(Finitely many minimal primes) + ($R_M$ domain for all maximal $M$) $\Rightarrow$ ($R$ = product of domains)

Question

We are doing a homework problem for our commutative algebra class, which asks us to prove:

Let $R$ be a commutative ring with $1$ containing finitely many minimal prime ideals $P_1, \dots, P_n$. If in addition $R$ satisfies that $R_M$ (the localization of $R$ at $M$) is a domain for all maximal ideals $M \subset R$, then $$ R \cong \frac{R}{P_1} \times \dots \times \frac{R}{P_n}. $$

If we could show that $R$ is Artinian, then we would be done by a result from class. But we think we don't have enough to show this.

Any help would be appreciated. Thanks.

Answer 1

If $i \neq j$, we deduce that $P_i+P_j$ is not contained in any maximal ideal: Assume $P_i+P_j \subset M$, then $R_M$ has at least two minimal primes - namely $P_i$ and $P_j$ - and is thus not a domain.

So $P_i+P_j=R$ for $i \neq j$. This is precisely what we need for the Chinese Remainder Theorem, hence we are done.

Note that $\bigcap_{1 \leq i \leq n} P_i$ is the nil-radical of $R$, but $R$ is reduced, since it is locally reduced.

By the way: Your conclusion, that there only finitely many maximal ideals, is plain wrong. For example, $R=\mathbb Z$ clearly satisfies all the assumptions and has infinitely many maximal ideals.