What is the connection between discriminants and Galois groups?

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The discussion centers on the relationship between discriminants and Galois groups in the context of field theory and polynomial roots. It establishes that the discriminant locus D in k^n identifies polynomials with fewer than n distinct roots, influencing the degree of mappings between solution spaces S, X, and Y. The Galois group of the normal field extension is equated to the group of deck transformations of S, with a focus on obtaining the Galois group of a specific polynomial from the general case. The conversation highlights the significance of inertia groups in this geometric framework.

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  • Understanding of field theory and algebraic closures
  • Familiarity with polynomial discriminants and their implications
  • Knowledge of Galois theory and Galois groups
  • Experience with algebraic geometry concepts, particularly affine spaces
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  • Study the properties of polynomial discriminants in field extensions
  • Explore the concept of Galois representations and their applications
  • Investigate inertia groups and their role in Galois theory
  • Learn about the relationship between monodromy and Galois groups in algebraic contexts
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Mathematicians, algebraic geometers, and students of Galois theory seeking to deepen their understanding of the connections between discriminants and Galois groups in polynomial equations.

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consider a field k and the affine n space Y = k^n, whose coordinates a1,...,an parametrize monic polynomials over k. and over it consider the space S of solutions of those polynomials, i.e. S = {(a,x): where a = (a1,...,an) is a polynomial, and (x1,...,xn) are the roots of a lying in an algebraic closure K of k}.

thus S is a subset of k^n x K^n.

Let D in k^n be th zero locus of the discriminant formula of polynomials of degree n, so that a is in D if and only if there are fewer than n distinct roots of the polynomial a.

Then, since the roots of each a are ordered, the degree of the map S--->Y is n!, but there are fewer than n! preimages precisely of the points a of D.

Now consider the map S-->>X from S onto a subspace X of k^n x K, sending (a,x1,...,xn) to (a,x1). Thus X is the space of all pairs (a,x) where x is a root of a. thus we have maps S-->X-->Y where the first (left) map has degree (n-1)! and the second (right) map has degree n.

Both maps S-->Y and X-->Y have the same discriminant locus D in Y. Then map S-->Y is the normal closure of the map X-->Y, in the sense that this is true of the induced map via pulback of functions, on their function fields.

Moreover the group of deck transformations of S should presumably equal the Galois group of this normal field extension, and in the complex case. also the monodromy representation of the fundamental group.

Now in the general case, this Galois group should still be S(n), but what happens if we focus on one point a, i.e. one polynomial. It would be nice to obtain the galois group of this polynomial from the general one.

supose we take the ring k[X1,...,Xn] of the space Y, and the maximal ideal m corresponding to the point a and localize at it, getting k of course. upstairs we have the ring of S i guess, and can consider all maximal ideals of that ring which pull back to m, via the ring inclusion of k[Y] into k.

now what? I suppose we want to consider the galois representation of the big galois group, no wait. we want some subgroup of the general group S(n). Maybe we should consider those automorphisms of the field k(S) that what? preserve the ring k? or that preserve one of the ideals over m?

hmmmmm... how do we get the Galois group of a from this geometry? I think it has something to do with "inertia groups" for those who know this stuff.

This is the level of galois theory one gets from Lang, as contrasted with other books, like mine.
 
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