ as $\gamma$, since these satisfy the same relation! I could write 
$$\prod_{\substack{1 \leq i \leq p-2\\ i\ \text{odd}}} \psi(\text{fish}^{i}) \equiv \prod_{\substack{1 \leq i \leq p-2\\ i\ \text{odd}}}\psi(\gamma^i) \bmod p$$
as long as I first specified that $\text{fish}$ is a primitive $(p-1)$th root of unity!)
