# A one-dimensional domain with infinitely many maximal ideals and non-zero Jacobson radical?

I'm looking for an example of a non-semi-local, non-Jacobson domain $A$, having $\dim(A)=1$.

A commutative ring $A$ is non-Jacobson if it has a prime ideal that is not an intersection of maximal ideals. For a one-dimensional domain, that comes down to the zero ideal $0$ being strictly contained in the Jacobson radical $rad(A)$, the intersection of all maximal ideals of $A$.

Such an $A$ is necessarily non-Noetherian, as is shown here: https://math.stackexchange.com/q/840896

Take for example $$A$$ to be the integral closure of $$\mathbb{Z}_{(p)}$$ in $$\overline{\mathbb{Q}}$$. Then $$A$$ is of dimension $$1$$ (because it is integral over something of dimension $$1$$) and a domain. Further, there are infinitely many prime ideals lying over $$(p)$$ (because there are algebraic field extensions over $$\mathbb{Q}$$ of arbitrary degree which are unramified over $$(p)$$), but every maximal ideal of $$A$$ contains $$p$$.
If you are more of a geometric minded person - like me - then, of course, the analogous example "take $$A$$ to be the integral closure of $$k[X]_{(X-a)}$$ in $$\overline{k(X)}$$" also works.
• For clarity, $p$ has order $a$ in the group $(\mathbb{Z}/n)^{*}$, where $n=p^{\, a}-1$. So $p$ splits into $\phi(n)/a$ different primes in $\mathbb{Z}[\zeta_{n}]$ by the cyclotomic decomposition law. Since this quantity tends to infinity as $a$ does, infinitely many primes of $A$ lie over $p$. – Matthé van der Lee Jun 6 '18 at 15:27