In general if I "add" to the field of rationals $\mathbb{Q}$ the square root of two, the extension will be algebraic and of the form:
$\mathbb{Q}(\sqrt{2})=\dfrac{\mathbb{Q}[X]}{(X^2-2)}=\left\{ q_0 +q_1\sqrt{2} \ \bigg| q_0,q_1 \in \mathbb{Q} \right\}$
The same kind of construction for the cubic root of two will yield the following extension:
$\mathbb{Q}(\sqrt[3]{2})=\dfrac{\mathbb{Q}[X]}{(X^3-2)}=\left\{ q_0 +q_1\sqrt[3]{2} + q_2(\sqrt[3]{2})^2 \ \bigg| q_0,q_1,q_2 \in \mathbb{Q} \right\}$
We easily see that $\mathbb{Q}(\sqrt{2}) \nsubseteq \mathbb{Q}(\sqrt[3]{2}), \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})= \mathbb{Q}(\sqrt[6]{2}) \supseteq \mathbb{Q}(\sqrt{2})$.
Now, if I want to describe $\mathbb{Q}(\sqrt[12]{2})$ I can proceed in this way:
$12=3 \times 4= 3 \times 2^2$
This implies that $\mathbb{Q}(\sqrt[12]{2}) = \mathbb{Q}(\sqrt[4]{2},\sqrt[3]{2})=\mathbb{Q}(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2})$
Now, with these examples in mind, I want to build, if it has a sense, the infinite extension of $\mathbb{Q}$, adding all the $n$-roots of $2$, where $n \in \mathbb{N}$.
$\mathbb{Q}(\sqrt{2},\sqrt[3]{2}, \sqrt[4]{2},\sqrt[5]{2},...,\sqrt[n]{2},...) $
I can describe this extension using only prime numbers and their power:
$\large \mathbb{Q}(\left\{ \sqrt[p^{\alpha}]{2} \right\}_{p \ \text{over} \ \textit{primes},\\ \alpha \in \mathbb{N}} )$
But actually I don't know if this is helpful.
Is there a way to give sense at this extensions?
P.s. I was also thinking about direct limits, and describe this extension as a direct limit of finite $n$th-extensions of the form $\left\{ \mathbb{Q}(\sqrt{2},\sqrt[3]{2}, \sqrt[4]{2},\sqrt[5]{2},...,\sqrt[n]{2}) \right\}_{n\in \mathbb{N}}$. But here i don't know properties of closure to direct limits of the category of fields $\mathcal{Fld}$
It's a union of the chain of subfields
$$Q\left(2^{1/n!}\right), \;n = 1,2,3, ...$$
of the field of algebraic numbers.