The Sequence

\xymatrix{0 \ar[r] & Br(k) \ar[r] & ⨁_{v} Br(k_{v}) \ar[r] & ℚ/ℤ \ar[r] & 0}

Let k be a number field. Finding a primitive solution over ℤ (or ℚ) of a homogenous polynomial over ℤ (or ℚ) relates via the norm map to the injectivity of \xymatrix{Br(k) \ar[r] & ⨁_{v} Br(k)} (which is the subject of the Brauer Hasse Noether theorem).

Example: Consider the polynomial ax^{2} + by^{2} + cz^{2} = 0 with a, b, c ∈ ℤ a, b, c ≠ 0. If we set z = 1 and divide by a we get x^{2} + \frac{b}{a}y^{2} = −\frac{c}{a}. Now

consider the field extension K = ℚ(\sqrt{−\frac{b}{a}}) over ℚ. The norm map for this ex-

tension is given by N(α + β\sqrt{−\frac{b}{a}}) = α^{2} + \frac{b}{a}β^{2} α, β ∈ ℚ. So the polynomial has a (primitive) solution with z ≠ 0 if and only if −\frac{c}{a} is a norm. From the Brauer group sequence one gets an injective map

\xymatrix{ℚ^{∗}/NK^{∗} \ar[r] & ∏_{l∪∞} ℚ_{l}/Nℚ_{l}^{∗}}  (later)

.

So one can check if −\frac{c}{a} is a norm in ℚ_{l} for every l and the injectivity allows a conclusion for ℚ. Explicitely, for the solutions this means that a tripel (α, β, 1) is a solution in ℚ if and only if it is a solution in all ℚ_{l} (or equivalently in all ℤ/l^{n}ℤ).

In this example one needs a special norm map which fits to the situation. The field extension was constructed the way that it fits. But if one just takes a field extension, one gets a norm map from which one can deduce a polynomial on which again one can apply this local-global method.

Now we want to relate the norm map of a cyclic extension L/k of a number field to the Brauer group sequence. For this we need a map

\xymatrix{k^{∗}/NL^{∗} \ar[r] & Br(k)}

But first:

Tate cohomology

Let G be a cyclic group and A, B and C be G-modules. Consider the functor [figure: A; A/(G − 1)A = A_{G}] Its left derived functor gives homology H_{i}(G, A).

One glues its long exact sequence with the one of the usual group cohomology

via a norm map \xymatrix{A_{G} \ar[r]^{N_{A}} & A^{G}} . From an exact sequence

\xymatrix{0 \ar[r] & A \ar[r] & B \ar[r] & C \ar[r] & 0}

we get

\xymatrix{ &  & Ĥ^{0}(G, A) & Ĥ^{0}(G, B) & Ĥ^{0}(G, C) \ar@{-->}[dr] \\  & 0 \ar[r] & A^{G} \ar[r] \ar[u] & B^{G} \ar[r] \ar[u] & C^{G} \ar[r] \ar[u] & H^{1}(G, A) \ar[r] & … \\ … \ar[r] & H_{1}(G, A) \ar@{-->}[dr] \ar[r] & A_{G} \ar[r] \ar[u]^{N_{A}} & B_{G} \ar[r] \ar[u]^{N_{B}} & C_{G} \ar[r] \ar[u]^{N_{C}} & 0 \\  &  & Ĥ_{0}(G, A) \ar[u] & Ĥ_{0}(G, B) \ar[u] & Ĥ_{0}(G, C) \ar[u]}

The Tate cohomology groups are then defined to be

Ĥ^{n}(G, A) = H^{n}(G, A)  n ≥ 1
Ĥ^{0}(G, A) = coker(N_{A})
Ĥ^{−}1(G, A) = ker(N_{A}) = Ĥ_{0}(G, A)
Ĥ^{−n}(G, A) = H_{n−1}(G, A)  n ≥ 2

