Bézout Domains

Introduction

Bézout's identity says that in the integers the greatest common divisor of $a$ and $b$ is a linear combination of them, $d = ax + by$; by the ideal-theoretic reading of GCD Domains, directly above this article, it says that the ideal $(a, b)$ is principal. This article promotes the identity from a theorem about $\mathbb{Z}$ to a definition: a Bézout domain is an integral domain in which every finitely generated ideal is principal. The rung lies strictly between the GCD domains, where the gcd exists but need not be a combination, and the principal ideal domains, which are the Bézout domains with a finiteness hypothesis added.

The rung has two uses. It isolates exactly what makes the gcd computable by an identity rather than by a factorisation, and it is the only rung of the chain that is not trivially Noetherian: the standard example, the ring $\overline{\mathbb{Z}}$ of all algebraic integers, is a Bézout domain in which a single element may fail to factor into irreducibles at all. Throughout, $R$ is an integral domain, and divisibility, associates, the gcd and the lcm are used in the sense of Integral Domains and GCD Domains, above this article.

Bézout's Identity as a Definition

The Definition and the Two-Generated Case

Definition. An integral domain $R$ is a Bézout domain if every finitely generated ideal of $R$ is principal.

Proposition. For an integral domain $R$ the following are equivalent.

(a) $R$ is a Bézout domain.

(b) Every ideal of $R$ generated by two elements is principal.

(c) For all $a, b \in R$ the greatest common divisor $\gcd(a,b)$ exists and is a Bézout combination, $(a,b) = (\gcd(a,b))$.

Proof. (a) $\Rightarrow$ (b) is immediate. (b) $\Rightarrow$ (c) is the proposition of GCD Domains identifying a gcd with the generator of a principal ideal $(a)+(b)$. (c) $\Rightarrow$ (a): if $I = (a_1, \ldots, a_n)$ is finitely generated, then $I = (a_1, \ldots, a_{n-1}) + (a_n)$ equals $(\gcd(a_1,\ldots,a_{n-1})) + (a_n) = (\gcd(\gcd(a_1,\ldots,a_{n-1}), a_n))$ by induction, once $\gcd$ is known to exist for $n - 1$ elements; the induction is on $n$ and starts at $n = 2$, so every finitely generated ideal of $R$ is principal.

The induction in the proof needs the gcd of more than two elements to exist; that follows from the two-element case, since $(\gcd(a,b),c)$ is principal and its generator is a gcd of $a, b, c$. The definition is thus genuinely a two-element condition.

Theorem (Bézout). Let $R$ be a Bézout domain and $a, b \in R$ not both zero. Then $\gcd(a,b)$ exists and there are $x, y \in R$ with

$$ \gcd(a,b) = a x + b y . $$

Proof. By the proposition, $(a, b)$ is principal, say $(a,b) = (d)$; the generator $d$ is a gcd by the ideal-theoretic characterisation of GCD Domains, and $d \in (a,b)$ means precisely that $d = ax + by$ for some $x, y$.

Corollary. Every principal ideal domain — the rung Principal Ideal Domains, below this article in this category — is a Bézout domain, and every Bézout domain is a GCD domain.

Proof. In a principal ideal domain every ideal is principal, so every finitely generated ideal is; and a Bézout domain has $(a,b)$ principal for all pairs, so gcds exist as in (c) of the proposition.

The gcd in a Bézout Domain

Proposition. Let $R$ be a Bézout domain and let $a_1, \ldots, a_n \in R$, not all zero.

(a) $\gcd(a_1, \ldots, a_n)$ exists and $(a_1, \ldots, a_n) = (\gcd(a_1,\ldots,a_n))$.

(b) $\gcd(a_1,\ldots,a_n)$ is a linear combination $\sum_i a_i x_i$ with $x_i \in R$.

(c) $a_1, \ldots, a_n$ are coprime, in the sense that their only common divisors are units, if and only if $(a_1,\ldots,a_n) = R$.

Proof. (a) and (b) follow by the induction of the proposition above, the combination statement from $\gcd \in (a_1,\ldots,a_n)$. (c) The common divisors are the divisors of the gcd $d$; they are all units exactly when $d$ is a unit, that is, when $(d) = R$.

