21.02.05 · number-theory / quadratic-forms-local-fields

Hilbert symbol $(a, b)_{v}$ and the product formula

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Anchor (Master): Serre Ch. III §2 (product formula via quadratic reciprocity); Cassels-Fröhlich 1967 *Algebraic Number Theory* Ch. VI–VII (class-field-theoretic embedding via the local invariant); Neukirch 1999 *Algebraic Number Theory* Ch. V §3 + Ch. VI (Galois-cohomological framing as cup product in $H^2(\mathbb{Q}_v, \mu_2)$); O'Meara 1963 *Introduction to Quadratic Forms* §63 (link to Hasse-Witt invariant); Tate 1976 *Invent. Math.* 36 (higher-degree symbols, explicit reciprocity laws)

Intuition Beginner

The Hilbert symbol is a single bit of information — a plus one or a minus one — that you compute from two numbers $a$ and $b$ at one place at a time. A place is just a way of measuring size on the rationals. The familiar absolute value on $Q$ is one place, written $v = \infty$ . For each prime number $p$ there is a second kind of size, the $p$ -adic size, which counts how many factors of $p$ a rational number has, and that gives one place per prime. So the places of $Q$ are one Archimedean place and one place per prime — countably many in total.

For each place $v$ and each pair of nonzero rationals $(a, b)$ , the rule is simple. You ask: does the equation $z^{2} = a x^{2} + b y^{2}$ have a solution where the numbers $x, y, z$ live in the completion of $Q$ at $v$ , and not all three are zero? If yes, the Hilbert symbol $(a, b)_{v}$ is $+ 1$ . If no, the Hilbert symbol is $- 1$ . That is all the symbol records.

The reason this matters: when you compute the symbol at every place and multiply all the answers together, the product is always $+ 1$ . Most of the factors are $+ 1$ — only finitely many places give $- 1$ — and the rule that the minus signs cancel out across all places is one of the cleanest examples in mathematics of a local-to-global principle. It is exactly equivalent to Gauss's quadratic reciprocity law, the rule from elementary number theory that connects $p mod q$ with $q mod p$ for two odd primes.

Visual Beginner

A schematic showing the row of places of the rational numbers, with one box for the Archimedean place $\infty$ and one box for each small prime ( $2, 3, 5, 7, 11$ ). Under each box is printed the value of the Hilbert symbol $(7, 11)_{v}$ , with the product of all the entries displayed below: $+ 1$ , $- 1$ , $+ 1$ , $+ 1$ , $+ 1$ , $- 1$ , and multiplying out to $+ 1$ as the product formula promises.

The picture captures the essential geometry of the product formula: across the entire infinite list of places, only two slots carry a minus sign, and they cancel. The product is forced to be positive, and that one fact is the source of quadratic reciprocity.

Worked example Beginner

Compute the Hilbert symbol $(7, 11)_{v}$ at every place of $Q$ and check that the product across all places is $+ 1$ .

Step 1. The Archimedean place. We ask whether the equation $z^{2} = 7 x^{2} + 11 y^{2}$ has a real solution with $(x, y, z)$ not all zero. Take $x = 1, y = 0$ , so the right side is $7$ . Then $z = 7$ is a real number, which gives $(x, y, z) = (1, 0, 7)$ , a real solution. So $(7, 11)_{\infty} = + 1$ . (More directly: whenever $a, b$ are both positive reals, you can write $a x^{2}$ as any positive number you like by adjusting $x$ , so the right side ranges over all positive reals and any positive number is the square of some real.)

Step 2. Most primes give $+ 1$ . For an odd prime $p$ that does not divide $2 \cdot 7 \cdot 11 = 154$ — that is, $p \neq = 2, 7, 11$ — there is a general rule that says $(7, 11)_{p} = + 1$ . The rule comes from the fact that the equation $z^{2} = 7 x^{2} + 11 y^{2}$ has a solution in $Q_{p}$ whenever none of the coefficients $7, 11, - 77$ is forced to be a non-square mod $p$ . For $p \neq = 2, 7, 11$ , all three of these are units mod $p$ and there is no obstruction. So $(7, 11)_{3} = (7, 11)_{5} = (7, 11)_{13} = \dots = + 1$ .

Step 3. The place $p = 7$ . Here we use the formula that $(7, 11)_{7}$ equals the Legendre symbol $(11/7)$ , the indicator of whether $11$ is a square mod $7$ . Reduce $11 mod 7$ : $11 = 7 + 4$ , so $11 \equiv 4 (mod 7)$ . Is $4$ a square mod $7$ ? Yes — $4 = 2^{2}$ already in the integers. So $(11/7) = + 1$ and $(7, 11)_{7} = + 1$ .

Step 4. The place $p = 11$ . By symmetry of the Hilbert symbol, $(7, 11)_{11} = (11, 7)_{11} = (7/11)$ , the Legendre symbol of $7$ mod $11$ . Check the squares mod $11$ : the seven squares of ${1, 2, 3, 4, 5}$ mod $11$ are $1, 4, 9, 5, 3$ . So the squares mod $11$ are ${1, 3, 4, 5, 9}$ . Is $7$ in this list? No. So $(7/11) = - 1$ and $(7, 11)_{11} = - 1$ .

Step 5. The place $p = 2$ . The formula at $2$ is more delicate and depends on the values of $7$ and $11$ mod $8$ : $7 \equiv - 1 (mod 8)$ and $11 \equiv 3 (mod 8)$ . The $2$ -adic formula then gives $(7, 11)_{2} = - 1$ .

Step 6. Multiply everything together. The product over all places is $(7, 11)_{\infty} \cdot (7, 11)_{2} \cdot (7, 11)_{3} \cdot (7, 11)_{5} \cdot (7, 11)_{7} \cdot (7, 11)_{11} \cdot (7, 11)_{13} \dots$ . All factors past $11$ are $+ 1$ . The remaining six factors are $+ 1, - 1, + 1, + 1, + 1, - 1$ . The product is $(+ 1) (- 1) (+ 1) (+ 1) (+ 1) (- 1) = + 1$ .

What this tells us: the two negative signs at $p = 2$ and $p = 11$ cancel exactly, leaving the product $+ 1$ . The same pattern works for any pair $a, b$ of nonzero rationals: minus signs come in matched pairs across the places, and the product always equals $+ 1$ . That is the product formula, and it is the heart of quadratic reciprocity.

Check your understanding Beginner

Formal definition Intermediate+

Let $Q$ denote the rational numbers and let $v$ denote a place of $Q$ . By Ostrowski's theorem, the places of $Q$ are the Archimedean place $v = \infty$ corresponding to the standard absolute value $∣ x ∣_{\infty} = ∣ x ∣$ , together with one finite place $v = p$ per prime, corresponding to the $p$ -adic absolute value $∣ x ∣_{p} = p^{- ord_{p} (x)}$ where $ord_{p} (x)$ is the exponent of $p$ in the prime factorisation of $x$ . The completion of $Q$ at the place $v$ is the field $Q_{v}$ : at $v = \infty$ this is $R$ , and at $v = p$ this is the field $Q_{p}$ of $p$ -adic numbers.

Definition (Hilbert symbol). Let $v$ be a place of $Q$ and let $a, b \in Q_{v}^{\times}$ . The Hilbert symbol $(a, b)_{v} \in {\pm 1}$ is defined by $$ (a, b)_v = \begin{cases} +1 & \text{if the equation } z^2 = a x^2 + b y^2 \text{ has a solution } (x, y, z) \in \mathbb{Q}_v^3 \setminus {(0, 0, 0)}, \ -1 & \text{otherwise}. \end{cases} $$

The symbol depends only on the square classes of $a$ and $b$ : $(a, b)_{v} = (a c^{2}, b)_{v} = (a, b c^{2})_{v}$ for any $c \in Q_{v}^{\times}$ . Equivalently, the symbol descends to a function on $Q_{v}^{\times} / (Q_{v}^{\times})^{2}$ in each argument.

Equivalent characterisation (norm form). The Hilbert symbol satisfies $(a, b)_{v} = 1$ iff $b$ lies in the image of the norm map $N_{K / Q_{v}} : K^{\times} \to Q_{v}^{\times}$ where $K = Q_{v} (a)$ is the (étale) quadratic algebra over $Q_{v}$ . When $a$ is already a square in $Q_{v}$ , the algebra $K = Q_{v} \times Q_{v}$ is split and the norm map is surjective, so $(a, b)_{v} = 1$ for every $b$ . When $a$ is a non-square, $K$ is a quadratic field extension, and the image $N (K^{\times})$ is an index-two subgroup of $Q_{v}^{\times}$ ; the symbol detects whether $b$ falls into this subgroup.

The two definitions agree because: if $b = N_{K / Q_{v}} (α)$ for some $α = u + w a \in K$ with $u, w \in Q_{v}$ , then $b = u^{2} - a w^{2}$ , giving the solution $(x, y, z) = (w, 1, u)$ to $z^{2} = a x^{2} + b y^{2}$ (substituting yields $u^{2} = a w^{2} + b$ , matching after the relabel). Conversely, a non-zero solution to $z^{2} = a x^{2} + b y^{2}$ with $y \neq = 0$ produces $b = (z / y)^{2} - a (x / y)^{2}$ , a norm from $K$ . The cases $y = 0$ or $a$ a square are handled by separate squaring arguments.

Properties of the Hilbert symbol. The symbol $(a, b)_{v}$ satisfies the following identities for every $a, b, a^{'}, b^{'} \in Q_{v}^{\times}$ :

(i) Bimultiplicativity. $(a a^{'}, b)_{v} = (a, b)_{v} \cdot (a^{'}, b)_{v}$ and $(a, b b^{'})_{v} = (a, b)_{v} \cdot (a, b^{'})_{v}$ .

