# Dimension of $\mathbb Z[\sqrt 5]$

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.

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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.

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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. :) – Peter Patzt Jan 10 '12 at 11:13

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

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

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

COROLLARY $\:$ A proper chain of prime ideals in $\rm\:R\:$ cannot contract to a shorter such chain in $\rm\:\mathbb Z\:.$

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

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Wonderful!!! ${}$ – Pierre-Yves Gaillard Jan 11 '12 at 5:49

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 $I\cap A$.

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.
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 $48$. – Bill Dubuque Jan 10 '12 at 18:13