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

Quadratic reciprocity via Gauss sums

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Anchor (Master): Gauss 1801 *Disquisitiones Arithmeticae* §§107-150 (six distinct proofs of quadratic reciprocity, originator); Eisenstein 1844 *J. reine angew. Math.* 28 (sine-product geometric proof); Eisenstein 1845 *J. reine angew. Math.* 29 (proof via $p$-th roots of unity and Gauss sums; the prototype of the present unit); Artin 1927 *Abh. Math. Sem. Univ. Hamburg* 5, 353-363 (general Artin reciprocity); Serre 1973 *A Course in Arithmetic* (GTM 7) Ch. I §3; Ireland-Rosen 1990 *A Classical Introduction to Modern Number Theory* (GTM 84) Ch. 5-6; Lang 1994 *Algebraic Number Theory* (GTM 110, 2nd ed.) Ch. IV §3; Lemmermeyer 2000 *Reciprocity Laws: From Euler to Eisenstein* (Springer Monographs); Hilbert 1897 *Zahlbericht* §64 on the product formula for the quadratic Hilbert symbol; Tate 1967 *Algebraic Number Theory* (Cassels-Fröhlich) on the Hilbert symbol and class field theory framing

Intuition Beginner

A natural question in number theory is whether a given integer $a$ is a perfect square modulo a prime $p$ . For $p = 7$ the squares modulo $7$ are $1, 2, 4$ (since $1^{2} = 1$ , $2^{2} = 4$ , $3^{2} = 2$ , and so on); the non-squares are $3, 5, 6$ . So $2$ is a square mod $7$ but $3$ is not. The Legendre symbol $(\frac{a}{p})$ packages this question into a single value: $+ 1$ when $a$ is a non-zero square mod $p$ , $- 1$ when it is a non-square, and $0$ when $p$ divides $a$ .

Quadratic reciprocity is one of the most surprising patterns in all of number theory. It says that the answer to "is $p$ a square mod $q$ ?" is essentially the same as the answer to "is $q$ a square mod $p$ ?", with at most a sign change controlled by the residues of $p$ and $q$ modulo $4$ . For two odd primes the rule is $(\frac{p}{q}) (\frac{q}{p}) = (- 1)^{(p - 1) (q - 1) /4}$ . The two sides have nothing to do with each other on the surface — one lives in arithmetic mod $p$ , the other in arithmetic mod $q$ — yet they conspire.

Gauss called this theorema aureum, the golden theorem, and produced six distinct proofs of it over his lifetime. The proof you will see uses a clever sum involving roots of unity, called a Gauss sum, whose square turns out to be $p$ up to a sign. Once you know the square, you can extract the reciprocity law by computing in two different ways inside a cyclotomic field.

Visual Beginner

A schematic showing two perpendicular axes labelled "mod $p$ " and "mod $q$ " for two odd primes $p$ and $q$ . On the mod- $p$ axis the residues $1, 2, \dots, p - 1$ are split into squares (filled circles) and non-squares (open circles), and similarly on the mod- $q$ axis. Two highlighted points $q$ on the mod- $p$ axis and $p$ on the mod- $q$ axis are joined by a curved arrow. The arrow is labelled with the sign $(- 1)^{(p - 1) (q - 1) /4}$ , which is $+ 1$ unless both $p$ and $q$ are congruent to $3$ modulo $4$ .

The picture captures the symmetry: the two questions "is $q$ a square mod $p$ ?" and "is $p$ a square mod $q$ ?" have the same answer when at least one of $p, q$ is congruent to $1$ mod $4$ , and opposite answers when both are congruent to $3$ mod $4$ .

Worked example Beginner

Compute $(\frac{3}{17})$ and $(\frac{5}{29})$ by reciprocity.

Step 1. Squares modulo $17$ . The non-zero squares mod $17$ are ${1, 2, 4, 8, 9, 13, 15, 16}$ . So $3$ is not in the list of squares mod $17$ , which means $(\frac{3}{17}) = - 1$ . The direct check confirms the answer; the point of reciprocity is that we can also reach it without enumerating squares.

Step 2. Reciprocity for $(\frac{3}{17})$ . The primes are $p = 3$ and $q = 17$ . Since $17 \equiv 1 (mod 4)$ , the reciprocity sign $(- 1)^{(p - 1) (q - 1) /4} = (- 1)^{2 \cdot 16/4} = (- 1)^{8} = + 1$ . So $(\frac{3}{17}) = (\frac{17}{3})$ .

Step 3. Reduce. $17 \equiv 2 (mod 3)$ . The non-zero squares mod $3$ are ${1}$ , so $2$ is not a square mod $3$ and $(\frac{2}{3}) = - 1$ . Therefore $(\frac{17}{3}) = - 1$ , and $(\frac{3}{17}) = - 1$ . Agrees with Step 1.

Step 4. Reciprocity for $(\frac{5}{29})$ . The primes are $p = 5$ and $q = 29$ . Both are congruent to $1$ mod $4$ , so the reciprocity sign is $(- 1)^{4 \cdot 28/4} = (- 1)^{28} = + 1$ . So $(\frac{5}{29}) = (\frac{29}{5})$ .

Step 5. Reduce $29 (mod 5)$ : $29 = 5 \cdot 5 + 4$ , so $29 \equiv 4 (mod 5)$ . The non-zero squares mod $5$ are ${1, 4}$ , and $4 = 2^{2}$ is a square. Therefore $(\frac{29}{5}) = + 1$ , and $(\frac{5}{29}) = + 1$ .

Step 6. Check by enumeration: the non-zero squares mod $29$ are ${1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28}$ . So $5$ is indeed a square mod $29$ ; in fact $5 = 1 1^{2} (mod 29)$ since $1 1^{2} = 121 = 4 \cdot 29 + 5$ . Both methods agree.

What this tells us: reciprocity reduces a question about residues modulo a large prime to a question about residues modulo a smaller one. Combined with the supplementary laws for $(\frac{- 1}{p})$ and $(\frac{2}{p})$ , every Legendre symbol becomes computable by repeated reduction. The cost of each step is a small remainder calculation, and the answer follows in a handful of steps regardless of how big the primes were to begin with.

Check your understanding Beginner

Formal definition Intermediate+

Let $p$ be an odd prime. The Legendre symbol is the function $$ \left(\frac{a}{p}\right) = \begin{cases} +1 & \text{if } a \text{ is a non-zero square in } \mathbb{F}_p, \ -1 & \text{if } a \text{ is a non-square in } \mathbb{F}_p^\times, \ 0 & \text{if } p \mid a. \end{cases} $$ The symbol depends only on $a$ modulo $p$ and is multiplicative: $(\frac{ab}{p}) = (\frac{a}{p}) (\frac{b}{p})$ for all integers $a, b$ . Multiplicativity reflects the fact that the squares form an index- $2$ subgroup of the cyclic group $F_{p}^{\times}$ , so the symbol is the unique non-identity homomorphism $F_{p}^{\times} \to {\pm 1}$ extended to $Z$ by zero on multiples of $p$ .

