# Quotients of a ring of integers

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 ?

Greg

• 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. – Jason Starr Jul 5 '19 at 14:17

## 1 Answer

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 $$\mathcal O_L$$.