$\sqrt{I}=R \iff I=R$ Let $R$ be a commutative ring with identity and $I$ and ideal. I have a proof that $\sqrt{I}=R \iff I=R$ and I am wondering if its correct:
"$\Leftarrow$":Obvious, since $I\subset \sqrt{I}$
"$\Rightarrow$":We know that $\sqrt{I}=\bigcap{p_i}$ where $p_i$ is prime and $p_i\supset I$ for all $i$. Because $\bigcap{p_i}=R$ we get that there isn't any proper prime ideal containing $I$ and because each proper ideal is contained in a (proper) maximal ideal we get that $I$ is maximal.
Let $x \in R\setminus I$. Then by maximality of $I$ we have $xI=R$. But this means that $x$ is invertible and therefore also $x^n$. Since $x \in \sqrt{I}\Rightarrow x^n \in I$  we have that $I=R$ since it contains an invertible ellement.
Thanks in advance !

It's much simpler to use the definition: if $\sqrt{I}=R$, then $1\in\sqrt{I}$, so $1^n\in I$, for some $n$.
Your proof is incorrect. From $\bigcap_i p_i=R$ you argue that $I$ is maximal, which is wrong. You should instead argue that there is no prime ideal containing $I$. Therefore $I$ is not a proper ideal.