Euler's criterion identifies the Legendre symbol with a power residue: $$ \left(\frac{a}{p}\right) \equiv a^{(p-1)/2} \pmod p \qquad \text{for } \gcd(a, p) = 1. $$ Both sides take values in ${\pm 1}$ when computed modulo $p$ , and the congruence is an equality of integers.

Fix $ζ_{p} = e^{2 π i / p}$ , a primitive $p$ -th root of unity in $C$ . The quadratic Gauss sum is $$ g_a = \sum_{x \in \mathbb{F}_p} \zeta_p^{a x^2} \in \mathbb{Z}[\zeta_p] $$ for $a \in F_{p}^{\times}$ , and we write $g = g_{1}$ . An equivalent expression is $$ g_a = \sum_{t \in \mathbb{F}_p} \left(\frac{t}{p}\right) \zeta_p^{a t}, $$ obtained by reorganising the first sum according to the value $t = x^{2}$ : each non-zero square contributes twice and each non-square contributes zero, so the corrected sum has Legendre-symbol weights.

The quadratic reciprocity law for two distinct odd primes $p$ and $q$ is the identity $$ \left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2} \cdot \frac{q-1}{2}}. $$ The sign on the right is $+ 1$ unless both $p$ and $q$ are congruent to $3$ modulo $4$ .

Write $p^{*} = (- 1)^{(p - 1) /2} p$ : this is $p$ when $p \equiv 1 (mod 4)$ and $- p$ when $p \equiv 3 (mod 4)$ . The arithmetic of $p^{*}$ is what the Gauss sum sees; the field $Q (p^{*})$ is the unique quadratic subfield of the cyclotomic field $Q (ζ_{p})$ , and this inclusion is what couples the two primes.

Counterexamples to common slips

The Legendre symbol $(\frac{a}{p})$ is only defined for $p$ odd; the symbol $(\frac{a}{2})$ has no useful meaning at the prime $2$ , and reciprocity at $p = 2$ is captured by the supplementary law for $(\frac{2}{p})$ instead.
Euler's criterion is a congruence modulo $p$ , not a literal equality of complex numbers; one must reduce $a^{(p - 1) /2}$ mod $p$ before reading off $\pm 1$ .
The Gauss sum $g_{a}$ depends on the choice of $ζ_{p}$ , but $g_{a}^{2} = p^{*}$ does not. Replacing $ζ_{p}$ by another primitive $p$ -th root of unity $ζ_{p}^{c}$ scales $g$ by $(\frac{c}{p})$ , and squaring removes the sign.

Key theorem with proof Intermediate+

Theorem (Euler's criterion). For an odd prime $p$ and $a \in F_{p}^{\times}$ , $$ a^{(p-1)/2} \equiv \left(\frac{a}{p}\right) \pmod p. $$

Proof. Fermat's little theorem gives $a^{p - 1} \equiv 1 (mod p)$ for every $a \in F_{p}^{\times}$ . Factoring as a difference of squares in $F_{p} [X]$ , the polynomial $X^{p - 1} - 1$ splits as $(X^{(p - 1) /2} - 1) (X^{(p - 1) /2} + 1)$ . So $a^{(p - 1) /2}$ is a root of one of the two factors, meaning $a^{(p - 1) /2} \equiv \pm 1 (mod p)$ .

If $a = b^{2}$ is a non-zero square, then $a^{(p - 1) /2} = b^{p - 1} \equiv 1 (mod p)$ by Fermat. So the squares lie in the $+ 1$ factor. There are $(p - 1) /2$ squares and the polynomial $X^{(p - 1) /2} - 1$ has at most $(p - 1) /2$ roots in the field $F_{p}$ , so the squares are exactly the roots of $X^{(p - 1) /2} - 1$ . The remaining $(p - 1) /2$ non-squares are forced into the other factor, giving $a^{(p - 1) /2} \equiv - 1 (mod p)$ for non-squares. Both cases combine into the single congruence $a^{(p - 1) /2} \equiv (\frac{a}{p}) (mod p)$ . $□$

Theorem (Gauss-sum identity). Let $p$ be an odd prime, $ζ_{p} = e^{2 π i / p}$ , and $g = \sum_{x \in F_{p}} ζ_{p}^{x^{2}}$ the quadratic Gauss sum. Then $$ g^2 = (-1)^{(p-1)/2} p = p^*. $$

Proof. Use the Legendre-weighted form $g = \sum_{t \in F_{p}} (\frac{t}{p}) ζ_{p}^{t}$ (the equivalence with the squared-exponent form is the substitution $t = x^{2}$ noted above; the $t = 0$ term contributes $0 \cdot ζ_{p}^{0} = 0$ on the Legendre side and one extra $ζ_{p}^{0} = 1$ on the squared side, but the Legendre side accumulates each $t \neq = 0$ either $+ 1$ or $- 1$ times via the Legendre symbol while the squared side accumulates each non-zero square twice and each non-square zero times, so the two sums differ only by the $x = 0$ contribution which is $1$ on the squared side; we adjust by writing the equivalent form $g = - 1 + 2 \sum_{x \neq = 0} ζ_{p}^{x^{2}}$ when convenient, or absorb the difference into the calculation).

Compute the product $g \cdot \overset{g}{ˉ}$ where $\overset{g}{ˉ}$ is the complex conjugate. Since $\overset{ˉ}{ζ}_{p} = ζ_{p}^{- 1}$ and $\overline{(\frac{t}{p})} = (\frac{t}{p})$ (Legendre symbols are real), $$ \bar g = \sum_{t} \left(\frac{t}{p}\right) \zeta_p^{-t} = \sum_{u} \left(\frac{-u}{p}\right) \zeta_p^{u} = \left(\frac{-1}{p}\right) g $$ substituting $u = - t$ and using multiplicativity of the Legendre symbol. So $g \overset{g}{ˉ} = (\frac{- 1}{p}) g^{2}$ and also $g \overset{g}{ˉ} = ∣ g ∣^{2} \geq 0$ , whence $(\frac{- 1}{p}) g^{2} = ∣ g ∣^{2}$ .

Now compute $∣ g ∣^{2}$ directly: $$ |g|^2 = g \bar g = \sum_{s, t} \left(\frac{s}{p}\right) \left(\frac{t}{p}\right) \zeta_p^{s - t} = \sum_{s, t} \left(\frac{s t}{p}\right) \zeta_p^{s - t} $$ by multiplicativity. Substitute $s = u t$ for $t \neq = 0$ (and note that the $t = 0$ terms vanish because $(\frac{0}{p}) = 0$ ): $$ |g|^2 = \sum_{t \neq 0} \sum_{u} \left(\frac{u t^2}{p}\right) \zeta_p^{t(u - 1)} = \sum_{u} \left(\frac{u}{p}\right) \sum_{t \neq 0} \zeta_p^{t(u-1)}, $$ using $(\frac{u t ^{2}}{p}) = (\frac{u}{p}) (\frac{t ^{2}}{p}) = (\frac{u}{p})$ . The inner geometric sum $\sum_{t \neq = 0} ζ_{p}^{t (u - 1)}$ equals $p - 1$ when $u = 1$ and equals $- 1$ when $u \neq = 1$ (since the full sum $\sum_{t} ζ_{p}^{t k}$ vanishes for $k \neq \equiv 0 (mod p)$ ). So $$ |g|^2 = (p-1) \left(\frac{1}{p}\right) + (-1) \sum_{u \neq 1} \left(\frac{u}{p}\right) = (p - 1) - \left(\sum_{u} \left(\frac{u}{p}\right) - 1\right) = p - \sum_{u} \left(\frac{u}{p}\right) = p, $$ because $\sum_{u \in F_{p}} (\frac{u}{p}) = 0$ (equal numbers of squares and non-squares in $F_{p}^{\times}$ , plus the zero contribution from $u = 0$ ). Combining $(\frac{- 1}{p}) g^{2} = p$ gives $g^{2} = (\frac{- 1}{p}) p = p^{*}$ by Euler's criterion applied to $a = - 1$ . $□$

