In this post we shall investigate an example which I am hoping will clarify some of the more subtle and potentially confusing behaviour in the theory of local fields (and its interaction with that of global fields).

Given a finite extension {L/K} of number fields, a prime {v} of {K} can extend to several different primes of {L}. If {L/K} is Galois, it is sufficiently symmetric that even though there may be several primes, {Gal(L/K)} acts transitively on them and each {\sigma \in Gal(L/K)} gives an isometry {L_{(w)} \rightarrow L_{(\sigma(w))}} (where we write {L_{(w)}} to mean the field {L} equipped with the norm coming from {w}). In particular, one can deduce that the ramification and inertia degree of each of these primes {w|v} are the same. However, if {L/K} is not Galois, the primes {w_1,...,w_k} over {v} could behave very differently from each other, and in general we do have the formula

\displaystyle e_1f_1+...+e_kf_k = [L:K]
but the {e_i} and {f_i} can vary as {i} varies.

Contrast this situation with that of local fields. If {L/K} is a finite extension of local fields, then each prime {v} of {K} extends to a unique prime of {L}, and {ef=[L:K]}. In particular, let us think about what happens if we take a global field {K} and complete it at a prime (obtaining something I’ll write as {K_v}). What happens to an extension {L} of {K}? If {L/K} is Galois, then we saw above that all the primes looked fairly similar, and the completions at all primes are isomorphic via the isometries we noted above.

When {L/K} is not Galois, things become complicated, and different completions of {L} are no longer (generally) isomorphic. Though it shouldn’t be if we think clearly enough about these things, this can be a source of great confusion. We are accustomed to writing things like {K(\sqrt[3]{2})} and assuming it means a well-defined thing (at least up to isomorphism of field extensions), but alas this no longer is so.

The easiest way to see this is probably to consider completion at the infinite place. It is true that as abstract fields (letting {\omega} be a 3rd root of unity) {{\mathbb Q}(\sqrt[3]{2}) \cong {\mathbb Q}(\omega \sqrt[3]{2})}, but if we complete with respect to the obvious archimedean norm (considering these things embedded in {{\mathbb C}}) one of the fields we get is {{\mathbb R}} and the other {{\mathbb C}}. Thinking about this more functorially, we had an abstract field {{\mathbb Q}[X]/(X^3-2)} and because it isn’t Galois, it is allowed to have different-looking archimedean places, and in fact it does: it has one real embedding and one pair of complex embeddings.

Exactly the same issue is a problem at the finite primes, except now life is in general much more complicated because the parameters {e} and {f} can vary so freely, so it is even more important to be very careful. Let us recycle the example above, but look at what happens at the prime {5}. It does not ramify, and splits into two primes, of inertial degrees 1 and 2. Why is this happening? Well, exactly as at the infinite place, there is one embedding into {{\mathbb Q}_5}, and a conjugate pair of embeddings into {{\mathbb Q}_{25}} (where by conjugate I mean there is an element of {G_{{\mathbb Q}_5}} that exchanges them). This corresponds to the fact that {X^3-2 = (X-3)(X^2+3X-1)} modulo {5}, and by Hensel’s lemma this factorisation lifts to a factorisation of {X^3-2} in {{\mathbb Q}_5}.

So what is the careful way to work out what is going on? As the above examples illustrate, if {L/K} an extension of number fields, {L=K(\alpha)} with {g} denoting the minimal polynomial of {\alpha} and {v} is a place of {K}, then the distinct places of {L} extending {v} are precisely the number of ways to embed {L} into an extension of {K_v}, modulo the action of {G_{K_v}}. In other words, the number of ways to realise {\alpha} (a generic root of {g}) as an element of an extension of {K_v}, modulo the {G_{K_v}} action. So actually this is just the number of irreducible factors of the polynomial {g} considered as an element of {K_v}.

Let’s write this more algebraically. Recall that {L=K[X]/(g(X))}, so

\displaystyle L\otimes_K K_v = K_v[X]/(g(X)) = \prod K_v[X](g_i(X)),
where {g=g_1...g_n}, where the {g_i} are distinct irreducible elements of {K_v[X]}. Note (being careful not to confuse what happens here with what happens with ramified primes when you pass to the residue field) that they are always actual irreducible polynomials, since {g} has distinct roots in {\bar{K}}, and of course each factor is a finite extension of a local field so gives a unique local field up to isomorphism (because primes of local fields extend uniquely).

So the upshot is that to see the different possible ways to localise your extension of global fields (i.e. the different places), it is essential to factorise the minimal polynomial of a generator of the global field considered as a polynomial in the appropriate local field (or at least find an irreducible factor), and specify which factor you are using (and of course by basic Galois theory this is equivalent to giving a {G_{K_v}}-orbit of embeddings {L \hookrightarrow \bar{K}_v}).

We finish with a short example, which I found somewhat confusing until I started writing this post. Consider the extension {{\mathbb Q}(\sqrt[3]{10})/{\mathbb Q}}. Let us try to compute the power of {3} dividing its discriminant. This is a scenario fairly similar to our examples above, except of course 3 is now ramifying. By Hensel’s lemma (and {10 \cong 1 \mod 9}), there is some {\alpha} a cube root of 10 in {{\mathbb Q}_3}, but there is no third root of unity, so in {{\mathbb Q}_3} we get {X^3-10 = (X-\alpha)(X^2 + \alpha X + \alpha^2)}. The only contribution to the different can come from the second factor {f(X)=X^2 + \alpha X + \alpha^2}, whose roots {\beta} generate the local ring of integers. Indeed a direct computation shows that {f'(\beta)} is a uniformiser, which tells us that the local different has valuation 1, and therefore (since the inertia degree is 1) only one power of {3} divides the discriminant of {{\mathbb Q}(\sqrt[3]{10})/{\mathbb Q}}. I suppose in general the moral is that it’s probably a bad idea to ever write anything like {{\mathbb Q}_3(\sqrt[3]{10})} unless it’s at least Galois.