Quadratic field
In algebraic number theory, a quadratic field is a finite field extension of the rational numbers \mathbb{Q} of degree two, obtained by adjoining a square root of a square-free integer d \neq 1 to \mathbb{Q} and denoted K = \mathbb{Q}(\sqrt{d}) = \{ x + y \sqrt{d} : x, y \in \mathbb{Q} \}.[1] These fields are the simplest nontrivial examples of number fields and form the foundation for studying more advanced extensions. Quadratic fields are classified into two types based on the sign of d: real quadratic fields when d > 0, which embed into the real numbers, and imaginary quadratic fields when d < 0, which do not.[2] The ring of integers \mathcal{O}_K of such a field K, consisting of the algebraic integers in K, has a explicit description that depends on d modulo 4: if d \equiv 2 or $3 \pmod{4}, then \mathcal{O}_K = \mathbb{Z}[\sqrt{d}]; if d \equiv 1 \pmod{4}, then \mathcal{O}_K = \mathbb{Z}\left[\frac{1 + \sqrt{d}}{2}\right].[1] This ring is a Dedekind domain, meaning that every nonzero ideal factors uniquely into prime ideals, even though elements may not factor uniquely—a phenomenon quantified by the finite ideal class group \mathrm{Cl}(K), whose order is the class number h(K). Key arithmetic features of quadratic fields include the norm N(\alpha) = \alpha \overline{\alpha} for \alpha \in K, where \overline{\cdot} denotes the nontrivial automorphism (conjugation), and the trace \mathrm{Tr}_{K/\mathbb{Q}}(\alpha) = \alpha + \overline{\alpha}, which facilitate the study of factorization and ramification of primes from \mathbb{Z} in \mathcal{O}_K.[1] Primes p \in \mathbb{Z} may remain inert, split into two distinct prime ideals, or ramify (factor as a square of a prime ideal) in \mathcal{O}_K, depending on the Legendre symbol \left( \frac{d}{p} \right).[3] The group of units \mathcal{O}_K^\times = \{ \alpha \in \mathcal{O}_K : N(\alpha) = \pm 1 \} is finite for imaginary quadratic fields (typically \{ \pm 1 \}, except for d = -1, -3) but infinite for real quadratic fields, generated by \pm 1 and a fundamental unit solving a Pell equation.[1] Quadratic fields play a central role in algebraic number theory as concrete models for broader phenomena, including the failure of unique factorization, Dirichlet's unit theorem, and the class number problem; for imaginary quadratic fields, the class number is tied to the Dedekind zeta function via an explicit formula involving the discriminant and L-functions. They also connect to quadratic forms, continued fractions, and modular forms, with applications in cryptography, Diophantine equations, and the distribution of primes.Fundamentals
Definition
In algebraic number theory, a quadratic field is defined as a finite field extension K of the rational numbers \mathbb{Q} having degree 2.[4] Explicitly, every quadratic field can be expressed as K = \mathbb{Q}(\sqrt{d}), where d is a square-free integer not equal to 0 or 1; the elements of K are thus all expressions of the form a + b\sqrt{d} with a, b \in \mathbb{Q}.[1] This construction ensures that adjoining \sqrt{d} to \mathbb{Q} yields a proper extension, as d being square-free guarantees the irreducibility of the relevant polynomial over \mathbb{Q}.[3] The degree of the extension [\mathbb{Q}(\sqrt{d}) : \mathbb{Q}] = 2 follows from the fact that the minimal polynomial of \sqrt{d} over \mathbb{Q} is x^2 - d.[5] This monic polynomial is irreducible over \mathbb{Q} precisely because d is square-free and not 0 or 1, as otherwise \sqrt{d} would already lie in \mathbb{Q}.[1] As a vector space over \mathbb{Q}, K admits the basis \{1, \sqrt{d}\}, which spans K and is linearly independent over \mathbb{Q}.[4] Quadratic fields are classified into two types depending on the sign of d: if d > 0, then K is a real quadratic field, embedded into the real numbers \mathbb{R} via two distinct real embeddings; if d < 0, then K is an imaginary quadratic field, with no real embeddings but two complex conjugate embeddings into \mathbb{C}.[6]Examples
