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.

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    $\begingroup$ 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. $\endgroup$ Jul 31 '21 at 22:10
  • $\begingroup$ @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. $\endgroup$
    – user736925
    Jul 31 '21 at 22:19
  • $\begingroup$ 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. $\endgroup$
    – Bernard
    Jul 31 '21 at 22:21
  • $\begingroup$ @user736925 - note that this is iff: Given two ideals $I,J \subseteq A$. It follows $I+J=(1)$ iff $\sqrt{I}+\sqrt{J}=(1)$. $\endgroup$
    – hm2020
    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.$

  • $\begingroup$ 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! $\endgroup$
    – user736925
    Jul 31 '21 at 22:30

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