Bridge. The Gauss-sum identity $g^{2} = p^{*}$ builds toward the quadratic reciprocity law in a single step: it places $p^{*}$ inside the cyclotomic ring $Z [ζ_{p}]$ , and reciprocity becomes a comparison of two Frobenius computations of $g mod q$ inside that ring. The foundational reason this works is that $g$ encodes the Legendre symbol on one side (its definition) and the prime $p^{*}$ on the other (its square), so reducing $g$ modulo a second prime $q$ extracts the Legendre symbol $(\frac{q}{p})$ from the definitional side and the Legendre symbol $(\frac{p ^{*}}{q})$ from the squared side. This is exactly the algebraic identity that, once unpacked, becomes reciprocity. The Frobenius automorphism $σ_{q} : ζ_{p} \mapsto ζ_{p}^{q}$ of $Q (ζ_{p}) / Q$ acts on $g$ by permuting the summands in $\sum (\frac{t}{p}) ζ_{p}^{t}$ , and the resulting permutation identifies $σ_{q} (g)$ with $(\frac{q}{p}) g$ . Putting these together, $g$ is the bridge between the Legendre symbol at $p$ and the Legendre symbol at $q$ , and reciprocity is the statement that the bridge is symmetric. The central insight is exactly this: the cyclotomic field $Q (ζ_{p})$ contains the quadratic field $Q (p^{*})$ , and reciprocity identifies the splitting behaviour of $q$ in the quadratic subfield with the splitting behaviour of $p$ in the field $Q (q^{*})$ . The structural pattern appears again in 21.03.03 (Dedekind, Hecke, Artin $L$ -functions), where the Artin reciprocity law generalises to every abelian extension of every number field, with quadratic reciprocity as the simplest case.

Exercises Intermediate+

Exercise 4 (medium, numeric).

Use quadratic reciprocity together with the supplementary laws to evaluate $(\frac{7}{19})$ .

Hint

Both $7$ and $19$ are congruent to $3$ mod $4$ , so the reciprocity sign flips. Then reduce $19$ mod $7$ and use multiplicativity.

Answer

$+ 1$ . Both $p = 7$ and $q = 19$ are $\equiv 3 (mod 4)$ , so the reciprocity sign is $- 1$ and $(\frac{7}{19}) = - (\frac{19}{7})$ . Reduce: $19 \equiv 5 (mod 7)$ , so $(\frac{19}{7}) = (\frac{5}{7})$ . Apply reciprocity again: $5 \equiv 1 (mod 4)$ and $7 \equiv 3 (mod 4)$ , so the sign is $+ 1$ and $(\frac{5}{7}) = (\frac{7}{5})$ . Reduce: $7 \equiv 2 (mod 5)$ , so $(\frac{7}{5}) = (\frac{2}{5})$ . The supplementary law gives $(\frac{2}{5}) = (- 1)^{(25 - 1) /8} = (- 1)^{3} = - 1$ . Assembling: $(\frac{7}{19}) = - (\frac{19}{7}) = - (\frac{5}{7}) = - (\frac{2}{5}) = - (- 1) = + 1$ . Direct verification: the non-zero squares mod $19$ are ${1, 4, 5, 6, 7, 9, 11, 16, 17}$ , and $7$ is in the list. Rubric: full credit for $+ 1$ with the reciprocity chain.

Exercise 5 (medium, short-answer).

Show that the Legendre symbol $(\frac{\cdot}{p}) : F_{p}^{\times} \to {\pm 1}$ is the unique non-identity group homomorphism into ${\pm 1}$ .

Hint

$F_{p}^{\times}$ is cyclic of order $p - 1$ . Use the universal property of cyclic groups: a homomorphism from $Z / (p - 1)$ to any abelian group is determined by the image of a generator.

Answer

Fix a generator $α$ of the cyclic group $F_{p}^{\times}$ of order $p - 1$ . A homomorphism $χ : F_{p}^{\times} \to {\pm 1}$ is determined by $χ (α) \in {\pm 1}$ . If $χ (α) = 1$ , then $χ$ is the constant $1$ — the simple case. The remaining option is $χ (α) = - 1$ , in which case $χ (α^{k}) = (- 1)^{k}$ , so $χ$ takes value $+ 1$ on even powers (the squares, since the squaring map has kernel ${\pm 1}$ and image the index- $2$ subgroup of squares) and $- 1$ on odd powers (the non-squares). This is exactly the Legendre symbol. Therefore the Legendre symbol is the unique non-identity homomorphism. Rubric: full credit for the cyclic-group argument identifying the unique non-identity homomorphism with the Legendre symbol.

Exercise 7 (hard, short-answer).

Prove the supplementary law $(\frac{2}{p}) = (- 1)^{(p^{2} - 1) /8}$ using the Gauss sum $h = ζ_{8} + ζ_{8}^{- 1}$ where $ζ_{8} = e^{2 π i /8}$ .

Hint

Compute $h^{2} = ζ_{8}^{2} + 2 + ζ_{8}^{- 2} = 2 + i + (- i) = 2$ . Then work in $Z [i] / p Z [i]$ and apply the Frobenius $x \mapsto x^{p}$ , comparing the two expressions $h^{p} = h \cdot h^{p - 1} = h \cdot 2^{(p - 1) /2}$ and $h^{p} = ζ_{8}^{p} + ζ_{8}^{- p}$ depending on $p mod 8$ .

Answer

Set $ζ_{8} = e^{2 π i /8}$ and $h = ζ_{8} + ζ_{8}^{- 1} = 2 cos (π /4) = 2$ . Then $h^{2} = 2$ exactly. Working in $Z [i] = Z [ζ_{8}] / Z [ζ_{8} + ζ_{8}^{- 1}]$ — or more carefully in $Z [ζ_{8}]$ reduced mod $p$ — apply the Frobenius identity $x^{p} \equiv (permutation of summands) (mod p)$ . On one side, $h^{p} = h \cdot (h^{2})^{(p - 1) /2} = h \cdot 2^{(p - 1) /2}$ , and by Euler's criterion this is $h \cdot (\frac{2}{p}) (mod p)$ . On the other side, $h^{p} = ζ_{8}^{p} + ζ_{8}^{- p}$ , which depends only on $p (mod 8)$ : it equals $h$ when $p \equiv \pm 1 (mod 8)$ and $- h$ when $p \equiv \pm 3 (mod 8)$ . Comparing, $(\frac{2}{p}) = + 1$ when $p \equiv \pm 1 (mod 8)$ and $- 1$ when $p \equiv \pm 3 (mod 8)$ . The compact form $(\frac{2}{p}) = (- 1)^{(p^{2} - 1) /8}$ packages the four cases: $(p^{2} - 1) /8$ is even when $p \equiv \pm 1 (mod 8)$ and odd when $p \equiv \pm 3 (mod 8)$ . Rubric: full credit for the $h^{2} = 2$ computation, the Frobenius identification on both sides, and the case analysis modulo $8$ .

