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Comaximal ideals in a commutative ring

On this webpage http://www.imsc.res.in/~kapil/geometry/caag/finite.html it's a stated fact that if $I$ and $J$ are coprime ideals of a ring $A$, then $I^k$ and $J^\ell$ are also coprime for any integers $k,\ell$.

There's no proof given. Why is this fact true?

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marked as duplicate by Bill Dubuque, Willie Wong Mar 6 '12 at 8:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4 Answers 4

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Let $R$ be your ring. I am assuming my ring is commutative. Saying $I,J$ are coprime ideals is the same as saying that there is no prime $q \subset R$ such that $I + J \subset q$. Well suppose there exists a prime $p \subset R$ such that $I^l + J^k \subset p$. Then $I^l \subset p$ and $J^k \subset p$. But this means $I \subset p$ and $J \subset p$ (why?), so $I + J \subset p$, contradicting the fact that $I,J$ are coprime.

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I hadn't thought of this. I like it! –  Dylan Moreland Mar 6 '12 at 3:32
    
I sometimes get confused when raising binomial or multi-nomial expressions to powers, so I like this one. Thanks @DylanMoreland. –  Rankeya Mar 6 '12 at 3:38
    
To fill in the gap you left me, if $a\in I$, then $a^l\in p$, but since $p$ is prime, necessarily $a\in p$. Same for $J$. I like this too! Thanks. –  Cye Mar 6 '12 at 4:37
    
Yup. The more general statement is this: if $A$ is a ring and $\alpha_1,...,\alpha_n \subset A$ are ideals, $p \subset A$ a prime ideal and $\alpha_1...\alpha_n \subset p$, then $\exists i \in \{1,...,n\}$ such that $\alpha_i \in p$. This is a useful result to keep in mind. There is an analogous statement with products replaced by finite intersections. –  Rankeya Mar 6 '12 at 4:46

You have $I + J = A$. Take each side to the $(k + \ell - 1)$-th power; then each term on the left is contained in either $I^k$ or $J^\ell$, and it follows that $I^k + J^\ell \supset A$. You could instead work with an expression $x + y = 1$, where $x \in I$ and $y \in J$, if you prefer.

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If you are familar with ideal radicals then 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\ $ which is true more generally if $\rm\: A+B\:$ is invertible (e.g. $(1)$ or any principal ideal), e.g.

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

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

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

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

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

For more see this prior answer.

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First prove that if $I$ is coprime to $J$ and $J'$ then $I$ is coprime to $JJ'$.

Then use induction to show $I$ and $J^\ell$ are coprime for any $\ell \ge 1$, and then use induction on $k$ in the same way.

Alternatively:

$$A=A^{k+\ell-1}=(I+J)^{k+\ell-1} \subseteq I^k+J^\ell$$

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