Proposition 2

— Let $A$ be a commutative ring of characteristic $p$, $K$ a subring of $A$, $S$ a subset of $A$ and $f$ a positive integer.

a) We have $K[S]^{p^f} = K^{p^f}[S^{p^f}]$, and if $A$ is a field, $K(S)^{p^f} = K^{p^f}(S^{p^f})$.

b) If the $K$-module $K[S]$ is generated by the family $(a_i)_{i \in I}$ of elements of $A$, then the $K$-module $K[S^{p^f}]$ is generated by the family $(a_i^{p^f})_{i \in I'}$

Since $K[S]$ is the subring of $A$ generated by $K \cup S$, its image $KISlp^f$ under the endomorphism $\pi : a \mapsto a^{p^f}$ of the ring $A$ is the subring of $A$ generated by the image $K^{p^f} \cup S^{p^f}$ of $K \cup S$ under $\pi$, whence $K[S]^{p^f} = K^{p^f}[S^{p^f}]$. The case of fields is treated similarly; this proves a).

It is clear that the family $(a_i^{p^f})_{i \in I}$ generates the $K^{p^f}$-module $K[S]^{p^f}$. The $K$-module $K[S^{p^f}]$ is generated by products of the form $x_1^{p^f} \cdots x_n^{p^f} = (x, \ldots x_n)^{p^f}$ with $x_1, \ldots, x_n$ arbitrary in $S$, hence also by the set $K[S]^{p^f}$. Assertion b) follows directly from this.

^1 Of course the set $S^{p^f}$ should not be confused with the set product of $p^f$ sets equal to $S$, nor with the set of products of $p^f$ elements belonging to $S$.