Integral domain ($ab=0\Rightarrow a=0\text{ or }b=0$)

An integral domain is a nonzero commutative ring without zero divisors. Equivalently, a product of two nonzero elements is always nonzero. The zero ring is excluded, and $\mathbb{Z}/6\mathbb{Z}$ fails because $2\cdot 3=0$. Every field is an integral domain, but the integers show that the converse fails: $2$ has no multiplicative inverse in $\mathbb{Z}$. See MIT's Basic Properties of Rings, definitions 15.9–15.10 and example 15.11.

The practical benefit is multiplicative cancellation. In the integers, $2x=6$ has at most one solution and indeed $x=3$ works. Yet $2x=1$ has no integer solution. Thus multiplicative cancellation guarantees uniqueness when a solution exists; it does not guarantee division by every nonzero element. Passing to the fraction field $\mathbb{Q}$ supplies the missing solution $1/2$.

The polynomial ring $\mathbb{Z}[X]$ is also an integral domain. For example, the product of $2X+1$ and $3X-4$ has leading coefficient $6$ and cannot vanish. Generally the leading coefficient of a product is the product of the two leading coefficients; an integral domain prevents that coefficient from vanishing. This connects integer arithmetic to polynomial arithmetic through one reusable principle.

Historically, this connection grew far beyond elementary examples. Keith Conrad identifies the ring of integers of a number field and a ring of polynomial functions on a smooth plane curve as early important examples of Dedekind domains. This shared structure connects number theory with algebraic geometry; see Conrad's Ideal Factorization, section 6.

In an integral domain, $ab=ac$ with $a\ne 0$ gives $a(b-c)=0$ by the distributive property. Since there are no zero divisors, $b-c=0$. Conversely, in a nonzero commutative ring, this cancellation property rules out zero divisors by comparing $ab=0$ with $a0=0$.

For an integral domain $A$ with fraction field $K$, a fractional ideal is a nonzero $A$-submodule $I\subseteq K$ for which $dI\subseteq A$ for some $0\ne d\in A$. A finitely generated submodule of $K$ has a common denominator and is therefore a fractional ideal. The product is generated by pairwise products of elements. An invertible fractional ideal has a multiplicative inverse and is finitely generated; arbitrary fractional ideals need not be finitely generated over a non-Noetherian domain. The familiar fractional ideals of a number field are the special case $A={\mathcal{O}}_K$. This general definition also applies to the ideals representing Cartier divisors on affine integral schemes.

Over an integral domain $A$ with fraction field $K$, an invertible fractional ideal is a finitely generated nonzero $A$-submodule $I\subseteq K$ for which another fractional ideal $J$ satisfies $IJ=A$. Here $I^{-1}=\{x\in K:xI\subseteq A\}$ equals $J$. Choose $1=\sum u_i{v}_i$ with $u_i\in I$, $v_i\in J$. On the distinguished open where $u_i{v}_i$ is invertible, $I$ is generated by $u_i$, since $x/u_i=(x{v}_i)/(u_i{v}_i)$. The maps $x\mapsto v_{i}x$ form a finite dual basis and prove that $I$ is a rank-one projective module. Conversely, a finitely generated rank-one projective module embeds in $K$ after choosing a basis over $K$; its dual identifies with $I^{-1}$, and local evaluation shows $I{I}^{-1}=A$.

A principal fractional ideal of an integral domain $A$ with fraction field $K$ is $aA$ for a nonzero $a\in K$. Its generators differ by a unit in a ring in $A$. For a number field, taking $A={\mathcal{O}}_K$ and quotienting nonzero fractional ideals by principal fractional ideals gives the ideal class group; requiring congruences and signs of the generators instead gives a ray class group.

An integral ideal is a fractional ideal of an integral domain $A$ that is contained in $A$. For a number field, take $A$ to be its ring of integers ${\mathcal{O}}_K$.

Multiplication by a nonzero element $a$ of an integral domain is injective, by multiplicative cancellation. When the domain is finite, this self-map is surjective, so some element $b$ satisfies $ab=1$. Commutativity gives $ba=1$, so every nonzero element has a multiplicative inverse and the domain is a field.

Even when the domain convention does not initially assume an identity element, finiteness produces one. Surjectivity supplies $e$ with $ae=a$; then $a(ex-x)=0$ forces $ex=x$ for every $x$. Commutativity makes $e$ a two-sided identity, after which the inverse argument applies.

A nonzero nonunit $p$ in an integral domain is prime when $p\mid ab$ implies $p\mid a$ or $p\mid b$.

A nonzero nonunit $p$ in an integral domain is irreducible when every factorization $p=ab$ has at least one unit factor.

