R a commutative ring and A an ideal in R with $A=m_1···m_r=n_1···n_s$ with $m_i$ distinct maximal ideals and all the $n_j$ distinct maximal ideals.

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.

• Do you want $m_i = n_{\sigma(i)}$ rather than $m_i = n_\sigma(i)$? Cheers! – Robert Lewis Jan 16 at 5:01
• @RobertLewis Yes! Sorry, it was a typo. – davidh Jan 16 at 5:06

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.