A ring R is simple if it has no two-sided ideal. A ring R is Abelian if each idempotent in R is central and R is domain if for each a,b in R, ab=0 implies that a=0 or b=0.
Dear Yaser Maleki
Consider Z6 = {[0], [1], [2], [3], [4], [5]}
Notice that [2]x[3] = [6] = [0],
but nether [2] nor [3] is zero.
Best regards
Dear Yaser Maleki
Consider Z6 = {[0], [1], [2], [3], [4], [5]}
Notice that [2]x[3] = [6] = [0],
but nether [2] nor [3] is zero.
Best regards
Issam Kaddoura
Since Z_{6} is commutative every principal left ideal of Z_{6} is two-sided ideal so its not simple.
Yaser Maleki
Yes, I didn't pay attention to the condition on the ring.
If R is a simple abelian ring, then R should be a field and hence an integral domain.
Therefore, there is no example of a simple abelian ring which is not a domain.
See
en.wikipedia.org › wiki › Simple_ring
Simple ring - Wikipedia
Do you include the existence of multiplicative identity 1 to axioms of ring? If so, then every simple Abelian ring is a domain.
Indeed, assume ring R is Abelian but is not a domain, i.e. there is a couple of non-zero elements a, b whose product is 0. Consider subset X_a of R consisting of all those x \in R, for which xa = 0. This X_a is a left ideal (if xa = 0, then for every y we have yxa = 0, if ua = 0 and va = 0 then (u + v)a = 0), by commutativity this ideal is two-sided, it is not the zero ideal (b \in X_a) and it is not the whole ring: 1 does not belong to X_a.
So we demonstrated that R contains a non-trivial two-sided ideal.
Vladimir Kadets Issam Kaddoura
Note that by the definition I've mentioned before a ring R is Abelian if each idempotent of R is central. there are Abelian rings which are not commutative rings. for example, every division ring is Abelian but not commutative.
Yaser Maleki
Thank you for pointing this. I missed the definition, sorry! In this case the reasoning from my previous post demonstrates the following weaker statement:
your ring R does not have idempotent divisors of zero. This is not too much ...
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