Frobenius powers of an ideal does not depend on the choice of a system of generators

Question

Let $I$ = $(x_1 , . . . , x_n )$ be an ideal of a ring $R$ of characteristic $p$. For each nonnegative integer $e$ we set $I^{[p^e]}$=$(x_1^{p^e},...x_n^{p^e}$)$R$. These ideals are called the Frobenius powers of $I$.

Claim: Frobenius powers of $I$ does not depend on the choice of generators for the ideal $I$.

Answer 1

This is pretty straight forward and heavily relies on the additivity of the Frobenius. Assume $(x_1, \dotsc, x_n ) = (y_1, \dotsc, y_m)$ and show $(x_1^{p^e}, \dotsc, x_n^{p^e}) = (y_1^{p^e}, \dotsc, y_m^{p^e})$.

For that you only have to check $y_1^{p^e} \in (x_1^{p^e}, \dotsc, x_n^{p^e})$. Everything else comes from symmetric reasons.