Let $I$ be an ideal of a ring $A$. Then $\sqrt{I}=A$ iff $I=A$
One side is obvious. If $I=A$, since $I\subset\sqrt{I}$ then $A\subset\sqrt{I}$. Therefore $\sqrt{I}=A$.
I'm having trouble with the other side though. I know that if $\sqrt{I}=A$ then for $a\in A$, $a\in \sqrt{I}$ so $a^n \in I$ for some $n\geq 1$ but I'm stuck on where to go from there. Any hints would be appreciated please.
Assume the ring has a unit $1$. If $\sqrt{I}=A$ then $1\in \sqrt{I}$ and this implies that for some $n$ one has $1^n=1\in I$. Then $1\in I$. One has $\forall a\in A, 1.a=a\in I$.
If the ring has no unit $A=2\Bbb{Z}$ and $I=(4)$ is a counter example