Endomorphism Ring: Proving f(R) = E

[Pietjuh]

I'm working on a problem about the endomorphism ring:
Let R be a ring. Show that R is isomorphic to E = { f in End(R+) | f(xr) = f(x)r for all x,r in R }

I'm trying to do it with the following theorem in our lecture notes:
Let R be a ring and R+ the additive group of R. Let for r in R, L_r: R+ --> R+ : x |--> rx be the left multiplication map. Then f: R --> End(R+): r |--> L_r is a injective homomorphism, and R is isomorphic to a subring of End(R+), which is f(R).

I had the following strategy in mind: If I can show that E is a subring of End(R+) and that f(R) = E, the statement is proved. I almost succeeded with it, but I had a small problem on the end proving the equality of f(R) and E.

To show that f(R) lies in E isn't that hard. Because L_r (xy) = rxy = L_r(x) y, so L_r satisfies the conditions. But I couldn't figure out how to show that E lies in f(R).

Could anyone help me with this?

 

[AKG]

EDIT: I had written a lot but I see most of it was unnecessary. You want to show that R is isomoprhic to E, and there is a natural function f (the one given in your post) which you know to be injective and homomorphic, you just need to show that f(R) = E. f(R) is the sub-ring of left-multiplications, so you want to show that E is the sub-ring of left-multiplications. Any element g in E satisfies the property that for all x, r in R, we have:

g(xr) = g(x)r

from this we can deduce:

g(x) = g(1x) = g(1)x

Let g(1) = a_g. Then we can say that E consists of all functions of the form:

g(x) = a_gx

where a_g is in R. Clearly, each g is a left-multiplication, and since E consists of all such functions (i.e. since all left-multiplications g satisfy g in End(R+), g(xr) = g(x)r for all x, r in R), we deduce that E is indeed the sub-ring of left-multiplications, so f(R) = E.