I know that every non-zero prime ideal in a Dedekind Domain is maximal. Since $\mathbb Z[\sqrt 5]$ is not integrally closed in its field of fractions $\mathbb Q[\sqrt 5]$, I wonder whether we can still show that non-zero primes are maximal.

More generally is it known for which of the square-free $D\in \mathbb Z$ there are non-zero primes of $\mathbb Z[\sqrt D]$ that are not maximal.


The ring $\mathbb{Z}[\sqrt{D}]$ is integral over $\mathbb{Z}$. Therefore a prime ideal $P$ of it is maximal if and only if the prime ideal $p:=P\cap\mathbb{Z}$ is maximal. Since all non-zero prime ideals of $\mathbb{Z}$ are maximal we get that all $P$ such that $P\cap\mathbb{Z}\neq 0$ are maximal. The latter condition is equivalent to $P\neq 0$, which proves the claim.

  • $\begingroup$ good ... i thought this property only holds for rings that are integrally closed over another ring but I looked it up, it is enough that it is integral. :) $\endgroup$ – Peter Patzt Jan 10 '12 at 11:13

Below is a very simple direct proof for any ring $\,R\,$ of algebraic integers.

Lemma $\ $ If $\ I \supsetneq P\ $ are ideals of $\,R,\,$ with $\,P\,$ prime then there is an integer $\,f_k\in I\,$ but $\,f_k\not\in P\,.$

Proof $\ $ Choose $\,\alpha\in I,\ \alpha\not\in P\,.\,$ Being an algebraic integer, $\,f(\alpha) = 0\,$ for a monic $\,f(x)\in \mathbb Z[x],\ $ $\,f(x) \, =\, x^n +\cdots + f_1\ x +\, f_0\,.\,$ Note $\,f_n = 1\not\in P\,.\,$ Let $\,k\,$ be least with $\,f_k\,\not\in P.\,$ $\ f(\alpha) = 0\in P\,$ $\, \Rightarrow\,$ $\,(\alpha^{n-k}+\cdots+f_k)\ \alpha^k\in P,\ \alpha\not\in P$ $\,\Rightarrow\,$ $\,\alpha^{n-k}+\cdots+f_k\in P\subset I\,.\,$ So $\ \alpha\in I$ $\,\Rightarrow\,$ $\,f_k \in I\,.\ $ QED

Corollary $\, $ A proper chain of prime ideals in $\,R\,$ cannot contract to a shorter such chain in $\,\mathbb Z\,.$

Remark $\ $ Alternatively, reduce to the simpler case $\,P = 0\,$ by way of factoring out the prime $\,P\,.\,$ Then $\,f_k\,$ is the constant term of a minimal polynomial for $\,\alpha\,$ over a domain, so $\,f_k\ne 0,\,$ i.e. $\,f_k\not\in P,\,$ which is explained in much detail in this post, as a generalization of rationalizing denominators.


New answer

After having read Bill Dubuque's beautiful answer, I'm rewriting mine.

To answer the question as asked, the simplest seems to prove this.

If $P$ and $I$ are ideals of $\mathbb Z[\sqrt5]$, and if $P$ is prime, then $$ P < I\quad\implies\quad\mathbb Z\cap P\ < \ \mathbb Z\cap I, $$ where $ < $ means "properly contained in".

To see this, set $x=a+b\sqrt5$ with $a,b\in\mathbb Z$, and let $x$ be in $I$ but not in $P$. Putting
$$ x':=a-b\sqrt5,\quad t:=x+x'=2a,\quad n:=xx'=a^2+5b^2, $$ we have $$ x^2-tx+n=0. $$ If $n$ is not in $P$, we are done because $n$ is in $\mathbb Z\cap I$. If $n$ is in $P$, then so is $x-t$, and $t$ is in $\mathbb Z\cap I$ but not in $P$.

More generally, if $B$ is a (commutative) ring, if $A$ is a subring, if $B$ is integral over $A$, if $P$ and $I$ are ideals of $B$, and if $P$ is prime, then $$ P < I\quad\implies\quad A\cap P\ < \ A\cap I. $$ Indeed, on taking the quotient by $P$, we can assume that $B$ is a domain, and that $P=0$. Let $x$ be a nonzero element of $I$, and let $f\in A[X]$ a monic polynomial of least degree such that $f(x)=0$. Then the constant term of $f$ is a nonzero element of $I\cap A$.

Old answer

In view of the second Corollary on page 7 of the text [1] by Mel Hochster, if a ring $B$ is integral over its subring $A$, then $A$ and $B$ have the same Krull dimension.

This shows in particular that if a ring $A$ is between $\mathbb Z$ and the ring of integers of a number field, then $A$ has Krull dimension $1$, that is, its nonzero primes are maximal.

[1] Integral extensions and integral dependence: pdf file --- html page.

  • 2
    $\begingroup$ More generally if the ring extension $\rm\: R \subset T\:$ satisfies GU (going-up) and INC (incomparable) then $\rm\:dim\ R\ =\ dim\ T\:.\:$ For a proof see Kaplansky, Commutative Rings, Theorem $48$. $\endgroup$ – Bill Dubuque Jan 10 '12 at 18:13

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