First, an example:
Consider the algebraic number field $\mathbb{Q}(\sqrt{-5})$. Its class number is two. That means there exist ideals of the ring of integers $\mathbb{Z}[\sqrt{-5}]$ of this field that are not principal, but the square of any ideal is principal. For instance, we have the decomposition
$$6 = 2 \cdot 3 = ( 1 + \sqrt{-5})\cdot (1-\sqrt{-5})$$
The ideals $\mathfrak{p}\colon =(3, 1+\sqrt{-5})$ is prime and of order $2$. That means that its square $\mathfrak{p}^2$ is principal. Indeed, we have from first principles
\begin{eqnarray}
\mathfrak{p}^2 = (\, 9,\, 3( 1 +\sqrt{-5}),\, (1+\sqrt{-5})^2)=\\
= (\, (2+\sqrt{-5})(2-\sqrt{-5}),\, (-1 + \sqrt{-5})(2 -\sqrt{-5}),\, -2(2 - \sqrt{-5})\, )= (2 - \sqrt{-5})
\end{eqnarray}
Let's consider the $\mathbb{Q}(\sqrt{2 - \sqrt{-5}}) $ of the field $\mathbb{Q}(\sqrt{5})$. In the ring of the integers of this number field we have the equality of ideals (extend the ideals)
$$(3, 1+\sqrt{-5})^2 =(2 - \sqrt{-5}) = ( \sqrt{2 - \sqrt{-5}})^2$$
Because of the unique decomposition of ideals of any ring of integers of an algebraic number field, the above equality implies
$$(3, 1+\sqrt{-5})= ( \sqrt{2 - \sqrt{-5}})$$
Let's check this equality. For comfort, fix an embedding of $\mathbb{Q}(\sqrt{2 - \sqrt{-5}}) $ into $\mathbb{C}$, $\sqrt{2 - \sqrt{-5}} \mapsto \frac{ \sqrt{5} - i }{\sqrt{2}}$ ( so $\sqrt{-5} \mapsto \sqrt{5}i$)
We have $3^2 = 9 = (2+ \sqrt{5}i) \cdot (2- \sqrt{5}i)$ so
$$
3= \frac{ \sqrt{5} + i }{\sqrt{2}}\cdot \frac{ \sqrt{5} - i }{\sqrt{2}}
$$
Note that $\frac{ \sqrt{5} + i }{\sqrt{2}}$ is in the ring of integers of $\mathbb{Q}(\frac{ \sqrt{5} - i }{\sqrt{2}})$. Indeed, the minimal polynomial of $\frac{ \sqrt{5} + i }{\sqrt{2}}$ is $x^4 - 4 x^2 + 9$ and moreover
$$\frac{ \sqrt{5} + i }{\sqrt{2}}= 4/3 \cdot \frac{ \sqrt{5} - i }{\sqrt{2}}-1/3 \cdot \left(\frac{ \sqrt{5} - i }{\sqrt{2}}\right)^3 $$
Also, from $(1+ \sqrt{5}i)^2 = -2 ( 2 - \sqrt{5}i)$ we get
$$1+ \sqrt{5}i = \sqrt{2} i \cdot \frac{ \sqrt{5} - i }{\sqrt{2}}$$
Again, $\sqrt{2}i$ is an algebraic integers and $\mathbb{Q}(\frac{ \sqrt{5} - i }{\sqrt{2}})$, since
$$\sqrt{2} i =1
/3 \cdot \frac{ \sqrt{5} - i }{\sqrt{2}}-1/3 \cdot \left(\frac{ \sqrt{5} - i }{\sqrt{2}}\right)^3 $$
So far, both $3$ and $1+ \sqrt{5}i$ are divisible by $\frac{ \sqrt{5} - i }{\sqrt{2}}$. Now we'll see that $\frac{ \sqrt{5} - i }{\sqrt{2}}$ is an (algebraic) integer combination of $3$ and $1+ \sqrt{5}i$. Indeed, we have
$$\frac{ \sqrt{5} - 3 i}{\sqrt{2}} \cdot (1 + \sqrt{5} i) + \frac{ \sqrt{5} - i }{\sqrt{2}}\cdot 3 = \frac{ \sqrt{5} - i }{\sqrt{2}}$$
And so it goes in general. Consider finitely many algebraic integers $(\alpha_s)$ . Let $K$ the algebraic number field generated by them, $\mathcal{O_K}$ its ring of integers. The ideal class of $K$ is finite. Therefore, the $m$ power of ideal of $\mathcal{O}_K$ generated by the $\alpha_s$ is principal for some integer $m$ ( for instance, if $m$ is the class number of $K$). There exists $\beta$ in $\mathcal{O}_K$ so that $(\alpha_s)^m_s = (\beta)$ ($\beta$ uniquely determined by this $m$ up to a unit element ( in $\mathcal{O}_K
^{\times}$). Extend $K$ by adding $\beta^{1/m}$ to $L = K(\beta^{1/m})$. We have the equality of $\mathcal{O}_L$ ideals
$$(\alpha_s)_s^m = (\beta)$$ and so
$$(\alpha_s)_s = (\beta^{1/m})$$
Hint for a and c:
Prove that $\alpha\in \overline{\mathbf Z}\;$ if and only the ring $\mathbf Z[\alpha]$ is a finitely generated $\mathbf Z$-module.
No irreducible elements: just show the square root of an algebraic integer is an algebraic integer.
b) Show that any element $\alpha\notin \mathfrak p$ (a non-zero prime ideal of $\overline{\mathbf Z}$) is a unit modulo $\mathfrak p$. For that, show that $\mathfrak p\cap\mathbf Z[\alpha] $ is a maximal ideal of $\mathbf Z[\alpha] $.
Sub-hint: Set $\mathfrak p\cap\mathbf Z=p\mathbf Z$ and show the quotient $\mathbf Z[\alpha]/\mathfrak p\cap\mathbf Z[\alpha]$ is a finite-dimensional $\mathbf Z/p\mathbf Z$-vector space.
c) Compute the minimal polynomials of each element.
Best Answer
Yes, $\mathcal{O}$ has lots of prime ideals (the axiom of choice is equivalent to every non-unit in any commutative ring being contained in a maximal ideal).
A concrete example is not so easy, but the point is this: for any finite field extension $L/\mathbb{Q}$, pick a prime $\mathfrak{p}_L$ of the ring of integers $L$, and do so compatibly, i.e. if $L \subset L'$ we need $\mathfrak{p}_{L'} \cap L = \mathfrak{p}_L$. The union of all the $\mathfrak{p}_L$ will then be a prime.