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That the endomorphism ring of a Drinfeld $A$-module of rank 1 is isomorphic to $A$ appears as a corollary to the more general result that a Drinfeld module of rank $r$ has endomorphism ring of rank at most $r^2$ (a result obtained by analyzing the map $\text{Hom}(\phi,\psi)\otimes_A A_{\mathfrak{p}}\to\text{Hom}(T_{\mathfrak{p}}(\phi),T_{\mathfrak{p}}(\psi))$). Without using this more general result or the uniformization theorem for Drinfeld modules, is there a way to get a more direct handle on $\text{End}(\phi)$ when $\phi$ is of rank 1?

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We can construct a new object from $O:=\text{End}(\phi)$ by looking at $\psi:O\to \mathscr{F} \{ \tau \}$ defined by $\psi_f = f$. This thing is not automatically a Drinfeld module, but it is close.

We can either relax our definition of Drinfeld modules or use an isogeny to assume that $\psi$ is a Drinfeld module (See the proof of Proposition 4.7.17 in Basic Structures of Function Field Arithmetic by Goss and Explicit Class Field Theory of Global Function Fields by Hayes).

Now, by looking at the torsion $\psi[a]$ for some $a\in A$ (we need to be a little careful here if $\psi$ is not really a Drinfeld module), we can see that the rank of $\psi$ is less than that of $\phi$ and also that $O$ is equal to $A$.

EDIT: I forgot to say that if $\text{End}(\phi)$ is not commutative you may have to look at a commutative subring, just like in Proposition 4.7.17 of Goss' book.

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