Localization, the Different, and the Discriminant

As I said before, the different and the discriminant are stable under localization, which is why it’s possible to use local techniques to show that a prime integer ramifies in the ring of integers of a field extension K iff it divides the discriminant d(K).

As usual, I’m going to assume the ground field is Q. This isn’t really necessary – the calculations extend trivially to any global field whose ring of integers is a PID, and only slightly less trivially to a general global field – but I think it makes a lot of calculations clearer.

First, the different. To show that the different D is stable under localization, it’s enough to show this about the inverse different I. It’s enough to show that if S is any multiplicatively closed subset of Z then tr(xy) is in S^(-1)Z for all y in S^(-1)O(K) iff x is in S^(-1)I. But x is in S^(-1)I iff for each y in O(K), tr(sxy) is in Z for some s in S, which holds iff tr(xy) is in S^(-1)Z for all y in O(K), which holds iff for each y in S^(-1)O(K), tr(sxy) is in S^(-1)Z for some s in S, which holds iff tr(xy) is in S^(-1)Z for all y in S^(-1)O(K).

Now, completion. Let R be a discrete valuation ring in Q (i.e. Z localized at some prime ideal (p)) and S be its integral closure in K. Using primes to denote completion, we want to have D’ = DS’, i.e. IS’ = I’ = {x in K’: tr(xy) is in R’ for all y in S’}.

I’ can’t be bigger than IS’, because otherwise it would intersect S in a bigger ideal than I, and for some x not in S, tr(xy) would be in R’ for all y in S’, i.e. tr(xy) would be in R for all y in S, a contradiction. It can’t be smaller, because it contains every element of I. To see why, note that for a fixed x, tr(xy) is a continuous function from K’ to Q(p), so for each y in S’, let y(i) be a sequence in S approaching y, and note that tr(xy(i)) is in R for every i, so that tr(xy) is in R’.

Second, the discriminant. Relative to S^(-1)Z, the ring of integers of K is just S^(-1)O(K), so any integral basis for O(K) over Z can also serve as an integral basis for S^(-1)O(K) over S^(-1)Z. This is enough to show that the ideal generated by the discriminant is stable under localization.

For completion, let’s use the same notation as above. It’s enough to show that if {x1, …, x(n)} is an integral basis for S over R, then it’s also an integral basis for S’ over R’. But {x1, …, x(n)} is a basis for K’ over Q(p), so every element of K can be written as a1x1 + … + a(n)x(n) where the a(i)’s are in Q(p). If the element is in S’ but some a(i) is not in R’ = Z(p), then by subtracting, we get an element b1x1 + … + b(n)x(n) in S’ where each b(i) is in Q and at least one is not in R. That element will then be in S, contradicting the assumption that {x1, …, x(n)}.

Third, we need to show that when K(P) is an extension of Q(p) with different D, K is ramified iff its unique prime divides D. But if K is unramified, then it’s generated by some primitive kth root of unity, z(k), where k is not divisible by p. To show that D = S = I, it’s enough to show that tr(a/p) is not in Z(p) for some a in K(P), or that tr(a) is not divisible by p.

If tr(1) = phi(k) is not divisible by p, then we are done. Otherwise, let us look at l, the maximal squarefree divisor of k. tr(z(l)) is phi(k)/phi(l) = k/l times the sum of the distinct conjugates of z(l). Every conjugate of z(l) can be uniquely written as the product consisting of one conjugate of each z(q) where q divides l, so the sum of the distinct conjugates of z(l) is the product of the sums of the distinct conjugates of each z(q). But that latter sum can be shown to be -1, so tr(z(l)) = (+/-)k/l, which doesn’t divide p.

Conversely, if K(P) is ramified, then it’s enough to show tr(1/q) is in Z(p) for every prime element q. This is because for a fixed q0, every x/q0 with x in O(K)(P) is either in O(K)(P) or of the form 1/q for some q. Ramification can only come from taking some root of a prime element in Z(p), say p‘. But tr(p‘^(1/e)) = (p‘^(1/e))(1 + z(e) + z(e)^2 + … + z(e)^(e-1)) = 0, and elements of that form generate O(K)(P). But tr(p‘^(1/e)/p) = tr(p‘^(1/e))/p = 0. This shows that the different is divisible not only by P, but even by P^(e-1); if (and only if) p divides e, it’s even divisible by P^e because tr(1/p) = ef/p.

And fourth, we need to show that N(D(K(P)) = d(K(P)) where the discriminant is taken as an ideal. If D is the unit ideal, then the extension is unramified, and the roots of unity are a suitable integral basis. But {z(k)^i: gcd(i, k) = 1 and 0 < i < k} has a fixed discriminant divisible only by primes dividing k. As p doesn’t divide k, it follows that d(K(P)) is the unit ideal as well.

In the ramified case, we can split K(P) via an intermediate field F containing all roots of unity of K(P) of order coprime to p. Then K(P)/F is generated by element of the form p‘^(1/e), and d(K(P)/Q(p)) is generated from matrices of bases of the form {(z(k)^i1)p‘^(i2/e): gcd(i1, k) = 1, 0 < i1 < k, and 0 <= i2 < e}. Replacing each p‘^(i2/e) with a 1 corresponds to dividing the discriminant by a factor of p^(e-1), which takes care of the case when p doesn’t divide e (which is called tamely ramified, since it’s not so problematic as the other case).

In the wildly ramified case, we can use the same trick with F to get an extension generated entirely by p^kth roots of unity, where k > 1 (when k = 1, the ramification is tame). In that case the discriminant ideal is generated by p^((pkk – 1)p^(k-1)), and the different can be shown by an explicit calculation to be the same power of P.

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