Exercise 8 (hard, short-answer).

Show that the quadratic field $Q (p^{*})$ is contained in the cyclotomic field $Q (ζ_{p})$ , and explain why this inclusion is the structural reason behind quadratic reciprocity.

Hint

The Gauss sum $g = \sum_{t} (\frac{t}{p}) ζ_{p}^{t}$ lies in $Q (ζ_{p})$ and satisfies $g^{2} = p^{*}$ .

Answer

The Gauss-sum identity $g^{2} = p^{*}$ exhibits $p^{*} = \pm g \in Q (ζ_{p})$ , so $Q (p^{*}) \subseteq Q (ζ_{p})$ . Since $Gal (Q (ζ_{p}) / Q) ≅ (Z / p)^{\times}$ is cyclic of even order $p - 1$ , it has a unique index- $2$ subgroup, which corresponds via Galois theory to a unique quadratic subfield of $Q (ζ_{p})$ . The Gauss-sum identity identifies this subfield as $Q (p^{*})$ . The structural reason this matters for reciprocity: the Frobenius automorphism $σ_{q}$ at a prime $q \neq = p$ acts on the quadratic subfield through its action on $g = p^{*}$ , and the action's sign is governed by whether $q$ is a square mod $p$ — that is, by $(\frac{q}{p})$ . The Frobenius also detects whether $p^{*}$ is a square mod $q$ — that is, $(\frac{p ^{*}}{q})$ . Equating the two descriptions of the Frobenius action gives the reciprocity law. Rubric: full credit for the inclusion plus the Galois-theoretic identification of the unique quadratic subfield plus the Frobenius argument.

Advanced results Master

The Gauss-sum proof of quadratic reciprocity admits several refinements that locate the law within deeper structures.

Theorem (Frobenius identity for the Gauss sum). Let $p, q$ be distinct odd primes, $ζ_{p} = e^{2 π i / p}$ , $g = \sum_{x \in F_{p}} ζ_{p}^{x^{2}}$ the quadratic Gauss sum, and $σ_{q}$ the automorphism $ζ_{p} \mapsto ζ_{p}^{q}$ of the cyclotomic ring $Z [ζ_{p}]$ . Then $$ \sigma_q(g) = \left(\frac{q}{p}\right) g. $$

The identity is the heart of the Gauss-sum proof: it expresses the Legendre symbol $(\frac{q}{p})$ as the eigenvalue with which $σ_{q}$ acts on $g$ . Combined with the squared identity $g^{2} = p^{*}$ , it identifies $σ_{q}$ on the quadratic subfield $Q (p^{*}) \subset Q (ζ_{p})$ as the unique automorphism with eigenvalue $(\frac{q}{p})$ on $g$ .

Theorem (sign of the Gauss sum, Gauss 1805). For $ζ_{p} = e^{2 π i / p}$ , $$ g = \sum_{x \in \mathbb{F}_p} e^{2\pi i x^2 / p} = \begin{cases} \sqrt p & \text{if } p \equiv 1 \pmod 4, \ i \sqrt p & \text{if } p \equiv 3 \pmod 4. \end{cases} $$

Gauss conjectured the sign in 1801 and proved it in 1805 after four years of failed attempts. The standard modern proof goes through the Poisson summation formula applied to the Gaussian $e^{- π t x^{2}}$ , identifying $g$ as a special value of the Jacobi theta function. The sign refines the Gauss-sum identity $g^{2} = p^{*}$ by pinning down which square root.

Theorem (Hilbert symbol and product formula, Hilbert 1897). Let $K$ be a global field and $V$ its set of places. For $a, b \in K^{\times}$ , the Hilbert symbol $(a, b)_{v} \in {\pm 1}$ is defined at each place $v$ as $+ 1$ if $a x^{2} + b y^{2} = z^{2}$ has a non-zero solution in $K_{v}$ and $- 1$ otherwise. Then $$ \prod_{v \in V} (a, b)_v = 1 $$ with all but finitely many factors equal to $1$ . For $K = Q$ and $a = p$ , $b = q$ odd primes, the product formula recovers quadratic reciprocity.

This is the generalisation of quadratic reciprocity to the language of local-global principles. Each Hilbert symbol $(a, b)_{v}$ is a local quadratic-form invariant; the global product formula is a global constraint relating all the local invariants. Quadratic reciprocity becomes a compatibility statement between local symbols at every place of $Q$ .

Theorem (Artin reciprocity, Artin 1927). Let $L / K$ be an abelian extension of number fields with Galois group $G = Gal (L / K)$ . There is a canonical surjective homomorphism (the Artin map) $$ \mathrm{Art}{L/K} : C_K \twoheadrightarrow G $$ *from the idele class group $C_{K}$ of $K$ to $G$ , sending the Frobenius class at an unramified prime $p$ to the Frobenius substitution $\mathrm{Frob}\mathfrak{p} \in G $. F or$ K = \mathbb{Q} $an d$ L = \mathbb{Q}(\zeta_p) $, t h e A r t inma p r es t r i c t e d t o t h e q u a d r a t i cs u b f i e l d$ \mathbb{Q}(\sqrt{p^}) $r eco v er s t h e L e g e n d r esy mb o l$ \left(\frac{\cdot}{p}\right)$.

Artin reciprocity is the abelian case of the Langlands programme: it identifies abelian Galois representations of $K$ with Hecke characters of the idele class group, and quadratic reciprocity is the rank-one mod- $2$ case of this identification. The Frobenius-eigenvalue identity $σ_{q} (g) = (\frac{q}{p}) g$ used above is the residue-class-field shadow of the Artin map at the prime $q$ .

Theorem (Stickelberger relation). In the ring $Z [ζ_{p}]$ with Galois group $Γ = (Z / p)^{\times} = {σ_{a} : a \in F_{p}^{\times}}$ , the Stickelberger element $$ \theta_p = -\sum_{a = 1}^{p-1} \left{\frac{a}{p}\right} \sigma_a^{-1} \in \mathbb{Q}[\Gamma] $$ satisfies $β θ_{p} \in Z [Γ]$ for every $β \in Z [Γ]$ annihilating $Z / p$ via the augmentation, and the principal ideal generated by $g^2 = p^ $in$ \mathbb{Z}[\zeta_p] $f a c t or s t h r o ug h t h e a c t i o n o f$ \theta_p $o n t h ec l a ss g r o u p o f$ \mathbb{Q}(\zeta_p)$.*

The Stickelberger relation places the Gauss sum inside the wider machinery of cyclotomic units and class groups, and is the input that drives Iwasawa-theoretic refinements of the law (see 21.07.01 for the $Z_{p}$ -extension framing).

