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An associator ring is a ring in which equality of the principal ideals generated by two elements forces them to be associate elements: $(a)=(b)$ implies $a=ub$ for a unit in a ring $u$. The reverse implication always holds, since $a=ub$ and $b=u^{-1}a$ give both ideal inclusions. The property separates equality of generated ideals from equality of generators up to an invertible factor.

Every integral domain is an associator ring. If $a\ne 0$ and $(a)=(b)$, write $a=rb$ and $b=sa$. Then $(1-rs)a=0$, so cancellation gives $rs=1$ and $a=rb$ with $r$ a unit. The zero case is immediate. The converse fails: $\mathbb{Z}/4\mathbb{Z}$ has generator classes $\{0\}$, $\{2\}$ and $\{1,3\}$ for its three ideals, and $1,3$ differ by a unit, although $2$ is a nonzero zero divisor.

Finite direct product of rings preserve this property. Equality of principal ideals is coordinatewise, and a tuple of units is a unit. Every quotient ring of a principal ideal domain is also an associator ring: first, modulo a prime power $p^k$, every nonzero class is a power $p^e$, $e<k$, times a unit, and distinct powers generate distinct ideals. Next factor a nonzero nonunit modulus into prime powers and apply the Chinese remainder theorem. The zero modulus gives the original integral domain, while a unit modulus gives the zero ring.

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.

The group ring of an infinite cyclic group is the Laurent polynomial ring $R=\mathbb{Z}[t,t^{-1}]$, with augmentation $\epsilon (t)=1$. Multiply any element in the kernel by a power of $t$ to obtain a polynomial vanishing at one. The factor theorem makes it divisible by $t-1$. Thus the augmentation ideal is $(t-1)R$. Since $R$ is an integral domain, multiplication by $t-1$ is an injective module map, so this ideal is a free $R$-module of rank one, even though its additive group has infinite rank.

For an ideal $I$ in $R[x]$, let $J_n$ contain the coefficients of $x^n$ in its elements of degree at most $n$. These are ideals of $R$ and form an ascending chain, since multiplying by $x$ embeds one into the next. If $R$ is a Noetherian ring, the chain stabilizes at $N$, and each $J_n$ for $0\le n\le N$ has finitely many generators. Choose corresponding polynomials in $I$. Their union generates $I$: for an element of degree $d$, subtract a combination of the chosen polynomials for $J_d$ if $d<N$, or the polynomials for $J_N$ multiplied by $x^{d-N}$ otherwise. This cancels its leading coefficient and lowers its degree. Induction completes the proof without assuming that $R$ is an integral domain.

For an integral domain $A$ with fraction field $K$, realize Cartier divisors as invertible fractional ideals, with local divisor equation $f$ corresponding to the ideal $fA$. This is the $\mathcal{O}(-D)$ convention. The maps in the displayed exact sequence are unit inclusion, $f\mapsto fA$, and $I\mapsto [I]$. A principal fractional ideal is trivial as a divisor exactly when its generator is a unit. An invertible fractional ideal is trivial in the Picard group of a ring exactly when it is principal. Every rank-one projective module embeds in $K$ and produces an invertible fractional ideal, proving surjectivity. Local multiplication identifies $I{\otimes }_{A}J$ with $IJ$, so these are group homomorphisms.

The positive characteristic of a ring which is an integral domain is a prime number. The characteristic cannot be one because $1\ne 0$. If its least positive value were $n=ab$ with $1<a,b<n$, minimality would make $a1$ and $b1$ nonzero, while $(a1)(b1)=n1=0$ would give zero divisors.

The analogous result holds for a nonzero commutative nonunital ring without zero divisors, using the characteristic of a rng. If $n=ab$ were composite and minimal, choose $r$ with $ar\ne 0$. For every $s$, $(ar)(bs)=n(rs)=0$, so cancellation forces $bs=0$ for every $s$, contradicting minimality of $n$.

