Steenrod Algebra as automorphisms of additive group Is there a direct way to see that the subalgebra of the mod-$p$ Steenrod algebra ${\mathcal A}_p$ generated by the reduced powers is isomorphic to the dual of the Hopf algebra ${\mathcal O}(\text{Aut}({\mathbb G}_a))$ of functions on the automorphisms of the additive formal group law over ${\mathbb F}_p$?
So far I only understand this by looking at their respective actions on the infinite polynomial ring over ${\mathbb F}_p$, but I wonder if there is more direct way to get from ${\mathbb F}_p$-cohomology operations to the additive group law over ${\mathbb F}_p$.
Thank you!
 A: Unless explicitly stated otherwise all cohomology will be taken with $\mathbb{F}_p$-coefficients and $p$ will be an odd prime.
Consider the exact sequence of groups 
$$0 \to \mathbb{Z}/p\mathbb{Z} \to U(1) \overset{(-)^p}{\to} U(1) \to 1$$
Since the classifying space functor $B(-)$ takes short exact sequence of groups to fiber sequences we get the fiber sequence 
$$ B\mathbb{Z}/p\mathbb{Z} \to BU(1) \to BU(1)$$
Applying the serre spectral sequence to this fibration we see that the cohomology $H^{\ast}(B\mathbb{Z}/p\mathbb{Z})$ must be generated as an algebra by two classes $x\in H^1, \beta(x) \in H^2$. Where $x$ exists to kill the cohomology of the base, i.e. $d_2(x) = u \in H^2(BU(1))$ where $H^{\ast}(BU(1)) = \mathbb{F}_p[[u]]$. The class $\beta(x)$ as the notation suggests is the bockstein of $x$ and it surives to $E_{\infty}$ and becomes a generator of the cohomology of the total space $H^{\ast}(BU(1)) = \mathbb{F}_p[[j^{\ast}\beta(x)]]$ (where $j :B \mathbb{Z}/p \mathbb{Z} \to BU(1)$). 
Alltogether (and carrying the extra structure over) this shows that the map of abelian groups $\mathbb{Z}/p\mathbb{Z} \to U(1)$ induce a map of bi-commutative completed (we will return to this point in a second) hopf algebras $j^{\ast}: H^{\ast}BU(1) \to H^{\ast}(B\mathbb{Z}/p \mathbb{Z})$ which identifies $H^{\ast}(BU(1))$ as the even subalgebra $\mathbb{F}_p[[\beta(x)]] \subset H^{\ast}(B\mathbb{Z}/p\mathbb{Z})$. By naturality this identification respects the action of the Steenrod algebra.
Completeness means the comultiplication lands in the completed tensor product $$H^{\ast}(BU(1)) \to H^{\ast}(BU(1)) \hat{\otimes}H^{\ast}(BU(1))$$
Since $H^{\ast}(BU(1))$ is a formal power series ring and because the compltiplication satisfies further conditions stemming from the abelian group axioms on $U(1)$ this map is equivalent to what is called a formal group law.
It is an easy exercise that the abelian group structure on $BU(1)$ coming from tensoring line bundles makes $H^{\ast}(BU(1),\mathbb{Z})$ into the completed hopf algebra corresponding to the additive formal group. It is a slightly less trivial observation that the Steenrod action of $\mathcal{A}_p$ on $H^{\ast}(B \mathbb{Z} /p \mathbb{Z})$ respects the hopf algebra structure. Now note that The even part of the Steenrod algebra (preserves and thus) acts on the even part of $H^{\ast}(B\mathbb{Z}/p\mathbb{Z})$ and this action also respect the hopf-structure and as such it "acts on the additive formal group over $\mathbb{F}_p$". 
To interpret the statement above more precisely one can take the even part $\mathcal{P} \subset \mathcal{A}_p$ and note that the coaction of $\mathcal{P}^{\vee}$ (the dual hopf algebra) on $H^{even}(B\mathbb{Z} /p \mathbb{Z})$ is the same thing as an action of the group scheme $\mathbb{G}_p := Spec \mathcal{P}^{\vee}$ on the formal group scheme $Spf H^{even}(B \mathbb{Z}/p \mathbb{Z}) = Spf H^{even}(BU(1))$. So by definition we get a homomorphism:
$$\mathbb{G}_p \to Aut(\widehat{\mathbb{G}_a})$$
Verifying that it is bijective onto the strict automorphisms is a staightforward computation (once the right hand side is known algebraically).