(ii) Symmetry. $(a, b)_{v} = (b, a)_{v}$ .

(iii) Negation identity. $(a, - a)_{v} = 1$ for every $a \in Q_{v}^{\times}$ .

(iv) Complementarity identity. $(a, 1 - a)_{v} = 1$ for every $a \in Q_{v} ∖ {0, 1}$ .

Bimultiplicativity in either argument, together with symmetry, makes the Hilbert symbol a symmetric bimultiplicative pairing $$ (\cdot, \cdot)_v : \mathbb{Q}_v^\times / (\mathbb{Q}_v^\times)^2 \times \mathbb{Q}_v^\times / (\mathbb{Q}_v^\times)^2 \to {\pm 1}. $$ For $v$ a finite place of odd residue characteristic, $Q_{v}^{\times} / (Q_{v}^{\times})^{2}$ is a Klein four-group, so the Hilbert symbol on $Q_{v}$ is a ${\pm 1}$ -valued pairing on $(Z /2)^{2} \times (Z /2)^{2}$ . For $v = 2$ , the square-class group $Q_{2}^{\times} / (Q_{2}^{\times})^{2}$ has order eight, and the pairing is correspondingly larger.

Counterexamples to common slips

The condition $(a, b)_{v} = 1$ is not the same as saying $a$ and $b$ are both squares in $Q_{v}$ . The symbol is $+ 1$ whenever either of them is a square in $Q_{v}$ , and also $+ 1$ in many cases where neither is. The exact criterion is that $b$ is a norm from $Q_{v} (a)$ .
The equation $z^{2} = a x^{2} + b y^{2}$ always has the all-zero solution. The Hilbert symbol asks about non-zero solutions in $Q_{v}^{3}$ , so the all-zero solution is excluded explicitly.
The symbol $(a, b)_{v}$ depends only on $a, b$ modulo squares in $Q_{v}^{\times}$ , but it does not depend only on the residue classes $\overset{a}{ˉ}, \overset{ˉ}{b} \in F_{p}$ at a finite place. At odd $p$ the residue class plus the $p$ -adic valuation determine the square class, and the explicit formula combines both pieces.

Explicit formula at the Archimedean place. For $a, b \in R^{\times} = Q_{\infty}^{\times}$ : $$ (a, b)_\infty = \begin{cases} -1 & \text{if } a < 0 \text{ and } b < 0, \ +1 & \text{otherwise}. \end{cases} $$ The reasoning: in $R^{3}$ , the equation $z^{2} = a x^{2} + b y^{2}$ has a non-zero solution iff at least one of $a, b$ is positive (then take $y = 0$ and $x = 1$ if $a > 0$ , or symmetrically), iff $a$ and $b$ are not both negative.

Explicit formula at an odd prime $p$ . Write $a = p^{α} u$ and $b = p^{β} w$ with $u, w \in Z_{p}^{\times}$ and $α, β \in Z$ . Then $$ (a, b)_p = (-1)^{\alpha \beta \varepsilon(p)} \cdot \left(\frac{\bar u}{p}\right)^\beta \cdot \left(\frac{\bar w}{p}\right)^\alpha, $$ where $ε (p) = (p - 1) /2 mod 2$ and $(\frac{x}{p})$ is the Legendre symbol of $x mod p$ . In particular, when $α = β = 0$ (so $a, b \in Z_{p}^{\times}$ are both $p$ -adic units), $(a, b)_{p} = + 1$ . So $(a, b)_{p} = 1$ whenever $p$ is odd and $p$ does not divide $2 ab$ .

Explicit formula at $p = 2$ . For $a, b \in Q_{2}^{\times}$ , write $a = 2^{α} u$ and $b = 2^{β} w$ with $u, w \in Z_{2}^{\times}$ — that is, $u, w$ odd $2$ -adic integers. Define $ε (u) = (u - 1) /2 mod 2 \in {0, 1}$ and $ω (u) = (u^{2} - 1) /8 mod 2 \in {0, 1}$ , depending on $u mod 8$ . Then $$ (a, b)_2 = (-1)^{\varepsilon(u) \varepsilon(w) + \alpha \omega(w) + \beta \omega(u)}. $$ This is the bilinear formula in $(α, ε (u), ω (u))$ versus $(β, ε (w), ω (w))$ that Serre records as Theorem 1 of Ch. III §1.

Key theorem with proof Intermediate+

Theorem (Hilbert product formula; Hilbert 1897, Hasse 1924). For every pair $a, b \in Q^{\times}$ of non-zero rationals, the product of the Hilbert symbols across all places of $Q$ equals $+ 1$ : $$ \prod_v (a, b)_v = 1, $$ the product being over the Archimedean place and all finite places. The product is well-defined: for fixed $a, b$ , the value $(a, b)_{p}$ is $+ 1$ for all but finitely many primes $p$ .

The proof proceeds in three steps. First, reduce to the case $a, b$ each a product of a sign and a square-free integer composed of distinct primes. Second, use bimultiplicativity to reduce to the four building blocks $(- 1, - 1)$ , $(- 1, p)$ , $(p, q)$ , $(2, p)$ for primes $p, q$ . Third, verify each building block by direct computation using the explicit local formulas of §1, and identify the product formula on each building block with one of the supplementary laws or with quadratic reciprocity itself.

Step 1: reduction to building blocks. By bimultiplicativity, the function $(a, b) \mapsto \prod_{v} (a, b)_{v}$ is bimultiplicative on $Q^{\times} \times Q^{\times}$ , and it descends to the square classes. The square-class group $Q^{\times} / (Q^{\times})^{2}$ is generated by ${- 1} \cup {p : p prime}$ , since every rational is a product of $\pm 1$ and prime powers and only the parity of each exponent matters. So the product formula reduces to verifying $\prod_{v} (x, y)_{v} = 1$ on every pair of generators $(x, y)$ .

The unordered pairs of generators fall into four families: (a) $(- 1, - 1)$ , (b) $(- 1, p)$ for $p$ prime, (c) $(2, p)$ for $p$ an odd prime, (d) $(p, q)$ for distinct odd primes $p, q$ , plus the diagonal cases $(p, p)$ which reduce by $(p, p) = (p, - 1)$ via the identity $(p, p) = (p, - p) \cdot (p, - 1) = 1 \cdot (p, - 1)$ .

Step 2: verification on $(- 1, - 1)$ . Compute at every place:

$(- 1, - 1)_{\infty} = - 1$ since both arguments are negative.
$(- 1, - 1)_{2}$ : applying the formula at $p = 2$ with $a = b = - 1$ gives $α = β = 0$ , $u = w = - 1 \equiv 7 (mod 8)$ , so $ε (- 1) = 1, ω (- 1) = 0$ . Then $(- 1, - 1)_{2} = (- 1)^{1 \cdot 1 + 0 + 0} = - 1$ .
$(- 1, - 1)_{p}$ for $p$ odd: applying the odd- $p$ formula with $α = β = 0$ and $u = w = - 1$ a unit, the formula gives $(- 1, - 1)_{p} = (- 1)^{0 \cdot ε (p)} = 1$ .

The product is $(- 1) (- 1) (1) (1) \dots = + 1$ . The product formula holds on $(- 1, - 1)$ .

Step 3: verification on $(- 1, p)$ for $p$ odd prime. At the relevant places:

$(- 1, p)_{\infty} = + 1$ since $p > 0$ .
$(- 1, p)_{2}$ : $a = - 1$ has $α = 0, u = - 1$ ; $b = p$ has $β = 0, w = p$ . The 2-adic formula gives $(- 1)^{ε (- 1) ε (p)} = (- 1)^{1 \cdot ε (p)} = (- 1)^{(p - 1) /2}$ .
$(- 1, p)_{p}$ : $a = - 1$ has $α = 0, u = - 1$ ; $b = p$ has $β = 1, w = 1$ . The odd- $p$ formula gives $(- 1, p)_{p} = (\frac{- 1}{p})^{1} \cdot (\frac{1}{p})^{0} = (\frac{- 1}{p}) = (- 1)^{(p - 1) /2}$ .
$(- 1, p)_{q}$ for $q \neq = 2, p$ : both arguments are $q$ -adic units, so the symbol is $+ 1$ .

Product: $1 \cdot (- 1)^{(p - 1) /2} \cdot (- 1)^{(p - 1) /2} \cdot 1 \dots = (- 1)^{p - 1} = + 1$ since $p - 1$ is even. The product formula holds.

The verification of the building block $(- 1, p)$ is the first supplementary law of quadratic reciprocity in disguise: it says $(\frac{- 1}{p}) = (- 1)^{(p - 1) /2}$ , which the calculation above derives from the requirement that $(- 1, p)_{2} \cdot (- 1, p)_{p} = + 1$ .

Step 4: verification on $(2, p)$ for $p$ odd prime.

$(2, p)_{\infty} = + 1$ since both are positive.
$(2, p)_{2}$ : $a = 2$ has $α = 1, u = 1$ ; $b = p$ has $β = 0, w = p$ . The 2-adic formula gives $(- 1)^{ε (1) ε (p) + 1 \cdot ω (p) + 0 \cdot ω (1)} = (- 1)^{ω (p)}$ , with $ω (p) = (p^{2} - 1) /8 mod 2$ .
$(2, p)_{p}$ : $a = 2$ has $α = 0, u = 2$ ; $b = p$ has $β = 1, w = 1$ . The odd- $p$ formula gives $(\frac{2}{p})^{1} \cdot (\frac{1}{p})^{0} = (\frac{2}{p})$ .
$(2, p)_{q}$ for $q \neq = 2, p$ : both are $q$ -adic units, so $+ 1$ .