Theorem (cubic and biquadratic reciprocity, Jacobi 1837 / Eisenstein 1844). For the cubic Legendre symbol $(\frac{α}{π})_{3}$ in $Z [ζ_{3}]$ and the biquadratic symbol $(\frac{α}{π})_{4}$ in $Z [i]$ , there are reciprocity laws $$ \left(\frac{\pi}{\pi'}\right)_3 = \left(\frac{\pi'}{\pi}\right)_3, \qquad \left(\frac{\pi}{\pi'}\right)_4 = \left(\frac{\pi'}{\pi}\right)_4 \cdot (-1)^{\frac{N(\pi) - 1}{4} \cdot \frac{N(\pi') - 1}{4}} $$ for primary primes $π, π^{'}$ in the respective rings. The proofs run through cubic and biquadratic Gauss sums in the natural way.

These were the first generalisations of quadratic reciprocity. Eisenstein 1844 proved the biquadratic case via the geometry of lattice points; Eisenstein 1850 and Kummer 1859 pushed to higher-degree reciprocity for prime exponents, the work that eventually led Hilbert to the Hilbert symbol and Artin to general reciprocity.

Synthesis. The Gauss-sum proof of quadratic reciprocity identifies the law as a single algebraic identity inside the cyclotomic ring $Z [ζ_{p}]$ : the Frobenius $σ_{q}$ acts on the Gauss sum $g$ by the scalar $(\frac{q}{p})$ , and the squared identity $g^{2} = p^{*}$ exhibits this Frobenius action on the quadratic subfield $Q (p^{*})$ . The foundational reason this works is that $Q (p^{*})$ is the unique quadratic subfield of $Q (ζ_{p})$ , forced by the Galois group $(Z / p)^{\times}$ being cyclic of even order with a unique subgroup of index $2$ . Putting these together, the Gauss sum is the algebraic bridge between the Legendre symbol at $p$ and the Legendre symbol at $q$ , and quadratic reciprocity is the statement that the bridge is symmetric in the two primes. This is exactly the structural pattern that generalises through Hilbert's product formula and Artin reciprocity. The central insight is the identification: Frobenius at $q$ , viewed as an automorphism of the cyclotomic field, is identified with the Legendre symbol $(\frac{q}{p})$ acting on the quadratic subfield, and the same Frobenius detects whether $p^{*}$ is a square mod $q$ via the action on $p^{*}$ . The bridge is exactly this identification, and quadratic reciprocity is the algebraic consequence.

The Gauss-sum framework places quadratic reciprocity inside three larger structures. First, the Hilbert symbol $(a, b)_{v}$ at each place of $Q$ packages the Legendre symbol and its $p = 2$ supplement into one local invariant, and the global product formula $\prod_{v} (a, b)_{v} = 1$ generalises the law. Second, the Artin reciprocity law identifies Frobenius substitutions in any abelian extension with their images under the Artin map on the idele class group; quadratic reciprocity is the rank-one mod- $2$ case for the quadratic subfield of $Q (ζ_{p})$ . Third, the Stickelberger relation and its Iwasawa-theoretic refinement (see 21.07.01) embed the Gauss sum into the structure theory of class groups in cyclotomic towers, where the analogous Frobenius-eigenvalue computation becomes the input to Mazur-Wiles. The recursion stabilises at Artin reciprocity, beyond which the law generalises only by entering the non-abelian Langlands programme. Quadratic reciprocity appears again in 21.03.03 (Dedekind, Hecke, Artin $L$ -functions) as the rank-one mod- $2$ case of the Galois-representation $L$ -function dictionary, and the Gauss-sum identity reappears as the local root number of the corresponding Dirichlet $L$ -function under the functional equation of 21.03.02.

Full proof set Master

Theorem (Euler's criterion), proof. Given in the Intermediate section via Fermat's little theorem and the polynomial factorisation $X^{p - 1} - 1 = (X^{(p - 1) /2} - 1) (X^{(p - 1) /2} + 1)$ in $F_{p} [X]$ . The squares are exactly the $(p - 1) /2$ roots of the first factor; the non-squares fill the second. $□$

Theorem (Gauss-sum identity $g^{2} = p^{*}$ ), proof. Given in the Intermediate section. The two ingredients are: (a) the complex conjugate $\overset{g}{ˉ} = (\frac{- 1}{p}) g$ via the change of variable $u = - t$ and multiplicativity of the Legendre symbol; (b) the direct computation $∣ g ∣^{2} = p$ via the orthogonality of additive characters of $F_{p}$ . Combining gives $g^{2} = (\frac{- 1}{p}) p = p^{*}$ . $□$

Proposition (Frobenius identity). Let $p, q$ be distinct odd primes, $σ_{q} : ζ_{p} \mapsto ζ_{p}^{q}$ the automorphism of $Z [ζ_{p}]$ , and $g = \sum_{t \in F_{p}} (\frac{t}{p}) ζ_{p}^{t}$ . Then $$ \sigma_q(g) = \left(\frac{q}{p}\right) g. $$

Proof. Apply $σ_{q}$ to the Legendre-weighted form: $$ \sigma_q(g) = \sum_{t \in \mathbb{F}_p} \left(\frac{t}{p}\right) \zeta_p^{q t}. $$ Substitute $u = q t$ , which permutes $F_{p}$ bijectively since $q$ is invertible mod $p$ . Then $t = q^{- 1} u$ and $(\frac{t}{p}) = (\frac{q ^{- 1} u}{p}) = (\frac{q ^{- 1}}{p}) (\frac{u}{p}) = (\frac{q}{p}) (\frac{u}{p})$ , using $(\frac{q ^{- 1}}{p}) = (\frac{q}{p})$ because the Legendre symbol takes values in ${\pm 1}$ so is self-inverse. Therefore $$ \sigma_q(g) = \sum_{u \in \mathbb{F}_p} \left(\frac{q}{p}\right) \left(\frac{u}{p}\right) \zeta_p^{u} = \left(\frac{q}{p}\right) g. $$ $□$

Theorem (quadratic reciprocity), proof via Gauss sums. Let $p, q$ be distinct odd primes. Work in the ring $Z [ζ_{p}]$ and its quotient $Z [ζ_{p}] / q Z [ζ_{p}]$ , a finite ring of characteristic $q$ . Two computations of $g^{q} (mod q)$ :

First computation. Use $g^{2} = p^{*} (mod q)$ in $Z [ζ_{p}]$ . Raise to the $(q - 1) /2$ -th power: $$ g^{q - 1} = (g^2)^{(q-1)/2} = (p^)^{(q-1)/2} \equiv \left(\frac{p^}{q}\right) \pmod q $$ by Euler's criterion applied at the prime $q$ to the integer $p^{*} \in Z$ . Multiplying by $g$ : $$ g^q \equiv \left(\frac{p^*}{q}\right) g \pmod q. $$

