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.

score 3 · Answer 1 · edited May 04 '18 at 17:53

3

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.

edited May 04 '18 at 17:53

Michael Hardy

1

answered May 04 '18 at 17:06

carmichael561

53,688

Actually the $(m+n-1)$th power suffices. – May 04 '18 at 17:36
@Jakob Werner: good point, though of course what's important is that some power works. – carmichael561 May 04 '18 at 17:43

rschwieb · Accepted Answer · 2018-05-04T17:54:41.000

2

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.

edited May 04 '18 at 17:54

answered May 04 '18 at 16:59

rschwieb

153,510

1

I've been trying this for a while now, and although your comment didn't solve the problem directly for me it somehow inspired me to find this solution: $1 \in rad(I) + rad(J) \implies 1^k=1\in I + J$. And then you're done. (I actually did find it before that other answer was posted haha) – Daven May 04 '18 at 17:07
@Daven Well, since I thought it should convince anyone nearly instantly, I've gone ahead and completed my argument as a spoiler. I think it's a rather useful way of solving the problem. It's certainly nice to have an element-free solution to the problem. (Nothing against elements, really.) – rschwieb May 04 '18 at 17:55
This seems like a nice solution, however it is not obvious to me why $I+J$ being in $M$ means that $rad(I)$ is in $M$. – Daven May 05 '18 at 12:08
@Daven Do you know that $rad(I)$ is equal to the intersection of all prime ideals containing $I$? M is a prime ideal containing both I and J – rschwieb May 05 '18 at 12:17

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

2 Answers2