So my old proof isn’t really working on wordpress for some reason, so I’ve taken it as a sign to do it in a different way. This method is far more complicated than the old way (in which I just call upon some theorems and look at orders and then am done), but I think it better gets at what is going on.
Anyways, since we were on the topic of Galois theory, here is a fact I found astonishing the first time I heard it (maybe it is quite obvious to you). Every finite group arises as the Galois group of a field extension, moreover we can choose the two fields to be number fields. Recall that a number field is just a subset of the complex numbers that is algebraic over .
Proof: Let G be a finite group. Then by Cayley’s theorem for some n. But then there is a prime p, such that n<p, meaning . Let’s find a Galois extension such that .
Our natural choice is the splitting field of , (note the sarcasm) where we chose our prime to be of the form . Thus we have that , it is irreducible by Eisenstein’s criterion, and it has exactly 2 roots in . Thus, . The details of this just amounts to playing around with cycles.
Now we can just invoke the Fundamental Theorem of Galois Theory. We have the subgroup which corresponds to the fixed field, say L, where and , which is what we wanted.
Now that this is typed out, I think the other way is better since I didn’t have to skip over the cycle argument. This method is just as mysterious overall, but in fact the other is probably more believable, since I only use things that are standard.
Lastly, let me just clear up some mystery about at least. The way we know it has real roots is that alternates signs between pairs , then , all the way to . Thus by the IVT, there is a real root between each of those. Thus there are 2 complex roots left over. We can even locate them to be near if p is sufficiently large by Rouche’s Theorem. Also, is irreducible iff is irreducible, and and here it is clearly seen that p divides all the coefficients except the constant term and does not divide the highest term (Eisenstein).
Oh, and while I’m at it, I might as well lead you in the right direction if you want to check that . By order, we know that (at least there is an isomorphic copy in there). So let the copy of act on . The action is transitive, so there is a 2-cycle, and since p divides the order there is a p-cycle. These three things should get you there (that it is inside , that there is a 2-cycle, and that there is a p-cycle).
So although they exist, they aren’t always the easiest to find. In fact, if we want our base field to be , then this is known as the “Inverse Galois Problem” and is still open. Some cases have been resolved. For instance, it is known that every finite solvable group arises as a Galois group of an algebraic extension of .