This is the lemma we shall use in the proof of the reciprocity law, to reduce the cyclic case to the cyclotomic case:

Lemma 4 (Artin)Let be a cyclic extension of degree and a prime of unramified in . Then we can find a field , a subextension of , with such that in the lattice of fields

we have:

1. splits completely in

2. , so that is cyclotomic

3. is unramified in

Moreover, we can choose such that it is divisible only by arbitrarily large primes.

The proof of this will use the previous number-theory lemmas and the basic tools of Galois theory.

So, first of all, we know that is a subextension of some . We don’t know what is, but pretend we do, and will start carrying out the proof. As we do so, we will learn more and more about what has to be like, and eventually choose it.

** 2.1. The lattice condition **

The first thing we will want is for and to satisfy , so there is a lattice

The reason is that we then have a nice description of (as a product of Galois groups ), and we will define as a fixed field of a certain subgroup.

I claim that this condition will happen once is divisible only by sufficiently large primes. For in this case, is unramified everywhere over (because has fixed, finite ramification and the cyclotomic part is ramified precisely at the primes dividing ). It is a theorem, depending on the Minkowski bound for the discriminant, that any number field different from is ramified somewhere, so it follows that . Now in the lattice of fields

we have that , the last equality in view of the lattice of fields

However, we also know (by the same reasoning) that . It follows that , as claimed, when is not divisible by the primes of ramified in .

** 2.2. Unramification in **

Since we are going to define as a subfield of , this one is easy: just choose not divisible by the rational prime prolongs.

** 2.3. Complete splitting **

We are now going to choose a subgroup , whose fixed field will be .

Now since splits completely in , we know that we are going to require to contain the decomposition group of some extension of it, i.e. the group generated by the Frobenius elements

This, however, is still not enough. We need and .

** 2.4. **

So is the fixed field of the subgroup of generated by and by . If we want , we will need

We can arrange this if we choose a generator of and some and add to , so that is generated by two elements. So do this. We will choose and later.

** 2.5. is cyclotomic; choice of **

This is the most subtle part of the proof, and where the number theory developed earlier will come in handy. We will prove that , so that is a subextension of a cyclotomic extension.

The field is the fixed field of the intersection of and , and we therefore want this intersection to be the trivial subgroup. We have defined to be generated by

where is yet to be chosen. Suppose this subgroup intersected with ; we’d then be able to write for some ,

which is to say that the cyclic groups in generated by (since this is how acts on ) and intersect.

OK. Now it’s starting to become clear how this intersects (groan) with our previous number-theoretical shenanigans.

So, let . Choose divisible only by super large primes and such that has order in dividing , and such that there also exists of order dividing such that the cyclic groups generated by respectively are disjoint. We can do this by yesterday’s lemma.

We have now chosen . For take it to correspond to . I claim now that , which will prove the last remaining claim in Artin’s lemma.

Indeed, if we had some element in the intersection, then first of all we’d be able to write

as mentioned earlier; this is because we’d have an expression for some in the intersection of the two groups. This implies that by independence and are consequently divisible by . But then, because has order , it follows that , and too. Thus the intersection of the two subgroups is the trivial subgroup as claimed. This proves Artin’s lemma.

June 22, 2010 at 10:38 am

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