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

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.

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.

• You are right about my statement about maximal ideals. I will correct this error. – Doug Apr 25 '16 at 6:09
• I'm sorry if this is obvious, but why does $P_i + P_j \subset M$ imply $P_i$ and $P_j$ are in $M$ (which is my interpretation of your statement, if I am correct that you mean the extensions of $P_i$ and $P_j$ are minimal primes in $R_M$)? – Doug Apr 25 '16 at 6:18
• $P_i \subset P_i+P_j$. – MooS Apr 25 '16 at 6:19
• Yes, I am being stupid, thank you. – Doug Apr 25 '16 at 6:21
• Oh, another question. So by the CRT, $$\frac{R}{P_1 \dots P_n} \cong \frac{R}{P_1} \times \dots \times \frac{R}{P_n}.$$ How do we know $R/P_1...P_n \cong R$? – Doug Apr 25 '16 at 6:25