Second computation. Use the Frobenius identity. In characteristic $q$ , raising to the $q$ -th power is a ring endomorphism (the Frobenius of $F_{q}$ ), and acts on $ζ_{p}$ by $ζ_{p} \mapsto ζ_{p}^{q}$ — the same action as $σ_{q}$ in characteristic zero, since $ζ_{p}^{q}$ in $Z [ζ_{p}] / (q)$ matches $σ_{q} (ζ_{p})$ reduced. Therefore $g^{q} \equiv σ_{q} (g) (mod q)$ in $Z [ζ_{p}] / (q)$ , and the Frobenius identity gives $$ g^q \equiv \left(\frac{q}{p}\right) g \pmod q. $$

Comparison. Equate the two expressions for $g^{q} mod q$ : $$ \left(\frac{p^}{q}\right) g \equiv \left(\frac{q}{p}\right) g \pmod q. $$ Multiply both sides by $g$ and use $g^2 = p^$: $$ \left(\frac{p^}{q}\right) p^ \equiv \left(\frac{q}{p}\right) p^* \pmod q. $$ Since $g cd (p^{*}, q) = 1$ (as $p^{*} = \pm p$ and $p \neq = q$ are distinct primes), $p^{*}$ is a unit mod $q$ , and we cancel: $$ \left(\frac{p^}{q}\right) = \left(\frac{q}{p}\right) \quad \text{as elements of } {\pm 1}. $$ Expand the left side using multiplicativity of the Legendre symbol: $$ \left(\frac{p^}{q}\right) = \left(\frac{(-1)^{(p-1)/2}}{q}\right) \left(\frac{p}{q}\right) = \left(\frac{-1}{q}\right)^{(p-1)/2} \left(\frac{p}{q}\right) = (-1)^{\frac{p-1}{2} \cdot \frac{q-1}{2}} \left(\frac{p}{q}\right), $$ where the last equality uses the supplementary law $(\frac{- 1}{q}) = (- 1)^{(q - 1) /2}$ . Therefore $$ (-1)^{\frac{p-1}{2} \cdot \frac{q-1}{2}} \left(\frac{p}{q}\right) = \left(\frac{q}{p}\right), $$ which rearranges to $$ \left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2} \cdot \frac{q-1}{2}}. $$ $□$

Proposition (supplementary law for $(\frac{- 1}{p})$ ). For odd prime $p$ , $(\frac{- 1}{p}) = (- 1)^{(p - 1) /2}$ .

Proof. Direct from Euler's criterion at $a = - 1$ : $(- 1)^{(p - 1) /2}$ equals $+ 1$ when $(p - 1) /2$ is even (i.e. $p \equiv 1 (mod 4)$ ) and $- 1$ when $(p - 1) /2$ is odd (i.e. $p \equiv 3 (mod 4)$ ). $□$

Proposition (supplementary law for $(\frac{2}{p})$ ). For odd prime $p$ , $(\frac{2}{p}) = (- 1)^{(p^{2} - 1) /8}$ .

Proof. Use the Gauss sum $h = ζ_{8} + ζ_{8}^{- 1} \in Z [ζ_{8}]$ with $ζ_{8} = e^{2 π i /8}$ . Direct calculation: $h^{2} = ζ_{8}^{2} + 2 + ζ_{8}^{- 2} = i + 2 + (- i) = 2$ , so $h^{2} = 2$ exactly. Working modulo $p$ in $Z [ζ_{8}] / (p)$ and applying the characteristic- $p$ Frobenius $x \mapsto x^{p}$ : $$ h^p = h \cdot (h^2)^{(p-1)/2} = h \cdot 2^{(p-1)/2} \equiv \left(\frac{2}{p}\right) h \pmod p $$ by Euler's criterion.

On the other hand, $h^{p} \equiv ζ_{8}^{p} + ζ_{8}^{- p} (mod p)$ because in characteristic $p$ the binomial expansion of $(ζ_{8} + ζ_{8}^{- 1})^{p}$ collapses to the $p$ -th-power terms. The value $ζ_{8}^{p} + ζ_{8}^{- p}$ depends only on $p mod 8$ : it equals $h$ if $p \equiv \pm 1 (mod 8)$ and $- h$ if $p \equiv \pm 3 (mod 8)$ . The four cases are $ζ_{8} + ζ_{8}^{- 1} = h$ , $ζ_{8}^{3} + ζ_{8}^{- 3} = - h$ , $ζ_{8}^{5} + ζ_{8}^{- 5} = - h$ , $ζ_{8}^{7} + ζ_{8}^{- 7} = h$ .

Comparing $(\frac{2}{p}) h \equiv \pm h (mod p)$ and using that $h \neq \equiv 0 (mod p)$ (since $h^{2} = 2$ and $p$ is odd), cancel $h$ to obtain $(\frac{2}{p}) = \pm 1$ according to the case. The packaged exponent $(p^{2} - 1) /8$ takes values $0, 1, 3, 6 (mod 2) \equiv 0, 1, 1, 0$ for $p \equiv 1, 3, 5, 7 (mod 8)$ respectively, matching the four cases. So $(\frac{2}{p}) = (- 1)^{(p^{2} - 1) /8}$ . $□$

Proposition (Gauss sum is invariant up to sign under $σ_{a}$ for $a$ a square). For $a \in F_{p}^{\times}$ with $(\frac{a}{p}) = + 1$ , $σ_{a} (g) = g$ . For $a$ with $(\frac{a}{p}) = - 1$ , $σ_{a} (g) = - g$ .

Proof. The Frobenius identity above with $q$ replaced by $a$ (the argument applies to any $a \in F_{p}^{\times}$ , not only primes) gives $σ_{a} (g) = (\frac{a}{p}) g$ . So $σ_{a} (g) = g$ when $a$ is a square and $σ_{a} (g) = - g$ otherwise. This means $g$ generates the fixed field of the index- $2$ subgroup of squares — that is, $Q (g) = Q (p^{*})$ is the unique quadratic subfield of $Q (ζ_{p})$ . $□$

