Commuting square in representation theory of affine group schemes I am reading Milne's 'Basic Theory of Affine Group Schemes', and in particular I am trying to understand his proof of the fact that linear representations of an affine group $G$ are in one-to-one correspondence with $\mathcal{O}(G)$-comodule structures (proposition 6.1).
One of his steps is to acknowledge that for a representation $r$, $g\in G(R)$ and $a\in G(\mathcal{O}(G))$ the universal object of the affine group, the morphism $r_R(g)$ is the unique $R$-linear map making the following square commute (Milne writes $A$ for the coordinate ring $\mathcal{O}(G)$):
$\hspace{5cm}$
I can show that this is indeed true by using the naturality of $r$ and explicitly computing both compositions, but the way Mile writes this proof makes it clear that this is not the intended way to come to this result (in particular, Milne later makes the same computation for another step). Is there a nice way to see why $r_R(g)$ makes this square commute?

I'm not sure what do you mean by computation, but it is a direct consequence of definition. having a group action means that for each $k$-algebra map $R\to R'$ the diagram

is commutative(see page 112 of the Milne book). now being the universal element means that for each $g\in G(R)$ there is a map $g:O(G)\to R$ which sends $a$ to $g$. the commutativity of the above diagram for the map $g:O(G)\to R$ implies the commutativity of the diagram you want.
