Is it true that given ideals $I_1,\dots ,I_n$ of a commutative ring we have $\sqrt{\sum_k I^{n_k}_k}\supset \sum_k I_k$? How can I prove this?
I think I can manage if the identity $\sqrt{I^m}=\sqrt I$ is true. But this seems strange to me. The inclusion $\sqrt{I^m}\subset \sqrt I$ follows from monotonicity, but tells us that if $r^n\in I^m$ for some $n$ then $r^k\in I$ for some $k$. This seems counter intuitive. For the opposite inclusion $\sqrt{I^m}\supset \sqrt I$, if $r^k\in I$ then $(r^k)^m\in I^m$. Is my proof correct?
For all $i$ you have $$I_i \subset \sqrt{I_i^{n_i}} \subset \sqrt{ \sum_kI_k^{n_k}}$$ so that $\sqrt{ \sum_kI_k^{n_k}}$ contains all the $I_i$s. Hence it contains their sum.