a submodule of $R^n$ Let $R$ be a commutative ring of positive characteristic p. If $M$ is a submodule of $R^n$, let $M^{[p]}$ be the submodule of $R^n$ generated by $(a_1^p,\cdots,a_n^p)$ where $(a_1,\cdots,a_n)\in M$. The question is: Is $M^{[p]}$ independent of the choice of the basis of $R^n$?
Thanks!
 A: Let $R$ be a commutative ring. If $M$ is a submodule of $R^n$ then let $M^{[m]}$ be the submodule of $R^n$ generated by the elements $(a_1^m, \dots, a_n^m)$ as $(a_1, \dots, a_n)$ varies in $M$.
I interpret your question in the following way:
Question. If $p$ is a prime, $R$ has characteristic $p$, $M$ is a submodule of $R^n$ and $\phi \colon R^n \to R^n$ is an automorphism of $R^n$, then $\phi(M^{[p]}) = \phi(M)^{[p]}$?
If $n = 1$, the answer is yes. A submodule of $R$ is an ideal $I$ of $R$ and an automorphism $\phi$ of $R$ is simply the multiplication by an invertible element $\mu$ of $R$. We have
$$
\phi(I^{[p]}) = \mu I^{[p]} = I^{[p]} = \mu^p I^{[p]} = \langle \mu^p a^p \mid a \in I \rangle = (\mu I)^{[p]} = \phi(I)^{[p]}.
$$
If $n \geq 2$, the answer is no. Consider the submodule $M$ of $R^2$ generated by $(0,1)$ and consider the automorphism $\phi$ of $R^2$ given by the matrix
$$
\phi = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix},
$$
where $t \in R$ is an element such that $t^p \neq t$ (for example you can pick $R = \mathbb{F}_p[t]$). Now we have:
$$
\phi(M^{[p]}) = R \begin{pmatrix} t \\ 1 \end{pmatrix}
$$
$$
\phi(M)^{[p]} = R \begin{pmatrix} t^p \\ 1 \end{pmatrix}.
$$
(Note that we haven't used the fact that $p$ is prime.)
