Remarks
In Part 3, we need not have
g(X)p = g(Xp)
if we replace Z/pZ by an arbitrary field K of
characteristic p.
For instance,
let K be the rational function field Z/pZ(Y).
Then g(X) =
YX 
K[X]
satisfies
g(X)p =
(YX)p = Yp Xp
Y Xp =
g(Xp).
If K is finite, of order say q,
it may happen that, for different powers r, s of p,
the maps x -> xr
and x -> xs are identical.
For instance, r = 1 = p0 and
s = q both represent the identity on K.