Product: $(- 1)^{ω (p)} \cdot (\frac{2}{p}) = + 1$ forces $(\frac{2}{p}) = (- 1)^{ω (p)} = (- 1)^{(p^{2} - 1) /8}$ , which is the second supplementary law of quadratic reciprocity. The product formula on $(2, p)$ encodes this law.

Step 5: verification on $(p, q)$ for distinct odd primes. This is the main case.

$(p, q)_{\infty} = + 1$ since both positive.
$(p, q)_{2}$ : $a = p$ has $α = 0, u = p$ ; $b = q$ has $β = 0, w = q$ . The 2-adic formula gives $(- 1)^{ε (p) ε (q)} = (- 1)^{(p - 1) /2 \cdot (q - 1) /2}$ .
$(p, q)_{p}$ : $a = p$ has $α = 1, u = 1$ ; $b = q$ has $β = 0, w = q$ . The odd- $p$ formula at $p$ gives $(\frac{1}{p})^{0} \cdot (\frac{q}{p})^{1} = (\frac{q}{p})$ .
$(p, q)_{q}$ : by symmetry, $(\frac{p}{q})$ .
$(p, q)_{r}$ for $r \neq = 2, p, q$ : both units, so $+ 1$ .

Product: $(- 1)^{(p - 1) /2 \cdot (q - 1) /2} \cdot (\frac{q}{p}) \cdot (\frac{p}{q}) = + 1$ , which rearranges to $$ \left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2} \cdot \frac{q-1}{2}}, $$ Gauss's quadratic reciprocity law. The product formula on the building block $(p, q)$ is precisely the main statement of quadratic reciprocity.

Steps 2–5 cover all the cases, so the product formula holds on every pair $(a, b)$ of generators, and bimultiplicativity extends the conclusion to every pair $a, b \in Q^{\times}$ . $□$

Bridge. The Hilbert product formula builds toward the entire edifice of class field theory, and it appears again in 21.02.08 (Hasse-Minkowski theorem), where the local-global theorem for quadratic forms over $Q$ relies on the same local invariants the Hilbert symbol packages. The foundational reason the product formula holds is exactly the bimultiplicative reduction to four building blocks $(- 1, - 1)$ , $(- 1, p)$ , $(2, p)$ , $(p, q)$ , each of which encodes one piece of quadratic reciprocity. The central insight is that quadratic reciprocity is the product formula on $(p, q)$ , and the supplementary laws are the product formula on $(- 1, p)$ and $(2, p)$ — a single local-global identity, computed once at each pair of generators, identifies the elementary number-theoretic reciprocity statement with a uniform statement across all places. The bridge is that this local-global pattern generalises: every Brauer class over $Q$ has local invariants in $Br (Q_{v})$ that sum to zero in $Q / Z$ , and the Hilbert product formula is the case of an order-two Brauer class. Putting these together, the Hilbert symbol identifies the elementary side of quadratic reciprocity with the cohomological side: $\prod_{v} (a, b)_{v} = 1$ is the case $n = 2$ of the global reciprocity exact sequence of class field theory.

Exercises Intermediate+

Exercise 3 (medium, numeric).

Compute the Hilbert symbol $(3, 5)_{v}$ at every place of $Q$ and verify that the product is $+ 1$ .

Hint

Both $3$ and $5$ are positive, so $(\cdot, \cdot)_{\infty} = + 1$ . At every prime $p \neq = 2, 3, 5$ , both are $p$ -adic units, so the symbol is $+ 1$ . At $p = 2, 3, 5$ , apply the explicit formulas.

Answer

Place-by-place: $(3, 5)_{\infty} = + 1$ (both positive); $(3, 5)_{3} = (\frac{5}{3}) = (\frac{2}{3}) = - 1$ since $2 \neq \equiv □ (mod 3)$ ; $(3, 5)_{5} = (\frac{3}{5})$ . The squares mod $5$ are ${1, 4}$ , so $3$ is not a square, giving $(\frac{3}{5}) = - 1$ . $(3, 5)_{2}$ : with $α = β = 0, u = 3, w = 5$ , $ε (3) = 1, ε (5) = 0$ , so the 2-adic formula gives $(- 1)^{ε (3) ε (5) + 0 + 0} = (- 1)^{1 \cdot 0} = + 1$ . At all other primes, both are units, so $+ 1$ . Product: $(+ 1) (+ 1) (- 1) (- 1) (+ 1) \dots = + 1$ . Rubric: full credit for the three non-identity place values plus the running product. (Note: this also verifies quadratic reciprocity for the pair $(3, 5)$ , $(\frac{3}{5}) (\frac{5}{3}) = (- 1)^{1 \cdot 2} = + 1$ , matching the product $(- 1) (- 1) = + 1$ .)

Exercise 4 (medium, symbolic).

Show that the equation $z^{2} = a x^{2} + b y^{2}$ has a non-zero solution in $Q_{v}^{3}$ iff $b$ is a norm from the quadratic algebra $Q_{v} (a)$ .

Hint

Split on whether $a$ is a square in $Q_{v}$ . If yes, the algebra splits and every $b$ is a norm. If no, $Q_{v} (a)$ is a quadratic extension and the norm of $u + w a$ is $u^{2} - a w^{2}$ .

Answer

Case 1: $a = c^{2}$ a square in $Q_{v}$ . Then $Q_{v} (a) = Q_{v} \times Q_{v}$ is the split étale algebra, and the norm map $(u, w) \mapsto u w$ is surjective onto $Q_{v}^{\times}$ , so every $b$ is a norm. Correspondingly, $z^{2} = a x^{2} + b y^{2} = (c x)^{2} + b y^{2}$ has the solution $z = c, x = 1, y = 0$ , so $(a, b)_{v} = + 1$ , matching $b$ -is-a-norm.

Case 2: $a$ is a non-square. Then $K = Q_{v} (a)$ is a quadratic field extension, and the norm of $α = u + w a$ is $N (α) = α \overset{α}{ˉ} = u^{2} - a w^{2}$ . Suppose $b = N (α) = u^{2} - a w^{2}$ for some $α$ with $α \neq = 0$ . Then $u^{2} = a w^{2} + b$ , so $(x, y, z) = (w, 1, u) \in Q_{v}^{3}$ is a solution to $z^{2} = a x^{2} + b y^{2}$ , and it is non-zero because the second coordinate is $1$ .

Conversely, suppose $z^{2} = a x^{2} + b y^{2}$ has a non-zero solution. If $y \neq = 0$ , dividing by $y^{2}$ gives $b = (z / y)^{2} - a (x / y)^{2} = N_{K / Q_{v}} (z / y - (x / y) a)$ , so $b$ is a norm. If $y = 0$ , then $z^{2} = a x^{2}$ with $(x, z) \neq = (0, 0)$ , forcing $a = (z / x)^{2}$ a square — contradicting Case 2. So $y \neq = 0$ in Case 2 and the conclusion holds.

The bijection $(b is a norm) \leftrightarrow (solution exists)$ is the equivalent definition of $(a, b)_{v}$ . Rubric: full credit for both directions of the case split.

Exercise 5 (medium, short-answer).

Prove the bimultiplicativity of the Hilbert symbol in the first argument: $(a a^{'}, b)_{v} = (a, b)_{v} \cdot (a^{'}, b)_{v}$ .

Hint

Use the norm-from- $Q_{v} (a)$ characterisation. The set of $b$ that are norms from $Q_{v} (a)$ is a subgroup of $Q_{v}^{\times}$ , and the assignment $a \mapsto N (Q_{v} (a)^{\times}) (mod (Q_{v}^{\times})^{2})$ has the right multiplicative structure modulo square classes.

Answer

Fix $b \in Q_{v}^{\times}$ . Define $χ_{b} : Q_{v}^{\times} \to {\pm 1}$ by $χ_{b} (a) = (a, b)_{v}$ . The claim is that $χ_{b}$ is a group homomorphism, i.e., $χ_{b} (a a^{'}) = χ_{b} (a) χ_{b} (a^{'})$ .

The set $N_{b} = {a \in Q_{v}^{\times} : (a, b)_{v} = + 1}$ is the kernel of $χ_{b}$ if $χ_{b}$ is a homomorphism. By the norm characterisation, $(a, b)_{v} = + 1$ iff $b \in N_{K_{a} / Q_{v}} (K_{a}^{\times})$ where $K_{a} = Q_{v} (a)$ . Equivalently, by symmetry, $(a, b)_{v} = + 1$ iff $a$ is a norm from $Q_{v} (b)$ . So $N_{b} = N_{K_{b} / Q_{v}} (K_{b}^{\times})$ when $b$ is a non-square, and $N_{b} = Q_{v}^{\times}$ when $b$ is a square.

The image of a norm map on $K_{b}^{\times}$ is a subgroup of $Q_{v}^{\times}$ , and the quotient $Q_{v}^{\times} / N (K_{b}^{\times})$ has order $2$ when $b$ is a non-square (by local class field theory, or by direct computation: the norm group is exactly $ker ((\cdot, b)_{v}) = ker (χ_{b})$ , which is index $2$ in $Q_{v}^{\times}$ ). So $χ_{b}$ is the unique non-identity character on $Q_{v}^{\times} / N (K_{b}^{\times})$ , which is a group homomorphism by construction. Multiplicativity $χ_{b} (a a^{'}) = χ_{b} (a) χ_{b} (a^{'})$ follows. The same argument with $a$ fixed gives multiplicativity in the second argument. Rubric: full credit for identifying the kernel with the norm group, plus the group-quotient argument.

Exercise 6 (medium, short-answer).

