Determine the full ring of quadratic integers

Try

The number ω=(1+√5)/2 is not in ℤ[√5], but it satisfies the monic equation ω²−ω−1=0. By contrast, √5/2 satisfies 4X²−5=0, which is not monic. Does having an integer polynomial equation make both numbers algebraic integers?

An element is integral over a ring A if it satisfies a monic polynomial with coefficients in A. Monic means leading coefficient 1. An algebraic integer is integral over ℤ. We will prove why this definition is stable under arithmetic and determine every algebraic integer in each quadratic field, rather than only checking some candidates.

Throughout, rings are commutative with identity and inclusions preserve identity. A module is finitely generated if a finite list spans it using coefficients in its scalar ring; this does not mean it has finitely many elements. Review modules and maps and Gauss’s lemma as needed.

Predict

If α satisfies a monic equation of degree n over A, predict why every power of α is an A-linear combination of 1,α,…,αⁿ⁻¹. Conversely, suppose multiplication by α preserves a finitely generated module containing 1. Can a determinant produce an equation even if the chosen generators are not a basis?

For a squarefree integer d other than 0 or 1, write K=ℚ(√d). Squarefree means no prime square divides d; negative d is allowed. Every element has a unique form a+b√d with a,b∈ℚ. Predict why d modulo 4 might determine whether half-integer coordinates are allowed. Do not assume that a and b must themselves be integers just because the whole element is integral.

Experiment

For α=a+b√d, its conjugate is ᾱ=a−b√d. Its trace and norm are T=2a and N=a²−db². Thus α satisfies X²−TX+N=0. Compute these values before checking the table.

The positive and negative examples use the same rational arithmetic. No order comparison involving √d is needed. The field is embedded in ℂ when d is negative, but the proofs below do not depend on a picture of its elements.

Understand

A finite module gives a monic equation

First, if α is integral over A, reducing its monic equation expresses αⁿ in lower powers. Repeated reduction expresses every higher power in 1,α,…,αⁿ⁻¹. Therefore A[α] is a finitely generated A-module.

For the converse let A⊆B, α∈B, and let M⊆B be an A-submodule containing 1, generated by m₁,…,mᵣ, such that αM⊆M. Choose coefficients cᵢⱼ∈A with αmᵢ=Σⱼ cᵢⱼmⱼ. Writing m as a column gives (αI−C)m=0. The adjugate identity, valid over any commutative ring, yields det(αI−C)mᵢ=0 for each i. Since 1 is an A-linear combination of the generators, det(αI−C)=0 in B. The polynomial det(XI−C) is monic of degree r with coefficients in A. Hence α is integral. The generators need not be independent, and no division or matrix inverse was used.

If 1=0 in B, the conclusion is immediate from the monic polynomial X and needs no nonempty generating list. Otherwise a module containing 1 cannot have an empty generating list. Thus α is integral exactly when such a module exists, and exactly when A[α] is finite as an A-module.

There is a related formulation for an abstract module M over A[α]: if M is finitely generated over A and faithful over A[α], the same determinant annihilates M and faithfulness makes it zero in A[α]. Faithful means that only the zero element of A[α] annihilates every member of M. Faithfulness merely over A is insufficient. Our submodule formulation avoids this ambiguity: containing 1 directly forces any annihilating scalar from B to be zero.

Closure and transitivity without a hidden field assumption

If α and β are integral over A with monic equations of degrees r and s, the products αⁱβʲ, 0≤i<r and 0≤j<s, span A[α,β] over A. Reduce each exponent using its monic equation. This finite module contains 1 and is preserved by multiplication by α+β, αβ and −α. The criterion proves all three integral. Elements of A are integral by X−a, so the integral elements in a fixed commutative extension form a subring.

Now suppose A⊆B⊆C, every element of B is integral over A, and γ∈C is integral over B. Only finitely many coefficients b₁,…,bₘ occur in one monic equation for γ. Iterating the preceding finite-generation argument makes A[b₁,…,bₘ] finite as an A-module. Adjoining γ is finite over this coefficient ring by monic reduction. Products of the two finite generating lists show A[b₁,…,bₘ,γ] is finite over A. It contains 1 and is preserved by γ, so γ is integral over A. We did not assume the whole ring B is a finite A-module.

Which trace and norm conditions are necessary?

A rational number u/v in lowest terms, with v>0, integral over ℤ must be an integer. Multiplying a monic degree-r equation by vʳ shows v divides uʳ; coprimality forces v=1. This also includes zero, whose reduced denominator is 1.

