$\sqrt{I}+\sqrt{J}=R$ implies $I+J=R$

Question

Let $R$ be a commutative ring with unity and $I,J$ ideals of $R$. Suppose that $$ \sqrt{I}+\sqrt{J}=R $$ I want to show that this implies $I+J=R$.

Take $r\in R$, then I can write $$ r=a+b, $$ for some $a\in \sqrt{I}, b\in \sqrt{J}$. This implies that there exist $n,m\in \mathbb{Z}_{>0}$ such that $a^n\in I, b^m\in J$. Consider $$r^{m+n}=(a+b)^{m+n}=\sum_{i=0}^{m+n} \binom{m+n}{i} a^i b^{m+n-i}.$$

Now, in $a^i b^{m+n-i}$ either $i\leq n$ and $b^{m+n-i}\in J$, so that $a^i b^{m+n-i}\in J$ or $i\geq n$ so that $a^i\in I$ and $a^i b^{m+n-i}\in I$. This means that $$ r^{m+n}=c+d, $$ where $c\in I, d\in J$. This decomposition should work for any $r\in R$, take $r=1$, then $$ 1=c+d $$ with $c\in I, d\in J$. Let $r\in R$, then $$ r=1\cdot r=(c+d)r=cr+dr \in I+J, $$ since $cr\in I, dr\in J$.

Is my proof correct?

Answer 1

Yes, this proof is correct. It can be shortened slightly by just using $1$ for the decomposition instead of an arbitrary $r$, but that's an issue of style.