# The radicals of two ideals are comaximal implies the ideals are comaximal.

I found this as a property on the Wikipedia page for the Radical of an Ideal, I found I can use it trivialize a result I wish to prove but I can’t prove the property itself! The property is as follows,

Let I and J be ideals of a commutative ring R, if $$\sqrt{I}$$ and $$\sqrt{J}$$ are comaximal then I and J are comaximal. Any suggestions/hints are appreciated.

• What have you tried? Do you understand the definitions and what you must show? Also, enclose math in dollar signs to have it render with mathjax. Jul 31 '21 at 22:10
• @paulblartmathcop hello, thanks just fixed the notation. Yes if two ideals are comaxmimal then there exists an x in I and and an y in J such that $x+y = 1.$ The radical of an ideal I is all x in R such that x^k is in I for any k. So the radicals of the ideals being comaximal tells us there exist $x \in \sqrt{I}, y \in \sqrt{J}$ such that $x + y = 1,$ but of course this is just the setup. I have found it hard to go from this seemingly stronger statement, since the radicals of the ideal contains the ideal itself, to the desired “weaker” statement about just the ideals. Jul 31 '21 at 22:19
• Hint: Let $x\in\sqrt I$, $y\in\sqrt J$ such that $x+y=1$. Also let $m,n$ integers such that $x^l\in I$, $y^n\in J$ respectively. Consider $\;(x+y)^{m+n}$ and expand it by the binomial formula. Jul 31 '21 at 22:21
• @user736925 - note that this is iff: Given two ideals $I,J \subseteq A$. It follows $I+J=(1)$ iff $\sqrt{I}+\sqrt{J}=(1)$. Aug 1 '21 at 9:02

If $$\sqrt{I}, \sqrt{J}$$ are comaximal, then there's some $$x \in \sqrt{I}, y \in \sqrt{J}$$ so that $$x+y=1.$$ Thus there's $$n, m \geq 1$$ so that $$x^n \in I, y^m \in J$$ and $$x+y=1.$$
Now, here's the kicker. Write $$1 = 1^{n+m} = (x+y)^{n+m} = x^{n+m} + \binom{n+m}{1}x^{n+m-1}y^1 + \cdots + y^{n+m}.$$
In each term in our binomial expansion, we'll have something like $$Cx^ay^b$$ where $$a+b = n+m,$$ and in particular either $$a \geq n$$ or $$b \geq m$$ since if both $$a < n$$ and $$b < m,$$ we'd have $$a + b < n + m.$$ But now this means every term on the right belongs to either $$I$$ or $$J,$$ so that a sum of (terms in $$I$$) + (terms in $$J$$) $$= 1.$$
• Ahhh Truthfully I was dreading the binomial expansion even though it certainly felt like a relevant tool but your answer has shed some light on what’s going on. Just to make sure I get it, the initial equalities are all given and we make the choice to raise $x+y$ to the $n+ m.$ Then we can see/show each term of the expansion is actually an element of our ideals and with some rearranging we have (sum of elements in I) + (sum of elements in J) and since ideals are additive subgroups we have found our desired $i \in I, j \in J$ such that $i+j = 1.$ Thanks so much! Jul 31 '21 at 22:30