Let $L$ be a number field, let $p$ be a prime number, and let $I$ be a ideal of $\mathcal{O}_L$ containing $p$. I am not assuming that $\mathcal{O}_L$ or that $I$ is prime. The quotient ring $\mathcal{O}_L/I$ has a natural structure of $\mathbb{F}_p$-algebra.

Question. Do we have an isomorphism of $\mathbb{F}_p$-algebras $$\mathcal{O}_L/I\simeq \mathbb{F}_p[X]/(f)$$ for some nonzero monic polynomial $f$ ?

I know that the answer if YES in several cases.

1) For example, it is true if $\mathcal{O}_L=\mathbb{Z}[\alpha]$ for some $\alpha$.

Indeed, since $I$ contains $p$, evaluation at $\alpha$ induces a morphism of $\mathbb{F}_p$-algebras $\mathbb{F}_p[X]\to \mathcal{O}_L/I$. This morphism is surjective since $\mathcal{O}_L=\mathbb{Z}[\alpha]$ . Its kernel is generated by a monic polynomial. Done.

2) If $L=K_1K_2,$ where $\mathcal{O}_{K_i}=\mathbb{Z}[\alpha_i]$, and the discriminants of $K_1$ and $K_2$ are coprime, and $p$ is totally ramified in $K_2$, and $I=\mathfrak{p}_2\mathcal{O}_L.$ where $\mathfrak{p}_2$ is the unique prime ideal of $\mathcal{O}_{K_2}$ lygin above $p$. In this case, one may show that $\mathcal{O}_L/I\simeq \mathbb{F}_p[X]/(\overline{\mu}_{\alpha_1,\mathbb{Q}})$.

Nevertheless, I suspect that I am missing obvious counterexamples...

About 2), i wonder if it is a particular case of 1), so there is a side question:

Side question. If $L=K_1K_2,$ where $\mathcal{O}_{K_i}=\mathbb{Z}[\alpha_i]$, and the discriminants of $K_1$ and $K_2$ are coprime, is $\mathcal{O}_L=\mathbb{Z}[\alpha]$ for some $\alpha$ ? For example, is $\alpha=\alpha_1+\alpha_2$ working ?

Any thoughts ?


  • 6
    $\begingroup$ Here is one counterexample. Let $K$ be $\mathbb{Q}$. Let $L$ be $\mathbb{Q}[s,t]/\langle s^2+s+2,t^2+t+4 \rangle$. Let $p$ be $2$, and let $I$ be $2\mathcal{O}_L$. Then $\mathcal{O}_L/I$ equals $\mathbb{F}_2[s,t]/\langle s^2+s,t^2+t \rangle$. This is isomorphic as an $\mathbb{F}_2$-algebra to a product of four copies of $\mathbb{F}_2$. Yet a monic polynomial has at most two $\mathbb{F}_2$-rational points. $\endgroup$ Jul 5 '19 at 14:17

The answer to your main question is no. Let $I = (p)$ where $p$ splits completely in $L$ and $r := [L:\mathbf Q] > p$. (Example: cubic field in which $p = 2$ splits completely.) Then $\mathcal O_L/(p) \cong \mathbf F_p^r$ but if $\mathcal O_L/(p) \cong \mathbf F_p[X]/(f)$ for monic $f$ in $\mathbf F_p[X]$ then $f$ is a product of $r$ distinct monic linear factors, which is impossible for $r > p$. This proves $\mathcal O_L$ is not $\mathbf Z[\alpha]$ for some $\alpha$ ($L$ is not monogenic).

Dedekind found the first example of this: $L = \mathbf Q(\theta)$ where $\theta$ is a root of $X^3-X^2-2X-8$. This is a cubic field in which $2$ splits completely. Thus $\mathcal O_L/(2) \cong \mathbf F_2^3$ and $\mathcal O_L/(2) \not\cong \mathbf F_2[X]/(f)$ for any $f$ in $\mathbf F_2[X]$. Also $\mathcal O_L \not= \mathbf Z[\alpha]$ for some $\alpha$ in $\mathcal O_L$.


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