Remark. The passage from divisibility to linear combinations is what makes the gcd effective: a common divisor $c$ of $a$ and $b$ divides $ax + by$, so any identity $ax + by = 1$ certifies that no non-unit common divisor exists, without any factorisation being exhibited. This is the form in which the gcd is used in the classical rings $\mathbb{Z}$ and $K[x]$, and it is the reason the rung is named after the identity rather than after the gcd.

Bézout Domains and GCD Domains

The Two Rungs

The two rungs are related by one implication and by one counterexample.

Theorem. Every Bézout domain is a GCD domain. The converse fails: the polynomial ring $\mathbb{Z}[x]$ is a GCD domain that is not a Bézout domain.

Proof. The implication is the corollary above. For the failure, $\mathbb{Z}[x]$ is a unique factorisation domain and hence a GCD domain, by GCD Domains, above this article, and Unique Factorisation Domains, below it in this category; but the ideal $(2, x)$ is not principal. Indeed, if $(2,x) = (d)$ then $d \mid 2$ and $d \mid x$; a common divisor of the constant $2$ and the polynomial $x$ is a unit of $\mathbb{Z}[x]$, by the description of the units in Integral Domains, above, since a non-unit divisor of $2$ is an associate of $2$ and $2 \nmid x$ in $\mathbb{Z}[x]$. A unit generates all of $\mathbb{Z}[x]$, whereas $(2,x)$ is a proper ideal: the evaluation map $\mathbb{Z}[x] \to \mathbb{Z}/2\mathbb{Z}$, $f \mapsto f(0) \bmod 2$, vanishes on $2$ and on $x$ and is not the zero map, since it sends $1$ to $1$. Hence $(2,x)$ is not principal and $\mathbb{Z}[x]$ is not a Bézout domain.

Example. In $\mathbb{Z}$ and in $K[x]$, $K$ a field (the field axioms are those of Fields, later in this category), every ideal is principal, so both are Bézout domains; the identity of the previous section is the classical Bézout identity in the first case and the gcd of polynomials in the second. In $K[x,y]$ the ideal $(x,y)$ is not principal, so the ring is not a Bézout domain; it is nevertheless a unique factorisation domain.

Corollary. A Bézout domain is integrally closed in its fraction field: any element $u \in \operatorname{Frac}(R)$ that is a root of a monic polynomial with coefficients in $R$ lies in $R$.

Proof. Write $u = a/b$ with $\gcd(a,b) \sim 1$, possible because the gcd calculus of the previous section is available. If $u^n + c_{n-1}u^{n-1} + \cdots + c_0 = 0$ with $c_i \in R$, then multiplying by $b^n$ gives

$$ a^n + c_{n-1}a^{n-1}b + \cdots + c_0 b^n = 0, $$

so $b \mid a^n$; since $a$ and $b$ are coprime, $a^n$ and $b$ are coprime as well by the gcd calculus above, so $b \mid 1$ and $b$ is a unit. Hence $u = ab^{-1} \in R$. The fraction field is Localization and the Fraction Field, below this article in this category.

The Divisibility Lattice

Theorem. Let $R$ be a Bézout domain. Then the set of associate classes of nonzero elements of $R$, ordered by divisibility, is a lattice: every pair has a meet and a join, namely $\gcd(a,b)$ and $\operatorname{lcm}(a,b)$.

Proof. The gcd exists by the proposition above, and the lcm exists by the identity $\gcd(a,b)\operatorname{lcm}(a,b) \sim ab$ of GCD Domains, above. In the divisibility order on classes, $d$ is the meet of $a$ and $b$ exactly when $d \mid a$, $d \mid b$ and every common divisor of $a$ and $b$ divides $d$, which is the definition of the gcd; the lcm is the join by the dual definition.

Corollary. In a Bézout domain the principal ideals form a lattice under $+$ and $\cap$: both $(a) + (b) = (\gcd(a,b))$ and $(a) \cap (b) = (\operatorname{lcm}(a,b))$ are principal. For finitely many nonzero elements,

$$ (a_1) + \cdots + (a_n) = (\gcd(a_1,\ldots,a_n)), \qquad (a_1) \cap \cdots \cap (a_n) = (\operatorname{lcm}(a_1,\ldots,a_n)) . $$

Proof. The first identity is the defining property of a Bézout domain; the second is the corollary of GCD Domains, above, identifying the intersection of two principal ideals with the principal ideal generated by the lcm; both extend to $n$ elements by induction, the $n$-element gcd being the gcd of a two-element gcd and the next element, and likewise for the lcm.

