Steenrod Operations an algebraic Approach Assume that $q=p^{r}$, where $p$ is a prime either 2 or odd and $\mathbb{F}_{q}$ is a Galois field and $V$ a finite dimensional $\mathbb{F}_{q}$-vector space. Then due to Larry Smith in this http://arxiv.org/pdf/0903.4997.pdf we can define the $\mathbb{F}_{q}$-algebras homomorphism ${\mathcal{P}}_{T}:\mathbb{F}_{q}[V] \rightarrow \mathbb{F}_{q}[V][T]$ such that ${\mathcal{P}}_{T}(l)=l+{l}^{q}T$ $\forall l \in V^{*}$, $l$ linear, and by extending that homomorphism linearly, we end up by a formula ${\mathcal{P}}_{T}(f)= \sum\limits_{i=1}^{\infty}\mathcal{P}^{i}(f) T^{i}$, $\forall f \in \mathbb{F}_{q}[V]$ and $\mathcal{P}^{i}$'s are exaclty the so-called Steenrod operations. So in the above notes, says that this algebra homomorphism commutes with the elements of $GL(V)$, but I can't see why! Can you help me please?
 A: I think that find something! Any comments or remarks are really acceptable. So, since we have the formula ${\mathcal{P}}_{T}(f)= \sum\limits_{i=1}^{\infty}\mathcal{P}^{i}(f) T^{i}$, $\forall f \in \mathbb{F}_{q}[V]$, it's only need to prove that Steenrod operations commute with the elements of $GL(V)$ i.e. $\mathcal{P}^{i}(A \thinspace f)=A \thinspace \mathcal{P}^{i}(f)$, $\forall f \in \mathbb{F}_{q}[V]$. But this construction is natural in the sense that, given an arbitrary $\phi:V \rightarrow V$, $\mathbb{F}_{p}$-linear homomorphism, gives rise to a homomorphism $\phi^{*}:\mathbb{F}_{p}[V] \rightarrow \mathbb{F}_{p}[V]$. In fact we have the following equality: $(\phi^{*} \circ \mathcal{P}^{i})(f)=\mathcal{P}^{i}(f) \circ \phi = \mathcal{P}^{i}(f \circ \phi)=\mathcal{P}^{i}(\phi^{*} \circ f)=(\mathcal{P}^{i} \circ \phi^{*}) (f)$. So that means that commutes with every element of $GL(V)$ as well. The last remark gives a answer to the former question. Am I right? Or do I miss something subtle here?
