I feel a bit ashamed to ask the following question here.
What is (actually, is there) Galois
theory for polynomials in
$n$-variables for $n\geq2$?
I am preparing a large audience talk on Lie theory, and decided to start talking about symmetries and take Galois theory as a "baby" example. I know that Lie groups are somehow to differential equations what discrete groups are to algebraic equations. But I nevertheless would expect Lie (or algebraic) groups to appear naturally as higher dimensional analogs of Galois groups.
Namely, the Galois group $G_P$ of a polynomial $P(x)$ in one variable can be defined as the symmetry group of the equation $P(x)=0$ (very shortly, the subgroup of permutations of the solutions/roots that preserves any algebraic equation satisfied by them).
Then one of the great results of Galois theory is that $P(x)=0$ is solvable by radicals if and only if the group $G_P$ is solvable (meaning that its derived series reaches $\{1\}$).
I was wondering what is the analog of the story in higher dimension (i.e. for equations of the form $P(x_1,\dots,x_n)=0$. I would naively expect algebraic group to show up…
I googled the main key words and found this presentation: on the last slide it is written that
the task at hand is to develop a
Galois theory of polynomials in two
variables
This convinced me to anyway ask the question
EDIT: the first "idea" I had
I first thought about the following strategy. Consider $P(x,y)=0$ as an polynomial equation in one variable $x$ with coefficients in the field $k(y)$ of rational functions in $y$, and consider its Galois group. But then we could do the opposite…what would happen?
Best Answer
(This should really be a comment I think, but I'm not highly rated enough to leave one, so please bear with me)
A Galois Theoretic condition for a polynomial in two variables to be solvable by radicals is found in the following paper: http://arxiv.org/abs/math/0305226. It seems to indicate that something similar can be done for higher variables. Perhaps I'll ask Jochen next time I see him about this.