Math 254A, UC Berkeley, Fall 2001 (Kedlaya) Addendum on the Relative Discriminant October 31, 2001

A lot of you had trouble with the homework problems on the relative discriminant last week. One big problem was typos on the problem set. The correct statement of problem 2 was: If L/K is a number field, prove that the ideal (Disc(L)/ Disc(K)^{[L:K]}) is the ideal generated by the absolute norm of the relative discriminant Disc(L/K). (I forgot the factor of [L : K] in the exponent. Mea maxima culpa.)

One approach to computing this is the one taken by Neukirch in Section 3.2, using the different. I had in mind a more direct argument, something like the following.

It is true (but I didn’t say this explicitly in class, and should have) that trace, norm and the discriminant can be computed two different ways for local fields just like for number fields. Namely, if L/K is an extension of finite extensions of ℚ_{p} of degree n, there will be exactly n embeddings τ_{1}, …, τ_{n} of L into \overline{K} (that much is true for any separable extension), and we can write

Trace_{L/K}(α) = ∑_{i} τ_{i}(α), Norm_{L/K}(α) = ∏_{i} τ_{i}(α), Disc(α_{1}, …, α_{n}) = (det(τ_{i}(α_{j}))_{i,j=1}^{n})^{2};

the equivalence of these definitions with the original definitions works the same way as in the number field case.

Using problem 1 and the relationship between the absolute discriminant and Neukirch’s form of the relative discriminant, problem 2 reduces to the following: if M/L and L/K are extensions of finite extensions of ℚ_{p}, then

Disc(M/K) = Disc(L/K)^{[M:K]} Norm_{L/K} Disc(M/L),  (1)

where Norm_{L/K} sends an ideal of 𝔬_{L} generated by α to the ideal (Norm_{L/K}(α)). Here’s one way to prove this.

Put m = [M : L] and n = [L : K], choose a basis α_{1}, …, α_{m} of 𝔬_{M} over 𝔬_{L} and a basis β_{1}, …, β_{m} of 𝔬_{L} over 𝔬_{K}, and choose an isomorphism f : \overline{L} → \overline{K} of fields once and for all. (Note: \overline{L} is a field together with a fixed map of L into it, so I’ll push elements of L into it without comment. Likewise, \overline{K} is a field together with a fixed map of K into it.)

Let τ_{1}, …, τ_{m} be the embeddings of M into \overline{L} over L, and let σ_{1}, …, σ_{n} be the embeddings of L into \overline{K} over K. Choose automorphisms ρ_{1}, …, ρ_{n} of \overline{K} such that ρ_{i} ∘ f = σ_{i} for each i. Now ρ_{i} ∘ τ_{j} gives all of the embeddings of M into \overline{K} over K, so using the basis α_{k}β_{l} of 𝔬_{M} over 𝔬_{K} (for k = 1, …, m and l = 1, …, n), I can write Disc(M/K) = (det C)^{2}, where the matrix C has rows and columns indexed by ordered pairs and is given by

C_{(i,j)(k,l)} = ρ_{i}(τ_{j}((α_{k}β_{l}))) = ρ_{i}(τ_{j}(α_{k})) · ρ_{i}(β_{l})).  (2)

(The τ_{j} dropped out in the second term because we have a fixed identification of L as a subfield of \overline{L}, and τ_{j} is L-linear.)

I’d like to factor this determinant using the identity det(A ⊗ B) = (det A)^{n}(det B)^{m} for the determinant of the tensor product of an m × m matrix A and an n × n matrix B. (The tensor product of A and B is the matrix with rows and columns indexed by pairs (i, j) and (A ⊗ B)_{(i,j)(k,l)} = A_{i}kB_{jl}. Why it’s called that is left to the reader.) Unfortunately, I haven’t quite got that yet, but I can modify the determinant in question to bring this about.

For ease of notation, let T be the matrix with T_{ij} = τ_{i}(α_{j}), so that C_{(i,j)(k,l)} = ρ_{i}(M_{jk})ρ_{i}(β_{l}). Now define the matrix D by

D_{(i,j)(k,l)} = \begin{cases}(T ρ_{i}(T)^{−1})_{jl} & i = k \\ 0 & i ≠ k. \end{cases}

We have

(DC)_{(i,j)(k,l)} = ∑_{(r,s)} D_{(i,j)(r,s)}C_{(r,s)(k,l)}

= ∑_{s} D_{(i,j)(i,s)}C_{(i,s)(k,l)}

= ∑_{s}(T ρ_{i}(T)^{−1})_{js}ρ_{i}(T)_{jk}ρ_{i}(β_{l})

= T_{jk}ρ_{i}(β_{l}).

That is, DC is the tensor product of the matrices M and N, where N_{il} = ρ_{i}(β_{l}). Since D is a block matrix, we can compute its determinant as

det(D) = ∏_{i} \frac{det T}{det ρ_{i}(T)}.

We conclude

det C = (det T^{−m} ∏_{i} det ρ_{i}(T)) det T^{m} det N^{m})

= (det N)^{m} ∏_{i} ρ_{i}(det T)

= (det N)^{m} Norm_{L/K}(det T).

To conclude, observe that (f^{−1}(det T))^{2} = (Disc(M/L)) and (det N)^{2} = (Disc(L/K)).