# Question about proof of Lemma in Niven's book about irrational numbers concerning minimal polynomial of a primitive nth root of unity

On page 35 in Niven's book on irrationals , during the proof of Lemma 3.6: That if $\omega$ is a primitive $n$th root of unity, and $f(x)$ is it's minimal polynomial, then for any prime $p$ such that $(p,n) = 1$ and any $\rho$ such that $f(\rho) = 0$ we have that $f(\rho^p) = 0,$ the author makes the following two statements:

1."By the fundamental theorem on symmetric polynomials, the $p$th powers of the roots of $f(x) = 0$ are roots of a polynomial equation which is also monic with integral coefficients, and so $\rho^p$ is an algebraic integer."

How exactly are we applying the fundamental theorem of symmetric polynomials here?

2."We now interpret these equations in $J_p[x]$." ($J_P[x]$ are the polynomials with coefficients in the field of integers modulo $p$).

Why exactly are we allowed to do that? Wouldn't that imply that the polynomials used in the aboive equations already have coefficients in $J_p[x]$?

1. If you write $f(X)= (X-\rho_1)\cdots (X-\rho_k)$, where $\rho_1=\rho$, then the coefficients of this polynomial are just elementary symmetric polynomials of $\rho_1,\ldots,\rho_n$, and by the assumption they are all integers. If you consider now $g(X)= (X-\rho_1^p)\cdots(X-\rho_k^p)$, the coefficients of it are clearly symmetric polynomials of $\rho_1,\ldots, \rho_k$, so by the fundamental theorem on symmetric polynomials they can be expressed as algebraic combinations of elementary symmetric polynomials, implying that they are also integers.
2. By interpreting it is meant that all coefficients are reduced modulo $p$. You can check that this is indeed ok (it preserves equations and so forth).
1. The elementary symmetric functions of the $\rho_i^p$, where the $\rho_i$ are the roots of $f$, are polynomials with integer coefficients in the elementary symmetric functions of the $\rho_i$, which are integers.
2. Any polynomial in $\mathbf Z[x]$ can have its coefficients reduced modulo $n$, thereby resulting in a polynomial in $\mathbf Z/n \mathbf Z[x]$.