The quadratic field \mathbb{Q}(\sqrt{2}) arises naturally in geometry, for instance, as the field generated by the length of the diagonal of a unit square.[7] Elements of this field take the form a + b\sqrt{2} with a, b \in \mathbb{Q}, and basic arithmetic operations follow from distributing over the basis \{1, \sqrt{2}\}; for example, addition yields (1 + \sqrt{2}) + (3 - \sqrt{2}) = 4, while multiplication gives (1 + \sqrt{2})( \sqrt{2} - 1) = (\sqrt{2} - 1 + 2 - \sqrt{2}) = 1.[7] Another real quadratic field is \mathbb{Q}(\sqrt{5}), which contains the golden ratio (1 + \sqrt{5})/2 and appears in problems involving pentagons and Fibonacci sequences.[7] Imaginary quadratic fields provide examples with complex elements. The field \mathbb{Q}(i) = \mathbb{Q}(\sqrt{-1}), known as the Gaussian rationals, consists of elements a + bi with a, b \in \mathbb{Q}.[7] Similarly, \mathbb{Q}(\sqrt{-3}), the Eisenstein rationals, includes elements of the form a + b \sqrt{-3} with a, b \in \mathbb{Q}, and is notable for its connections to equilateral triangles and cubic residues.[7] Early studies of specific quadratic fields, such as \mathbb{Q}(\sqrt{-1}), were pursued by Pierre de Fermat and Leonhard Euler in the context of sums of two squares; Fermat stated that an odd prime p is a sum of two squares if and only if p \equiv 1 \pmod{4}, a result later proved by Euler using infinite descent and linked to factorization in Gaussian integers.[8] Fermat also challenged contemporaries with Pell equations like x^2 - 2y^2 = \pm 1, tied to units in real quadratic fields such as \mathbb{Q}(\sqrt{2}), though the general solution method was developed later.[9] These examples illustrate the distinction between real quadratic fields, which embed into the reals, and imaginary ones, which do not; their discriminants, such as 8 for \mathbb{Q}(\sqrt{2}) and -4 for \mathbb{Q}(i), are detailed subsequently.[7]Integral Structure
Ring of Integers
In a quadratic field K = \mathbb{Q}(\sqrt{d}), where d is a square-free integer not equal to 0 or 1, the ring of integers \mathcal{O}_K is defined as the integral closure of \mathbb{Z} in K, consisting of all elements in K that are roots of monic polynomials with coefficients in \mathbb{Z}.[10] This ring serves as the maximal order in K, meaning it is the largest subring of K that is finitely generated as a \mathbb{Z}-module and integrally closed in K.[11] The explicit form of \mathcal{O}_K depends on the congruence class of d modulo 4. If d \equiv 2 or $3 \pmod{4}, then \mathcal{O}_K = \mathbb{Z}[\sqrt{d}] = \{ a + b\sqrt{d} \mid a, b \in \mathbb{Z} \}. If d \equiv 1 \pmod{4}, then \mathcal{O}_K = \mathbb{Z}\left[\frac{1 + \sqrt{d}}{2}\right] = \left\{ a + b \frac{1 + \sqrt{d}}{2} \mid a, b \in \mathbb{Z} \right\}.[10][11] To establish this form, consider the basis elements. For d \equiv 2 or $3 \pmod{4}, \sqrt{d} satisfies the minimal polynomial X^2 - d = 0, which is monic with integer coefficients, so \sqrt{d} is integral over \mathbb{Z}, and \mathbb{Z}[\sqrt{d}] is the full integral closure. For d \equiv 1 \pmod{4}, the element \rho = \frac{1 + \sqrt{d}}{2} satisfies the minimal polynomial X^2 - X + \frac{1 - d}{4} = 0; since d \equiv 1 \pmod{4}, \frac{1 - d}{4} is an integer, making the polynomial monic with integer coefficients, hence \rho is integral over \mathbb{Z}. This polynomial is irreducible over \mathbb{Q} because its discriminant d is square-free (hence not a perfect square), ensuring \mathbb{Z}[\rho] generates the full ring. Any larger ring would contradict the \mathbb{Z}-rank 2 of \mathcal{O}_K.