On p. 643 of Lang's Algebra, revised 3rd edition, he writes
Remark: Let $E$ be a 1-dimensional vector space over a division ring $D$, and let $\{v\}$ be a basis. For each $a \in D$, there exists a unique $D$ linear map $f_a: E \to E$ such that $f_a(v) = av.$ Then we have the rule $$f_af_b=f_{ba}.$$
I just cannot understand what he means. It would make sense to me if he said $f_a(v) = va$ (this would then be a left module homomorphism). I doubt it is a typo, since it's been through three editions and this would be a pretty glaring one...also I did not find it in the errata.
What am I missing?
OK, I figured it out with the help of a bright math professor.The issue is not the way that function composition is written, nor is there a typo.
The function $f_a$ is not defined by "left multiply the argument by $a$". Rather, it is defined by the fact that it maps the generator $v \mapsto av$, and from there we extend left-linearly. The same is true of $f_b$; therefore, $f_a(bv):= b f_a(v)= bav$.
Why go to all this trouble instead of just defining a left module homomorphism by $v \mapsto va$? Because, for a general left module $E$ over $D$, there is no right multiplication by a scalar. The special case $E=D$ does have a right multiplication defined, though, and we do know that $E \cong D$ as left $D$ modules, so we can work with that.
This is all by way of showing that $D^{op} \cong \text{End}_D(E)$ as rings, when $E$ is one-dimensional. I believe they cannot be made into $D$ algebras since $a \phi$ is not necessarily in $\text{End}_D(E)$ if $\phi$ is.