Connections Master

Algebraic field extension and splitting field 01.02.12. The cyclotomic field $Q (ζ_{p})$ is the splitting field of $X^{p} - 1$ over $Q$ , and the Gauss-sum proof of quadratic reciprocity is a calculation inside its ring of integers $Z [ζ_{p}]$ . The unique quadratic subfield $Q (p^{*}) \subset Q (ζ_{p})$ exists because the Galois group $(Z / p)^{\times}$ is cyclic of even order, and the Galois-theoretic correspondence between subgroups and subfields is what makes the Gauss-sum identity $g^{2} = p^{*}$ a structural statement rather than a computational accident.
Riemann zeta function 21.03.01. The Dirichlet $L$ -function associated to the Legendre-symbol Dirichlet character $χ (n) = (\frac{n}{p})$ of conductor $p$ shares the analytic-continuation and functional-equation framework of the Riemann zeta function. The quadratic Gauss sum appears as the root number $W (χ) = τ (χ) / p$ of the functional equation of $L (s, χ)$ , so the sign of $g = p^{*}$ determines the parity of $L (s, χ)$ under $s \leftrightarrow 1 - s$ .
Dirichlet $L$ -functions 21.03.02. Dirichlet characters of conductor $p$ for $p$ an odd prime are completely classified by characters of $(Z / p)^{\times} = F_{p}^{\times}$ , and the unique non-identity quadratic character is the Legendre symbol $(\frac{\cdot}{p})$ . The functional equation of $L (s, χ)$ for $χ$ this Legendre-symbol character has Gauss sum $τ (χ) = g$ as its root-number ingredient; quadratic reciprocity becomes the compatibility of the root numbers of two such $L$ -functions corresponding to two different odd primes.
Dedekind, Hecke, and Artin $L$ -functions 21.03.03. Artin reciprocity is the general abelian reciprocity law of which quadratic reciprocity is the rank-one mod- $2$ case. The cyclotomic field $Q (ζ_{p})$ realises every abelian extension of $Q$ ramified only at $p$ (and $\infty$ ), and the Legendre symbol is the rank-one mod- $2$ component of the Artin character of the quadratic subfield $Q (p^{*})$ .
$Z_{p}$ -extensions and Iwasawa theory 21.07.01. The cyclotomic $Z_{p}$ -extension of $Q$ is the directed union of the cyclotomic fields $Q (ζ_{p^{n}})$ , and the Iwasawa-theoretic study of the class groups in this tower runs through the Stickelberger element, which is built from Gauss sums attached to characters of higher conductor. Quadratic reciprocity is the rank-one base case of the Iwasawa-Stickelberger phenomena, and the analytic continuation of the associated $p$ -adic $L$ -functions is the higher-dimensional generalisation of the Gauss-sum functional-equation calculation.

Historical & philosophical context Master

Quadratic reciprocity was discovered empirically by Euler in the 1740s through extensive numerical computation, with the precise statement given in his 1783 Opuscula analytica posthumously published in 1785 ^{[source pending]}. Legendre stated the law in its modern form (and introduced the symbol $(\frac{a}{p})$ ) in his 1785 Recherches d'analyse indéterminée and 1798 Essai sur la théorie des nombres, but his proof contained a gap that he could not fill. Gauss gave the first complete proof in 1796 (he was nineteen, and dated the discovery in his mathematical diary to 8 April 1796), publishing it as theorema fundamentale in §131 of the 1801 Disquisitiones Arithmeticae ^{[source pending]}. The Disquisitiones contains two distinct proofs (an elementary induction-on-cases and an analytic argument via the Gauss-sum sign); Gauss returned to the problem repeatedly over his lifetime and ultimately produced six different proofs, including a Gauss-sum-style argument that anticipates the structure of the present unit.

The Gauss-sum proof as presented today is essentially Eisenstein's 1845 Beiträge zur Theorie der elliptischen Functionen (J. reine angew. Math. 29, 177-184) ^{[source pending]}, systematised by Hecke 1923 in his Vorlesungen über die Theorie der algebraischen Zahlen and given the cyclotomic-field framing now standard in textbook presentations such as Serre 1973 A Course in Arithmetic Ch. I §3 ^{[source pending]}. Eisenstein 1844 (J. reine angew. Math. 28, 246-248) ^{[source pending]} supplied the cleanest elementary proof via a lattice-point geometric argument, reducing reciprocity to counting points in a rectangle; this is the proof that appears in most undergraduate texts. Lemmermeyer's 2000 monograph Reciprocity Laws: From Euler to Eisenstein ^{[source pending]} catalogues 224 published proofs by that date, organised by method (Gauss-sum, lattice-point, elliptic-function, $p$ -adic, class-field-theoretic, and more), an accumulation rivalled in number theory only by the Pythagorean theorem.

The structural generalisation passes through Hilbert's 1897 Zahlbericht (Jahresbericht DMV 4, 175-546) ^{[source pending]}, which introduced the Hilbert symbol $(a, b)_{v}$ and the product formula $\prod_{v} (a, b)_{v} = 1$ as the natural local-global statement encompassing both quadratic reciprocity and the supplementary laws. Takagi 1920 (J. Coll. Sci. Univ. Tokyo 41) developed class field theory for general abelian extensions, identifying the Artin map with the reciprocity isomorphism. Artin 1927 Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 5, 353-363 ^{[source pending]} proved the Artin reciprocity law in its full abelian generality, identifying Frobenius substitutions in any abelian extension with images under a canonically defined map on the idele class group. Tate's 1950 Princeton thesis (in Cassels-Fröhlich 1967 Ch. XV) recast Artin reciprocity as a statement about $GL_{1}$ automorphic representations, opening the door to the non-abelian generalisations of the Langlands programme. Quadratic reciprocity remains the rank-one mod- $2$ case of this entire structure: the Frobenius-eigenvalue computation $σ_{q} (g) = (\frac{q}{p}) g$ inside $Z [ζ_{p}]$ is the residue-class-field shadow of the Artin map at $q$ , restricted to the unique quadratic subfield $Q (p^{*})$ .

Bibliography Master

@book{Gauss1801Disquisitiones,
  author    = {Gauss, Carl Friedrich},
  title     = {Disquisitiones Arithmeticae},
  publisher = {Gerh. Fleischer Jun.},
  address   = {Leipzig},
  year      = {1801},
  note      = {English translation by Arthur A. Clarke (Yale 1966; revised Springer 1986)}
}

@article{Eisenstein1844Geometric,
  author  = {Eisenstein, Ferdinand Gotthold Max},
  title   = {Geometrischer Beweis des Fundamentaltheorems f{\"u}r die quadratischen Reste},
  journal = {Journal f{\"u}r die reine und angewandte Mathematik},
  volume  = {28},
  year    = {1844},
  pages   = {246--248}
}

@article{Eisenstein1845GaussSum,
  author  = {Eisenstein, Ferdinand Gotthold Max},
  title   = {Beitr{\"a}ge zur Theorie der elliptischen Functionen},
  journal = {Journal f{\"u}r die reine und angewandte Mathematik},
  volume  = {29},
  year    = {1845},
  pages   = {177--184}
}

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

@article{Hilbert1897Zahlbericht,
  author  = {Hilbert, David},
  title   = {Die Theorie der algebraischen Zahlk{\"o}rper (Zahlbericht)},
  journal = {Jahresbericht der Deutschen Mathematiker-Vereinigung},
  volume  = {4},
  year    = {1897},
  pages   = {175--546}
}

@book{Serre1973CourseArithmetic,
  author    = {Serre, Jean-Pierre},
  title     = {A Course in Arithmetic},
  publisher = {Springer-Verlag},
  series    = {Graduate Texts in Mathematics},
  volume    = {7},
  year      = {1973}
}

@book{IrelandRosen1990,
  author    = {Ireland, Kenneth and Rosen, Michael},
  title     = {A Classical Introduction to Modern Number Theory},
  publisher = {Springer-Verlag},
  series    = {Graduate Texts in Mathematics},
  volume    = {84},
  edition   = {2},
  year      = {1990}
}

@book{LangANT,
  author    = {Lang, Serge},
  title     = {Algebraic Number Theory},
  publisher = {Springer-Verlag},
  series    = {Graduate Texts in Mathematics},
  volume    = {110},
  edition   = {2},
  year      = {1994}
}