[11][10] As a ring of integers in a number field, \mathcal{O}_K is a Dedekind domain, meaning every nonzero prime ideal is maximal and ideals factor uniquely into primes. It is a principal ideal domain (and thus a unique factorization domain) if and only if the class number of K is 1, which occurs for specific values of d, such as the imaginary quadratic fields with d = -1, -2, -3, -7, -11, -19, -43, -67, -163. For other d, like d = -5, \mathcal{O}_K = \mathbb{Z}[\sqrt{-5}] fails to be a PID due to non-unique factorization of elements like 6.[10][11] Non-maximal orders, such as \mathbb{Z}[\sqrt{d}] when d \equiv 1 \pmod{4}, are subrings of \mathcal{O}_K with finite index; the conductor of such an order measures this index and relates to how ideals factor differently in the order versus \mathcal{O}_K.[10][11]Discriminant
The discriminant \Delta_K of the ring of integers \mathcal{O}_K in a quadratic field K = \mathbb{Q}(\sqrt{d}), where d is a square-free integer not equal to 0 or 1, is defined as the determinant of the trace form matrix with respect to a \mathbb{Z}-basis of \mathcal{O}_K.[12] Specifically, if \{e_1, \dots, e_n\} is such a basis, then \Delta_K = \det(\operatorname{Tr}_{K/\mathbb{Q}}(e_i e_j)), where \operatorname{Tr}_{K/\mathbb{Q}} denotes the field trace.[12] The computation of \Delta_K depends on the congruence class of d modulo 4. If d \equiv 1 \pmod{4}, then \mathcal{O}_K = \mathbb{Z}\left[\frac{1 + \sqrt{d}}{2}\right] and \Delta_K = d; otherwise, \mathcal{O}_K = \mathbb{Z}[\sqrt{d}] and \Delta_K = 4d.[12] Here, d is taken to be the square-free part defining the field, making \Delta_K the fundamental discriminant associated to K.[12] Key properties of \Delta_K include its congruence modulo 4: \Delta_K \equiv 0 \pmod{4} or \Delta_K \equiv 1 \pmod{4}, reflecting the structure of the ring of integers.[12] The sign of \Delta_K distinguishes real quadratic fields (where d > 0, so \Delta_K > 0) from imaginary quadratic fields (where d < 0, so \Delta_K < 0).[12] The discriminant relates to the different ideal \mathfrak{D}_K, whose norm equals |\Delta_K|, providing a measure of ramification in the extension K/\mathbb{Q}.[12] In quadratic fields, the prime ideals dividing \mathfrak{D}_K are precisely the ramified primes, with the exponent in the factorization indicating the ramification index minus one.[12] For example, in K = \mathbb{Q}(\sqrt{2}), where d = 2 \equiv 2 \pmod{4}, the discriminant is \Delta_K = 4 \cdot 2 = 8.[12] In K = \mathbb{Q}(\sqrt{-3}), where d = -3 \equiv 1 \pmod{4}, the discriminant is \Delta_K = -3.[12]Arithmetic Properties
Norm and Trace
In quadratic fields, the trace and norm are fundamental linear and multiplicative maps from the field K = \mathbb{Q}(\sqrt{d}) to the base field \mathbb{Q}, where d is a square-free integer. These maps arise from the structure of K as a degree-2 extension of \mathbb{Q} and play a central role in the arithmetic of elements in K. They can be defined using the two embeddings of K into \mathbb{C}, which send \sqrt{d} to \pm \sqrt{d} (assuming d > 0) or to \pm i\sqrt{|d|} (if d < 0); the trace is the sum of the images under these embeddings, while the norm is their product.[13] For an element \alpha = a + b\sqrt{d} with a, b \in \mathbb{Q}, the trace is given by \operatorname{Tr}_{K/\mathbb{Q}}(\alpha) = 2a, which is the sum of \alpha and its Galois conjugate \overline{\alpha} = a - b\sqrt{d}. This map is \mathbb{Q}-linear and equals the trace of the matrix representing multiplication by \alpha on the basis \{1, \sqrt{d}\}.[13] The norm is N_{K/\mathbb{Q}}(\alpha) = a^2 - d b^2 = \alpha \overline{\alpha}, the product of \alpha and its conjugate, or equivalently the determinant of the same multiplication matrix. Unlike the trace, the norm is multiplicative: N_{K/\mathbb{Q}}(\alpha \beta) = N_{K/\mathbb{Q}}(\alpha) N_{K/\mathbb{Q}}(\beta) for all \alpha, \beta \in K. For example, in K = \mathbb{Q}(\sqrt{2}), N_{K/\mathbb{Q}}(1 + \sqrt{2}) = 1^2 - 2 \cdot 1^2 = -1. Elements \varepsilon \in K with N_{K/\mathbb{Q}}(\varepsilon) = \pm 1 are units in the ring of integers of K.[13] The norm extends to ideals in the ring of integers \mathcal{O}_K: for a principal ideal (\alpha), the ideal norm N((\alpha)) equals |N_{K/\mathbb{Q}}(\alpha)|, which counts the index [\mathcal{O}_K : (\alpha)]. This ideal norm is completely multiplicative over ideal multiplication and positive for nonzero ideals.[14]Units
