Suppose $K$ is a number field and we have finitely many elements $\alpha_{1}, \ldots, \alpha_{k} \in K$ and a polynomial $f(x)\in K[x]$ of degree $\geq 2$. Given $m \geq 1$, we define $K_{m}(\alpha_{i}) = K(f^{-m}(\alpha_{i}))$, which is the field obtained upon adjoining the $m^{th}$ pre-images of $\alpha_{i}$. We suppose that $k \geq 3$.
Suppose that for every $1 \leq i \leq k$, we have that $K_{m}(\alpha_{i})$ is not contained in the compositum extension of $\prod_{j\neq i} K_{m}(\alpha_{j})$ corresponding to the rest of $\alpha_{j}$'s. This to ensure that each of the fields are actually necessary. Again, we may as well suppose that none of the $\alpha_{i}$ have finite backward orbit under $f$ either. This to ensure that as $m$ varies, $K_{m}(\alpha_{i})$ actually gives bigger and bigger extensions.
I want to see whether this is enough to guarantee that linear-disjointness of the following kind holds: (If not, I am wondering what other additional conditions yield the conclusion).
For every finite extension $E$ of $K$, there exists a large enough $N$ such that for every $m \geq N$, there is a field extension $L = L^{(m)}$ of $E$ such that the fields $L_{m}(\alpha_{i})$'s are all linearly disjoint over $L_{m}$.
Of course, we want to choose $L = L^{(m)}$ such that none of the $L_{m}(\alpha_{i})$ properly contain $L$, so that the each of the extensions $L_{m}(\alpha_{i})$ are non-trivial.
For $f(x) = x^{2}$ and $\alpha_{i}\in\{2, 3, 5\}$ and $K = \mathbb{Q}$, we have that $\mathbb{Q}_{m}(\alpha_{i}) = \mathbb{Q}(\alpha_{i}^{1/2^{m}}, \mu_{2^{m}}) $. In this case, given any $E$, we can take $L^{(m)} = \mathbb{Q}(\mu_{2^{m}})$ for any $m$. We don't need to consider $\alpha_{i} \in \{2, 3, 6\}$, because in this case $\mathbb{Q}_{m}(6) = \mathbb{Q}_{m}(2) \mathbb{Q}_{m}(3)$. Now, upon doing some literature review, I stumbled upon terms in the vicinity of "arboreal-Galois-representations"; however, those words hardly mean anything to me. I would be grateful for any help or for accessible references that ponder about such things.
