Hilbert symbol: $(a,b)=1$ if $z^2-ax^2-by^2=0$ has a non-trivial solution and $-1$ else.

Norm map: $N_{K/F}(\alpha)$ is the determinant of the map $x\mapsto \alpha x$. For example, If $K$ is an extension of $F$ by adjoining $\sqrt{\beta}$ then $N(z+\sqrt{\beta}y) = z^2-\beta y^2.$

Proposition. $(a,b)=1$ iff $a\in Nk[\sqrt{b}]^{*}$

Proof. If $b$ is a square this is clear. If $b$ is a non-square then it is also true.

Proposition. The hilbert symbol satisfies: $(a,b)=(b,a)$, \[(a,c^2)=(a,-a) = (a,1-a) = 1,\] \[(a,b) =1 \implies (aa',b)=(a',b),\] \[(a,b)=(a,-ab) = (a, (1-a)b).\]

Proof. The implication $(a,b)=1\implies (aa',b)=(a',b)$ seems interesting. It’s true because we characterized $\left\{ x\; : \;(x,b)=1 \right\}$ as $Nk[\sqrt{b}]^{*}$. So, because $a\in Nk[\sqrt{b}]^{*}$ we have $aa'\in Nk[\sqrt{b}]^{*} \iff a'\in Nk[\sqrt{b}]^{*}$.

Lemma. $(aa',b)=(a,b)(a',b).$

Hilbert symbol is somehow related to legendre symbol.

main theorem: some relationship between hilbert symbol and legendre symbol

Theorem. In reals this is easy to compute. As always there is something different in the case that $p=2$. We just state for odd $p$.

Write $a=p^{\alpha}u, b = p^{\beta}v.$ In the $p$-adics we have $(a,b) = (-1)^{\alpha \beta \varepsilon(p)} \genfrac{(}{)}{}{}{u}{p}^{\beta} \genfrac{(}{)}{}{}{v}{p}^{\alpha}$

where $\varepsilon(u) = \frac{u-1}{2}\mod 2$, and where legendre symbol of element of $\mathbb{Z}_p^{*}$ actually means first take the element’s reduction modulo $p$ and then take the legendre symbol.

Corollary. If $b$ is not a square then group $Nk_b^{*}$ is a subgroup of index $2$ in $K^{*}$.

Recall: a point in $\mathbb{Z}_p^{m}$ is primitive if one of the $x_i$’s in the vector is invertable (not div by $p$).

recall lemma: it you have non-trivial zero then you can find primitive zero.

now we have:

Lemma. Let $v\in U$. if $z^2-px^2-vy^2=0$ has nontrivial solution in $\mathbb{Q}_p$ then it has one with $z,y\in U, x\in \mathbb{Z}_p$.

Proof. First get a primitive zero. Then take stuff mod p it can’t work unless $z,y$ are units.

The proof is kind of interesting. suppose our solution didn’t have $y,z$ being units. Then at least one of them is div by $p$. But taking reduction $\mod p$ we find that both must be $0$. But then $px^2\equiv 0 \mod p^2$ so $x,y,z$ all div by $p$ contradicting the primitiveness of the zero.

Now we prove the theorem

Proof. Split to three cases. The reason there are only three cases is by symmetry and the fact that multiplying stuff by squares doesn’t change the hilbert symbol.

Case 1: $\alpha =\beta = 0$. Then we want to prove $(u,v) = 1$. The equation $z^2-ux^2-vy^2 = 0$ is a quadratic form in 3 vars so has a solution modulo $p$. Furthermore, the discriminant of the quadratic form is a $p$-adic unit: $uv$. Thus by that other lemma “newton’s method” this lifts to a $p$-adic solution.

Case 2: $\alpha =0, \beta = 1$. Here the claim is \[ (pu,v) = \genfrac{(}{)}{}{}{v}{p} \] Using a non-trivial fact from earlier in this blog post (the first two propositions) we have in this case that $(pu,v)=(p,v)$. If $v$ is a square then both $(p,v)$ and $\genfrac{(}{)}{}{}{v}{p}$ are $1$ clearly. Else, if $v$ is not a square then we had this theorem earlier that said $p^{n}u$ is a square iff $n$ is even and $u\mod p$ is a square in $\mathbb{F}_p$. So then $\genfrac{(}{)}{}{}{v}{p} = -1$.

Now using that lemma from earlier that somewhat conspicuously had a $z^2-px^2-vy^2=0$ going on in it, an indication that it would be useful for understanding $(v,p)$, we see that if $(v,p)\neq -1$ then we would get units $z,y$, and p-adic integer $x$ such that $z^2-px^2-vy^2=0$. But this is ridiculous because then $v \equiv (zy^{-1})^2 \mod p$, while we know $v$ is not a square.

Case 3: $\alpha =\beta = 1$.

We use another interesting formula, $(a,b)=(a,-ab)$ to reduce this to case 2.

And the bilinearity thing follows from the fact that the legendre symbols and other stuff in our formula is bilinear.

section 2

We use $\mathbb{Q}_\infty$ to denote $\mathbb{R}$. We count $\infty$ as a prime number. Let $(a,b)_p$ be the hilbert symbol in $\mathbb{Q}_p$.

Theorem. $(a,b)_p = 1$ for all but finitely many $p$ and also \[ \prod_p (a,b)_p =1\]

Proof.

