Hilbert symbol

The following three properties follow directly from the definition, by choosing suitable solutions of the diophantine equation above:

• If a is a square, then (a, b) = 1 for all b.
• For all a,b in K^×, (a, b) = (b, a).
• For any a in K^× such that a−1 is also in K^×, we have (a, 1−a) = 1.

The (bi)multiplicativity, i.e.,

(a, b₁b₂) = (a, b₁)·(a, b₂)

for any a, b₁ and b₂ in K^× is, however, more difficult to prove, and requires the development of local class field theory.

The third property shows that the Hilbert symbol is an example of a Steinberg symbol and thus factors over the second Milnor K-group ${\displaystyle K_{2}^{M}(K)}$, which is by definition

K^× ⊗ K^× / (a ⊗ (1−a), a ∈ K^× \ {1})

By the first property it even factors over ${\displaystyle K_{2}^{M}(K)/2}$. This is the first step towards the Milnor conjecture.

The Hilbert symbol can also be used to denote the central simple algebra over K with basis 1,i,j,k and multiplication rules ${\displaystyle i^{2}=a}$, ${\displaystyle j^{2}=b}$, ${\displaystyle ij=-ji=k}$. In this case the algebra represents an element of order 2 in the Brauer group of K, which is identified with -1 if it is a division algebra and +1 if it is isomorphic to the algebra of 2 by 2 matrices.

For a place v of the rational number field and rational numbers a, b we let (a, b)_v denote the value of the Hilbert symbol in the corresponding completion Q_v. As usual, if v is the valuation attached to a prime number p then the corresponding completion is the p-adic field and if v is the infinite place then the completion is the real number field.

Over the reals, (a, b)_∞ is +1 if at least one of a or b is positive, and −1 if both are negative.

Over the p-adics with p odd, writing ${\displaystyle a=p^{\alpha }u}$ and ${\displaystyle b=p^{\beta }v}$, where u and v are integers coprime to p, we have

${\displaystyle (a,b)_{p}=(-1)^{\alpha \beta \epsilon (p)}\left({\frac {u}{p}}\right)^{\beta }\left({\frac {v}{p}}\right)^{\alpha }}$, where ${\displaystyle \epsilon (p)=(p-1)/2}$

and the expression involves two Legendre symbols.

Over the 2-adics, again writing ${\displaystyle a=2^{\alpha }u}$ and ${\displaystyle b=2^{\beta }v}$, where u and v are odd numbers, we have

${\displaystyle (a,b)_{2}=(-1)^{\epsilon (u)\epsilon (v)+\alpha \omega (v)+\beta \omega (u)}}$, where ${\displaystyle \omega (x)=(x^{2}-1)/8.}$

It is known that if v ranges over all places, (a, b)_v is 1 for almost all places. Therefore, the following product formula

${\displaystyle \prod _{v}(a,b)_{v}=1}$

makes sense. It is equivalent to the law of quadratic reciprocity.

The Hilbert symbol on a field F defines a map

${\displaystyle (\cdot ,\cdot ):F^{*}/F^{*2}\times F^{*}/F^{*2}\rightarrow \mathop {Br} (F)}$

where Br(F) is the Brauer group of F. The kernel of this mapping, the elements a such that (a,b)=1 for all b, is the Kaplansky radical of F.[2]

The radical is a subgroup of F^*/F^*2, identified with a subgroup of F^*. The radical is equal to F^* if and only if F has u-invariant at most 2.[3] In the opposite direction, a field with radical F^*2 is termed a Hilbert field.[4]

The Hilbert symbol is (multiplicatively) bilinear:

(ab,c) = (a,c)(b,c)

(a,bc) = (a,b)(a,c)

skew symmetric:

(a,b) = (b,a)⁻¹

nondegenerate:

(a,b)=1 for all b if and only if a is in K*ⁿ

It detects norms (hence the name norm residue symbol):

(a,b)=1 if and only if a is a norm of an element in K(ⁿ√b)

It has the "symbol" properties:

(a,1–a)=1, (a,–a)=1.

Hilbert's reciprocity law states that if a and b are in an algebraic number field containing the nth roots of unity then[6]

${\displaystyle \prod _{p}(a,b)_{p}=1}$

where the product is over the finite and infinite primes p of the number field, and where (,)_p is the Hilbert symbol of the completion at p. Hilbert's reciprocity law follows from the Artin reciprocity law and the definition of the Hilbert symbol in terms of the Artin symbol.

If K is a number field containing the nth roots of unity, p is a prime ideal not dividing n, π is a prime element of the local field of p, and a is coprime to p, then the power residue symbol (^a
_p) is related to the Hilbert symbol by[7]

${\displaystyle {\binom {a}{p}}=(\pi ,a)_{p}}$

The power residue symbol is extended to fractional ideals by multiplicativity, and defined for elements of the number field by putting (^a
_b)=(^a
_(b)) where (b) is the principal ideal generated by b. Hilbert's reciprocity law then implies the following reciprocity law for the residue symbol, for a and b prime to each other and to n:

${\displaystyle {\binom {a}{b}}={\binom {b}{a}}\prod _{p|n,\infty }(a,b)_{p}}$