Is there some irreducible $F \in \mathbb{Z}[x]$ such that $\mathbb{Z}[x]/(F)$ has no principal maximal ideal? Equivalently, is it possible that the $1$-dimensional integral domain $\mathbb{Z}[x]/(F)$ has no prime element?

The maximal ideals have the form $(p,f)$, where $p \in \mathbb{Z}$ is a prime and $f \in \mathbb{Z}[x]$ is a monic polynomial such that $f \bmod p$ is an irreducible factor of $F \bmod p$. In particular, $F$ should be reducible modulo every prime number. See here for a classification of biquadratic polynomials with this property. But this does not suffice, as $F=x^4+1$ shows. This is irreducible, reducible modulo every prime number, but $(2,x+1)=(x+1)$ in $\mathbb{Z}[x]/(x^4+1)$. A better candidate seems to be $F=x^4-10x^2+1$, I have checked $(p,f) \neq (f)$ for some primes $p$.

I suspect that the question is connected with the class group of the curve $V(F) \subseteq \mathbb{A}^1_{\mathbb{Z}}$?

Background: The question is equivalent to the question if $\mathbb{Z}[x]$ wins in the game of integral domains, which is a simplification of the game of rings by Will Sawin.

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    $\begingroup$ It seems to me that Cebotarev density will imply that there are infinitely many maximal ideals which are principal (this will certainly be true if $A={\mathbb Z}[x]/(F)$ is the ring of integers in the number field ${\mathbb Q}[x]/(F)$). The issue may be to prove Cebotarev for these orders $A$. $\endgroup$ Mar 18, 2013 at 12:03
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    $\begingroup$ @Aakumadula: To complete your thought, for any number field $K$ and order $A$ in the ring of integers $O_K$, the finiteness of the index of $A$ in $O_K$ as an abelian group implies that maximal ideals of $O_K$ whose residue characteristic does not divide $[O_K:A]$ lie entirely inside $A$. Thus, principal maximal ideals of $O_K$ whose residue characteristic does not divide $[O_K:A]$ do the job (i.e., they are also principal maximal ideals of $A$). $\endgroup$
    – user30180
    Mar 18, 2013 at 13:02
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    $\begingroup$ Why the downvote - looks like an attractive question to me! (Now compensated by an upvote) $\endgroup$ Mar 18, 2013 at 13:08
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    $\begingroup$ @ayanta, I first thought you were right, but I am now confused. If $P\subset O_K$ is a maximal ideal whose residue char does not divide $[O_K:A]$, how can it lie in $A$? If $P$ lies in $A$, it means that $[O_K:P]$ is in fact a multiple of $[O_K:A]$. @martin brandenburg: $Z[x]/(F)$ is definitely an order (since its ${\mathbb Q}$ span is the number field and this is a subring of $O_K$. $\endgroup$ Mar 18, 2013 at 15:25
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    $\begingroup$ @ayanta thanks for this. I think you may want to put this up as an answer since it answers the question completely (in the negative). $\endgroup$ Mar 19, 2013 at 1:12

1 Answer 1


Let $A$ be an order in the ring of integers $O_K$ of a number field $K$. We claim that there are infinitely many principal maximal ideals $P$ of $A$. By using localization at rational primes, we have a bijection between the sets of maximal ideals of $A$ and $O_K$ with residue characteristic relatively prime to $N = [O_K:A]$ via $P \mapsto P \cap A$ and $P' \mapsto P'O_K$.

In this way, we see that it is harmless to replace $A$ by a sub-order, so we may assume $A = \mathbf{Z} + M O_K$ for an integer $M > 0$. For $x \in 1 + MO_K$, we have $A \cap xO_K = xA$. Indeed, if $y \in O_K$ and $xy = c + Mt$ for $t \in O_K$ then we have to show that $y \in \mathbf{Z} + M O_K$, but this is clear since $xy \equiv c \bmod M O_K$ and $x \equiv 1 \bmod M O_K$. Hence, if $xO_K$ is a prime ideal of $O_K$ then $xA$ is a prime ideal of $A$, so it suffices to construct infinitely many maximal ideals $P$ of $O_K$ admitting a generator congruent to 1 modulo $M O_K$.

This latter formulation does not mention the order at all, and is a special case of the more general fact that for any nonzero ideal $J$ of $O_K$ whatsoever, $O_K$ has infinitely many principal maximal ideals $P$ admitting a generator $x \equiv 1 \bmod J$. The existence of infinitely many such $P$ follows from the method of proof of the "abelian" case of the Chebotarev Density Theorem (using generalized ideal class characters in the role of Dirichlet characters in the proof of Dirichlet's theorem on primes in arithmetic progressions). So tacitly here we are using the basic analytic properties of $L$-functions attached to characters of generalized ideal class groups (which can be proved in various ways, such as using $\zeta$-functions and class field theory if one wants to be ahistorical).

  • $\begingroup$ I sort of agree, but it is not clear that for the order $A={\mathbb Z[x]/(F)$, the intersection of the principal maximal ideal with $A$ is actually principal even if it is so for the smaller ring ${\mathbb Z}+MO_K$. . $\endgroup$ Mar 20, 2013 at 7:37
  • $\begingroup$ Dear Aakumadula: Here is the argument I had in mind. Since "ideal generated by" is inverse to the operation of "intersection" when considering maximal ideals relatively prime to the index (so to speak), we just use a bit of transitivity: if $A'' \subset A' \subset A$ is a containment of orders and $P$ is a maximal ideal of $A$ whose residue characteristic doesn't divide $[A:A'']$ then the prime ideals $P' = A' \cap P$ and $P'' = P' \cap A'' = P \cap A''$ satisfy $P'' A' = P'$ (!) so if $P \cap A''$ is principal then such a generator in $A''$ also generates $P'$ in $A'$. $\endgroup$
    – user30180
    Mar 23, 2013 at 1:16
  • $\begingroup$ Thanks a lot for your answer, ayanta. Unfortunately I don't have enough background in algebraic number theory. I didn't imagine that such a simple question requires so much theory. There are many number theorists on MO, perhaps one of them can confirm the proof? $\endgroup$ Mar 24, 2013 at 3:00
  • $\begingroup$ Dear Martin: The advantage of being at a department as strong as Muenster is that it has a lot of people who know number theory quite well both at this level and way beyond, so you can also ask any of them in person. Proceeding in that way you may give more understanding of the reasoning than by waiting for someone on MO to declare "Yes, I agree with this argument" (assuming I have not made a blunder, which I do not believe I have). $\endgroup$
    – user30180
    Mar 24, 2013 at 4:12

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