Hasse invariant of an algebra

Let K be a local field with valuation v and D a K-algebra. We may assume D is a division algebra with centre K of degree n. The valuation v can be extended to D, for example by extending it compatibly to each commutative subfield of D: the value group of this valuation is (1/n)Z.^[1]

There is a commutative subfield L of D which is unramified over K, and D splits over L.^[2] The field L is not unique but all such extensions are conjugate by the Skolem–Noether theorem, which further shows that any automorphism of L is induced by a conjugation in D. Take γ in D such that conjugation by γ induces the Frobenius automorphism of L/K and let v(γ) = k/n. Then k/n modulo 1 is the Hasse invariant of D. It depends only on the Brauer class of D.^[3]

The Hasse invariant is thus a map defined on the Brauer group of a local field K to the divisible group Q/Z.^[3][4] Every class in the Brauer group is represented by a class in the Brauer group of an unramified extension of L/K of degree n,^[5] which by the Grunwald–Wang theorem and the Albert–Brauer–Hasse–Noether theorem we may take to be a cyclic algebra (L,φ,π^k) for some k mod n, where φ is the Frobenius map and π is a uniformiser.^[6] The invariant map attaches the element k/n mod 1 to the class. This exhibits the invariant map as a homomorphism

${\displaystyle {\underset {L/K}{\operatorname {inv} }}:\operatorname {Br} (L/K)\rightarrow \mathbb {Q} /\mathbb {Z} .}$

The invariant map extends to Br(K) by representing each class by some element of Br(L/K) as above.^[3][4]

For a non-Archimedean local field, the invariant map is a group isomorphism.^[3][7]

In the case of the field R of real numbers, there are two Brauer classes, represented by the algebra R itself and the quaternion algebra H.^[8] It is convenient to assign invariant zero to the class of R and invariant 1/2 modulo 1 to the quaternion class.

In the case of the field C of complex numbers, the only Brauer class is the trivial one, with invariant zero.^[9]