Give an example of a proper ideal I of a ring R such that the radical of I is equal to R i.e. √I=R.

Question

I am having trouble with finding such example. In commutative case with identity such example does not exist as 1∈R=√I implies 1∈I and so I=R. Is there any in noncommutative case or a ring without identity?

Answer 1

Let’s $G$ be any abelian group with a nontrivial subgroup, and make $G$ into a ring using the operation $ab=0$ for all $a,b\in G$.

Every subgroup is an ideal of this ring, so any nontrivial ideal has radical $G$ ( in the usual commutative sense: all squares are zero, so everything squared gets inside of all ideals.)