A “common slot” counterexample in degree 3

Notation: For a, b nonzero elements in a field F containing a primitive cube root of unity ω, the symbol (a, b) denotes the element of the Brauer group of F represented by the F -algebra generated by elements α, β subject to

α^{3} = a,  β^{3} = b,  βα = ωαβ.

Let a_{1}, b_{1}, a_{2} ∈ F^{×}. If there exist x, y ∈ F^{×} such that

(a_{1}, b_{1}) = (a_{1}, x) + (a_{1}, y), (a_{1}, x) = −(a_{2}, x) and (a_{1}, y) = (a_{2}, y), (∗)

then the additivity of symbols yields (a_{1}, b_{1}) = (a_{2}, x^{−1}y). However, the next example shows that when (a_{1}, b_{1}) is split by F ( ^{3} \sqrt{a_{2}}), there need not exist elements x, y satisfying (∗).

Example: A global field F containing a primitive cube root of unity and elements a_{1}, b_{1}, a_{2}, b_{2} such that (a_{1}, b_{1}) = (a_{2}, b_{2}), but no couple of elements x, y satisfying (∗). In particular (taking x = 1), the field F does not contain any element y such that

(a_{1}, b_{1}) = (a_{1}, y) = (a_{2}, y) = (a_{2}, b_{2}).

Let F = 𝔽_{7}(t), where t is an indeterminate, a_{1} = t and a_{2} = t(1 − t). Note that (a_{1}, a_{2}) = 0. Therefore, for all place v of F, the local invariant (a_{1}, a_{2})_{v} is trivial. It follows that in the completion F_{v} of F at v we have either a_{1} ∈ F_{v}^{×3} or a_{1} ≡ a_{2} mod F_{v}^{×3} or a_{1} ≡ a_{2}^{2} mod F_{v}^{×3} or a_{2} ∈ F_{v}^{×3}, since the (generalized) Hilbert symbol ( , )_{v}: (F_{v}^{×}/F_{v}^{×3}) × (F_{v}^{×}/F_{v}^{×3}) → \frac{1}{3}ℤ/ℤ is a nondegenerate alternating pairing.

Consider in particular v_{1} the t-adic place and v_{2} the (t + 3)-adic place. Since a_{1}, a_{2} are uniformizing parameters at v_{1}, we have a_{1}, a_{2} ∉ F_{v_{1}}^{×3}; but a_{1} ≡ a_{2} mod F_{v_{1}}^{×3}. On the other hand, a_{1} and a_{2} have non-cube residues at v_{2}, hence a_{1}, a_{2} ∉ F_{v_{2}}^{×3} but a_{1} ≡ a_{2}^{−1} mod F_{v_{2}}^{×3}.

Let now A be the central simple F -algebra with local invariants 1/3 at v_{1}, 2/3 at v_{2} and 0 everywhere else. If v is a place of F where a_{1} ∈ F_{v}^{×3}, then v ≠ v_{1}, v_{2} hence [A]_{v} = 0. It follows that A is split by F ( ^{3} \sqrt{a_{1}}), hence we may find b_{1} ∈ F^{×} such that [A] = (a_{1}, b_{1}) in the Brauer group of F. Similarly, A is split by F ( ^{3} \sqrt{a_{2}}) hence we may find b_{2} ∈ F^{×} such that [A] = (a_{2}, b_{2}); thus,

(a_{1}, b_{1}) = (a_{2}, b_{2}).

Suppose now x, y ∈ F^{×} satisfy (∗). Since a_{1} ≡ a_{2} mod F_{v_{1}}^{×3}, the relation (a_{1}, x)_{v_{1}} = −(a_{2}, x)_{v_{1}} implies (a_{1}, x)_{v_{1}} = 0. On the other hand, since a_{1} ≡ a_{2}^{−1} mod F_{v_{2}}^{×3}, it follows from (a_{1}, y)_{v_{2}} = (a_{2}, y)_{v_{2}} that (a_{1}, y)_{v_{2}} = 0, hence (a_{1}, x)_{v_{2}} = (a_{1}, b_{1})_{v_{2}} = 2/3.

For v ≠ v_{1}, v_{2}, we consider four cases, according to the relation between a_{1} and a_{2} in the group of cube classes:

• if a_{1} ∈ F_{v}^{×3}, then clearly (a_{1}, x)_{v} = 0.

• if a_{1} ≡ a_{2} mod F_{v}^{×3}, then (a_{1}, x)_{v} = 0 as for v = v_{1} above.

• if a_{1} ≡ a_{2}^{−1} mod F_{v}^{×3}, then (a_{1}, x)_{v} = (a_{1}, b_{1})_{v} as for v = v_{2} above, hence (a_{1}, x)_{v} = 0.

• if a_{2} ∈ F_{v}^{×3}, then (a_{1}, x)_{v} = 0 follows from (a_{1}, x) = (a_{2}, x^{−1}). Thus, the invariants of (a_{1}, x) are:

(a_{1}, x)_{v_{2}} = 2/3, and (a_{1}, x)_{v} = 0 for v ≠ v_{2},

a contradiction to the reciprocity law.

Jean-Pierre Tignol, June 1996.