A User's Guide to Serre's Arithmetic/Hilbert Symbol

Local Properties[edit | edit source]

For a fixed (local) field $k=\mathbb {Q} _{p},\mathbb {R}$ the Hilbert symbol of two $a,b\in k^{*}$ is defined as

(a,b)_{p}={\begin{cases}1&{\text{if }}ax^{2}+by^{2}=z^{2}{\text{ for some }}(x,y,z)\in k^{3}-\{(0,0,0)\}\\-1&{\text{otherwise}}\end{cases}}

If we replace $a,b$ by $ac^{2},bd^{2}$ , then

z^{2}=ac^{2}x^{2}+bd^{2}y^{2}=a(cx)^{2}+b(dy)^{2}

showing that if we multiply, $a,b$ by squares, then their Hilbert symbols does not change. Hence the Hilbert symbol factors as

(\cdot ,\cdot )_{p}:{\frac {k^{*}}{(k^{*})^{2}}}\times {\frac {k^{*}}{(k^{*})^{2}}}\to \mathbb {F} _{2}

Serre goes on to prove that this is in fact a bilinear form over $\mathbb {F} _{2}$ in the next subsection.

After the definition he gives a method for computing the Hilbert Symbol in the proposition: It states that there is a short exact sequence

1\to Nk_{b}^{*}\to k^{*}{\xrightarrow {(\cdot ,a)_{p}}}\{\pm 1\}\to 1

where $k_{b}=k({\sqrt {(}}b))$ and

N:k_{b}^{*}\to k^{*}

sends

x+y{\sqrt {b}}\mapsto (x+y{\sqrt {b}})(x-y{\sqrt {b}})=x^{2}-by^{2}

He then goes on to prove/state some identities useful for computation: