It is known in basic ring theory that any ring R (with identity) can be embedded in its own endomorphism ring End(R) (This is the analogue of the Cayley theorem in group theory). Is there a characterization or classification for rings R such that R is isomorphic to the whole endomorphism ring End(R)? ; the ring of integers Z is an example of such rings.
Peter Breuer
I mean the endomorphism ring of the additive group (R,+) considered as a ring with homomorphism addition and multiplication defined by composition of maps. This endomorphism ring End(R) is known to contain a copy of R. In the case when R=Z, End( Z)=Z . I wonder if rings which share this property with Z have been classified. (In your reply f :Z---Z is a ring homomorphism ; it must be either the identity or the zero map but that is different from what I meant.) Regards
Consider the ring R as an R-module and let f: R-->End( R) defined by:
for every r \in R, f(r)=f_r, such that for every x\in R, f_r (x)=rx (multiplication by r).
This map is a ring homomorphism such that f_rr'=f_r °f_r'. This homo can be iso only if R is a divisible ring (divisible module).
What do you say dear Ahmed in this analysis?
Dear Peter Breuer
Are you saying that it is true for any ring R (just as for Z) that R iso E(R) ?
Can we rectify the answer to this one: "the case of R ≠ Z, then R is a divisible ring"?
You said all dear Peter :) I have to add this point only:
The homomorphism f: R---->End(R), such that f(r)=f_r is 1-1, yes. We look for the condition that makes it onto (surjective).
Let f_r \in End(R), such that f(r)=f_r. So we look for the existence of nonzero r such that for for all x, f_r(x)=rx.
If such r exists, then it satisfies: f_r(1)=r.1=r.
Yes I know, you will say: but f_r(1)=1 because f_r is an homo. The key in all this problem is that f(rs)=f_r°f_s, but not f_s°f_r (there is a reverse order in the composition, so we can not naturally have f_r(1)=1).
Dear Peter and Ahmed what do say? correct to me please.
Dears Peter and Ahmed, if the reasoning in my previous answer is right, then we have: rx=y, rx'=y' and so on...
So can we say for all y \in R there is x \in R, such that rx=y (in other word, f_r is onto) which gives rR=R? The last result meant that r is left invertible.
Conclusion: If R ≠ Z, then R is iso to End(R) in case that R is a skew field as Peter concluded.
Dear Peter Breuer
Dear Hanifa Zekraoui
I am enjoying the discussion and I appreciate all the valuable information I get from you. The connection with divisible (not division!) rings mentioned by Hanifa is so relevant (I think) but I still can not figure how to use it to show that the map f: R----E(R) (r----f(r)) is surjective under the condition that R is divisible ring (we know already that this map is injective when R has 1). Is is an imitation of the case when R=Z (in which case Z is ring-isomorphic to E(Z))
Dear Peter Breuer
Dear Hanifa Zekraoui
I quote from Peter Breuer answer
"If you mean instead to ask when f is onto all the endomorphisms of R+ (the additive group of R), then you are asking when every endomorphism g=fr for some r. That is, g(x)=fr(x)=rx for any endomorphism g "
Yes Peter this is the question. This is true for R=Z and the question is what other rings share this property with Z.
Ahmed Ali Khammash
Dear Ahmed,
The article bellow gave definition of a divisible module but no definition of a divisible ring. If we consider the same definition for the ring as a module on itself, then we can apply it in our problem:
" to show that the map f: R----E(R) (r----f(r)) is surjective "
As the ring is divisible, then for all y \in R and for all nondivisor of zero r \in R, there is x \in R, such that rx=y .
https://www.math.uaic.ro/~martar/pdf/articole/articol%202.pdf
Conclusion: If R ≠ Z, then R is isomorphic to End(R) in case that R is a skew field or a divisible ring.
Note: A divisible ring is note necessary a skew field.
The problem now is reduced to the definition of a divisible ring in bold above is it right or wrong?
There is a mistake in application in my previous answer.
For surjectivity of f, We need to prove this one:
for all x and y \in R there is a nonzero divisor r \in R such that rx=y.
Really, it is difficult to benefit from a divisible ring in the proof.
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