In quadratic number fields, the structure of the unit group of the ring of integers is governed by Dirichlet's unit theorem, which asserts that the group is finitely generated with rank equal to the number of real embeddings plus half the number of complex embeddings minus one.[15] For imaginary quadratic fields \mathbb{Q}(\sqrt{d}) with d < 0 square-free, there are no real embeddings, yielding rank zero; thus, the unit group is finite and torsion, consisting solely of roots of unity in the field.[16] In most cases, these units are simply \{\pm 1\}.[15] However, exceptions occur for d = -1 and d = -3: in \mathbb{Q}(\sqrt{-1}), the ring of integers is \mathbb{Z} and the units are \{\pm 1, \pm i\}, the fourth roots of unity; in \mathbb{Q}(\sqrt{-3}), the ring of integers is \mathbb{Z}\left[\frac{-1 + \sqrt{-3}}{2}\right] and the units are the six sixth roots of unity \{\pm 1, \pm \omega, \pm \omega^2\}, where \omega = \frac{-1 + \sqrt{-3}}{2}.[15][17] For real quadratic fields \mathbb{Q}(\sqrt{d}) with d > 0 square-free, there are two real embeddings, yielding rank one; the unit group is therefore isomorphic to \mathbb{Z} \times \{\pm 1\}, generated by -1 and a fundamental unit \varepsilon > 1.[16] This fundamental unit \varepsilon is the smallest element greater than 1 with multiplicative inverse also in the ring of integers, and it satisfies a Pell equation of the form x^2 - d y^2 = \pm 1 (or \pm 4 if the ring basis involves halves).[15] The full unit group is then \{\pm \varepsilon^n \mid n \in \mathbb{Z}\}.[16] A representative example is \mathbb{Q}(\sqrt{2}), where the ring of integers is \mathbb{Z}[\sqrt{2}] and the fundamental unit is \varepsilon = 1 + \sqrt{2}, satisfying x^2 - 2 y^2 = -1.[16] Units in quadratic fields are precisely the elements of norm \pm 1.[15] Algorithms for computing the fundamental unit in real quadratic fields rely on the continued fraction expansion of \sqrt{d}, which is purely periodic with period length related to the regulator of the unit group.[18] The expansion \sqrt{d} = [a_0; \overline{a_1, \dots, a_\ell}] produces convergents p_k / q_k; the fundamental unit arises from the convergent immediately preceding the period repetition, where p_k^2 - d q_k^2 = \pm 1 yields the minimal solution to the Pell equation.[18] This method efficiently bounds the search by the period length \ell, often small for small d, and underpins computational number theory tools for unit groups.[18]Embeddings and Galois Theory
Embeddings
A quadratic field K = \mathbb{Q}(\sqrt{d}), where d is a square-free integer not equal to 0 or 1, admits exactly two distinct embeddings into the complex numbers \mathbb{C}, as the degree of the extension is 2.[19] These embeddings are field homomorphisms \sigma: K \to \mathbb{C} that fix \mathbb{Q} pointwise and are determined by the image of \sqrt{d}.[20] Specifically, the two embeddings send \sqrt{d} to the two roots of the minimal polynomial x^2 - d = 0 in \mathbb{C}, namely \pm \sqrt{d}, where \sqrt{d} denotes the principal square root in \mathbb{C} (the non-negative real value if d > 0, or i times the positive real square root of |d| if d < 0).[19] The nature of these embeddings depends on the sign of d. For real quadratic fields, where d > 0, both embeddings are real, meaning \sigma_1 and \sigma_2 map K into \mathbb{R} \subseteq \mathbb{C}.