Hilbert Symbols when $K =$ the $p$-adic numbers

Question

How can I show that the Hilbert Symbol is bimultiplicative, when the local field is the $p$-adic numbers? Everything I can find just sort of asserts bimultiplicativity without much proof, so I'm guessing it's pretty straight forward, but I haven't done much work with the $p$-adics so I'm a little unclear.

Moreover, for what primes $p$, is it the case that there exists an element $z$ of the $p$-adics such that $(-1, z) = -1$. That is, the the Hilbert symbol acts on $-1$ and $z$ and evaluates to $-1$.

Answer 1

For the multiplicativity, there is a proof in Chapter III of J.-P. Serre's, A course in arithmetic.
It is also proved in that book that the Hilbert symbol is nondegenerate, meaning that if $a$ is not a square, then there is some $b$ such that $(a,b)=-1$.

When $a$ is a square, you have $(a,b)=1$, for all $b$, so the answer to your second question is there is such $z$ if and only if $-1$ is not a square in $\mathbb{Q}_p$, and this happens precisely when $p\equiv 3\bmod{4}$.

Answer 2

For a first viewing on $p$-adics and Hilbert symbols, I prefer the approach of defining the Hilbert symbol for an odd prime $p$ by

$$\langle a,b\rangle_p=\left(\frac{r}{p}\right)$$

where $r=(-1)^{v_p(a)v_p(b)}\frac{a^{v_p(b)}}{b^{v_p(a)}}$, and for $p=2$, set

$$ \langle a,b\rangle_2=(-1)^{(r^2-1)/8}\cdot (-1)^{(u-1)(v-1)/4}. $$

You can then develop properties of this symbol algebraically (in particular, bi-multiplicativity is now relatively trivial), and then later use them to show that this is equivalent to the definition in terms of existence of solutions to quadratic forms (this equivalence is not particularly hard). This is the path taken by Kato et al.'s wonderful "Number Theory 1."

You can answer your second question without invoking non-degeneracy (which is not to say that this wasn't a good, or even better, argument) by observing that

$$ (-1,p)_p= \left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}, $$ which returns $-1$ if $p$ is 3 mod 4.