I arrived at a confusion about finiteness of prime ideals in Dedekind domains. I am unable to see where is misunderstanding of mine.

Ref. Janusz's book on Algebraic Number Fields, page 10-11 (see here).

3.6. Let $B$ be commutative noetherian ring in which every prime ideal is maximal. Then every ideal of $B$ contains a product of prime ideals.

3.7. Let $B$ be as in 3.6. Then there exists distinct prime ideals $P_1,\cdots ,P_n$ in $B$ and positive integers $a_1,\cdots,a_n$ such that $P_1^{a_1}\cdots P_n^{a_n}=0.$

3.9. Let $B$ be as in 3.6 and let $P_1,\cdots,P_n$ be distinct prime ideals of $B$ such that $P_1^{a_1}\cdots P_n^{a_n}=0$. Then $B\cong B/P_1^{a_1}\oplus \cdots \oplus B/P_n^{a_n}$ as a ring.

3.10. With the assumptions of 3.9, the ideals $P_1,\cdots, P_n$ are all prime ideals of $B$.

Suppose $B$ is a Dedekind domain. Then each of the above statement applies to $B$, which concludes (finally from 3.10) that $B$ has only finitely many prime ideals. How can this be possible if $B$ is not PID? I didn't get my mistake in understanding the statements.

As in Ref.

  • An ideal $P$ of a ring is prime if $ab\in P$ implies $a\in P$ or $b\in P$. (We should also add that $P$ is proper ideal; isn't it? Let's see if needed)

  • 3.6 is correct for Noetherian and hence Dedekind domains. (I think, the extra condition in 3.6 -prime ideal is maximal - is not necessary.)

  • If $R$ is Dedekind domain, then $0$ (prime ideal) contains a product of prime ideals, hence 3.7.

  • For 3.9, since prime ideals are maximal, so distinct prime ideals are pairwise coprime. It can be shown that powers of distinct prime ideals are also pairwise coprime (its 3.8 in Ref.), and so CRT is applicable in addition to $P_1^{a_1}\cdots P_r^{a_1}=P_1^{a_1}\cap \cdots \cap P_2^{a_r}$ hence 3.9.

  • So it seems that either statement 3.10 is incorrect or (I am misunderstanding something.)

  • $\begingroup$ If $B$ is a Dedekind domain, then $B$ does not satisfy 3.6. $\endgroup$ Mar 10, 2018 at 7:34
  • $\begingroup$ Dedekind domain is a domain which is - by one definition- (1) Noetherian (2) Prime ideal of it is maximal (3) It it integrally closed $\endgroup$
    – Beginner
    Mar 10, 2018 at 8:11
  • $\begingroup$ You (or your source) haven't got that quite right. Please attend very carefully to the definition of prime ideal. $\endgroup$ Mar 10, 2018 at 8:14

1 Answer 1


A Dedekind domain does not satisfy 3.6 (unless it is a field). In a Dedekind domain, only the nonzero prime ideals are required to be maximal, but in 3.6 every prime ideal is required to be maximal. In any domain which is not a field, the zero ideal is a nonmaximal prime ideal.

  • $\begingroup$ Thats great; some misunderstanding of mine is now getting clear now (I have not completely understood summary of above results, but, your points make many things much clear. Thanks for pointing it. $\endgroup$
    – Beginner
    Mar 10, 2018 at 8:34

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