Let $R$ be a noncommutative ring. We know that every nil ideal (left or right) is contained in Rad($R$), for example Nil($R$). My question is: "Is Rad($R$) necessarlily a nil ideal? Of course if $R$ is an Artinian Ring the answer is yes because Nil($R$) is a nil ideal and Rad($R$)=Nil($R$). Thanks for any hint or an answer.
Rad($R$) is a nil ideal?
2
$\begingroup$
ring-theory
-
0I may have misunderstood what you meant, but I'll say something anyway: $Nil(R)=\{x\mid \exists n, x^n=0\}$ as defined in commutative rings is not usually an ideal. If you were thinking of $Nil^\ast(R):=$ sum of nil ideals of $R$, or of $Nil_\ast(R):=$ intersection of all prime ideals of $R$, then yes, those are both nil ideals of $R$. (The main body of your question was already taken care of below, of course :) ) – 2013-04-22
2 Answers
2
No, the Jacobson radical isn't necessarily nil. For a counterexample take the ring $R[[x]]$ of formal power series over a noncommutative division ring $R$. This ring is local and the maximal ideal (which is equal to the radical) is generated by $x$. Obviously, if $R$ had a non-nilpotent element $r$, we get a non-nilpotent series $rx$.
-
0@DylanMoreland: Thanks for hint. – 2012-01-27
1
In the non-commutative rings, the nil ideal not necessarily in the radical ideal.