Verify the negation identity $(a, - a)_{v} = 1$ at every place $v$ of $Q$ for $a \in Q^{\times}$ .

Hint

The equation $z^{2} = a x^{2} + (- a) y^{2} = a (x^{2} - y^{2})$ has the explicit solution $(x, y, z) = (1, 1, 0)$ for every $a$ .

Answer

Take $x = y = 1$ and $z = 0$ . Then $z^{2} = 0 = a (1) + (- a) (1) = a x^{2} + (- a) y^{2}$ , with $(x, y, z) = (1, 1, 0) \neq = (0, 0, 0)$ . The solution exists in $Q_{v}^{3}$ for every place $v$ (the values $1, 1, 0$ live in any field), so $(a, - a)_{v} = + 1$ at every place.

This identity has a conceptual reading: $- a = a \cdot (- 1)$ and the symbol satisfies $(a, - a)_{v} = (a, a)_{v} \cdot (a, - 1)_{v}$ . The first factor $(a, a)_{v}$ also evaluates to $(a, - 1)_{v}$ (from the identity $(a, a) = (a, - a) \cdot (a, - 1)$ and the present claim), so the two reduce consistently. Rubric: full credit for either the direct $(1, 1, 0)$ solution or the bimultiplicative reduction.

Exercise 7 (hard, short-answer).

Verify the complementarity identity $(a, 1 - a)_{v} = 1$ for $a \in Q^{\times}$ with $a \neq = 1$ , at every place $v$ .

Hint

The equation $z^{2} = a x^{2} + (1 - a) y^{2}$ has a non-zero solution for $(x, y, z) = (1, 1, 1)$ .

Answer

Substitute $x = y = z = 1$ : $z^{2} = 1$ , and $a x^{2} + (1 - a) y^{2} = a + (1 - a) = 1$ . The equation is satisfied, and $(1, 1, 1) \neq = (0, 0, 0)$ in any field, so $(a, 1 - a)_{v} = + 1$ at every place $v$ .

The complementarity identity $(a, 1 - a) = 1$ together with bimultiplicativity, symmetry, and the negation identity gives the Steinberg-style relations that characterise the Milnor $K_{2}$ of a field: $K_{2} (F)$ is generated by Steinberg symbols ${a, b}$ with the relations ${a, 1 - a} = 0$ and the multiplicativity-symmetry axioms, and the Hilbert symbol is the universal ${\pm 1}$ -valued Steinberg symbol on $Q_{v}$ . Rubric: full credit for the explicit solution; bonus context for naming the $K_{2}$ connection.

Exercise 8 (hard, short-answer).

Show that the product formula $\prod_{v} (a, b)_{v} = 1$ implies the supplementary law $(\frac{2}{p}) = (- 1)^{(p^{2} - 1) /8}$ for every odd prime $p$ .

Hint

Apply the product formula to the pair $(a, b) = (2, p)$ . Use the explicit local formulas: $(2, p)_{\infty} = + 1$ (both positive), $(2, p)_{q} = + 1$ for $q \neq = 2, p$ , the 2-adic formula gives $(- 1)^{(p^{2} - 1) /8}$ , and the $p$ -adic formula gives $(\frac{2}{p})$ .

Answer

The places contributing a possible $- 1$ to $\prod_{v} (2, p)_{v}$ are $\infty, 2, p$ , since for any other prime $q$ both arguments are $q$ -adic units and the symbol is $+ 1$ .

At $\infty$ : $(2, p)_{\infty} = + 1$ since $2, p > 0$ .

At $p$ : $a = 2$ has $α = 0, u = 2$ ; $b = p$ has $β = 1, w = 1$ . The odd- $p$ formula gives $(2, p)_{p} = (- 1)^{0 \cdot 1 \cdot ε (p)} \cdot (\frac{2}{p})^{1} \cdot (\frac{1}{p})^{0} = (\frac{2}{p})$ .

At $2$ : $a = 2$ has $α = 1, u = 1$ ; $b = p$ has $β = 0, w = p$ . Then $ε (1) = 0, ω (1) = 0, ε (p) = (p - 1) /2 mod 2, ω (p) = (p^{2} - 1) /8 mod 2$ . The 2-adic formula gives $(2, p)_{2} = (- 1)^{ε (1) ε (p) + 1 \cdot ω (p) + 0 \cdot ω (1)} = (- 1)^{ω (p)} = (- 1)^{(p^{2} - 1) /8}$ .

The product formula now reads $$ (+1) \cdot (-1)^{(p^2-1)/8} \cdot \left(\frac{2}{p}\right) \cdot (+1)(+1)\cdots = +1, $$ so $(\frac{2}{p}) = (- 1)^{(p^{2} - 1) /8}$ , the second supplementary law. Rubric: full credit for the three place values plus the product manipulation.

Exercise 9 (hard, short-answer).

For an odd prime $p$ , classify the square classes of $Q_{p}^{\times} / (Q_{p}^{\times})^{2}$ and compute the full multiplication table of the Hilbert symbol $(\cdot, \cdot)_{p}$ on the four square classes.

Hint

By Hensel's lemma, $u \in Z_{p}^{\times}$ is a square in $Q_{p}$ iff $\overset{u}{ˉ}$ is a square in $F_{p}^{\times}$ . So $Z_{p}^{\times} / (Z_{p}^{\times})^{2} = F_{p}^{\times} / (F_{p}^{\times})^{2} ≅ Z /2$ , plus a factor of $Z /2$ from the $p$ -adic valuation. The four classes are ${1, u, p, p u}$ where $u \in Z_{p}^{\times}$ is a non-square unit.

Answer

For odd $p$ , the square classes are represented by ${1, u, p, p u}$ where $u$ is any non-residue mod $p$ (e.g., $u =$ a non-residue lifted to $Z_{p}$ ). The group $Q_{p}^{\times} / (Q_{p}^{\times})^{2} ≅ (Z /2)^{2}$ . The Hilbert symbol restricted to these four classes is a symmetric ${\pm 1}$ -valued bilinear pairing. The table depends on $ε (p) = (p - 1) /2 mod 2$ ; for definiteness take $p \equiv 1 (mod 4)$ , so $ε (p) = 0$ and $- 1$ is a square.

The table (rows and columns indexed by $1, u, p, p u$ ):

$(\cdot, \cdot)_{p}$	$1$	$u$	$p$	$p u$
$1$	$+ 1$	$+ 1$	$+ 1$	$+ 1$
$u$	$+ 1$	$+ 1$	$- 1$	$- 1$
$p$	$+ 1$	$- 1$	$+ 1$	$- 1$
$p u$	$+ 1$	$- 1$	$- 1$	$+ 1$

The signs follow from the odd- $p$ formula: $(u, p)_{p} = (\frac{u}{p})^{1} = - 1$ since $u$ is a non-residue; $(p, p)_{p} = (p, - p)_{p} \cdot (p, - 1)_{p} = + 1 \cdot (\frac{- 1}{p}) = + 1$ when $p \equiv 1 (mod 4)$ . The case $p \equiv 3 (mod 4)$ (where $ε (p) = 1$ and $- 1$ is a non-residue) flips some entries — specifically $(p, p)_{p} = (\frac{- 1}{p}) = - 1$ in that case.

The Hilbert symbol is non-degenerate on $Q_{p}^{\times} / (Q_{p}^{\times})^{2}$ for odd $p$ : every non-identity class pairs to $- 1$ with some other class. Rubric: full credit for the table; partial for the four classes.

Exercise 10 (hard, short-answer).

Use the Hilbert symbol to decide whether $- 3$ is a norm from $Q (- 2)$ at every place of $Q$ . (This is the local-global question whose global answer is delivered by the Hasse-Minkowski theorem.)

Hint

$- 3$ is a norm from $Q_{v} (- 2)$ iff $(- 2, - 3)_{v} = + 1$ . Compute at the places $\infty, 2, 3$ (the relevant primes), and use the unit-unit rule at all other places.

Answer

Compute $(- 2, - 3)_{v}$ at every place.

At $\infty$ : $- 2, - 3$ both negative, so $(- 2, - 3)_{\infty} = - 1$ — not a norm.

At $p = 2$ : $- 2 = 2 \cdot (- 1)$ has $α = 1, u = - 1$ ; $- 3 = 1 \cdot (- 3)$ has $β = 0, w = - 3$ . The $2$ -adic unit $- 1$ reduces to $7 mod 8$ , giving $ε (- 1) = 1$ and $ω (- 1) = 0$ ; the $2$ -adic unit $- 3$ reduces to $5 mod 8$ , giving $ε (- 3) = 0$ and $ω (- 3) = 1$ . The 2-adic formula gives $(- 2, - 3)_{2} = (- 1)^{ε (- 1) ε (- 3) + α \cdot ω (- 3) + β \cdot ω (- 1)} = (- 1)^{1 \cdot 0 + 1 \cdot 1 + 0 \cdot 0} = - 1$ .

At $p = 3$ : $- 2 \equiv 1 (mod 3)$ is a unit, $α = 0$ ; $- 3 = 3 \cdot (- 1)$ has $β = 1, w = - 1$ . The odd- $p$ formula gives $(- 2, - 3)_{3} = (\frac{- 2}{3})^{1} \cdot (\frac{- 1}{3})^{0} = (\frac{1}{3}) = + 1$ .

At every other prime $p \neq = 2, 3$ : $- 2, - 3$ both units, so $(- 2, - 3)_{p} = + 1$ .

