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Let $A$ be a Noetherian commutative ring. Given two ideals $I$ and $J$ of $A$, I would like to know whether $I^2\cap J^2 = IJ$. More generally speaking, does $I^d\cap J^d = \sum_{k=1}^{d-1} I^k J^{d-k}$ hold? If not, what conditions on $A$ do I need for this to be true?

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No. For any ring $A$ and ideal $I$ such that $I^3 \neq I^4$, let $J = I^2$. You want conditions on $I, J$, not conditions on $A$. – Qiaochu Yuan Jul 7 '11 at 16:19
Why would you want an equality such as $I^2 \cap J^2 = I.J$ holds ? The Chinese remainder theorem states that if $I+J=A$ then $I \cap J = I.J$. If $A= \mathbb{Z}$, then $(n)^2 \cap (m)^2 = ppcm(n^2,m^2)$, so equality never holds except for trivial cases. – user10676 Jul 7 '11 at 16:24
Urk, that was truly a stupid question. – Jesko Hüttenhain Jul 7 '11 at 16:40
up vote 1 down vote accepted

The answer is no. The ring $\mathbb{Z}$ is a Noetherian commutative ring. If $I = p\mathbb{Z}$ and $J = q\mathbb{Z}$ where $p$ and $q$ are different primes, then $I^2 \cap J^2 = p^2 q^2 \mathbb{Z}$, while $IJ = pq\mathbb{Z}$.

As for the second part of the question, it might be more reasonable to ask when we have $I \cap J = IJ$ for ideals $I$ and $J$. In Dedekind domains (if you have seen these-they are a kind of ring where each proper non-zero ideal factors uniquely as a product of prime ideals), this happens if $I$ and $J$ have no common prime ideal divisor. In the case of $\mathbb{Z}$, which is a Dedekind domain, if $I$ and $J$ are proper non-zero ideals, then there are positive integers $m$ and $n$ such that $I = m\mathbb{Z}$ and $J = n\mathbb{Z}$. In that case, $I \cap J = IJ$ if and only if $m$ and $n$ are relatively prime (ie ${\rm gcd}(m,n) = 1$).

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I see my answer essentially duplicates Qiaochu's comments, which weren't there when I started writing. – Geoff Robinson Jul 7 '11 at 16:33

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