Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be a Krull ring. According to Theorem 12.3 in Matsumura's Commutative Ring Theory, the family of localizations of $A$ at height 1 prime ideals of $A$ forms a defining family of $A$.

Question: Why such family exists? In other words, why does a Krull ring have at least one height 1 prime ideal?

Remark: By definition a Krull ring is the intersection of DVRs and each DVR has dimension 1, hence its maximal ideal has height 1. However, if we contract this maximal ideal to $A$, it is not necessary that the prime ideal we get will have height 1.

Edit: I realized that the definition of a Krull ring given in Wikipedia is quite different from the one given in Matsumura. In fact, the Wikipedia definition trivially answers my question. Matsumura's definition is: an integral domain is called Krull if it is the intersection of a family of DVRs and every non-zero element in the domain is nonzero in only a finite number of corresponding discrete valuations. How to obtain that such a ring contains a height 1 ideal is not obvious to me.

share|cite|improve this question
up vote 1 down vote accepted

Let $R=\cap_{\lambda\in\Lambda}R_{\lambda}$ with $R_{\lambda}$ DVRs (as in Matsumura's definition of Krull domains). Assume that the intersection is irredundant, that is, if $\Lambda'\subsetneq\Lambda$ then $\cap_{\lambda\in\Lambda}R_{\lambda}\subsetneq\cap_{\lambda'\in\Lambda}R_{\lambda'}$.

Let's prove that $m_{\lambda}\cap R$ is a prime ideal of height one, for all $\lambda\in\Lambda$. First note that $m_{\lambda}\cap R\neq (0)$. If the height of some $m_{\alpha}\cap R$ is at least $2$, then there exists a nonzero prime $p\subsetneq m_{\alpha}\cap R$. From Kaplansky, Commutative Rings, Theorem 110, there exists $m_{\alpha'}\cap R\subseteq p$ (obviously $\alpha'\neq\alpha$). Let $x\in\cap_{\lambda\ne\alpha}R_{\lambda}$, $x\notin R$ (so $x\notin R_{\alpha}$), and $y\in m_{\alpha'}\cap R$, $y\neq 0$. One can choose $m,n$ positive integers such that $z=x^my^n$ is a unit in $R_{\alpha}$. Since $x\in R_{\alpha'}$ and $y\in m_{\alpha'}\cap R$ we get $z\in m_{\alpha'}$. Obviously $z\in R_{\lambda}$ for all $\lambda\ne \alpha, \alpha'$, so $z\in R$, $z$ is invertible in $R_{\alpha}$ and not invertible in $R_{\alpha'}$, a contradiction with $m_{\alpha'}\cap R\subseteq m_{\alpha}\cap R$.

(This argument is adapted from Kaplansky's proof of Theorem 114. Furthermore, using again Theorem 110 one can see that $m_{\lambda}\cap R$ are the only height one prime ideals of $R$.)

share|cite|improve this answer

Any finite-dimensional ring (indeed, any ring with a prime of finite, nonzero height) has prime ideals of height 1, basically by definition of the dimension: if $P$ has finite height $n$, there's a maximal chain $P_0 \subseteq P_1 \subseteq \dotsb \subseteq P_n = P$, and $P_1$ can't contain any more than one prime ideal, since otherwise we could make a longer chain. Indeed, so does any Noetherian ring, since it satisfies the descending chain condition on prime ideals.

If you're really interested in the non-Noetherian case, things get confusing -- for example, there are domains where every nonzero prime has infinite height. If $R$ has this property, though, then all localizations of $R$ have this property (the primes of any localization of $R$ corresponding to a downward-closed subset of the primes of $R$). In particular, no localization of $R$ can be a DVR, since DVRs have finite dimension, so $R$ can't be a Krull ring, even by Matsumura's definition. This should prove the equivalence of the two definitions, too.

share|cite|improve this answer
Thanks for your answer +1. A first question: lets work with Matsumura's definition. Thus $A=\cap_{\lambda \in \Lambda} R_{\lambda}$, where $R_{\lambda}$ is a DVR of the field of fractions $K$ of $A$. Why must we have that some localization of $A$ is DVR? – Manos May 30 '13 at 20:55
Each $R_\lambda$ is contained in $K$ and contains $A$, so it's a localization of $A$ -- not necessarily at a prime ideal, but just in the sense of inverting some elements of $A$. For any multiplicative set $S$, the primes of $S^{-1}A$ correspond to the primes of $A$ that don't meet $S$. – Paul VanKoughnett May 30 '13 at 22:38

Your question is answered e.g. in Samuel's Lectures on UFDs.

The short version is: $R$ is a Krull ring if and only if $\mathrm{Div}(R)$ (= the semigroup of divisorial fractional ideals, its neutral element is the submodule $R \subseteq Q(R)$) is (as a partially ordered group) isomorphic via some iso $\phi$ to a direct sum $\bigoplus_{i \in I}{\mathbb{Z}}$ equipped with the canonical partial order. If $R$ is a Krull ring but not a field, then $\mathrm{Div}(R) \neq 0$ and the canonical basis vectors $e_i, i \in I$ form a positive basis (Here is the existence statement!). Their preimages $\mathfrak{p}_i , i \in I$ under $\phi$ form a positive basis of $\mathrm{Div}(R)$ and they are in fact precisely the prime ideals of height one in $R$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.