[20] Thus, there are two real embeddings, corresponding to two archimedean (infinite) places of K, each of which is a real place.[19] In contrast, for imaginary quadratic fields, where d < 0, \sqrt{d} = i \sqrt{|d|} with i = \sqrt{-1}, so both embeddings are non-real complex embeddings that form a conjugate pair under complex conjugation.[20] Here, there are no real embeddings, and the two complex embeddings contribute to a single complex infinite place.[19] These embeddings encode key arithmetic invariants of elements in K. For \alpha \in K, the trace \operatorname{Tr}_{K/\mathbb{Q}}(\alpha) is the sum of the images under the two embeddings, \operatorname{Tr}_{K/\mathbb{Q}}(\alpha) = \sigma_1(\alpha) + \sigma_2(\alpha), while the norm N_{K/\mathbb{Q}}(\alpha) is their product, N_{K/\mathbb{Q}}(\alpha) = \sigma_1(\alpha) \sigma_2(\alpha).[19] For example, in the basis \{1, \sqrt{d}\}, the trace of a + b \sqrt{d} is $2a and the norm is a^2 - d b^2.[20] The infinite places of K thus decompose into r_1 = 2 real places and r_2 = 0 complex places for real quadratic fields, or r_1 = 0 real places and r_2 = 1 complex place for imaginary quadratic fields, where the total number of infinite places is r_1 + r_2.[19]Galois Group
A quadratic extension K = \mathbb{Q}(\sqrt{d}) of the rationals, where d is a square-free integer not equal to 0 or 1, is a Galois extension because it is the splitting field of the separable irreducible polynomial x^2 - d \in \mathbb{Q} (assuming characteristic not 2).[21] Thus, its Galois closure over \mathbb{Q} is K itself, as the extension is both separable and normal of degree 2.[20] The Galois group \Gal(K/\mathbb{Q}) is isomorphic to \mathbb{Z}/2\mathbb{Z}, the cyclic group of order 2.[20] It is generated by the unique non-trivial automorphism \sigma, known as the conjugation map, which sends \sqrt{d} to -\sqrt{d} and fixes \mathbb{Q} pointwise.[20] This action corresponds to the two embeddings of K into \mathbb{C}, where \sigma swaps the real and complex conjugates if applicable.[22] By the fundamental theorem of Galois theory, the fixed field of \Gal(K/\mathbb{Q}) is precisely \mathbb{Q}, as there are no proper subfields between \mathbb{Q} and K.[20] Quadratic fields provide prototypical examples of abelian extensions of \mathbb{Q}, with the Galois group being abelian.[20] In the context of class field theory, Artin reciprocity describes such extensions, associating the Galois group to a quotient of the idele class group modulo the norms from K.[23] The discriminant of K plays a key role in the conductor-discriminant formula for abelian extensions, where the conductor equals the absolute value of the discriminant, linking ramification to the extension's structure.[23]Ideal Theory
Prime Ideal Factorization
In quadratic fields, the factorization of a rational prime ideal (p) into prime ideals in the ring of integers \mathcal{O}_K is determined by the behavior of the minimal polynomial of a primitive element modulo p, leading to three possibilities: splitting, inertia, or ramification.[24] For an odd prime p, this behavior is governed by the Legendre symbol \left( \frac{\Delta_K}{p} \right), where \Delta_K is the discriminant of K.[25] Specifically, (p) splits into two distinct prime ideals if \left( \frac{\Delta_K}{p} \right) = 1 (the prime splits completely), remains prime (inert) if \left( \frac{\Delta_K}{p} \right) = -1, and ramifies as a square of a prime ideal if \left( \frac{\Delta_K}{p} \right) = 0, i.e., if p divides \Delta_K.[25] In formula terms, for an odd prime p, (p) = \mathfrak{p} \mathfrak{q}, \quad \mathfrak{p}^2, \quad \text{or remains prime}, respectively, where \mathfrak{p} and \mathfrak{q} are distinct prime ideals of norm p, and \mathfrak{p} has norm p in the ramified case.