Extensions of the $p$-adic valuation to $\overline{\mathbb{Q}}$

Question

I'm trying to find all the possible non-archimedean places (absolute value equivalence classes) on $\overline{\mathbb{Q}}$. I know that every valid absolute value on $\overline{\mathbb{Q}}$ must also be a valid absolute value on $\mathbb{Q}$, and I know that the every non-archimedean absolute value on $\mathbb{Q}$ is equivalent to the $p$-adic absolute value for some prime $p$, so I figured the best way to start would be trying to extend the $p$-adic absolute value to $\overline{\mathbb{Q}}$.

I know every place on a field is equivalent to a choice of a valuation ring, and the most natural extension of the $p$-adic valuation ring I thought of was the integral closure of $\mathbb{Q}\cap\mathbb{Z_p}$ in $\overline{\mathbb{Q}}$. ($\mathbb{Q}\cap\mathbb{Z_p}$ is the set of rational $p$-adic integers for some prime $p$, or the rational numbers whose denominator is not divisible by $p$ in lowest form)

My questions:

1. Is this a valid valuation ring? I have absolutely no idea how to go about proving whether this satisfies the definition. All I can think to do is try to find a counterexample, and I haven't found one. For every element of $\overline{\mathbb{Q}}$ that isn't in the integral closure of $\mathbb{Q}\cap\mathbb{Z_p}$ that I checked, the inverse is in the integral closure of $\mathbb{Q}\cap\mathbb{Z_p}$.
2. If it is a valid valuation ring, does it correspond to a real absolute value? I've read somewhere that some equivalent conditions for valuation ring corresponding to a real absolute value are: it has Krull dimension 1, it is a maximal proper subring of its field of fractions, and it is completely integrally closed (Note: I haven't seen a proof for any of these conditions). Again, I have no idea where to begin in proving this.
3. If the above are true, how do you actually define the valuation based on the choice of valuation ring? For some cases, like for a complex root of a rational number, it's easy: The absolute value of the $n$th root is the $n$th root of the absolute value. But in the general case (i.e the root of a general polynomial with rational coefficients, or some expression with radicals) I don't know how you would go about this. Ideally for rational numbers the absolute value should correspond to the standard $p$-adic absolute value ($|x|_p=p^{-v_p(x)}$)
4. Are all the non-archimedean places on $\overline{\mathbb{Q}}$ of this form?

Any help would be greatly appreciated.