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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

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Take for example $A$ to be the integral closure of $\mathbb{Z}_{(p)}$ in $\bar{\mathbb{Q}}$. 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 $\bar{k(X)}$" also works.

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    $\begingroup$ Very nice, thanks! $\endgroup$ – Matthé van der Lee Jun 6 '18 at 11:26
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    $\begingroup$ 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 $p$. $\endgroup$ – Matthé van der Lee Jun 6 '18 at 15:27

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