[26] The case p = 2 requires special treatment due to the possible forms of \mathcal{O}_K. If \Delta_K \equiv 0 \pmod{4}, then $2 ramifies as (2) = \mathfrak{p}^2 for some prime ideal \mathfrak{p}; if \Delta_K \equiv 1 \pmod{8}, then $2 splits as (2) = \mathfrak{p} \mathfrak{q}; and if \Delta_K \equiv 5 \pmod{8}, then $2 remains inert.[26] These conditions can be unified using the Kronecker symbol \left( \frac{\Delta_K}{2} \right), which extends the Legendre symbol and yields $1 for splitting, -1 for inertia, and $0 for ramification.[25] The ramified primes are precisely those dividing the discriminant \Delta_K, which for quadratic fields K = \mathbb{Q}(\sqrt{d}) (with d square-free) are the primes dividing d if d \equiv 1 \pmod{4}, or dividing $4d otherwise.[26] For example, in K = \mathbb{Q}(\sqrt{-5}) with \Delta_K = -20, the prime $2 ramifies as (2) = \mathfrak{p}^2 where \mathfrak{p} = (2, 1 + \sqrt{-5}), the prime $5 ramifies as (5) = \mathfrak{q}^2 where \mathfrak{q} = (5, \sqrt{-5}), and the prime $3 splits as (3) = \mathfrak{p}_3 \mathfrak{q}_3 where \mathfrak{p}_3 = (3, 1 + \sqrt{-5}) and \mathfrak{q}_3 = (3, 1 - \sqrt{-5}).[26] This factorization is explicitly given by Dedekind's theorem: for K = \mathbb{Q}(\alpha) with minimal polynomial f(T) \in \mathbb{Z}[T] and prime p not dividing the index [\mathcal{O}_K : \mathbb{Z}[\alpha]], if f(T) \equiv \prod_i \pi_i(T)^{e_i} \pmod{p} into distinct monic irreducibles \pi_i(T) over \mathbb{F}_p, then (p) = \prod_i \mathfrak{p}_i^{e_i} where \mathfrak{p}_i = (p, \pi_i(\alpha)) and each \mathfrak{p}_i is prime with residue degree \deg \pi_i.[24] In quadratic fields, f(T) is quadratic, so the factorization modulo p directly yields linear factors (splitting), a repeated linear factor (ramification), or irreducibility (inertia).[24]Class Group
In quadratic fields, the ideal class group Cl_K of a number field K = \mathbb{Q}(\sqrt{d}) is the group of fractional ideals of the ring of integers \mathcal{O}_K modulo the principal fractional ideals, forming a finite abelian group whose order is the class number h_K.[27] The finiteness of Cl_K follows from Minkowski's geometry of numbers, which bounds the norms of ideals representing each class.[28] Specifically, every class contains an ideal of norm at most the Minkowski bound, approximately \frac{\sqrt{|\Delta_K|}}{2} for real quadratic fields and \frac{2}{\pi} \sqrt{|\Delta_K|} for imaginary ones, where \Delta_K is the discriminant.[29] The class number h_K equals 1 if and only if \mathcal{O}_K is a principal ideal domain (and hence a unique factorization domain).[27] To compute Cl_K, one generates candidate ideals as prime ideals of norm below the Minkowski bound, factors rational primes into these using the prime ideal factorization in quadratic fields, and determines relations among them to find the group structure.[29] Gauss's genus theory further aids computation by determining the 2-rank of Cl_K, which equals the number of distinct prime factors of the discriminant minus one (for odd primes) plus adjustments for the ramified 2-adic part.[27] Class number formulas differ for real and imaginary quadratic fields, both involving Dirichlet L-functions L(s, \chi) associated to the Kronecker symbol \chi modulo |\Delta_K|. For imaginary quadratic fields, the formula is h_K = \frac{w \sqrt{|\Delta_K|}}{2\pi} L(1, \chi), where w is the number of roots of unity in K (typically 2, 4, or 6).[30] For real quadratic fields, it becomes h_K R_K = \sqrt{\Delta_K} L(1, \chi), with R_K the regulator from the unit group.[27] These express h_K analytically, highlighting growth like \sqrt{|\Delta_K|}. The Baker-Heegner-Stark theorem identifies all imaginary quadratic fields with h_K = 1: those with discriminants \Delta_K = -3, -4, -7, -8, -11, -19, -43, -67, -163.[31] For small |\Delta_K|, class numbers remain modest; the table below lists examples for fundamental discriminants up to 100.| Discriminant \Delta_K | Field Type | Class Number h_K |