Remark. In a general integral domain the meet of two associate classes need not exist, as $\mathbb{Z}[\sqrt{-5}]$ shows in GCD Domains, above; in a Bézout domain it always does, and the join as well. The divisibility order of a Bézout domain is therefore a lattice, and the map $a \mapsto (a)$ is an order-isomorphism from that lattice onto the lattice of principal ideals ordered by reverse inclusion.

The Noetherian Case: Principal Ideal Domains

Definition. A commutative ring $R$ is Noetherian if every ideal of $R$ is finitely generated; equivalently, every ascending chain of ideals of $R$ is eventually constant. The definition, the equivalence of the two forms and the theory of such rings are Noetherian and Artinian Rings, below this article in this category.

Theorem. An integral domain is a principal ideal domain if and only if it is a Bézout domain and is Noetherian.

Proof. A principal ideal domain is Bézout by the corollary above, and it is Noetherian because every ideal is generated by one element, hence finitely generated. Conversely, in a Bézout domain every finitely generated ideal is principal; if in addition every ideal is finitely generated, then every ideal is principal, which is the definition of a principal ideal domain.

Corollary. A Bézout domain is Noetherian if and only if it contains no infinite strictly increasing chain of principal ideals.

Proof. If $R$ is Noetherian then no chain of ideals is infinite and strictly increasing. Conversely, if $R$ is not Noetherian, let $I$ be an ideal that is not finitely generated and choose elements $a_1 \in I$ and recursively $a_{n+1} \in I \setminus (a_1, \ldots, a_n)$; the set subtracted is nonempty, since otherwise $I = (a_1,\ldots,a_n)$ would be finitely generated. In a Bézout domain each $(a_1, \ldots, a_n)$ is principal, say $(a_1, \ldots, a_n) = (d_n)$; then $(d_n) \subsetneq (d_{n+1})$ for every $n$, because $a_{n+1} \in (d_{n+1})$ while $a_{n+1} \notin (a_1, \ldots, a_n) = (d_n)$. So $(d_1) \subsetneq (d_2) \subsetneq (d_3) \subsetneq \cdots$ is an infinite strictly increasing chain of principal ideals.

Theorem. Let $R$ be a Bézout domain. Then $R$ is a unique factorisation domain if and only if $R$ is a principal ideal domain.

Proof. A principal ideal domain is a unique factorisation domain by Unique Factorisation Domains, below this article in this category. Conversely, if the Bézout domain $R$ is a unique factorisation domain, then every nonzero non-unit is a product of irreducibles, and an infinite strictly increasing chain of principal ideals $(a_1) \subsetneq (a_2) \subsetneq \cdots$ is impossible, since $(a_n) \subseteq (a_{n+1})$ gives $a_n = a_{n+1} b_{n+1}$ and hence a strictly shrinking multiset of prime factors of the fixed element $a_1$, which cannot occur infinitely often; this is the ascending chain condition on principal ideals, and it is stated in Unique Factorisation Domains, below this article in this category. By the corollary above, $R$ is Noetherian, hence a principal ideal domain by the theorem above.

The hypothesis "Bézout" is essential to the last theorem: $\mathbb{Z}[x]$ is a unique factorisation domain that is not a principal ideal domain, and it is not Bézout.

The Ring of All Algebraic Integers

Definition

Definition. An element $\alpha \in \mathbb{C}$ is an algebraic integer if it is a root of a monic polynomial with coefficients in $\mathbb{Z}$:

$$ \alpha^n + a_{n-1}\alpha^{n-1} + \cdots + a_0 = 0, \qquad a_i \in \mathbb{Z}. $$

The set of algebraic integers is written $\overline{\mathbb{Z}}$ and called the ring of all algebraic integers.

Proposition. $\overline{\mathbb{Z}}$ is a subring of $\mathbb{C}$ containing $\mathbb{Z}$, and it is an integral domain.

