Following Scott’s post, here is another post about some basic number theory that I haven’t seen written well in one place. First, a list of puzzles. In these puzzles, “polynomial” always means “polynomial with integer coefficients”:
Everyone of these sounds like a problem about fields but, with a little bit of knowledge, they are actually problems about finite group actions. By the way, although every problem is solvable as written, they become easier if you allow yourself a finite number of exceptions instead requiring them to actually work for all primes, and I will only explain how to solve them with this proviso. I do assume that you know the basics of Galois theory.
Here is the main idea. Let f be a polynomial with integer coefficients and without repeated roots. To make my life simpler, I’ll assume that f has leading coefficient one, so that the roots of f are algebraic integers, although actually all I need to do in the more general case is to exclude those primes that divide the leading term of f. Let K be the splitting field of f, that is to say, the smallest extension of in which f has all of its roots. Let the roots of f be , …, . Define R to be the subring of K generated (over ) by the roots of f. Let G be the Galois group of K over . Then G permutes the and hence acts on R. (Note for experts: the standard thing to do is to take the integral closure of inside K, rather than use R. This is undoubtedly the right thing to do, but I can avoid a lot of technical explanation by avoiding it.)
By definition, an -point of is a ring homomorphism from R to the algebraic closure of . (Yes, we are fixing an algebraic closure of .) I think that this is far too complicated a name for this idea so, in this blog post, I will call such a homomorphism a -widget. If I really need to be explicit, I’ll call it a -widget of f.
Now, is an integer (since it is Galois invariant) and is nonzero (since we assumed that f doesn’t have repeated roots). There are only finitely many primes which divide ; for any prime p which does not divide we know that f continues to have distinct roots after reduction modulo p. I’ll call the primes dividing the bad primes, and call all the others the good primes. For those trying to make the translation to standard terminology, “good” implies, but is not implied by, “unramified”.
What’s good about the good primes? Well, for one thing, let be any p-widget. If is good then for . Why is this? Well, is a ring homomorphism, so . But the right hand side is just the image in of an integer which we assumed not divisible by p, so every factor on the right must be nonzero. Therefore, if p is a good prime, then G acts faithfully on -widgets. It turns out, although I don’t see an easy proof, that G also acts transitively. I believe that this latter fact is due to Frobenius.
Now, there is also another action on -widgets. Instead of acting on the ring R, we can act on the field . Specifically, there is a map from to itself, called the Frobenius and we can turn one -widget into another by post composing with the Frobenius. We’ll denote the Frobenius by . If p is a good prime, and is a -widget, then for some unique g in G, as we saw that G acts transitively and faithfully on the -widgets. If, instead of , we chose some other -widget , this would change the resulting value of to . In other words, the conjugacy class of g is determined by uniquely by p, and is known as the Frobenius class of p and denoted .
Now, here are the parts that I think could be laid out more explicitly in the books I’ve read. As those of you who are familiar with Galois theory know, the irreducible factors of f (over the rationals) correspond (in number and size) to the G-orbits in the set . Now, similarly, F generates the Galois group of any finite extension of . Thus, the irreducible factors of f over correspond to the orbits of on the same set. Note that, although the actual orbits depend on the choice of a representative for , their number and size do not.
It shouldn’t be too hard to convince yourself of the following point, as well. If K is both the splitting field for f and for another polynomial g, and p is good with respect to both f and g, then is the same element of G whether it is computed using f or g. (I will concede that this is one thing which is easier when one uses the standard approach in terms of integral closure.)
You now know enough to solve puzzles (1), (2) and (4). For example, puzzle (1) asks us to find a group G acting on a finite set S such that every element of G fixes a point of S, but no point of S is fixed by all of G. The simplest example is where S has three orbits, each stabilized by a different 2-element subgroup of G. Technically, to solve the puzzle we must actually find a field extension with this Galois group — the easiest solution is to take so , which has this Galois group. Puzzle (2) is only a little harder. For puzzle (4), it will help you to have the following hint: f and g should have the same splitting field.
To get puzzle (3), you need some elementary group theory (which I leave to you) plus a result that says that every conjugacy class of G occurs as for some p. This is
Chebotarev’s Density Theorem : Let C be a conjugacy class in G. Then the density of primes p such that is .
Finally, let me make one more connection to standard terminology. The ring is a so-called Dedekind domain, so every ideal factors uniquely as a product of prime ideals. (Again, I am inverting in order to smear over my failure to take integral closures.) For a good prime p, the prime ideal factors of p correspond to the orbits of the -widgets under the action of Frobenius. If is a -widget, the corresponding prime ideal is the kernel of . If I had chosen R correctly to begin with, then R itself would be a Dedekind domain and I could define the following standard objects for you: if is the kernel of a -widget , then the decomposition group of is the stabilizer of and the inertia group is the stabilizer of .