# a case where contraction of a principal ideal is principal

Let $K$ be a field and $R_1,\cdots,R_n$ DVRs of $K$ with $m_i$ the maximal ideal of $R_i$. Define $A=\cap R_i$. Then $A$ is semilocal with maximal ideals $p_i=m_i \cap A$. Also, $A_{p_i} = R_i$.

Question: How can we see that $p_i$ is principal?

My effort: certainly $m_i$ is principal and in view of the isomorphism $A_{p_i} \cong R_i$ we have $p_i A_{p_i} = m_i = x_i A_{p_i}$ and we can take $x_i$ to be in $A$. Then $x_i$ is inside $p_i$. Now for every $y \in p_i$ we have $y s = a x_i$ with $s$ not inside $p_i$. That's as far as i got.

Reference: Theorem 12.2 in Matsumura. Note: Matsumura's argument is unclear to me.

• Which book by Matsumura? There are two candidates. – Martin Brandenburg May 23 '13 at 1:53
• @MartinBrandenburg I just found it in Comm. Ring Theory, pg. 87. – Corey Harris May 23 '13 at 1:56
• Yup, commutative ring theory. – Manos May 23 '13 at 2:09

Note: This is edited to correct a blunder pointed out by YACP in comments below.

Let $J := p_1 \cap \cdots \cap p_n$; this is the Jacobson radical of $A$. Note that the inclusion $J p_1 \subset p_1^2 \cap p_2 \cap \cdots \cap p_n$ is an equality (check after localizing at each $p_i$). Thus the Chinese Remainder Theorem shows that $p_1/J p_1 \cong p_1 /p_1^2 \times A/p_2 \times \ldots \times A/p_n,$ which is $\cong m_1/m_1^2 \times A/p_2 \times \ldots \times A/p_n$, and hence is a cyclic module. (If $C_i$ is a collection of cyclic modules over the rings $A_i$, then $\prod C_i$ is cyclic over $\prod A_i$ (a cyclic generator is given by taking a product of cyclic generators). By Nakayama, we see that $p_1$ itself is cyclic, i.e. a principal ideal. The same argument applies to each $p_i$.

Some more details added at OP's request: $p_1$, $p_2$, etc. are distinct maximal ideals, thus each pair $p_i, p_j$ ($i \neq j$) generates the unit ideal. Then $p_1^2, p_2,\ldots, p_n$ have the same property, and so CRT gives $A/Jp_1 = A/(p_1^2 \cap p_2 \cap \cdots \cap p_n) = A/p_1^2 \times A/p_2 \times \cdots A/p_n.$

Thus the image of $p$ in $A/ J p_1,$ which is equal to $p_1/J p_1$, is equal to the ideal generated by $p_1$ in the product. In each $A/p_i$ ($i > 1$) we have $p_1$ generates the unit ideal, and so we see that $p_1 / J p_1$ equals $p_1/p_1^2 \times A/p_2\times \cdots \times A/p_n$, as claimed.

A variation of this is as follows: a Dedekind domain with only finitely many prime ideals is necessarily a PID, by the same Chinese Remainder Theorem argument. Note that this gives a proof of the infinitude of primes: if it were false, then by general Dedekind domain theory, each ring of algebraic integers would again only have finitely many prime ideals, and hence would be a PID. But e.g. $\mathbb Z[\sqrt{-5}]$ is easily checked to not be a PID, and so there must actually be infinitely many primes! (This proof is due to Larry Washington, see here.)

• Dear Matt, thanks for your answer. Could you please explain how exactly you use the Chinese Remainder theorem to deduce that $p_1/Jp_1 \cong p_1/{p_1}^2$? – Manos May 26 '13 at 19:11
• Dear Matt, one more question please: i can see that $(p_1 A_{p_1})/(p_1^2 A_{p_1}) \cong m_1/m_1^2$, since $A_{p_1} \cong R_1$. But why is it true that $p_1/p_1^2 \cong m_1/m_1^2$? (i assume that this is what you mean in your third line) – Manos May 27 '13 at 16:17
• @Manos: Dear Manos, Because $p_1/p_1^2 = p_1 A_{p_1}/(p_1 A_{p_1})^2$ when $p_1$ is maximal, and by assumption $A_{p_1} = R_1,$ so that $p_1 A_{p_1} = m_1$. Regards, – Matt E May 28 '13 at 2:07
• Dear Matt, let me try and prove your statement: When $p$ is maximal, $A/p = A_p/(pA_p)$. Now $p/p^2$ is an $A/p$-vector space and hence an $A_p/(pA_p)$-vector space. Hence $p/p^2$ is an $A_p$-module. Now, $(pA_p)/(p^2A_p)=(p/p^2) \otimes_A A_p=(p/p^2) \otimes_{A_p} A_p = p/p^2$. Am i correct? – Manos May 29 '13 at 0:32
• @Manos: Dear Manos, Yep, that looks right. Cheers, – Matt E May 29 '13 at 4:08