. Now the Brauer group of k (still a number field) is Br(k) = H^{2}(Gal(k̄/k), k̄^{∗}). If we have a tower of Galois extensions k̄ ⊇ L ⊇ k where L is cyclic over k , and

G = Gal(k̄/k), I = Gal(k̄/L) and Γ = G/I

then from the Hochschild Serre spectral sequence

E_{2}^{r,s} = H^{r}(Γ, H^{s}(I, k̄^{∗})) ⇒ H^{r+s}(G, k̄^{∗})

we obtain a sequence of lower terms

\xymatrix{H^{0}(Γ, H^{1}(I, k̄^{∗})) \ar[r] & H^{2}(Γ, L^{∗}) \ar[r] & ker(H^{2}(G, k̄^{∗}) ⟶ H^{2}(I, k̄^{∗})) \ar[r] & H^{1}(Γ, H^{1}(I, k̄^{∗}))}

.

Hilbert 90 implies H^{2}(Γ, H^{1}(I, k̄^{∗})) = 0 = H^{1}(Γ, H^{1}(I, k̄^{∗})) so one gets an injection

\xymatrix{H^{2}(Γ, L^{∗}) \ar[r] & Br(k)}

For an arbitrary cyclic group Γ with generator s and a Γ-module A one has a simple projective resolution of A. Namely one uses the resolution of ℤ

\xymatrix{… \ar[r]^{σ−1} & ℤ[Γ] \ar[r]^{N} & ℤ[Γ] \ar[r]^{σ−1} & ℤ \ar[r] & 0}

and tensors with − ⨂_{ℤ[Γ]} A for homology and takes Hom_{ℤ[Γ]}(−, A) for cohomology. We then can see that Ĥ^{n}(Γ, A) ≅ Ĥ^{n+2}(G, A) for all n ∈ ℤ.

For n = 0 one obtains

A^{Γ}/NA ≅ Ĥ^{0}(Γ, A) ≅ Ĥ^{2}(Γ, A) = H^{2}(Γ, A)

In our situation we have L^{∗} for A and Gal(L/k) as Γ. So this gives us an injection

\xymatrix{k^{∗}/NL^{∗} \ar[r] & Br(k)}

We want to have the same map for the local fields of k. First we note that for two places w and w′ over v, NL_{w}^{∗} = NL_{w′}^{∗} because they are conjugate under Gal(L/k)(= Γ). Then since Gal(L_{w}/k_{v}) ⊆ Gal(L/k) the group Gal(L_{w}/k_{v}) is also cyclic so the same way as above we get

\xymatrix{k_{v}^{∗}/NL_{w}^{∗} \ar[r] & Br(k_{v})}

And finaly putting everything together we obtain a commutative diagram:

\xymatrix{k^{∗}/NL^{∗} \ar@{^{(}->}[r] \ar[d] & Br(k) \ar[d] \\ ∏_{v} k_{v}^{∗}/NL_{w}^{∗} \ar@{^{(}->}[r] & ⨁_{v} Br(k_{v})}

And the injectivity on the right side implies the one on the left side.

Next we want to take a closer look onto the Brauer group sequence itself.

Theorem. If k is a finite extension of ℚ_{p} then

Br(k) ≅ ℚ/ℤ .

Proof. We use that Br(k) ≅ H^{2}(Gal(k̄/k), k̄^{∗},). Now consider the tower k̄ ⊇ k^{nr} ⊆ k with the Galois groups G = Gal(k̄, /k,), I = Gal(k̄, /k^{nr},) and Γ = G/I (I ◁ G) where k^{nr} is the maximal unramified extension of k.