Proof. That $\overline{\mathbb{Z}}$ is closed under addition and multiplication is the standard stability of integral elements under the ring operations: if $\alpha$ and $\beta$ satisfy monic equations of degrees $m$ and $n$, then every element of the $\mathbb{Z}$-module generated by the $mn$ products $\alpha^i \beta^j$ with $0 \leq i < m$, $0 \leq j < n$ is a $\mathbb{Z}$-combination of them, so the module is preserved by multiplication by $\alpha$ and by $\beta$ and hence by any polynomial in them, and $\alpha \pm \beta$ and $\alpha\beta$ are integral; a proof belongs to Integral Extensions and Krull Dimension, below this article in this category. As a subring of the field $\mathbb{C}$ it is an integral domain.

Example. $\sqrt{2}$ and $\sqrt{-5}$ are algebraic integers, roots of $x^2 - 2$ and $x^2 + 5$; so are $\tfrac{1+\sqrt{-19}}{2}$ (root of $x^2 - x + 5$) and every $n$-th root of an integer. The element $\tfrac12$ is not an algebraic integer: if it satisfied a monic equation $\alpha^n + a_{n-1}\alpha^{n-1} + \cdots + a_0 = 0$ with integer coefficients, then multiplying by $2^n$ would give $1 + 2(\text{integer}) = 0$, which is impossible. Hence $\overline{\mathbb{Z}} \cap \mathbb{Q} = \mathbb{Z}$.

The Standard Non-Noetherian Bézout Domain

Theorem. The ring $\overline{\mathbb{Z}}$ of all algebraic integers is a Bézout domain: every finitely generated ideal of $\overline{\mathbb{Z}}$ is principal.

Proof. This is a standard theorem of the literature, cited here rather than reproduced; the argument uses the valuation theory of the algebraic integers, which lies beyond this rung. The companion example is the ring of analytic functions on a non-compact Riemann surface, which is a Bézout domain by Helmer's theorem.

Theorem. $\overline{\mathbb{Z}}$ is not Noetherian, not a principal ideal domain and not a unique factorisation domain.

Proof. The elements $2^{1/2^n}$ are algebraic integers, roots of $x^{2^n} - 2$, and

$$ (2) \subsetneq (2^{1/2}) \subsetneq (2^{1/4}) \subsetneq (2^{1/8}) \subsetneq \cdots $$

is a strictly increasing chain of principal ideals: $2 = (2^{1/2})^2$ gives $(2) \subseteq (2^{1/2})$, and in general $2^{1/2^n} = (2^{1/2^{n+1}})^2$ gives $(2^{1/2^n}) \subseteq (2^{1/2^{n+1}})$; the inclusion is strict, because equality would give $2^{1/2^n} = 2^{1/2^{n+1}} \cdot r$ with $r \in \overline{\mathbb{Z}}$, and cancelling the nonzero factor $2^{1/2^{n+1}}$ would exhibit $2^{1/2^{n+1}}$ as a unit of $\overline{\mathbb{Z}}$, whose inverse $2^{-1/2^{n+1}}$ is not in $\overline{\mathbb{Z}}$, since its $2^{n+1}$-st power $\tfrac12$ is not an algebraic integer. So there is an infinite strictly increasing chain of principal ideals, and $\overline{\mathbb{Z}}$ is not Noetherian. A principal ideal domain is Noetherian, and a unique factorisation domain satisfies the ascending chain condition on principal ideals by Unique Factorisation Domains, below this article in this category, so $\overline{\mathbb{Z}}$ is neither.

Corollary. The Bézout domain $\overline{\mathbb{Z}}$ is not a principal ideal domain, so a Bézout domain need not be Noetherian.

Remark. The ring $\overline{\mathbb{Z}}$ also shows that a Bézout domain can be very far from a unique factorisation domain: the element $2$ has no factorisation into irreducibles, since $2 = (2^{1/2})^2$, $2^{1/2} = (2^{1/4})^2$ and so on indefinitely, so no factor in the chain can be irreducible. Consequently no element of the chain is prime, although every gcd in $\overline{\mathbb{Z}}$ exists and is a linear combination. The rung is therefore neither vacuous nor automatically Noetherian: it is the one rung of the chain whose standard example is not Noetherian.

