Why is $\text{End}_K(K) = K$, or, more generally, $\text{End}_D(D) = D^{\text{op}}$?

Question

Let $K$ be a field and a vector space over itself, $D$ a division ring with a modul over itself, with its opposite ring $D^{\text{op}}$.

Why is then $\text{End}_K(K) = K$? Or more general, why is $\text{End}_D(D) = D^{\text{op}}$?

Answer 1

Let's say we treat $D$ as a left module over itself.

Consider the isomorphism $D^{op} \rightarrow \mathrm{End}_D(D)$ given by $d \mapsto [-\cdot d]$, where $[-\cdot d]$ is the map that takes $x \in D$ to $xd$.

It is not hard to see that

1. $[-\cdot d]$ is an endomorphism of $D$ as a left $D$-module,
2. $[-\cdot d_1]+[-\cdot d_2]=[-\cdot (d_1+d_2)]$,
3. $[-\cdot d_1] \circ [- \cdot d_2]=[- \cdot (d_2d_1)]$, $[-\cdot 1]=\mathrm{id}_D,$

i.e. the map above is a homomorphism of rings (notice that the condition 3. corresponds to the fact that the domain is $D^{op}$ and not $D$).

To check injectivity, look at the value of general $[-\cdot d]$ at $1$. To check surjectivity, take a general endomorphism $\varphi$ of $D$ over itself and compare $\varphi $ and $[-\cdot \varphi(1)]$ (here is danger of confusion: $\varphi(1)$ is not necessarily $1$...).

Also notice that the assumption that $D$ is a division ring wasn't needed anywhere, the whole thing will work for arbitrary ring.

Also warning. Similarly, you can prove that if one considers $D$ as a right module over itself (now with the map $d \mapsto [d\cdot-]$), then $D \simeq \mathrm{End}_D(D)$, without any "op". The seeming asymmetry comes from the fact that one needs to choose some convention regarding the order of composition of maps, i.e. regarding the multiplication in the ring $\mathrm{End}_D(D)$.