|---|---|---|
| -3 | Imaginary | 1 |
| -4 | Imaginary | 1 |
| -7 | Imaginary | 1 |
| -8 | Imaginary | 1 |
| -11 | Imaginary | 1 |
| -15 | Imaginary | 2 |
| -19 | Imaginary | 1 |
| -20 | Imaginary | 2 |
| -23 | Imaginary | 3 |
| -24 | Imaginary | 2 |
| 5 | Real | 1 |
| 8 | Real | 1 |
| 12 | Real | 1 |
| 13 | Real | 1 |
| 17 | Real | 1 |
| 21 | Real | 1 |
| 29 | Real | 1 |
| 33 | Real | 1 |
| 37 | Real | 1 |
| 41 | Real | 1 |
Special Constructions
Quadratic Subfields of Prime Cyclotomic Fields
For an odd prime p, the p-th cyclotomic field \mathbb{Q}(\zeta_p), where \zeta_p is a primitive p-th root of unity, contains a unique quadratic subfield K. This subfield arises as the fixed field of the unique subgroup of index 2 in the Galois group \mathrm{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q}) \cong (\mathbb{Z}/p\mathbb{Z})^\times.[32][33] Explicitly, K = \mathbb{Q}\left(\sqrt{(-1)^{(p-1)/2} p}\right). When p \equiv 1 \pmod{4}, this yields the real quadratic field \mathbb{Q}(\sqrt{p}); when p \equiv 3 \pmod{4}, it yields the imaginary quadratic field \mathbb{Q}(\sqrt{-p}). The discriminant \Delta_K of K is (-1)^{(p-1)/2} p, as the square-free part is congruent to 1 modulo 4.[34][32][35] The square root of the discriminant admits an explicit construction via the quadratic Gauss sum g = \sum_{k=0}^{p-1} \left( \frac{k}{p} \right) \zeta_p^k, where \left( \frac{\cdot}{p} \right) denotes the Legendre symbol; this sum satisfies g^2 = (-1)^{(p-1)/2} p, and thus \mathbb{Q}(g) = K.[34][33] Representative examples include p=5, where K = \mathbb{Q}(\sqrt{5}) with \Delta_K = 5, and p=7, where K = \mathbb{Q}(\sqrt{-7}) with \Delta_K = -7. Historically, Carl Friedrich Gauss employed the quadratic subfield \mathbb{Q}(\sqrt{17}) of the 17th cyclotomic field in his 1796 proof of the constructibility of the regular 17-gon using ruler and compass.[32][36]Quadratic Subfields of Other Cyclotomic Fields
In the cyclotomic field \mathbb{Q}(\zeta_n), where n > 2 is composite or a higher prime power, the quadratic subfields are the fixed fields of the index 2 subgroups of the Galois group \mathrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\times.[37] These subgroups exist whenever the order \phi(n) is even, which holds for all n > 2, and the number of distinct quadratic subfields equals the number of such subgroups, potentially exceeding one when \phi(n)/2 > 1.[37] For n = p^k with odd prime p and k > 1, the Galois group is cyclic of order p^{k-1}(p-1), admitting a unique index 2 subgroup and thus a unique quadratic subfield, identical to that of \mathbb{Q}(\zeta_p).[37] Specifically, when p \equiv 3 \pmod{4}, this subfield is \mathbb{Q}(\sqrt{-p}).[37] For p = 2 and k = 2, \mathbb{Q}(\zeta_4) = \mathbb{Q}(i) = \mathbb{Q}(\sqrt{-1}), which is itself quadratic.[37] For n = 3, \mathbb{Q}(\zeta_3) = \mathbb{Q}(\sqrt{-3}).[37] When n has multiple distinct prime factors, such as n = 12 = 2^2 \cdot 3, the Galois group \mathrm{Gal}(\mathbb{Q}(\zeta_{12})/\mathbb{Q}) is isomorphic to C_2 \times C_2, yielding three index 2 subgroups and thus three quadratic subfields: \mathbb{Q}(\sqrt{-1}), \mathbb{Q}(\sqrt{3}), and \mathbb{Q}(\sqrt{-3}).[38] Similarly, for n = 15 = 3 \cdot 5, \mathbb{Q}(\zeta_{15}) contains the three quadratic subfields \mathbb{Q}(\sqrt{5}), \mathbb{Q}(\sqrt{-3}), and \mathbb{Q}(\sqrt{-15}), including both real (like \mathbb{Q}(\sqrt{5})) and imaginary examples.[39] These arise as the composita of the quadratic subfields from the prime power components, with multiple possibilities when \phi(n)/2 > 1. By the Kronecker-Weber theorem, every quadratic field \mathbb{Q}(\sqrt{d}) (for square-free integer d) embeds as a quadratic subfield of some cyclotomic field \mathbb{Q}(\zeta_m), where m is chosen based on the discriminant of \mathbb{Q}(\sqrt{d}).[40] For instance, the minimal such m is the absolute value of the discriminant when it is congruent to 0 or 1 modulo 4.[41] In cases where a quadratic subfield arises as the compositum-derived field from distinct prime factors, such as \mathbb{Q}(\sqrt{-15}) from the factors 3 and 5 in n=[15](/page/15), its discriminant is the product of the individual prime discriminants (up to units in the ring of integers), reflecting the ramification at those primes.[41]Non-Maximal Orders
