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Let R be a commutative ring and A be an ideal in R satisfying $A=m_1···m_r =n_1···n_s$ with all the $m_i$ distinct maximal ideals and all the $n_j$ distinct maximal ideals. Show that $r = s$ and there exists a $σ ∈ S_r$ satisfying $m_i = n_{σ(i)}$ for all i.

I know that maximal ideal implies it being prime, and the product is contained in $m_i$ and $n_j$ for all $i$ and $j$, but I'm not sure how to proceed further.

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  • $\begingroup$ Do you want $m_i = n_{\sigma(i)}$ rather than $m_i = n_\sigma(i)$? Cheers! $\endgroup$ – Robert Lewis Jan 16 at 5:01
  • $\begingroup$ @RobertLewis Yes! Sorry, it was a typo. $\endgroup$ – davidh Jan 16 at 5:06
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Your thinking is alright and you should try to expand on it. You can probably continue like this:

Obviously we have that $n_1 \cdots n_s = m_1 \cdots m_r \subseteq m_1$. So as $m_1$ is a prime ideal too we must have that some $n_i \subseteq m_1$. However as $n_i$ is a maximal ideal we must have that $n_i = m_1$ as $m_1 \not = R$. Now in a similar manner you can find a corresponsing ideal $n_j$ for every $m_i$. It's obvious that no two ideals $m_i$ would be paired with the same ideal $n_j$

Now you can do the same for every $n_j$, pairing it with an ideal $m_i$. So what we have done with these is just form a bijection from $\{m_1,m_2,\dots,m_r\}$ to $\{n_1,n_2,\dots,n_s\}$. From here it's easy to conclude that $r=s$. Hence the proof.

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