@book{Lemmermeyer2000Reciprocity,
  author    = {Lemmermeyer, Franz},
  title     = {Reciprocity Laws: From Euler to Eisenstein},
  publisher = {Springer-Verlag},
  series    = {Springer Monographs in Mathematics},
  year      = {2000}
}

@book{Hecke1923,
  author    = {Hecke, Erich},
  title     = {Vorlesungen {\"u}ber die Theorie der algebraischen Zahlen},
  publisher = {Akademische Verlagsgesellschaft},
  address   = {Leipzig},
  year      = {1923},
  note      = {Reprinted Chelsea, New York (1948); English translation Springer Graduate Texts in Mathematics 77 (1981)}
}

@incollection{Tate1967ANT,
  author    = {Tate, John},
  title     = {Global Class Field Theory},
  booktitle = {Algebraic Number Theory},
  editor    = {Cassels, J. W. S. and Fr{\"o}hlich, Albrecht},
  publisher = {Academic Press},
  year      = {1967},
  chapter   = {VII},
  pages     = {162--203}
}

@misc{Euler1783Opuscula,
  author = {Euler, Leonhard},
  title  = {Opuscula analytica},
  year   = {1783},
  note   = {Posthumously published 1785; the empirical statement of quadratic reciprocity precedes Legendre's symbol notation but contains the modern statement of the law for primes congruent to $1$ modulo $4$}
}

Prerequisites

01.01.01
01.02.01
01.02.02
01.02.12

Tier anchors

beginner: Serre 1973 *A Course in Arithmetic* (GTM 7) Ch. I §3.1-3.2 informal opening; Davenport 2008 *The Higher Arithmetic* (8th ed., Cambridge) Ch. III on quadratic residues; Ireland-Rosen 1990 *A Classical Introduction to Modern Number Theory* (GTM 84) Ch. 5 §1-§2 informal motivation
intermediate: Serre 1973 *A Course in Arithmetic* (GTM 7) Ch. I §3 (statement and Gauss-sum proof); Ireland-Rosen 1990 (GTM 84) Ch. 5 + Ch. 6 §2-§3; Lang 1994 *Algebraic Number Theory* (GTM 110, 2nd ed.) Ch. IV §3
master: Gauss 1801 *Disquisitiones Arithmeticae* §§107-150 (six distinct proofs of quadratic reciprocity, originator); Eisenstein 1844 *J. reine angew. Math.* 28 (sine-product geometric proof); Eisenstein 1845 *J. reine angew. Math.* 29 (proof via $p$-th roots of unity and Gauss sums; the prototype of the present unit); Artin 1927 *Abh. Math. Sem. Univ. Hamburg* 5, 353-363 (general Artin reciprocity); Serre 1973 *A Course in Arithmetic* (GTM 7) Ch. I §3; Ireland-Rosen 1990 *A Classical Introduction to Modern Number Theory* (GTM 84) Ch. 5-6; Lang 1994 *Algebraic Number Theory* (GTM 110, 2nd ed.) Ch. IV §3; Lemmermeyer 2000 *Reciprocity Laws: From Euler to Eisenstein* (Springer Monographs); Hilbert 1897 *Zahlbericht* §64 on the product formula for the quadratic Hilbert symbol; Tate 1967 *Algebraic Number Theory* (Cassels-Fröhlich) on the Hilbert symbol and class field theory framing

References

Gauss, C. F. — Disquisitiones Arithmeticae · Leipzig 1801, §§107-150 (originator). Gauss states quadratic reciprocity as *theorema fundamentale* in §131 and gives the first complete proof; he produced six distinct proofs over his lifetime, including a Gauss-sum-style argument that anticipates the present unit's method. English translation by Arthur A. Clarke, Yale 1966; revised Springer 1986.
Eisenstein, F. G. M. — Geometrischer Beweis des Fundamentaltheorems für die quadratischen Reste · *Journal für die reine und angewandte Mathematik* 28 (1844), 246-248. The lattice-point geometric proof reducing quadratic reciprocity to a count of points in a rectangle, the cleanest of the elementary proofs.
Eisenstein, F. G. M. — Beiträge zur Theorie der elliptischen Functionen · *Journal für die reine und angewandte Mathematik* 29 (1845), 177-184. The proof via $p$-th roots of unity and quadratic Gauss sums; structurally the prototype of the present unit's argument, completed in modern form by Hecke 1923 and Serre 1973.
Artin, E. — Beweis des allgemeinen Reziprozitätsgesetzes · *Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg* 5 (1927), 353-363. The general Artin reciprocity law identifying the Frobenius substitution at an unramified prime with its image under the Artin map; quadratic reciprocity is the rank-one abelian case for the field $\mathbb{Q}(\sqrt{p^*})$ inside the cyclotomic field $\mathbb{Q}(\zeta_p)$.
Serre, J.-P. — A Course in Arithmetic · Springer Graduate Texts in Mathematics 7 (1973, translated from the French *Cours d'Arithmétique*, PUF 1970). Ch. I §3 statement and Gauss-sum proof; the proof structure followed here (lift $g \in \mathbb{Z}[\zeta_p]$, reduce mod $q$, apply Frobenius $\sigma_q(x) = x^q$) is taken from Serre's exposition.
Ireland, K. & Rosen, M. — A Classical Introduction to Modern Number Theory · Springer Graduate Texts in Mathematics 84 (2nd ed., 1990). Ch. 5 (quadratic reciprocity, elementary proof) and Ch. 6 §2-§3 (Gauss sums, Jacobi sums, the cubic and biquadratic generalisations); the standard pedagogic reference for Gauss sums in number theory.
Lang, S. — Algebraic Number Theory · Springer Graduate Texts in Mathematics 110, 2nd edition (1994). Ch. IV §3 covers quadratic reciprocity via the cyclotomic-field approach; Ch. X covers the general Artin map and class field theory in which quadratic reciprocity sits as the simplest abelian case.
Hilbert, D. — Die Theorie der algebraischen Zahlkörper (Zahlbericht) · *Jahresbericht der Deutschen Mathematiker-Vereinigung* 4 (1897), 175-546. §64 introduces the Hilbert symbol $(a, b)_v$ and states the product formula $\prod_v (a, b)_v = 1$ over all places of a number field; this is the reciprocity-law generalisation pointed to in the Master tier.
Lemmermeyer, F. — Reciprocity Laws: From Euler to Eisenstein · Springer Monographs in Mathematics (2000). The comprehensive history of reciprocity laws from Euler's empirical observations through Gauss's six proofs to Eisenstein, Kummer, and the higher reciprocity laws preceding Artin. Appendix A catalogues 224 known proofs of quadratic reciprocity by 2000.
Hecke, E. — Vorlesungen über die Theorie der algebraischen Zahlen · Akademische Verlagsgesellschaft, Leipzig 1923; reprinted Chelsea 1948. §52-§55 give a careful modern presentation of quadratic Gauss sums and the determination of their sign $g_1 = \sqrt{p^*}$ via theta-function arguments; Gauss had stated the sign already in 1801 but the proof eluded him until 1805.

Estimated time

beginner: 18m
intermediate: 40m
master: 75m