The multiplicative field norm $N(m+n\sqrt{-7})=m^2+7n^2$ is an integer. Units have field norm one and are exactly $\pm 1$. A nontrivial factorization of 2 would force both factors to have field norm two, which the displayed quadratic form cannot represent. Thus 2 is an irreducible element. It is not a prime element, since it divides $(1+\sqrt{-7})(1-\sqrt{-7})=8$ but neither factor. This is an example in a nonmaximal quadratic order, rather than the full ring of integers of $\mathbb{Q}(\sqrt{-7})$; the precise ring matters for factorization.

An atomic domain is an integral domain in which every nonzero nonunit is a finite product of irreducible elements. Every unique factorization domain is atomic, but atomicity alone does not imply uniqueness.

A greatest-common-divisor domain is an integral domain in which every two nonzero elements have a greatest common divisor, defined up to multiplication by a unit.

If $\gcd (a,b)=1$ and $a\mid bc$ in a greatest-common-divisor domain, then $a\mid c$. Indeed, the least common multiple of $a$ and $b$ is associate to $ab$ and divides every common multiple; applying this to $bc$ and cancelling $b$ proves the claim.

The fraction field of an integral domain $R$ consists of fractions $a/b$ with $a,b\in R$ and $b\ne 0$, modulo the usual equivalence relation.

A principal ideal domain is an integral domain whose every ideal has one generator.

Let $\pi$ be an irreducible element of a principal ideal domain $R$, and suppose $\pi \nmid a$. The ideal $(\pi ,a)=(d)$ has generator $d$ dividing $\pi$. Irreducibility makes $d$ a unit or an associate of $\pi$; the latter would imply $\pi \mid a$. Hence $(\pi ,a)=R$, giving $u\pi +va=1$ for some $u,v\in R$. If $\pi \mid ab$, multiply this identity by $b$: both $u\pi b$ and $vab$ are divisible by $\pi$, so $\pi \mid b$. This is precisely the prime element property.

For a prime number $p$, the ideal $(p,X)$ in the polynomial ring $\mathbb{Z}[X]$ is proper: the constant coefficient of every member is divisible by $p$. If $(p,X)=(g)$, then $g$ divides the nonzero constant $p$, so the additivity of polynomial degree over an integral domain makes $g$ constant. Since $g$ also divides $X$, it must divide its leading coefficient $1$; thus $g=\pm 1$. This would make the ideal the whole ring, a contradiction. Therefore $\mathbb{Z}[X]$ is not a principal ideal domain, and hence not a Euclidean domain.

Over a field $F$, the ideal $(X,Y)$ in a two-variable polynomial ring is not principal. A proposed generator would divide both $X$ and $Y$. Any nonconstant divisor of the degree-one polynomial $X$ is a scalar multiple of $X$, which does not divide $Y$; therefore the generator would have to be a unit. But every element of $(X,Y)$ vanishes at $(0,0)$, so the ideal cannot contain one. This contrasts with $F[X]$, where the division algorithm makes every ideal principal.

The ascending chain condition supplies existence of factorizations into irreducible elements. The Bezout identity makes every irreducible element prime, which supplies uniqueness up to order and multiplication by units.

Every submodule of a finite-rank free module over a principal ideal domain is free. Induction on rank splits off a generator of the image under one coordinate projection and applies the result to the kernel.

A Euclidean domain admits division with remainder measured by a decreasing Euclidean function.

For a field $F$, polynomial division makes the one-variable polynomial ring $F[X]$ a Euclidean domain with Euclidean function equal to the polynomial degree. For a divisor $g\ne 0$, subtract a suitable scalar multiple of $X^{\mathrm{deg}f-\mathrm{deg}g}g$ to cancel the leading coefficient of the current remainder. Its degree strictly decreases. Iteration terminates at $f=qg+r$, where $r=0$ or $\mathrm{deg}r<\mathrm{deg}g$. Division by the leading coefficient is valid because it is nonzero in a field.

Let $R$ be a Euclidean domain with Euclidean function $\nu$, and let $I$ be a nonzero ideal. Choose $b\in I\setminus \{0\}$ minimizing the nonnegative integer $\nu (b)$. For $a\in I$, divide $a=qb+r$ with $r=0$ or $\nu (r)<\nu (b)$. Since $r=a-qb\in I$, the latter case contradicts minimality. Hence every $a$ is a multiple of $b$, giving $I=(b)$. The zero ideal is $(0)$, so every ideal is a principal ideal and $R$ is a principal ideal domain.