Orders of Small Discriminant
In quadratic fields, non-maximal orders provide important examples where arithmetic properties, such as unique factorization, deviate further from those of the rationals compared to the maximal order. An order \mathcal{O} in a quadratic field K is defined as a subring of the ring of integers \mathcal{O}_K that contains \mathbb{Z} and is finitely generated as a \mathbb{Z}-module of rank equal to [K:\mathbb{Q}]. Every such order admits a unique expression \mathcal{O} = \mathbb{Z} + f \mathcal{O}_K for a positive integer f \geq 1, where f is the conductor of \mathcal{O}. The index [\mathcal{O}_K : \mathcal{O}] equals f, and the discriminant of \mathcal{O} is given by f^2 \Delta_K, where \Delta_K is the discriminant of K.[42] The structure \mathcal{O} = \mathbb{Z} + f \mathcal{O}_K implies that \mathcal{O} is generated over \mathbb{Z} by f times a basis for \mathcal{O}_K. For imaginary quadratic fields, the unit group \mathcal{O}^\times consists of the units of \mathcal{O}_K that are congruent to $1 modulo f \mathcal{O}_K, resulting in a proper subgroup of \mathcal{O}_K^\times when f > 1. Thus, non-maximal orders have strictly smaller unit groups than the maximal order. In real quadratic fields, the situation is analogous, though the infinite units lead to a Dirichlet unit theorem adapted to the conductor, yielding fewer fundamental units.[43] The ideal class group of a non-maximal order, known as the ring class group, extends the ideal class group of \mathcal{O}_K and incorporates the conductor f through the ring class field, an abelian extension of K whose Galois group is isomorphic to the ring class group. This group fits into an exact sequence relating it to the class group of \mathcal{O}_K, the units modulo f, and the ray class group modulo f in \mathcal{O}_K. Computations for small conductors often reveal larger class numbers than in the maximal case, highlighting obstructions to unique factorization.[43] Non-maximal orders arise naturally in examples illustrating the failure of unique factorization. In the imaginary quadratic field \mathbb{Q}(\sqrt{-5}) with \Delta_K = -20, the suborder \mathcal{O} = \mathbb{Z}[2\sqrt{-5}] of conductor f=2 has discriminant -80. Here, elements like $21 = 3 \cdot 7 = (1 + 2\sqrt{-5})(1 - 2\sqrt{-5}) (up to units) demonstrate non-unique factorization, extending the known failure in the maximal order \mathcal{O}_K = \mathbb{Z}[\sqrt{-5}]. Similar phenomena occur in other small-discriminant fields, where suborders amplify arithmetic complexities.[42] The following table presents representative non-maximal orders in quadratic fields with small field discriminants |\Delta_K| < 30, focusing on conductors f=2 to keep order discriminants |\operatorname{disc}(\mathcal{O})| < 100. These examples illustrate the scaling of discriminants and typical generators for the orders.| Field K | \Delta_K | Conductor f | Order \mathcal{O} | \operatorname{disc}(\mathcal{O}) |
|---|---|---|---|---|
| \mathbb{Q}(i) | -4 | 2 | \mathbb{Z}[2i] | -16 |
| \mathbb{Q}(\sqrt{-3}) | -3 | 2 | \mathbb{Z}[\sqrt{-3}] | -12 |
| \mathbb{Q}(\sqrt{-2}) | -8 | 2 | \mathbb{Z}[2\sqrt{-2}] | -32 |
| \mathbb{Q}(\sqrt{-7}) | -7 | 2 | \mathbb{Z}[\sqrt{-7}] | -28 |
| \mathbb{Q}(\sqrt{5}) | 5 | 2 | \mathbb{Z}[1 + \sqrt{5}] | 20 |
| \mathbb{Q}(\sqrt{-5}) | -20 | 2 | \mathbb{Z}[2\sqrt{-5}] | -80 |