- #1

Pietjuh

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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?