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Let $R$ be a commutative ring and $I_1, \dots, I_n$ pairwise comaximal ideals in $R$, i.e., $I_i + I_j = R$ for $i \neq j$. Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r \in\mathbb N$) also comaximal?

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This is a standard exercise. The standard hint is to take powers of the relation I_i + I_j = R. – Qiaochu Yuan Nov 15 '10 at 14:27
up vote 14 down vote accepted

It is sufficient to prove this for the case two comaximal ideals, say $I,J$. Need to show, $I^m+J^n=R$ for any positive integers $m,n$. Now, $R=R^{m+n-1}=(I+J)^{m+n-1}\subseteq I^m+J^n$ since in the binomial expansion of $(I+J)^{m+n-1}$, in every term, either the power of $I$ is at least $m$ or the power of $J$ is at least $n$ by pigeonhole principle.

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If you are familar with ideal radicals then, as I mentioned on sci.math, the proof is a one-liner:

$$\rm rad\ (I^m +\: \cdots\: + J^n) \ \supset\ I +\:\cdots\:+ J\: =\: 1\ \ \Rightarrow\ \ I^m +\:\cdots\: + J^n\: =\: 1 $$

Alternatively, and much more generally, it may be viewed as an immediate consequence of the Freshman's Dream binomial theorem $\rm\ (A + B)^n = A^n + B^n\ $. This theorem is true for both arithmetic of GCDs and (invertible) ideals simply because, in both cases, multiplication is cancellative and addition is idempotent, i.e. $\rm\ A + A = A\ $ for ideals and $\rm\ (A,A) = A\ $ for GCDs. Combining this with the associative, commutative, distributive laws of addition and multiplication we obtain the following very elementary high-school-level proof of the Freshman's Dream:

$\rm\qquad\quad (A + B)^4 \ =\ A^4 + A^3 B + A^2 B^2 + AB^3 + B^4 $

$\rm\qquad\quad\phantom{(A + B)^4 }\ =\ A^2\ (A^2 + AB + B^2) + (A^2 + AB + B^2)\ B^2 $

$\rm\qquad\quad\phantom{(A + B)^4 }\ =\ (A^2 + B^2)\ \:(A + B)^2 $

So $\rm\quad\ {(A + B)^2 }\ =\ \ A^2 + B^2\ $ if $\rm\ A+B\ $ is cancellative, e.g. if $\rm A+B = 1$

The same proof works generally since, as above

$\rm\qquad\quad (A + B)^{2n}\ =\ A^n\ (A^n + \:\cdots\: + B^n) + (A^n +\:\cdots\: + B^n)\ B^n $

$\rm\qquad\quad\phantom{(A + B)^{2n}}\ =\ (A^n + B^n)\ (A + B)^n $

In the GCD case $\rm\ A+B\ := (A,B) = \gcd(A,B)\ $ for $\rm\:A,B\:$ in a GCD-domain, i.e. a domain where $\rm\: \gcd(A,B)\:$ exists for all $\rm\:A,B \ne 0,0\:$. So the Dream is true since $\rm\:(A,B)\:$ is cancellable, being nonzero in a domain.

In a domain, nonzero principal ideals are cancellable, so Dream is true for ideals in a PID (e.g. $\mathbb Z\:$), or f.g. (finitely generated) ideals in a Bezout domain. More generally, Dream also holds true in any Dedekind domain (e.g. any number ring) since nonzero ideals are invertible hence cancellable.

In fact this "Freshman's Dream" is true for all f.g. ideals in domain $\rm\:D\:$ iff every nonzero f.g. ideal is invertible. Such domains are known as Prufer domains. They're non-Noetherian generalizations of Dedekind domains. Moreover they form an important class of domains because they may also be equivalently characterized by a large number of other important properties, e.g. they are precisely the domains satisfying CRT (Chinese Remainder Theorem); $\ $ Gauss's Lemma: the content ideal $\rm\ \ c(fg) = c(f)\ c(g)\:$;$\ $ nonzero f.g. ideals are cancellable; $\ $ f.g. ideals satisfy contains $\Rightarrow$ divides; $\: $ etc. It's been estimated that there are close to a hundred such characterizations known. See my post here for about thirty such characterizations.

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A slight variation on the radical one-liner: Note that two ideals $A$ and $B$ are comaximal (i.e. $A+B=R$) if and only if the ideal $A+B$ is not contained in any maximal ideal of the domain. Now take the ideal $A^{m} + B^{n}$. Claim: $A^{m} + B^{n} = R$. For if not then $A^{m} + B^{n}$ must be contained in a maximal ideal $M$. But as $A^{m}$, $B^{n}$ are contained in $A^{m} + B^{n}$, we have $A^{m}$, $B^{n}$ contained in $M$. Since $M$ is a prime we get $A$, $B$ contained in $M$, a contradiction. Similar reasoning will show that $A^{m}$, $B^{n}$ comaximal implies that $A$, $B$ are comaximal. Muhammad

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As it suffices to prove that the claim holds modulo any given maximal ideal, we are reduced to the easy case where our ring is a field.

(See this MO answer of Georges Elencwajg and the accompanying comments.)

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+1 I agree that its worth explicitly mentioning this local view of the standard proof, i.e. that in MZ's answer (and in one of my posts in said sci.math thread Oct 6, 2008). Thanks much for the link to the interesting Mathoverflow discussion. – Bill Dubuque Sep 11 '11 at 18:16

In the comments here has been raised the question if such property holds for commutative rings without unity.

The answer is negative as shows the following example: $R=\mathbb Z$ with zero multiplication, that is, $a*b=0$ for any $a,b\in\mathbb Z$, and $I=2\mathbb Z$, $J=3\mathbb Z$. Clearly $I+J=R$, and since $R^2=0$ we have $I^m+J^n=0$ for any $m,n\ge2$.

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This follows by induction from the more general fact: If I + J = I + K = R, then I + JK = R. This is seen by noting that (I+J)(I+K) is contained in I + JK. (Just multiply it out.)

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Alternatively: Any maximal ideal containing JK contains either J or K, hence cannot contain I. – Jason Juett Feb 9 '15 at 10:45

A slight variation on other proofs. Suppose $I+J=R$, so $a+b=1$ for some $a\in I, b \in J$. I want to replace $a$ by $a^n$, so I write $a^n + (a-a^n) + b = 1$. If I knew that $a-a^n$ is in $J$, I could group it together with $b$ and get $a^n+b' = 1$, so $I^n+J = R$. But $a-a^n = a(1-a^{n-1})$ is divisible by $1-a = b \in J$, so it is in fact in $J$. From $I^n+J=R$ the general case $I^n+J^m=R$ follows by another application of the same argument.

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