The following problem is from Michael Artin's Algebra, chapter 12, M.6, unstarred:
Let $R$ be a domain, and let $I$ be an ideal that is a product of distinct maximal ideals in two ways, say $I=P_1\dotsb P_r=Q_1\dotsb Q_s$. Prove that the two factorizations are the same, except for the ordering of the terms.
Well, following the proof of the fundamental theorem of arithmetic, it seems that we should prove the theorem in the following way:
- Show that $P_1=Q_j$ for some $j$.
- Cancel $P_1,Q_j$ from both sides and reduce it to the induction hypothesis.
The first bulletin is relatively easy. It follows from the fact that maximal ideals are prime, therefore if $M$ is maximal, and $M\supset AB$, then $M\supset A$ or $M\supset B$, thus there's some $Q_j\subset P_1$, therefore $Q_j=P_1$ following from the maximality of $Q_j$.
However, the second one seems hard. There seems no cancellation laws among ideals. There's a counterexample even for maximal ideals, since maximal ideals could be zero. It's not like prime number, which are always positive. I should note that the distinctness of $P_k$,$Q_j$ and the fact that $R$ is a domain isn't used.
So how can we proceed next? Any help? Thanks!
Let $I=P_1 ... P_r = Q_1 ... Q_s$, and suppose $P_1=Q_1$. Let $J=P_2 ... P_r$ and $J'=Q_2...Q_s$. We want to show $J=J'$. For any $P=P_i$, $i=2,...,r$, we have $P \supset I$. Since $P$ and $P_1$ are distinct, $P\not\supset P_1$ by maximality of $P_1$. Therefore $P\supset J'$ since $P$ is prime. Since $P_2,...,P_r$ are maximal ideals, we have $P_2 \cap ... \cap P_r = P_2 ... P_r$ (Recall that $IJ = I \cap J$ if the ring $R$ is commutative and I+J=R). We have $P \supset J'$ for $P=P_2,...,P_r$, therefore $J=P_2 ... P_r = P_2 \cap ... \cap P_r \supset J'$. The same argument shows $J'\supset J$, so $J=J'$.
Thus we can "cancel" maximal ideals on both sides of $P_1 ... P_r = Q_1 ... Q_s$. By doing this repeatedly, we get $r=s$ and $\{P_1,...,P_r\} = \{Q_1,...,Q_s\}$.