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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.

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.

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 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.

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.

Is there some irreducible $F \in \mathbb{Z}[x]$ such that $\mathbb{Z}[x]/(F)$ has no principal maximal ideal?

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.

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 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.