Consider the following extension $L$ of $\mathbb{Q}$: $L = \mathbb{Q}(\alpha_1, \alpha_2, \ldots)$, where $\alpha_t$ is the positive number such that $\alpha_t ^{2^t} = 3$, $t \in\mathbb{N}$. Show that the ring of algebraic integers $L_\mathbb{Z}$ in $L$ is integrally closed and that every prime ideal of $L_\mathbb{Z}$ is maximal.
Clearly, $\alpha_n \in L_\mathbb{Z}$ for all $n\in\mathbb{N}$. Consequently, $L_{\mathbb{Z}} = \mathbb{Z}[\alpha_1, \alpha_2, \ldots]$. Indeed, $L_{\mathbb{Z}}$ is a ring not containing rational numbers. It is not hard to argue that $L_\mathbb{Z}$ is integrally closed in its field of fractions as if $f/g \in L$, then we can plug it in any polynomial with coefficients in $L_{\mathbb{Z}}$ and obtain a contradiction after multiplying through by a suitable power of $g$. However, I am having troubles to prove that every prime ideal of $L_\mathbb{Z}$ is maximal. This is true for $\mathbb{Z}$ because it is a principal ideal domain, but I am not sure how to extend this to $L_\mathbb{Z}$.
Let $L_n$ denote the subfield $\mathbb Q(\alpha_1, \dotsc, \alpha_n)$ of $L$, and $L_{n, \mathbb Z}$ denote the ring of integers of $L_n$.
For the first statement: if $L_\mathbb Z$ is not integrally closed, then there is an element $x$ in $L\backslash L_\mathbb Z$, such that $x$ is integral over $L_\mathbb Z$. Since a polynomial equation $f(x)=0$ only involves finitely many elements of $L$, there is an $n$ such that $L_n$ contains all the coefficients of $f$, as well as $x$. This identity then contradicts the fact that $L_{n, \mathbb Z}$ is integrally closed.
For the second statement: obviously, the correct statement is that every non-zero prime ideal is maximal.
If $P$ is a non-zero prime ideal of $L_\mathbb Z$, then we may look at the intersection $P_n=P \cap L_{n, \mathbb Z} $ for each $n$, which must be a non-zero prime ideal of $L_{n, \mathbb Z}$, hence a maximal ideal.
If $Q$ is any proper ideal containing $P$, then we define $Q_n$ similarly and see that we must have $Q_n=P_n$, by maximality of $P_n$. Therefore we have $P= \bigcup P_n = \bigcup Q_n = Q$.
The above argument is a more elementary way of saying that $L_{\mathbb Z}/P $ is the inductive limit of $L_{n, \mathbb Z}/P_n$, and an inductive limit of fields is still a field.