Let $\mathcal{O}$ be the ring of all algebraic integers: elements of $\mathbb{C}$ which occur as zeros of monic polynomials with coefficients in $\mathbb{Z}$.

It is known that $\mathcal{O}$ is a Bezout domain: any finitely generated ideal is a principal ideal.

In addition, $\mathcal{O}$ has no irreducible elements, since any $x \in \mathcal{O}$ which is not a unit can be written as $x = \sqrt{x}\cdot\sqrt{x}$, where $\sqrt{x}$ is also not a unit in $\mathcal{O}$.

My question is:

Does $\mathcal{O}$ have any prime ideal other than $(0)$?

  • 5
    $\begingroup$ Of course: it has nonzero maximal ideals, and any maximal ideal is prime. $\endgroup$ Jun 9, 2012 at 21:00
  • 2
    $\begingroup$ @Cocopuffs It may be that "normed polynomials" is actually correct, but I think the usual, standard name is "monic polynomial" = a pol. with main coefficient equal to $\,1\,$ $\endgroup$
    – DonAntonio
    Jun 9, 2012 at 21:01
  • $\begingroup$ @DonAntonio Yes, that's what I mean. $\endgroup$
    – Cocopuffs
    Jun 9, 2012 at 21:03
  • $\begingroup$ If $R\rightarrow S$ is an integral extension of rings, then the induced map $\mathrm{Spec}(S)\rightarrow\mathrm{Spec}(R)$ is surjective. Concretely, given a prime $\mathfrak{p}$ of $R$, there exists a prime $\mathfrak{P}$ of $S$ with $\mathfrak{P}\cap R=\mathfrak{p}$. Ultimately this fact too boils down to the existence of maximal ideals in non-zero rings (with identity). $\endgroup$ Jun 9, 2012 at 21:20
  • $\begingroup$ I wonder: is it possible to show that $\mathcal{O}$ has nonzero prime ideals without somewhere invoking the Axiom of Choice? Both the arguments in the comments above and @countinghaus 's argument below seem to need countable choice. $\endgroup$ Jun 10, 2012 at 11:18

3 Answers 3


Yes, $\mathcal{O}$ has lots of prime ideals (the axiom of choice is equivalent to every non-unit in any commutative ring being contained in a maximal ideal).

A concrete example is not so easy, but the point is this: for any finite field extension $L/\mathbb{Q}$, pick a prime $\mathfrak{p}_L$ of the ring of integers $L$, and do so compatibly, i.e. if $L \subset L'$ we need $\mathfrak{p}_{L'} \cap L = \mathfrak{p}_L$. The union of all the $\mathfrak{p}_L$ will then be a prime.

  • 1
    $\begingroup$ Thanks - I'm kicking myself that I didn't see it. A constructive proof is more of what I was looking for, and you answered that nicely as well. $\endgroup$
    – Cocopuffs
    Jun 9, 2012 at 21:09

We don't need the axiom of choice. We can write down a maximal ideal.

First, we observe that $R$ is countable: For every monic $p \in \mathbb{Z}[x]$ let $V(p)$ be the set of its complex roots. It has a canonical enumeration, when we order the roots using the lexicographic order on $\mathbb{R} \times \mathbb{R}$. For every $n \geq 1$ the set of these $p$ of degree $n$ identifies with $\mathbb{Z}^n$ and is therefore countable, with an explicit enumeration. It follows that also $\mathcal{O}_n = \cup_{\mathrm{deg}(p)=n} V(p)$ has an explicit enumeration, and then also $\mathcal{O}= \cup_n \mathcal{O}_n$. This enumeration is complicated, but it is computable.

Now let $R \neq 0$ be any countable ring. Any enumeration $R=\{a_0,a_1,\dotsc\}$ produces a maximal ideal as follows: Define an increasing chain of proper ideals $I_k$ as follows: Let $I_0=0$. If $I_k + \langle a_k \rangle$ is proper, let $I_{k+1} = I_k + \langle a_k \rangle$. If not, call $a_k$ bad, and let $I_{k+1}=I_k$. Then $I:=\cup_k I_k$ is a maximal ideal: By construction $1 \notin I$. Now let $a_k \in R \setminus I$. Then $a_k$ has to be bad (otherwise $a_k \in I_{k+1} \subseteq I$), so that $I_k + \langle a_k \rangle = R$ and hence $I+\langle a_k \rangle = R$. $\square$

More generally, let $R$ be a ring $\neq 0$ whose underlying set is well-orderable. Every enumeration $R=\{a_{\alpha} : \alpha < \kappa\}$ with some limit ordinal $\kappa$ produces a maximal ideal (without using the axiom of choice): Let $I_0=0$, and construct $I_{\alpha+1}$ from $I_{\alpha}$ as above. For limit ordinals $\lambda<\kappa$, let $I_{\lambda}=\cup_{\alpha<\lambda} I_{\alpha}$. Then $I=\cup_{\alpha<\kappa} I_{\alpha}$ is maximal: If $\alpha<\kappa$ and $a_{\alpha} \notin I$, then $a_{\alpha}$ is bad (otherwise $\alpha+1<\kappa$, $a_{\alpha} \in I_{\alpha+1} \subseteq I$), hence $I_{\alpha}+\langle a_{\alpha} \rangle = R$ and $I+\langle a_{\alpha} \rangle = R$.

Even more generally, the proof of "well ordering principle $\Rightarrow$ Zorn's Lemma" in ZF actually shows that every partial order, in which every chain has an upper bound, and whose underlying set is well-orderable, has a maximal element.

  • $\begingroup$ Very nice observation. I can't believe I've never seen this before! A non-constructive question: does the analogous construction by transfinite induction give a maximal ideal from any well-ordering of a commutative ring? I can't see any immediate obstruction. $\endgroup$ May 20, 2013 at 14:08
  • $\begingroup$ You are right. I've added this. $\endgroup$ May 20, 2013 at 20:50

This is not a "constructive" proof, but is certainly a very short one: if an integral domain has $(0)$ as its only prime ideal, then it is a field. But $\mathcal O$ is obviously not a field: for instance, $2$ is not invertible in $\mathcal O$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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