Summary: $(- 2, - 3)_{\infty} = - 1$ , $(- 2, - 3)_{2} = - 1$ , and $(- 2, - 3)_{p} = + 1$ for every other place. The product is $(- 1) (- 1) (+ 1) (+ 1) \dots = + 1$ , matching the product formula. The local-global conclusion: $- 3$ is a norm from $Q_{v} (- 2)$ at every finite place $p \neq = 2$ , but fails at $p = 2$ and at $\infty$ ; globally $- 3$ is therefore not a norm from $Q (- 2)$ , since global-norm membership requires norm membership at every place. Rubric: full credit for the place-by-place computation and the product-formula check.

Advanced results Master

Theorem (non-degeneracy of the local pairing; Serre Ch. III §1). At every place $v$ of $Q$ , the Hilbert symbol $(\cdot, \cdot)_{v} : Q_{v}^{\times} / (Q_{v}^{\times})^{2} \times Q_{v}^{\times} / (Q_{v}^{\times})^{2} \to {\pm 1}$ is a non-degenerate bilinear pairing of finite-dimensional $F_{2}$ -vector spaces: for every $a \in Q_{v}^{\times}$ that is not a square, there exists $b$ with $(a, b)_{v} = - 1$ .

The proof: at the Archimedean place, $R^{\times} / (R^{\times})^{2} ≅ Z /2$ generated by $- 1$ , and $(- 1, - 1)_{\infty} = - 1$ . At an odd $p$ , the four-element group $Q_{p}^{\times} / (Q_{p}^{\times})^{2} ≅ (Z /2)^{2}$ has the non-degenerate pairing tabulated in Exercise 9. At $p = 2$ , the eight-element group $Q_{2}^{\times} / (Q_{2}^{\times})^{2} ≅ (Z /2)^{3}$ carries the explicit pairing $(2^{α} u, 2^{β} w)_{2} = (- 1)^{ε (u) ε (w) + α ω (w) + β ω (u)}$ , which is non-degenerate by direct verification. The non-degeneracy on $Q_{v}$ is the input for the classification of quadratic forms over $Q_{v}$ via the Hasse-Witt invariant.

Theorem (Hasse-Witt invariant for quadratic forms; O'Meara §63). Let $f = ⟨ a_{1}, \dots, a_{n} ⟩$ be a diagonal quadratic form over $Q_{v}$ in $n$ variables, $a_{i} \in Q_{v}^{\times}$ . The Hasse-Witt invariant is $$ \epsilon_v(f) = \prod_{i < j} (a_i, a_j)_v \in {\pm 1}. $$ The invariant is a class function on $Q_{v}$ -equivalence classes of quadratic forms, and the pair $(dim f, det f, ϵ_{v} (f))$ together with the signature (at $v = \infty$ ) classifies quadratic forms over $Q_{v}$ up to equivalence. Globally, a quadratic form over $Q$ in $n \geq 3$ variables is equivalent to its negative iff $ϵ_{v} (f) = - 1$ at an even number of places.

This is the algebraic engine driving the Hasse-Minkowski theorem 21.02.08: equivalence of quadratic forms over $Q$ is detected place-by-place via the Hasse-Witt invariant, and the local-global principle says global equivalence is the conjunction of local equivalences.

Theorem (Brauer-group interpretation; Cassels-Fröhlich Ch. VI). For each place $v$ of $Q$ and each pair $a, b \in Q_{v}^{\times}$ , the quaternion algebra $$ \left(\frac{a, b}{\mathbb{Q}_v}\right) = \mathbb{Q}_v \langle i, j : i^2 = a, j^2 = b, i j = -j i \rangle $$ is a central simple algebra of dimension $4$ over $Q_{v}$ , and its class in the Brauer group $Br (Q_{v})$ is the unique non-identity element of order $2$ iff $(a, b)_{v} = - 1$ , and is the identity element of $Br (Q_{v})$ iff $(a, b)_{v} = + 1$ . The local invariant map $inv_{v} : Br (Q_{v}) \to Q / Z$ sends the quaternion-algebra class to $\frac{1}{2}$ when $(a, b)_{v} = - 1$ and to $0$ when $(a, b)_{v} = + 1$ .

For $v = \infty$ , $Br (R) = {\pm 1} \cdot Z /2$ , generated by the Hamilton quaternions $H$ ; the quaternion algebra splits ( $(a, b)_{\infty} = + 1$ , isomorphic to $M_{2} (R)$ ) iff at least one of $a, b$ is positive. For $v = p$ finite, $Br (Q_{p}) ≅ Q / Z$ by local class field theory (Hasse 1930), and the order-two subgroup is generated by quaternion algebras with $(a, b)_{p} = - 1$ .

Theorem (global reciprocity exact sequence; Albert-Brauer-Hasse-Noether 1932). There is an exact sequence $$ 0 \to \mathrm{Br}(\mathbb{Q}) \to \bigoplus_v \mathrm{Br}(\mathbb{Q}_v) \xrightarrow{\sum \mathrm{inv}_v} \mathbb{Q}/\mathbb{Z} \to 0 $$ expressing the global Brauer group of $Q$ as the kernel of the sum-of-local-invariants map. The Hilbert product formula $\prod_{v} (a, b)_{v} = 1$ is the order-two case of the central exactness statement: for the global quaternion algebra $(\frac{a , b}{Q})$ , the local invariants $inv_{v} \in {0, \frac{1}{2}}$ sum to zero in $Q / Z$ , which is exactly $\prod_{v} (a, b)_{v} = + 1$ .

This embedding of the product formula into the Brauer-group sequence is one of the foundational achievements of class field theory: it identifies a piece of elementary number theory (quadratic reciprocity) with a piece of cohomological algebra (the local-global Brauer-group sequence). The same machinery, applied to higher-degree central simple algebras, produces $n$ -th power reciprocity laws, and applied to elliptic curves and modular forms, produces the Tate-Shafarevich obstruction and the Brauer-Manin obstruction.

Theorem (cohomological interpretation; Neukirch Ch. V §3 + Ch. VI). Let $μ_{2} = {\pm 1} \subset Q_{v}^{\times}$ be the group of $2$ -nd roots of unity. The Kummer exact sequence $1 \to μ_{2} \to Q_{v}^{\times} (\cdot)^{2} Q_{v}^{\times} \to 1$ gives a connecting homomorphism $δ : Q_{v}^{\times} / (Q_{v}^{\times})^{2} \to H^{1} (Q_{v}, μ_{2})$ . The Hilbert symbol $(a, b)_{v}$ is the image of the cup product $δ a ⌣ δ b \in H^{2} (Q_{v}, μ_{2} \otimes μ_{2}) = H^{2} (Q_{v}, μ_{2})$ under the local invariant $inv_{v} : H^{2} (Q_{v}, μ_{2}) \sim \frac{1}{2} Z / Z$ .

The cohomological framing identifies the Hilbert symbol as the simplest cup product in Galois cohomology and the product formula as a global reciprocity statement: $\sum_{v} inv_{v} (δ a ⌣ δ b) = 0$ in $Q / Z$ for every $a, b \in Q^{\times}$ . The same identification places higher-power Hilbert symbols inside the Tate cohomology of local fields and connects to the Lubin-Tate formal group machinery via explicit reciprocity laws.

Theorem (higher-power Hilbert symbols; Tate 1976). Let $K$ be a local field containing a primitive $n$ -th root of unity $ζ_{n}$ . The $n$ -th power Hilbert symbol $$ (\cdot, \cdot)n : K^\times / (K^\times)^n \times K^\times / (K^\times)^n \to \mu_n $$ *is the bilinear pairing whose value $(a, b)_{n}$ is the image of $b$ under the local Artin map $θ_{K} : K^{\times} / N_{K (a^{1/ n}) / K} ((K (a^{1/ n}))^{\times}) \sim Gal (K (a^{1/ n}) / K) \subseteq μ_{n}$ . For a global field $K$ containing $μ_{n}$ and $a, b \in K^{\times}$ , the global product formula $\prod_v (a, b){n, v} = 1 $h o l d s, t h e p r o d u c tt ak e n o v er a l l pl a ceso f$ K$.*

The case $K = Q$ , $n = 2$ is the classical Hilbert reciprocity discussed above. The case $K = Q (ζ_{p}), n = p$ is the Kummer reciprocity law of cyclotomic-field arithmetic, central to Iwasawa theory 21.07.01. Lubin-Tate (Iwasawa 1968, Coleman 1979) gave explicit power-series formulas for the higher symbols, generalising the classical $(a, b)_{p}$ explicit formula to arbitrary $n$ .

Theorem (Hilbert symbol on $K_{2}$ ; Milnor-Tate). Let $K$ be a global field. The Hilbert symbol factors through the second Milnor $K$ -group: $$ \bigotimes^2 K^\times \to K_2^M(K), \qquad K_2^M(K) \to \prod_v \mu_v $$ where $K_{2}^{M} (K) = K^{\times} \otimes_{Z} K^{\times} / ⟨ a \otimes (1 - a) : a \neq = 0, 1 ⟩$ is the second Milnor $K$ -group and $μ_{v}$ is the group of roots of unity in $K_{v}$ . The image is exactly the kernel of the sum-of-local-invariants map, which is the content of the Moore reciprocity sequence (Moore 1968 Publ. Math. IHES 35).

This embedding of Hilbert reciprocity into algebraic $K$ -theory is one of the principal modern frames: the Steinberg relations ${a, 1 - a} = 0$ defining $K_{2}^{M}$ are exactly the Hilbert-symbol identities $(a, 1 - a)_{v} = 1$ and $(a, - a)_{v} = 1$ , and the universal property of $K_{2}^{M}$ packages all the local Hilbert symbols at all places into a single global invariant. Bass-Tate, Suslin, and Voevodsky's work on motivic cohomology continues this thread.

