# Non-zero prime ideals in the ring of all algebraic integers

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 $(0)$?

• Of course: it has nonzero maximal ideals, and any maximal ideal is prime. Jun 9, 2012 at 21:00
• @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 $\,1\,$ Jun 9, 2012 at 21:01
• @DonAntonio Yes, that's what I mean. Jun 9, 2012 at 21:03
• 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). Jun 9, 2012 at 21:20
• 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. Jun 10, 2012 at 11:18

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.

• 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. 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.

• 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. May 20, 2013 at 14:08
• You are right. I've added this. 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 $\mathcal O$.