# Prove that $m_1m_2\ldots m_r=n_1n_2\ldots n_s$ implies $r=s$ for distinct maximal ideals

Let $R$ be a commutative ring where $m_1,m_2,\ldots,m_r$ and $n_1,n_2,\ldots,n_s$ are maximal ideals such that $m_1m_2\ldots m_r=n_1n_2\ldots n_s$ and $m_i \neq m_j$, $n_i \neq n_j$ if $i \neq j$. Prove that $r=s$ and there the exists $\sigma \in S_r$ such that $m_i=n_{\sigma(i)}$ for all $i$.

So far I have proved that $\lbrace m_i \rbrace$ are pairwise comaximal, i.e., $m_i+m_j=R$ and $m_1m_2 \ldots m_r=m_1 \cap m_2 \cap \ldots \cap m_r$. But I have no idea how to proceed to prove that $r=s$. Any hints or ideas?

• $n_{1}\cdots n_{s}=m_{1}\cdots m_{r}\subseteq m_{1}$, $n_{i}\subseteq m_{1}$ for some $i$. $n_{i}=m_{1}$. Jan 14, 2016 at 4:01
• how did you get $n_i \subseteq m_1$? thanks Jan 14, 2016 at 4:48
• It is an exercise to show if $IJ\subseteq P$, $I,J$ ideals, $P$ prime ideal, then $I$ or $J\subseteq P$. This is the definition of prime ideal in Hungerford. Jan 15, 2016 at 2:05

Without a loss of generality, assume that $r \leqslant s$.
Since every maximal ideal is a prime ideal, note that $$m_1m_2\ldots m_r=n_1n_2\ldots n_s\subset n_i,i=1,\ldots ,s$$ then there exist a $\sigma(i)\in \{ 1,\dots,r\}$ such that $m_{\sigma(i)}\subset n_1$. But since $m_{\sigma(i)}$ is maximal, and $n_1$ is proper, $m_{\sigma(i)}=n_i$.
Note that $\sigma(i)\not=\sigma(j)$ for $i\not=j$ since $n_i\not=n_j$. It follows that $r=s$ and $\sigma$ is a permutation in $S_r$.
• Could you explain why $m_1m_2 \ldots m_r \subset n_1$ implies that $m_{\sigma(1)} \subset n_1$? thank you! Jan 14, 2016 at 5:04
• Just check out the definition of prime ideal: A prime ideal is an ideal $I$ such that if $AB \subset I$, then either $A \subset I$ or $b \subset I$. And by a simple induction you can freely replace the $AB$ in the original definition by any product of finitely terms. Jan 14, 2016 at 5:08
• I think there's a little work to do, but it is standard/simple. Let $A$ and $B$ are ideals and $P$ is a prime ideal with $AB\subset P$. If $A$ is not contained in $P$ then there is an $a\in A$ with $a\notin P$. Let $b$ be any element of $B$. Then $ab\in P$ since $AB\subset P$. Since $P$ is prime and $a\notin P$, that shows $b\in P$. So $B \subset P$. That shows that if $AB\subset P$, then $A\subset P$ or $B\subset P$ Jan 14, 2016 at 5:24