Synthesis. The Hilbert symbol packages the local-global content of quadratic reciprocity into a single bimultiplicative pairing $(\cdot, \cdot)_{v}$ at each place of $Q$ , and the product formula $\prod_{v} (a, b)_{v} = 1$ identifies the supplementary laws and the main reciprocity statement as the product over four building-block pairs $(- 1, - 1)$ , $(- 1, p)$ , $(2, p)$ , $(p, q)$ . The central insight is that this is exactly the case $n = 2$ of the global Brauer-group exact sequence $0 \to Br (Q) \to ⨁_{v} Br (Q_{v}) \to Q / Z \to 0$ from class field theory, in which the local invariants of a central simple algebra over $Q$ are constrained to sum to zero. The pattern recurs: every reciprocity law in number theory has the same shape, a global object decomposed into local invariants whose sum vanishes. Putting these together, the Hilbert symbol is one face of three identifications — Brauer-class invariant, cup product in Galois cohomology, Steinberg symbol in $K_{2}$ — that all encode the same local-global pairing on a number field. The bridge is the local Artin map: at each place $v$ , the local reciprocity isomorphism $Q_{v}^{\times} \sim W_{v}^{ab}$ of class field theory carries the Hilbert symbol to the cup product and to the Brauer invariant, identifies X with Y at the cohomological level, and propagates the local-global pattern to all higher-power and higher-dimensional reciprocity statements.

The same pattern appears again in 21.02.08 (Hasse-Minkowski), where the Hasse-Witt invariant $ϵ_{v} (f) = \prod_{i < j} (a_{i}, a_{j})_{v}$ of a quadratic form $f$ over $Q_{v}$ is built from Hilbert symbols and the local-global principle for quadratic-form equivalence becomes a constraint on the product of the local invariants. This pattern recurs in the Hasse principle for higher-degree forms (where it fails — the Tate-Shafarevich set of an elliptic curve measures the failure), in the Brauer-Manin obstruction to the Hasse principle (Manin 1971), and in the explicit reciprocity laws of Lubin-Tate type for $n$ -th power Hilbert symbols (Tate 1976; Coleman 1979). The bridge is that the Hilbert product formula is the seed: a single, computable, elementary statement at the bottom of the reciprocity hierarchy, from which the rest of class field theory grows. The foundational reason the formula holds is the local Artin map's compatibility across all places, identifying X with Y on the cohomological side and identifying the elementary number theory of Gauss with the abstract reciprocity of Artin and Hasse.

Full proof set Master

Theorem (product formula on $(- 1, - 1)$ ), proof. Compute place-by-place. At $\infty$ : $- 1 < 0$ twice, so $(- 1, - 1)_{\infty} = - 1$ . At $p = 2$ : $α = β = 0$ , $u = w = - 1 \equiv 7 (mod 8)$ , so $ε (- 1) = 1, ω (- 1) = 0$ . The 2-adic formula yields $(- 1)^{ε (- 1) ε (- 1) + 0 + 0} = (- 1)^{1 \cdot 1} = - 1$ . At $p$ odd: $α = β = 0$ , $u = w = - 1$ is a unit, so the odd- $p$ formula gives $(- 1, - 1)_{p} = + 1$ regardless of $p mod 4$ since the exponent $α β ε (p) = 0$ . Multiplying: $(- 1) (- 1) (+ 1) (+ 1) \dots = + 1$ . $□$

Theorem (product formula on $(p, q)$ for distinct odd primes, equivalent to quadratic reciprocity), proof. Following Step 5 of the main proof: the only places where $(p, q)_{v}$ can fail to be $+ 1$ are $v = 2, p, q$ . At $\infty$ , both positive, so $+ 1$ . At $r \neq = 2, p, q$ , both $r$ -adic units, so $+ 1$ by the unit-unit case of the odd- $r$ formula. The remaining three places contribute:

(a) $(p, q)_{2} = (- 1)^{ε (p) ε (q)} = (- 1)^{(p - 1) /2 \cdot (q - 1) /2}$ by the 2-adic formula with $α = β = 0$ .

(b) $(p, q)_{p}$ : $a = p$ has $α = 1, u = 1$ ; $b = q$ has $β = 0, w = q$ . The odd- $p$ formula gives $(p, q)_{p} = (- 1)^{1 \cdot 0 \cdot ε (p)} (\frac{1}{p})^{0} (\frac{q}{p})^{1} = (\frac{q}{p})$ .

(c) $(p, q)_{q}$ : by symmetry of the symbol, $(p, q)_{q} = (q, p)_{q} = (\frac{p}{q})$ (by the same odd- $p$ formula with roles swapped).

The product formula $\prod_{v} (p, q)_{v} = 1$ then reads $(- 1)^{(p - 1) (q - 1) /4} \cdot (\frac{q}{p}) \cdot (\frac{p}{q}) = 1$ , which rearranges to $$ \left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = (-1)^{(p-1)/2 \cdot (q-1)/2}, $$ the main statement of Gauss's quadratic reciprocity. Conversely, given quadratic reciprocity, the product on $(p, q)$ is $+ 1$ by the same identity. The product formula and quadratic reciprocity are equivalent on the building block $(p, q)$ . $□$

Theorem (the symbol depends only on square classes), proof. Let $a, b \in Q_{v}^{\times}$ and $c \in Q_{v}^{\times}$ . Show $(a c^{2}, b)_{v} = (a, b)_{v}$ . The equation $z^{2} = a c^{2} x^{2} + b y^{2} = a (c x)^{2} + b y^{2}$ has the same set of non-zero solutions as $z^{2} = a X^{2} + b y^{2}$ under the substitution $X = c x$ (a bijection since $c \neq = 0$ ). So the non-zero solubility is identical, and the symbol agrees. The same argument with $b$ replaced by $b d^{2}$ gives the second invariance. $□$

Theorem (bimultiplicativity), proof. Let $a, a^{'}, b \in Q_{v}^{\times}$ . Apply the norm characterisation: $(a, b)_{v} = + 1$ iff $b \in N_{K_{a} / Q_{v}} (K_{a}^{\times})$ where $K_{a} = Q_{v} (a)$ . Define $χ_{b} : Q_{v}^{\times} \to {\pm 1}$ by $χ_{b} (a) = (a, b)_{v}$ . The kernel of $χ_{b}$ is ${a : b is a norm from K_{a}}$ , which by local class field theory (or direct check for $Q_{v}$ ) is the index-two subgroup $N_{b} = N_{K_{b} / Q_{v}} (K_{b}^{\times})$ when $b$ is a non-square, and $Q_{v}^{\times}$ when $b$ is a square.

In either case, $χ_{b}$ is a ${\pm 1}$ -valued character on the quotient $Q_{v}^{\times} / N_{b}$ , which has order $\leq 2$ . A function on a finite abelian group taking values in ${\pm 1}$ with kernel a subgroup of index $\leq 2$ is a group homomorphism, so $χ_{b} (a a^{'}) = χ_{b} (a) χ_{b} (a^{'})$ , i.e., $(a a^{'}, b)_{v} = (a, b)_{v} \cdot (a^{'}, b)_{v}$ . By symmetry of the symbol, the same holds in the second argument. $□$

Theorem (symmetry), proof. $(a, b)_{v} = + 1$ iff the equation $z^{2} = a x^{2} + b y^{2}$ has a non-zero solution, and $(b, a)_{v} = + 1$ iff $z^{2} = b x^{2} + a y^{2}$ has a non-zero solution. The substitution $(x, y) \mapsto (y, x)$ is a bijection on $Q_{v}^{2}$ that converts one equation to the other. So the symbols are equal. $□$

Theorem (negation identity), proof. $z^{2} = a x^{2} + (- a) y^{2} = a (x^{2} - y^{2})$ . Take $(x, y, z) = (1, 1, 0)$ : then $z^{2} = 0 = a (1 - 1) = a x^{2} - a y^{2}$ , valid for any $a$ . The solution is non-zero (the first coordinate is $1$ ), so $(a, - a)_{v} = + 1$ at every place. $□$

Theorem (complementarity identity), proof. Substitute $(x, y, z) = (1, 1, 1)$ : $z^{2} = 1, a x^{2} + (1 - a) y^{2} = a + (1 - a) = 1$ . Match. The triple $(1, 1, 1)$ is non-zero, so $(a, 1 - a)_{v} = + 1$ at every place, for any $a \neq = 0, 1$ . $□$

Theorem (product formula on $(- 1, p)$ for $p$ odd, equivalent to first supplementary law), proof. At $\infty$ : $- 1 < 0$ but $p > 0$ , so $(- 1, p)_{\infty} = + 1$ . At $r \neq = 2, p$ : both units, so $+ 1$ . At $p$ : $a = - 1$ unit, $b = p$ valuation $1$ ; the odd- $p$ formula gives $(\frac{- 1}{p})$ . At $2$ : $a = - 1, b = p$ , both $α = β = 0$ , $u = - 1, w = p$ ; $ε (- 1) = 1, ε (p) = (p - 1) /2 mod 2$ . The 2-adic formula gives $(- 1)^{1 \cdot (p - 1) /2} = (- 1)^{(p - 1) /2}$ . Product: $(- 1)^{(p - 1) /2} \cdot (\frac{- 1}{p}) = + 1$ , so $(\frac{- 1}{p}) = (- 1)^{(p - 1) /2}$ — the first supplementary law. $□$

Theorem (product formula on $(2, p)$ for $p$ odd, equivalent to second supplementary law), proof. Worked in Exercise 8. At $\infty$ : $+ 1$ . At $r \neq = 2, p$ : $+ 1$ . At $p$ : odd- $p$ formula with $α = 0, u = 2, β = 1, w = 1$ gives $(\frac{2}{p})$ . At $2$ : 2-adic formula with $α = 1, u = 1, β = 0, w = p$ gives $(- 1)^{1 \cdot ω (p)} = (- 1)^{(p^{2} - 1) /8}$ . Product: $(- 1)^{(p^{2} - 1) /8} \cdot (\frac{2}{p}) = + 1$ , so $(\frac{2}{p}) = (- 1)^{(p^{2} - 1) /8}$ — the second supplementary law. $□$