The ring of integers $\mathbb{Z}[(1+\sqrt{-7})/2]$ is a Euclidean domain for $N(z)=z\overline{z}$. In its lattice choose the nearest row, whose imaginary-coordinate error is at most $\sqrt{7}/4$, then the nearest point on that row, whose real-coordinate error is at most $1/2$. Every complex number is thus within squared distance $11/16<1$ of a lattice point, giving Euclidean division. The suborder $\mathbb{Z}[\sqrt{-7}]$ has the nonprincipal ideal $(2,1+\sqrt{-7})$ and consequently is not a Euclidean domain for any Euclidean function.

A Euclidean function on an integral domain $R$ assigns a nonnegative integer to every nonzero element so that, for $a,b\in R$ with $b\ne 0$, there are $q,r\in R$ satisfying $a=bq+r$ and either $r=0$ or $d(r)<d(b)$.

A Gaussian integer has the form $a+bi$ with $a,b\in \mathbb{Z}$. Its multiplicative norm is $N(a+bi)=a^2+b^2$, which makes $\mathbb{Z}[i]$ a Euclidean domain.

The product of a Gaussian integer with its complex conjugate. It is multiplicative: $N(zw)=N(z)N(w)$, which gives the Brahmagupta-Fibonacci identity.

An odd Gaussian integer is primary when it satisfies the displayed congruence. The four units have distinct images in $(\mathbb{Z}[i]/(1+i{)}^3{)}^{\times }$, a group of order four; hence every odd ideal has a unique primary generator. Indeed $v_{1+i}(-2)=2$ and $v_{1+i}(i-1)=v_{1+i}(-i-1)=1$.

If $\alpha =a+bi$ is a Gaussian integer and $P$ is a nonzero prime ideal of $R=\mathbb{Z}[\alpha ]$, then $R/P$ is a finite field. Complex conjugation preserves $R$, so a nonzero $\beta \in P$ gives the positive integer $\beta \bar{\beta }\in P$. Primality supplies a rational prime $p\in P$. Since $\alpha$ satisfies the monic polynomial $t^2-2at+a^2+b^2$, the ring $R/P$ is a quotient of ${\mathbb{F}}_p[t]/(\bar{F})$ and has at most $p^2$ elements. It is a finite integral domain, hence a field. The proof does not require $R$ to be a principal ideal domain.

For a primitive cube root of unity $\omega$, the Eisenstein integers form the triangular latticeTheir multiplicative norm is $N(a+b\omega )=a^2-ab+b^2$.

For an Eisenstein integer $z=a+b\omega$, with $\omega =(-1+i\sqrt{3})/2$, the norm is $N(z)=z\overline{z}=a^2-ab+b^2$. It is a positive integer for $z\ne 0$, since $N(z)=(a-b/2{)}^2+3b^2/4$, and is multiplicative by the multiplicativity of the complex modulus. An element is a unit exactly when its norm is one: one direction follows from multiplicativity and integrality, and the other from the inverse $z^{-1}=\overline{z}$.

Let $p\ge 3$ be a prime number dividing $n^2+3$, and set $\omega =(-1+i\sqrt{3})/2$. The case $p=3$ is $N(1-\omega )=3$. For $p>3$, useThe Eisenstein integer $p$ divides this product, but divides neither factor: divisibility by the rational integer $p$ in the basis $1,\omega$ requires both integer coefficients to be divisible by $p$, whereas the second coefficients are $2$ and $-2$. Thus $p$ is not a prime element of $\mathbb{Z}[\omega ]$.

The Euclidean norm on the Eisenstein integers makes this ring a principal ideal domain. Since irreducible elements of a principal ideal domain are prime, $p$ admits a factorization $p=uv$ into two nonunits. The positive integral Eisenstein norm is multiplicative, and its value is one exactly on units: when $N(u)=1$, its complex conjugate is an inverse. Thus $N(u),N(v)>1$ and $N(u)N(v)=p^2$, forcing both norms to be $p$. Writing $u=a+b\omega$ proves the displayed representation.

Every complex number can be written uniquely as $x+y\omega$ with real $x,y$, where $\omega =(-1+i\sqrt{3})/2$. Round $x,y$ separately to integers $m,n$. The error is $s+t\omega$ with $|s|,|t|\le 1/2$, andFor Eisenstein integers $\alpha ,\beta$ with $\beta \ne 0$, apply this to $\alpha /\beta$ to choose $q=m+n\omega$. The remainder $r=\alpha -q\beta$ is an Eisenstein integer and $N(r)\le 3N(\beta )/4<N(\beta )$ if nonzero. Thus the Eisenstein norm is a Euclidean function and these integers form a Euclidean domain. Coordinate rounding suffices; no closest-point theorem is needed.

A Bézout domain is an integral domain in which every finitely generated ideal is principal. Equivalently, every pair $a,b$ has a greatest common divisor $d$ satisfying a Bézout identity $d=ra+sb$. A Noetherian Bézout domain is a principal ideal domain because all its ideals are finitely generated.