Construct an isomorphism End$_K (K) \cong K$ of rings. For a commutative ring $K$ regarded as a $K$-module, construct an isomorphism End$_K (K) \cong K$ of rings. End$_K (K)$ denotes the ring of $K$-linear endomorphisms of $K$.
I am trying to find some property that is unique to each linear map in End$_K (K)$, so I can then use the property to construct the isomorphism. (I was originally thinking about the kernels of each endomorphism but now I'm not so convinced...) How should I approach problems like these?
 A: In general one way to approach this is to find a nice generating set of your module and then look at how endomorphisms must behave on your generating set. For instance, as a $K$-module, $K$ is generated by the element $1$. Now if $\phi$ is a $K$-linear endomorphism from $K$ to itself and $k$ is an element of $K$, we must have $$\phi(k)=\phi(k\cdot 1)=\dots$$
Do you see where we use $K$ linearity? How many choices do we have to completely determine $\phi$ uniquely? This should help with your isomorphism.
A: You are trying to define an isomorphism $\Phi:K\to End_K(K)$, so you should think about how to construct a map $\Phi(k)=\phi_k:K\to K$ for each $k\in K$. A natural map (depending on $k$) from $K\to K$ is $\phi_k(x)=kx$. Okay, so what do we like to know?


*

*$\phi_k$ is $K$-linear. That is, $\phi_k(k'x)=k'\phi_k(x)$. Well, that is obvious (why?)

*We need to see that $\Phi$ is a homomorphism. That is $\phi_{kk'}=\phi_k\phi_{k'}$ and $\phi_{k+k'}=\phi_k+\phi_{k'}$. Again, pretty clear.
That's a good start, so what about showing that $\Phi$ is bijective? Well, it is certainly injective since, if $\phi_k=\phi_{k'}$, we have $$k=\phi_{k}(1)=\phi_{k'}(1)=k'.$$
And this shows us how to prove the map is surjective. If $\phi\in End_K(K)$, and $\phi(1)=k$, then $\phi=\phi_k$. Indeed, since $\phi$ is $K$-linear, we have for any $x\in K$,
$$\phi(x)=\phi(x1)=x\phi(1)=xk=kx=\phi_k(x).$$