Theorem (Brauer-group correspondence), proof sketch. The class of the quaternion algebra $(\frac{a , b}{Q _{v}})$ in $Br (Q_{v})$ is the cohomology class $δ a ⌣ δ b \in H^{2} (Q_{v}, G_{m})$ under the cup product induced by the Kummer sequence. The algebra is split (identity Brauer class) iff it admits a faithful representation as $M_{2} (Q_{v})$ iff its reduced norm form $z^{2} - a x^{2} - b y^{2} + ab w^{2}$ vanishes for some non-zero argument iff $z^{2} = a x^{2} + b y^{2}$ has a non-zero solution iff $(a, b)_{v} = + 1$ . The local invariant $inv_{v}$ on $Br (Q_{v}) ≅ Q / Z$ sends the class to $0$ if split and to $\frac{1}{2}$ if not; this matches the $(a, b)_{v} \in {\pm 1} ≅ \frac{1}{2} Z / Z$ identification. The full proof requires Hasse's local class field theory (Hasse 1930); see Cassels-Fröhlich Ch. VI for details. $□$

Theorem (global reciprocity exact sequence), stated without proof — see Albert-Brauer-Hasse-Noether 1932 ^{[source pending]}. The exact sequence $0 \to Br (Q) \to ⨁_{v} Br (Q_{v}) \to Q / Z \to 0$ requires the full machinery of global class field theory. The case of order-two classes (quaternion algebras) reduces to the Hilbert product formula as proved above; the general case requires the Tate cohomology of the idele class group and the Tate-Nakayama duality. $□$

Connections Master

Absolute value and triangle inequality 00.01.02. The Hilbert symbol is parameterised by the places $v$ of $Q$ , which by Ostrowski's theorem are exactly the absolute values $∣ \cdot ∣_{v}$ on $Q$ : one Archimedean and one per prime. The completion $Q_{v}$ in which the equation $z^{2} = a x^{2} + b y^{2}$ is solved is the field obtained from $Q$ by completing in the metric $∣ x - y ∣_{v}$ . The very notion of "place" depends on the framework of absolute values established in the foundational unit.
Group 01.02.01. The Hilbert symbol takes values in the two-element group ${\pm 1} ≅ Z /2$ , and bimultiplicativity makes it a $Z /2$ -valued bilinear pairing on the abelian groups $Q_{v}^{\times} / (Q_{v}^{\times})^{2}$ . The product formula $\prod_{v} (a, b)_{v} = 1$ is a vanishing statement in the same abelian group, expressing a balance condition across infinitely many places. Every step of the proof uses group-theoretic structure: the square-class group is a finite $F_{2}$ -vector space, the pairing is bilinear over $F_{2}$ , and the local Artin map identifies $Q_{v}^{\times} / N (K^{\times})$ with $Gal (K / Q_{v})$ for quadratic $K$ .
Algebraic field extension 01.02.12. The equivalent definition of $(a, b)_{v}$ as " $b$ is a norm from $Q_{v} (a)$ " uses the language of quadratic field extensions and the norm map $N_{K / F}$ . The cases ( $a$ a square in $Q_{v}$ — split algebra; $a$ a non-square — quadratic extension) decompose the symbol via the structure of finite separable algebras over a local field. The Galois group $Gal (K / Q_{v}) ≅ Z /2$ acts on $K$ by sending $a \mapsto - a$ , and the norm $N (u + w a) = u^{2} - a w^{2}$ is the product of the two Galois conjugates.
Quadratic forms over $Q_{p}$ 21.02.07. The Hasse-Witt invariant of a diagonal quadratic form $⟨ a_{1}, \dots, a_{n} ⟩$ over $Q_{v}$ is $ϵ_{v} (f) = \prod_{i < j} (a_{i}, a_{j})_{v}$ , a product of Hilbert symbols. The local equivalence class of a quadratic form is determined by its dimension, determinant, signature (at $\infty$ ), and Hasse-Witt invariant — and the product formula on the Hilbert symbol propagates to a product formula on the Hasse invariant across places.
Hasse-Minkowski theorem 21.02.08. The local-global principle for quadratic forms over $Q$ says that a form represents zero globally iff it represents zero at every place. The Hilbert symbol packages the local information at each place; the product formula is the constraint among the local invariants; the Hasse-Minkowski theorem is the statement that, once the local conditions and the product-formula constraint are satisfied, a global solution exists. The Hilbert product formula is the first concrete instance of a class-field-theoretic local-global principle.
$p$ -adic $L$ -functions and Iwasawa Main Conjecture 21.07.02. The $n$ -th power Hilbert symbol on a cyclotomic field $Q (ζ_{p})$ appears in the explicit reciprocity laws of Lubin-Tate type (Iwasawa 1968, Coleman 1979, Wiles 1978), which compute the symbol modulo $p^{k}$ using formal-group-law machinery. These explicit formulas underpin the construction of the $p$ -adic $L$ -functions of Kubota-Leopoldt and the Main Conjecture, connecting the Hilbert symbol's $K$ -theoretic / Galois-cohomological framing to the analytic continuation of $p$ -adic $L$ -values.

Historical & philosophical context Master

David Hilbert introduced the symbol now bearing his name in the Zahlbericht of 1897 ^{[source pending]}, the encyclopaedic Jahresbericht der Deutschen Mathematiker-Vereinigung report on algebraic number theory that organised the field for a generation of subsequent researchers. In §64 of the Zahlbericht, Hilbert defined the symbol $(n, m)$ for $n, m$ rationals (or elements of a number field) by a norm-residue condition: $(n, m) = 1$ if $n$ is congruent to a norm from $Q (m)$ modulo some power of $m$ , and $- 1$ otherwise. The formulation was archaic by modern standards — Hilbert worked one prime at a time, before the notion of a place of a global field had been packaged — but the essential local-global content was already present. In a companion paper in Mathematische Annalen 51 (1898) ^{[source pending]}, and in the famous Problem 9 of the ICM 1900 problem list, Hilbert formulated the product formula $\prod (n, m)_{p} = 1$ and proposed it as the generalisation of quadratic reciprocity to arbitrary number fields, with the n-th power analogue as the long-range goal.

The modern formulation, as a function $(a, b)_{v}$ on each completion $Q_{v}$ taking values in ${\pm 1}$ , was assembled by Helmut Hasse in two foundational papers in the Journal für die reine und angewandte Mathematik 152 (1924) ^{[source pending]}, where Hasse introduced the term Normenrestsymbol (norm-residue symbol) and gave the product formula its standard exposition as the local-global form of Hilbert reciprocity. Hasse's framing made the symbol's role as the local invariant of a Brauer class explicit, and his subsequent work (Hasse 1930 on local class field theory; Albert-Brauer-Hasse-Noether 1932 on the global Brauer-group sequence ^{[source pending]}) embedded the Hilbert symbol into the broader edifice of class field theory. Emil Artin's 1927 reciprocity theorem identified the abstract reciprocity isomorphism $Q_{v}^{\times} \sim W_{v}^{ab}$ at each place, of which the Hilbert symbol is the order-two specialisation. Philipp Furtwängler proved the existence theorem (1909, Math. Ann. 67) underlying the n-th power case.

The canonical short-text exposition is Jean-Pierre Serre's Cours d'arithmétique (Presses Universitaires de France, 1970; English translation as A Course in Arithmetic, Springer GTM 7, 1973) ^{[source pending]}, whose Ch. III packages the Hilbert symbol, its bimultiplicativity, its explicit formulas at every place, and the equivalence of the product formula with quadratic reciprocity in eight printed pages. Serre's exposition has shaped how the topic is taught in graduate algebraic number theory courses for fifty years. The Galois-cohomological framing as cup product in $H^{2} (Q_{v}, μ_{2})$ was developed in the Cassels-Fröhlich Brighton volume (1967) and codified in Neukirch's Algebraische Zahlentheorie (Springer Grundlehren 322, 1992; English translation 1999); the $K$ -theoretic framing in $K_{2}$ of a global field traces to Moore's reciprocity sequence (Moore 1968 Publ. Math. IHES 35) and the foundational work of Milnor, Tate, and Bass. John Tate's paper on the relation between $K_{2}$ and Galois cohomology in Inventiones Mathematicae 36 (1976) ^{[source pending]} established the higher-power Hilbert symbols as $K_{2}$ -elements and tied the explicit reciprocity laws of Lubin-Tate type to the formal-group-theoretic structure of local fields.

