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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

Bernard
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Matthé van der Lee
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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.

user26857
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Louis
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    Very nice, thanks! – Matthé van der Lee Jun 06 '18 at 11:26
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    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 06 '18 at 15:27