Hello everybody;
Let $\displaystyle R$ be commutative ring with unity and let $\displaystyle I$ and $\displaystyle J$ be two ideals of $\displaystyle R$ Show that
$\displaystyle \sqrt{I+J}=\sqrt{\sqrt{I}+\sqrt{J}}$
$\displaystyle \sqrt{I+J} \subseteq \sqrt{\sqrt{I}+\sqrt{J}}$ is obvious. for the other inclusion, let $\displaystyle a \in \sqrt{\sqrt{I}+\sqrt{J}}$. then there exists a positive integer $\displaystyle n$ such that $\displaystyle a^n = b+ c$, for some $\displaystyle b \in \sqrt{I}$ and $\displaystyle c \in \sqrt{J}$. so $\displaystyle b^r \in I$ and $\displaystyle c^s \in J$, for some positive integers $\displaystyle r, s$. now show that $\displaystyle a^{n(r+s)} \in I + J$ and thus $\displaystyle a \in \sqrt{I+J}$.
Thank you very much, I catch the point. In order to show that $\displaystyle a^{n(r+s)}\in I+J$, we do the following:
$\displaystyle (b+c)^{r+s}=\sum_{k=0}^{r+s}\binom{r+s}{k}\times b^{r+s-k}\times c^{k} \in I+J$
This shows that $\displaystyle a^{n(r+s)} \in I+J$