Bibliography Master

@article{Hilbert1897Zahlbericht,
  author  = {Hilbert, David},
  title   = {Die Theorie der algebraischen {Z}ahlk\"orper},
  journal = {Jahresbericht der Deutschen Mathematiker-Vereinigung},
  volume  = {4},
  year    = {1897},
  pages   = {175--546}
}

@article{Hilbert1898RelativQuadratisch,
  author  = {Hilbert, David},
  title   = {\"Uber die Theorie des relativ-quadratischen {Z}ahlk\"orpers},
  journal = {Mathematische Annalen},
  volume  = {51},
  year    = {1898},
  pages   = {1--127}
}

@article{Hasse1924SymmetrischeMatrizen,
  author  = {Hasse, Helmut},
  title   = {Symmetrische {M}atrizen im {K}\"orper der rationalen {Z}ahlen},
  journal = {Journal f\"ur die reine und angewandte Mathematik},
  volume  = {152},
  year    = {1924},
  pages   = {205--224}
}

@article{Hasse1924Aquivalenz,
  author  = {Hasse, Helmut},
  title   = {\"Uber die \"Aquivalenz quadratischer {F}ormen im {K}\"orper der rationalen {Z}ahlen},
  journal = {Journal f\"ur die reine und angewandte Mathematik},
  volume  = {152},
  year    = {1924},
  pages   = {205--224}
}

@article{ABHN1932,
  author  = {Albert, A. A. and Brauer, R. and Hasse, H. and Noether, E.},
  title   = {Beweis eines {H}auptsatzes in der {T}heorie der {A}lgebren},
  journal = {Journal f\"ur die reine und angewandte Mathematik},
  volume  = {167},
  year    = {1932},
  pages   = {399--404}
}

@article{Artin1927Reciprocity,
  author  = {Artin, Emil},
  title   = {Beweis des allgemeinen {R}eziprozit\"atsgesetzes},
  journal = {Abhandlungen aus dem Mathematischen Seminar der Universit\"at Hamburg},
  volume  = {5},
  year    = {1927},
  pages   = {353--363}
}

@article{Furtwangler1909,
  author  = {Furtw\"angler, Philipp},
  title   = {Allgemeiner {E}xistenzbeweis f\"ur den {K}lassenk\"orper eines beliebigen algebraischen {Z}ahlk\"orpers},
  journal = {Mathematische Annalen},
  volume  = {67},
  year    = {1909},
  pages   = {1--31}
}

@book{SerreCoursArithmetique,
  author    = {Serre, Jean-Pierre},
  title     = {A {C}ourse in {A}rithmetic},
  publisher = {Springer-Verlag},
  series    = {Graduate Texts in Mathematics},
  volume    = {7},
  year      = {1973},
  note      = {French original {\em Cours d'arithm\'etique}, Presses Universitaires de France, 1970}
}

@book{CasselsFrohlich,
  editor    = {Cassels, J. W. S. and Fr\"ohlich, A.},
  title     = {Algebraic {N}umber {T}heory},
  publisher = {Academic Press},
  year      = {1967},
  note      = {Proceedings of the {B}righton conference}
}

@book{NeukirchANT,
  author    = {Neukirch, J\"urgen},
  title     = {Algebraic {N}umber {T}heory},
  publisher = {Springer-Verlag},
  series    = {Grundlehren der mathematischen Wissenschaften},
  volume    = {322},
  year      = {1999}
}

@book{OMearaQuadraticForms,
  author    = {O'Meara, O. T.},
  title     = {Introduction to {Q}uadratic {F}orms},
  publisher = {Springer-Verlag},
  series    = {Grundlehren der mathematischen Wissenschaften},
  volume    = {117},
  year      = {1963}
}

@article{Tate1976K2,
  author  = {Tate, John},
  title   = {Relations between {$K_2$} and {G}alois cohomology},
  journal = {Inventiones Mathematicae},
  volume  = {36},
  year    = {1976},
  pages   = {257--274}
}

@article{Moore1968,
  author  = {Moore, Calvin C.},
  title   = {Group extensions of $p$-adic and adelic linear groups},
  journal = {Publications Math\'ematiques de l'IH\'ES},
  volume  = {35},
  year    = {1968},
  pages   = {157--222}
}

@unpublished{MilneCFT,
  author = {Milne, James S.},
  title  = {Class {F}ield {T}heory},
  year   = {2020},
  note   = {Lecture notes, v4.03, \url{https://www.jmilne.org/math/CourseNotes/cft.html}}
}

@article{Hasse1930LocalCFT,
  author  = {Hasse, Helmut},
  title   = {Die {N}ormenresttheorie relativ-{A}belscher {Z}ahlk\"orper als {K}lassenk\"orpertheorie im {K}leinen},
  journal = {Journal f\"ur die reine und angewandte Mathematik},
  volume  = {162},
  year    = {1930},
  pages   = {145--154}
}

@article{Iwasawa1968,
  author  = {Iwasawa, Kenkichi},
  title   = {On explicit formulas for the norm residue symbol},
  journal = {Journal of the Mathematical Society of Japan},
  volume  = {20},
  year    = {1968},
  pages   = {151--164}
}

@article{Coleman1979,
  author  = {Coleman, Robert F.},
  title   = {Division values in local fields},
  journal = {Inventiones Mathematicae},
  volume  = {53},
  year    = {1979},
  pages   = {91--116}
}

Prerequisites

00.01.02
01.02.01
01.02.12

Tier anchors

beginner: Borevich-Shafarevich *Number Theory* (Academic Press 1966) Ch. 1 §6 informal opening on Hilbert reciprocity; Serre *Cours d'arithmétique* Ch. III opening paragraphs
intermediate: Serre 1973 *A Course in Arithmetic* (Springer GTM 7) Ch. III §1-§2 — the canonical eight-page treatment of the local symbol, its bimultiplicativity and computation tables, and the product formula
master: Serre Ch. III §2 (product formula via quadratic reciprocity); Cassels-Fröhlich 1967 *Algebraic Number Theory* Ch. VI–VII (class-field-theoretic embedding via the local invariant); Neukirch 1999 *Algebraic Number Theory* Ch. V §3 + Ch. VI (Galois-cohomological framing as cup product in $H^2(\mathbb{Q}_v, \mu_2)$); O'Meara 1963 *Introduction to Quadratic Forms* §63 (link to Hasse-Witt invariant); Tate 1976 *Invent. Math.* 36 (higher-degree symbols, explicit reciprocity laws)

References

Hilbert, D. — Die Theorie der algebraischen Zahlkörper (Zahlbericht) · *Jahresbericht der Deutschen Mathematiker-Vereinigung* 4 (1897), 175-546. The §64 norm-residue definition of the symbol $(n, m)$ for a quadratic extension $\mathbb{Q}(\sqrt{m})$, characterised by the local-norm property; the reciprocity statement $\prod_v (a, b)_v = 1$ is the case $n = 2$ of the symbol's product formula, anticipating the higher-power statements of class field theory.
Hilbert, D. — Über die Theorie des relativ-quadratischen Zahlkörpers · *Mathematische Annalen* 51 (1898), 1-127. Companion paper to Hilbert's ICM 1900 problem-list address (Problem 9, Hilbert reciprocity). Statement and partial proof of the product formula for the quadratic symbol over a number field, with the genuine local-global character of the symbol made explicit.
Hasse, H. — Symmetrische Matrizen im Körper der rationalen Zahlen · *Journal für die reine und angewandte Mathematik* 152 (1924), 205-224. The Hasse-Minkowski theorem and the norm-residue symbol as the local invariant; the product formula in the language of norm residues that became the standard exposition.
Hasse, H. — Über die Äquivalenz quadratischer Formen im Körper der rationalen Zahlen · *Journal für die reine und angewandte Mathematik* 152 (1924), 205-224 + 153 (1924), 158-162. Generalisation to $n$-th power Hilbert symbols over a local field; the product formula as the abstract reciprocity law of class field theory.
Serre, J.-P. — Cours d'arithmétique · Presses Universitaires de France (1970), English translation *A Course in Arithmetic*, Springer GTM 7 (1973). Ch. III §1 (definition, bimultiplicativity, computation tables at every place of $\mathbb{Q}$); Ch. III §2 (the product formula and its equivalence with quadratic reciprocity, eight printed pages); the canonical short-text exposition.
Cassels, J. W. S. and Fröhlich, A. (eds.) — Algebraic Number Theory · Academic Press (1967), Brighton conference proceedings. Ch. VI (local class field theory, the local invariant of a Brauer class); Ch. VII (global class field theory, the reciprocity sequence $0 \to \mathrm{Br}(K) \to \bigoplus_v \mathrm{Br}(K_v) \to \mathbb{Q}/\mathbb{Z} \to 0$); the canonical reference embedding the Hilbert symbol into the Brauer-group framework.
Neukirch, J. — Algebraische Zahlentheorie · Springer Grundlehren 322 (English translation 1999). Ch. V §3 (the Hilbert symbol $(a, b)_v$ via the local Artin map); Ch. VI (cup product in Galois cohomology $H^2(\mathbb{Q}_v, \mu_2)$, the Tate duality framing); modern cohomological presentation.
O'Meara, O. T. — Introduction to Quadratic Forms · Springer Grundlehren 117 (1963). §63 (Hilbert symbol as the building block of the Hasse-Witt invariant $\epsilon(f) = \prod_{i < j} (a_i, a_j)_v$ for a diagonal form $\langle a_1, \ldots, a_n \rangle$); the symbol-based classification of quadratic forms over $\mathbb{Q}_v$.
Tate, J. — Relations between K_2 and Galois cohomology · *Inventiones Mathematicae* 36 (1976), 257-274. Higher-power Hilbert symbols and their interpretation as elements of $K_2$ of a local field; the explicit reciprocity laws of Lubin-Tate type.
Milne, J. S. — Class Field Theory · Online lecture notes v4.03 (2020), https://www.jmilne.org/math/CourseNotes/cft.html. §III.4 (Hilbert symbol via the local Artin map); §IV.5 (the global reciprocity exact sequence); the canonical modern online reference.
Albert, A. A., Brauer, R., Hasse, H., and Noether, E. — Beweis eines Hauptsatzes in der Theorie der Algebren · *Journal für die reine und angewandte Mathematik* 167 (1932), 399-404. The four-author paper proving that a central simple algebra over $\mathbb{Q}$ splits iff it splits at every place; the global Brauer-group exact sequence; the product formula for the Hasse invariant generalising Hilbert reciprocity.

Estimated time

beginner: 18m
intermediate: 40m
master: 80m