Suffices to consider case that $a,b$ are $-1$ or primes. Due to bilinearity.

Case 1: $a=-1,b=-1$. I.e., $x^2+y^2+z^2=0.$
We have $(-1,-1)_\infty = (-1,-1)_2 = -1$ (clear in $\mathbb{R}$, look at it $\mod 4$ to deduce the thing about $\mathbb{Q}_2$).
and, $(-1,-1)_p = 1$ for $p\neq 2$. This is because the quadratic form $x^2+y^2+z^2$ has a non-trivial solution modulo p by that one lemma that we really like, and then that solution lifts to a p-adic solution because you can’t have all the partial derivatives be zero at the solution: the derivative is just $2(x+y+z)$.
Case 2: $a=-1, b=2$
that is, $x^2+y^2-2z^2 =0$

Then $(-1,2)_2 = 1$. You can see this by one of the op quadratic form lemmas. For instance $(1,1,1)$ is a solution modulo $8$, and the matrix defining the quadratic form has non-zero determinant modulo $4$.

And clearly $(-1,2)_p = 1$: op quadratic form stuff.

Case 3: $a=-1, b=p\neq 2$.
Case 4:
this is basically just quadratic reciprocity

Existence of rationals with given hilbert symbols

Now the section I present for the class!

Theorem. Let $a_i$ be a finite fam of elts in $\mathbb{Q}^{*}$ and let $\varepsilon_{i,p}$ be a fam of signs $\pm 1$.

Suppose that there was $x\in \mathbb{Q}^{*}$ such that \[ (a_i, x)_v = \varepsilon_{i,v} \] for all $i,v$.

By the theorem above, we know that in order for this to possibly happen we need the following conditions:

Almost all $\varepsilon_{i,v}$ are $1$.
For all $i$ we have $\prod \varepsilon_{i,v} = 1$.
For all $v\in V$ there exists $x_v\in \mathbb{Q}_v^{*}$ such that $(a_i, x_v)_v = \varepsilon_{i,v}$ for all $i$.

It turns out these are sufficient as well.

Remark. The reason it’s interesting to have $I$ is that we are finding a single $x$ that works for all of them.

We need some lemmas.

Lemma. Chinese remainder theorem:

Let $m_1,\ldots, m_n$ be rel prime. Then for any set of $a_i$’s there is a solution to the simultaneous set of equations \[ a\equiv a_i \mod m_i. \]

Proof. $\left\{ m_1z\bmod m_2\; : \;z\in \mathbb{Z} \right\} = \mathbb{Z}/m_2\mathbb{Z}.$ Then do induction.

Lemma. Approximation theorem

Proof. Let $S$ be a finite subset of $V=\text{primes}\cup \infty$. The image of $\mathbb{Q}$ in $\prod_{v\in S} \mathbb{Q}_v$ is dense in the product topology.

Proof. Assume wlog that $\infty\in S$. Fix $\varepsilon>0$ and $x_\infty, x_1,x_2,\ldots, x_n \in \prod \mathbb{Q}_v$. Our goal is to find $x\in \mathbb{Q}$ such that $|x-x_v|_v < \varepsilon$ for each $v\in S$.

It’s nicer to work with $x_1,\ldots, x_n$ being $p$-adic integers. We can accomplish this by scaling them by a large number integer. This isn’t gonna make it harder to get a $\mathbb{Q}$-approximation to them. Fix $k\in \mathbb{N}$ such that $2^{-k}<\varepsilon$. Now, we can solve by CRT the simultaneous equations $z \equiv x_i \mod p_i^{k}$. Let $z_0$ be the solution. This is a good approx to the p-adic dudes but a lousy approx to the real dude.

Let $q\perp p_i$ for all $i$. Then we take $x = z_0+\frac{a}{q^m}p_1^{k}\cdots p_n^{k}$ where $m$ is large enough that $\frac{1}{q^{m}}<\varepsilon$. This is still a great p-adic approx to all the dudes. But now it is also a good real approx to the dudes, for appropriate choice of $a$.

Lemma. Dirichlet’s Theorem. Let $a\perp m$. Then there are infinitely many primes in the AP $a+m\mathbb{Z}$.

Proof. Dirichlet characters :O

Finally, proof of the theorem

Proof. We start by multiplying the $a_i$’s by the square of integers to make the $a_i$’s into integers. This is wlog because square of integer doesn’t impact hilbert symbol.

Let $S$ be $2,, $ and the prime factors of the $a_i$’s. Let $T$ be the set of $v\in V$ such that for some $i$ we have $\varepsilon_{i,v} = -1.$ Note that these sets are finite by assumption.

Case 1: $S \cap T = \varnothing$.

Define $a = \prod_{t\in T \setminus \infty} t$. Define $m = 8 \prod_{s\in S\setminus \left\{ 2,\infty\right\}} s.$

We have $a\perp m$.

Let $p$ be a prime not contained in $S\cup T$; exists by Dirichlet’s theorem.

Claim: $x=ap$ has the desired property! We do some QRrocity and use the product formula for the Hilbert Symbol.

Take $v\in S$. We must check $(a_i, x)_v = 1$.
First, $(a_i, x)_\infty = 1$: can take square-roots because $x>0$.

TODO