If $R$ is a ring and $I$ and $J$ are ideals, then if $\operatorname{rad}(I) + \operatorname{rad}(J) = R$, then $I + J = R$

Question

How can I show that if $R$ is a ring and $I$ and $J$ are ideals, then if $\operatorname{rad}(I) + \operatorname{rad}(J) = R$, then $I + J = R$? This has had me stumped for some time, which is frustrating as I imagine it has a fairly elementary solution.

Assume $R$ is commutative with an identity.

Answer 1

Hint: If $I+J\neq R$, then $I+J$ is contained in a maximal ideal $M$, hence...

$I, J $ are contained in $M$, hence $rad(I), rad(J)$ and $rad(I)+rad(J)$ are contained in $M$. So, $rad(I)+rad(J)\neq R$ and we've proven the contrapositive.

The nice part about this is that it works even if the radical is the Jacobson radical, and you don't have to do any binomial expansion.

Answer 2

If $\operatorname{rad}(I)+\operatorname{rad}(I)=R$, then we can write $$1=x+y$$ with $x\in \operatorname{rad}(I)$ and $y\in\operatorname{rad}(J)$, say $x^m\in I$ and $y^n\in J$. Then consider what happens if we raise both sides of $1=x+y$ to the $(m+n)$th power.