To compute the Brauer group of k it is sufficient to only consider the field k^{nr}. Namely one has Br(k) ≅ H^{2}(Γ, k^{nr^{∗}}): We consider the Hochschild Serre spectral sequence again

E_{2}^{r,s} = H^{r}(Γ, H^{s}(I, k̄^{∗})) ⇒ H^{r+s}(G, k̄^{∗})

Using that H^{1}(I, k̄^{∗}) = 0 (Hilbert 90) and that H^{2}(I, k̄^{∗}) = Br(k^{nr}) = 0 (k^{nr} is a C_{1} field) we find E_{2}^{n,1} = 0 and E^{n,2} = 0 for all n. So

\xymatrix{0 = E_{2}^{0,1} \ar[r]^{d} & E_{2}^{2,0} \ar[r]^{d} & 0}

gives

E_{2}^{2,0} ≅ E_{∞}^{2,0} ≅ F^{2}/F^{3} ≅ F^{2}

and also

\begin{matrix}E^{1,1} = 0 ⇒ F^{1} ≅ F^{2} & and \\ E^{0,2} = 0 ⇒ F^{0} ≅ F^{1} \end{matrix}

. Finaly we get E^{2} = F^{0} = F^{1} = F^{2} = E_{2}^{2,0} which means Br(k) = H^{2}(G, k̄^{∗}) = H^{2}(Γ, H^{0}(I.k̄^{∗})) = H^{2}(Γ, k^{nr^{∗}}).

Now we have a valuation

\xymatrix{k^{nr^{∗}} \ar@{->>}[r] & ℤ}

which is G equivariant. If we apply the cohomology functor we get a morphism

\xymatrix{Br(k) \ar[r]^{φ} & H^{2}(Γ, ℤ)}

. At last we want to see that H^{2}(Γ, ℤ) ≅ ℚ/ℤ and that φ is an isomorphism. From the corresponding short exact sequence we have

\xymatrix{H^{1}(Γ, ℚ) \ar[r] & H^{1}(Γ, ℚ/ℤ) \ar[r] & H^{2}(Γ, ℤ) \ar[r] & H^{2}(Γ, ℚ)}

But H^{i}(Γ, ℚ) = 0 ∀i > 0 and thus using that Γ ≅ ℤ̂ we obtain H^{2}(Γ, ℤ) ≅ H^{1}(ℤ̂, ℚ/ℤ). Since ℤ̂ = lim_{→}(ℤ/nℤ) this reduces to H^{2}(Γ, ℤ) ≅ lim_{→} H^{1}(ℤ/nℤ, ℚ/ℤ) ≅ lim_{→} ℤ/nℤ ≅ ℚ/ℤ. So we have a Morphism

\xymatrix{Br(k) \ar[r] & ℚ/ℤ} .

Consider the exact sequence of Γ-modules

\xymatrix{0 \ar[r] & A^{∗} \ar[r] & k^{nr^{∗}} \ar[r] & ℤ \ar[r] & 0}

where A is just the valuation ring of this valuation. This sequence splits so k^{nr^{∗}} ≅ A^{∗} ⊕ ℤ The cohomology functor is additiv, therefor we have an exact sequence

\xymatrix{0 \ar[r] & H^{2}(Γ, A^{∗}) \ar[r] & Br(k) \ar[r] & ℚ/ℤ \ar[r] & 0}

and it only remains to show, that H^{2}(Γ, A^{∗}) = 0.

From Milne’s Étale Cohomology I 4.5 + IV 1.6 we have an injection \xymatrix{Br(A) \ar@{^{(}->}[r] & Br(A/m)} . But A/m is a finite field and the Brauer group of a finite field is zero. So Br(A) = H^{2}(Γ, A^{∗}) must also be zero. Thus we obtain an isomorphism \xymatrix{Br(k) \ar[r]^{≃} & ℚ/ℤ} .

□

Using the fact that Br(ℝ) = ℤ/2ℤ and Br(ℂ) = 0 the sequence rewrites as

\xymatrix{0 \ar[r] & Br(k) \ar[r] & ⨁_{l} ℚ/ℤ ⊕ ⨁ ℤ/2Z \ar[r] & ℚ/ℤ \ar[r] & 0}

where the last map is given by

\xymatrix{(a_{1}, …; b_{1}, …, b_{n}) \ar[r] & Σa_{i} + Σb_{i}} .