If α=a+b√d is not rational, b≠0 and its monic minimal polynomial over ℚ is X²−2aX+(a²−db²). Indeed it vanishes at α and a nonrational element of the degree-two field cannot have a linear minimal polynomial. If α is integral, this minimal polynomial has integer coefficients. Here is the Gauss step: the monic minimal polynomial divides a monic integer polynomial over ℚ. Write it as g/c, where g is primitive in ℤ[X] with positive leading coefficient c, and similarly write the monic quotient as h/e. Their product gives gh=ceF for a monic integer F. Gauss’s lemma makes gh primitive; since F is primitive, its content is 1, forcing ce=1. Thus c=e=1 and the minimal polynomial lies in ℤ[X]. Consequently T=2a and N=a²−db² are integers.

For rational α=a, integrality already implies a∈ℤ, so these same T,N are integers. Conversely, for any a,b∈ℚ, integer T and N supply the displayed monic quadratic equation. Therefore α is integral if and only if 2a and a²−db² are integers. This criterion includes rational elements; it does not silently identify a reducible quadratic with their minimal polynomial.

Derive every allowed denominator

Write 2a=m∈ℤ, and write b=r/s in lowest terms with s>0. If N is integral, multiplying m²/4−dr²/s²=N by 4s² gives

m²s²−4dr²=4Ns².

Hence s² divides 4dr². Coprimality of r and s implies s² divides 4d. For each odd prime dividing s, its exponent in s² is at least two, while its exponent in 4d is at most one by squarefreeness, impossible. At the prime 2 the exponent of 4d is at most three, so the exponent of 2 in s is at most one. Therefore s divides 2. This argument also works for negative d by taking absolute values for divisibility, and for b=0 in its reduced form 0/1.

We have a=m/2 and b=n/2 for integers m,n. The remaining norm condition is

m²−dn² ≡ 0 modulo 4.

Squares modulo 4 are 0 and 1. If n is even, m must be even. If n is odd, m²≡d modulo 4. A squarefree d has residue 1,2 or 3 modulo 4, so this requires d≡1 and m odd. These cases exhaust all possibilities.

When d≡2 or 3 modulo 4, both coordinates are integers and the integral elements form ℤ[√d]. When d≡1 modulo 4, the coordinates are either both integers or both half-integers with odd numerators. Put ω=(1+√d)/2. Then

(m+n√d)/2 = (m−n)/2+nω,

and the parity condition makes (m−n)/2 integral. Conversely ω satisfies X²−X+(1−d)/4=0, so all elements of ℤ[ω] are integral by closure. We have therefore proved the complete descriptions, not just inclusions:

O = ℤ[√d] if d≡2 or 3 modulo 4; O = ℤ[(1+√d)/2] if d≡1 modulo 4.

Each displayed pair 1,√d or 1,ω is a ℤ-basis: spanning was proved, and independence follows from irrationality of √d. These are integral bases in this quadratic setting.

Why this ring is integrally closed

The fraction field of O is K: it is contained in K, and its fraction field contains ℚ and √d, since √d is in O in either description. If γ∈K is integral over O, transitivity and the fact that every element of O is integral over ℤ make γ integral over ℤ. The classification puts γ back in O. Thus O is integrally closed in its fraction field.

This conclusion is not automatic for every ring generated by a quadratic integer. For d=5 the proper subring ℤ[√5] omits ω, even though ω lies in its fraction field and satisfies a monic polynomial with coefficients in ℤ⊆ℤ[√5]. Hence that subring is not integrally closed. Finite generation and integrality of its generators do not repair the omission.

Apply

Replace 5 by −7. Determine the full quadratic integer ring, decide whether (3+√−7)/2 is integral, and write its coordinates in an integral basis. Compare it with (1+√−7)/4. Supply both a successful monic equation and an obstruction; decimal approximations are not certificates.

Check the changed negative quadratic field

Since −7≡1 modulo 4, O=ℤ[ω] with ω=(1+√−7)/2 and ω²−ω+2=0. The first element is 1+ω. Its trace is 3 and norm (9+7)/4=4, so it satisfies X²−3X+4=0. Its basis coordinates are (1,1). For the second element the trace is 1/2, not an integer; necessity excludes it. Merely clearing its denominators to get an integer polynomial cannot make that polynomial monic.

Continue

Return to Gauss’s lemma for the primitive-factor argument, and integer presentations for the difference between a finite generating set and a finite quotient. The integral basis now supplies exact integer coordinates for later ideal-lattice and inverse-ideal calculations. No general theorem of unique ideal factorization or class-number calculation has been proved here.