Question Is there a connection between Abel and Galois theories of polynomial equations?
Recall that for every polynomial $p(x)\in \mathbb{Q}[x]$ (say, without the free coefficient), Abel considered the monodromy group of the Riemann surface of the analytic function $w(z)$ defined by $p(w(z))+z=0$.. There is an expression of $w(z)$ in radicals if the monodromy group is solvable (is it an "if and only if" statement?).
On the other hand, for every polynomial $p(x)\in \mathbb{Q}[x]$, Galois considered the automorphism group (the Galois group) of the splitting field of $p$. The roots of $p(x)$ are expressed in radicals if and only if the Galois group is solvable.
The question asks whether there are any known connections between the monodromy group of $w(z)$ and the Galois group of $p(x)+z$ (considered as polynomial in $x$).
I am pretty sure this is well known. I just cannot find it in the literature.
Update 1. What I called "Abel's proof" of Abels' theorem is in fact Arnold's proof written by Alexeev (an English translation can be found here). Abel's proof was based on different ideas, see this text. So some instances of the word "Abel" above should be replaced by "Arnold". Added "Arnold" to the title.
Update 2. I found a very nice book by Askold Khovanskii,
"Topological Galois Theory", where Arnold's proof and its strong generalizations to other types of equations, including differential equations, are explained in detail. Highly recommended.
Best Answer
The action of the monodromy group of $w(z)$ on the fiber $p^{-1}(a)$ for a non-critical value $a$ of $p$ (that is $|p^{-1}(a)|=\deg p$) is the same as the action of the Galois group of $p(x)+z$ over $\mathbb C(z)$ on the roots of $p(x)+z$ in some splitting field. One can see this by comparing each of these groups with the deck transformation group of the cover $X\to\mathbb P^1\mathbb C$, where $X$ is the normal hull of the cover $P^1\mathbb C\to P^1\mathbb C$, $x\mapsto p(x)$. (Remark: Though it doesn't make a difference, it is slightly more convenient to work with $p(x)-z$ instead of $p(x)+z$.)