Worked Ideals in the Ring of Algebraic Integers

Example. In $\overline{\mathbb{Z}}$ the ideal generated by $2$ and $2^{1/2}$ is principal: $2 = (2^{1/2})^2$, so $(2, 2^{1/2}) = (2^{1/2})$, and the gcd of $2$ and $2^{1/2}$ is $2^{1/2}$ up to units. Likewise $(2^{1/2}, 2^{1/4}) = (2^{1/4})$, and in general $(2^{1/2^n}, 2^{1/2^{n+1}}) = (2^{1/2^{n+1}})$. Every one of these ideals has two generators and one generator, which is what Bézout-ness asserts.

Example. The chain of the previous section shows that in $\overline{\mathbb{Z}}$ the ideal generated by all the elements $2^{1/2^n}$, $n \geq 1$, is not finitely generated. Indeed, a finite generating set $g_1, \ldots, g_k$ lies in the union of the chain, so each $g_i$ lies in some $(2^{1/2^{N_i}})$; taking $N = \max_i N_i$, all the generators lie in $(2^{1/2^N})$, and the ideal they generate is contained in $(2^{1/2^N})$, which is a proper subset of the union because $2^{1/2^{N+1}}$ lies in the union and not in $(2^{1/2^N})$. So $\overline{\mathbb{Z}}$ is a Bézout domain in which a genuinely infinite ideal-theoretic construction is needed.

The Rungs Compared

Ring	Bézout	GCD	UFD	PID
$\mathbb{Z}$	yes	yes	yes	yes
$K[x]$, $K$ a field	yes	yes	yes	yes
$\mathbb{Z}[x]$	no	yes	yes	no
$K[x,y]$	no	yes	yes	no
$\overline{\mathbb{Z}}$	yes	yes	no	no

The first two rows are principal ideal domains, so every column is positive. The next two rows are unique factorisation domains, so they are GCD domains, but $(2,x)$ in $\mathbb{Z}[x]$ and $(x,y)$ in $K[x,y]$ are not principal, so they are not Bézout domains and hence not principal ideal domains. The last row is the Bézout domain that is neither Noetherian nor a unique factorisation domain. The table fixes the two strictnesses at this rung: GCD domains need not be Bézout, and Bézout domains need not be Noetherian.

Summary

A Bézout domain is an integral domain in which every finitely generated ideal is principal; equivalently, every ideal generated by two elements is principal; equivalently, every pair has a greatest common divisor that is a linear combination of the pair. Every Bézout domain is a GCD domain, and the converse fails: $\mathbb{Z}[x]$ is a unique factorisation domain, hence a GCD domain, but $(2,x)$ is not principal. In a Bézout domain gcds of finite sets exist, are linear combinations, and the gcd being a unit is equivalent to the ideal generated by the set being the whole ring.

A Bézout domain is a principal ideal domain exactly when it is Noetherian, so the principal ideal domains are the Noetherian Bézout domains. The rung is strict in both directions: $\mathbb{Z}$ and $K[x]$ are principal ideal domains, while the ring $\overline{\mathbb{Z}}$ of all algebraic integers is a Bézout domain that is not Noetherian, not a principal ideal domain and not a unique factorisation domain, because $2$ has no factorisation into irreducibles in $\overline{\mathbb{Z}}$.

Summary of Notation

Symbol	Meaning
$R$	Integral domain
$R^{\times}$	Group of units
$a \mid b$	$a$ divides $b$
$a \sim b$	Associates
$(a)$, $(a, b)$, $(a_1,\ldots,a_n)$	Principal ideal; ideals generated by the listed elements
$\gcd(a,b)$, $\operatorname{lcm}(a,b)$	Greatest common divisor, least common multiple
Bézout domain	Domain in which every finitely generated ideal is principal
Bézout combination	$d = ax + by$ for $d = \gcd(a,b)$
Noetherian	Every ideal finitely generated
PID	Principal ideal domain: Bézout and Noetherian
$\overline{\mathbb{Z}}$	Ring of all algebraic integers: roots of monic polynomials over $\mathbb{Z}$
$K$	A field
$\operatorname{Frac}(R)$	Fraction field of $R$