This is a part of page 214 in Ireland and Rosen's "A Classical Introduction to Modern Number Theory". $p$ is a prime coprime to $m$, $\zeta_m = e^{\frac{2\pi i}m}$, $D_m$ is the ring of integers of $\mathbb Q(\zeta_m)$, $f$ is the least positive integer for which $p^f = 1 \mod m$, $g(P)$ is the gauss sum belonging to the $F = D_m / P$ characters $\chi (\gamma + P) = 1/(\gamma/P)_m, \psi (t) = \zeta_p ^{t + t^p + t^{p^2} + \dots + t^{p^{f-1}}}$, $\Phi(P) = g(P)^m$, and for all $1\le t < m$ coprime to $m$ $\sigma_t \in G(\mathbb Q (\zeta_m), \mathbb Q)$ is the automorphism that maps $\zeta_m$ to $\zeta_m ^t$.
Why can we talk about $g(P)^{\sigma_{t'}}$ if $g(P)$ doesn't belong to $\mathbb Q(\zeta_m)$? Assuming we can, why is $\sigma_{t'} (\psi (r))$ equal to $\psi (r)$? I assume it should follow from $\sigma _{t'} (\psi (r)) = \psi (r) ^{t'}$, but I don't see why that should be the case given that $\psi(r)$ is a $p$th root of unity (and not an $m$th one).