# Does every Krull ring have a height 1 prime ideal?

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.

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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 $R$.)

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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.

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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

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 $R$.