Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The direction $\sqrt{\sqrt{I}+\sqrt{J}}\supset \sqrt{I+J}$ is trivial as $\sqrt{I}+\sqrt{J} \supset I+J$ since $\sqrt{K} \supset K$ for any ideal $K$. Is the following correct?

My attempt: Suppose $f\in \sqrt{\sqrt{I}+\sqrt{J}}$ then $f^n \in \sqrt{I}+\sqrt{J}$ which means that $f^n$ can be written as the sum $g+h$ where $g^p\in I$ and $h^q \in J$. If we raise $f^n$ to the $t=p+q$ then we get $f^{nt}=(g+h)^t=g^t+tg^{t-1}h+\cdots+tgh^{t-1}+h^t$. The first $p$ terms have a factor $g^{p'}$ where $p'\geq p$ which means they are in $I$ and the next (and final) $q$ terms have a factor $h^{q'}$ where $q'\geq q$. Thus $f^{nt}$ is in $I+J$.

share|cite|improve this question
up vote 1 down vote accepted

This is roughly the right answer, but you need to correct your expansion of $(g+h)^t$ and pick the right $t$. ($t=2$ isn't gonna work.)

share|cite|improve this answer
Sorry, it was supposed to say $t=p+q$. That works, right? – Steven-Owen Feb 25 '13 at 21:40
Yes, except your expansion is wrong. For example $(g+h)^2\neq g^2 + gh + h^2$. – Thomas Andrews Feb 25 '13 at 21:40
Woops, I am making typos all over the place. Thanks for your help, Thomas! – Steven-Owen Feb 25 '13 at 21:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.