The characteristic of a ring $R$ is the least positive integer $n$ satisfying $n\cdot 1_R=0$, or zero if no such integer exists. A nonzero ring of positive characteristic has prime characteristic when it is an integral domain.

Let a finitely generated module over an integral domain have a finite generating set $S$ and a maximal independent subset $T$. For each $s\notin T$, dependence of $T\cup \{s\}$ gives a nonzero coefficient $a_s$ with $a_{s}s\in ⟨T⟩$. The product of those coefficients is nonzero and carries the whole module into $⟨T⟩$.

If $A$ is a finite-type integral domain over an algebraically closed field and $\dim A=d$, every maximal ideal $\mathfrak{m}$ has height of a prime ideal $d$. Choose a Noether normalization $k[z_1,\dots ,z_d]\subset A$. The contraction of $\mathfrak{m}$ is a maximal ideal of the polynomial ring and has height $d$; going-down theorem lifts its full chain since the polynomial ring is integrally closed domain. The upper bound is $\dim A$. A principal open subset containing a closed point consequently has dimension $d$, because the whole chain under that point survives localization.

For an ideal $K\subseteq R[t]$, let $A_n(K)$ consist of the coefficients at $t^n$ of elements of $K$ of polynomial degree at most $n$, including the zero polynomial. Each $A_n(K)$ is an ideal of $R$, and multiplication of representing polynomials by $t$ shows that $A_n(K)\subseteq A_{n+1}(K)$.

If $R$ is a Noetherian ring, this chain stabilizes at some $N$. Choose finite generating sets of an ideal for $A_0(K),\dots ,A_N(K)$, and for each nonzero generator choose a representing polynomial $f_{nj}\in K$ with that leading coefficient and polynomial degree $n$. These finitely many polynomials generate $K$: for any nonzero $f\in K$ of polynomial degree $d$, use the chosen generators at level $\min (d,N)$, multiplied by suitable powers of $t$, to cancel its leading coefficient. Induction on polynomial degree finishes the argument. This proves the Hilbert basis theorem without assuming that $R$ is an integral domain.

Commutative algebra studies commutative rings, ideals, modules, and the spectrum of a ring. Its basic question is which familiar rules of integer and polynomial arithmetic survive in a new setting. The distinction between units, zero divisors, and integral domains makes that question concrete.

A polynomial consisting only of a constant term. Over an integral domain, a nonzero constant polynomial has polynomial degree zero; it is a unit of the polynomial ring exactly when its value is a unit of the coefficient ring.

For an affine algebraic set $X\subseteq {\mathbb{A}}_k^n$, its coordinate ring isThe set $X$ is irreducible exactly when $k[X]$ is an integral domain.

The homomorphismhas kernel $(y^2-x^3)$ and image $k[t^2,t^3]$. The quotient is therefore an integral domain, so the curve is irreducible. In characteristic zero its gradient $(-3x^2,2y)$ vanishes at the origin, making the origin its unique singular point of an algebraic variety.

For affine varieties $X,Y$ over an algebraically closed field $k$, their product in a category is affine, with coordinate ring $A{\otimes }_{k}B$, where $A=k[X]$ and $B=k[Y]$. Presenting $A=k[x]/J$ and $B=k[y]/L$ identifies this ring with $k[x,y]/(J,L)$. If both varieties are irreducible, this is an integral domain: express a nonzero tensor $F$ using linearly independent second factors. The points $x$ where its specialization $F_x$ is nonzero form a nonempty open subset of $X$. For two nonzero tensors $F,G$, choose $x$ in the intersection of these open subsets; then $F_x{G}_x\ne 0$ in the domain $B$. The Hilbert Nullstellensatz identifies $(J,L)$ with the defining ideal of $X\times Y$. The universal property follows from the universal property of the tensor product of modules as a tensor product of commutative algebras.

The degree of a nonzero polynomial is the largest exponent of a monomial with a nonzero coefficient. Thus $X^3-3X^2+4$ has degree three, while any nonzero constant polynomial has degree zero. The zero polynomial is treated separately in degree statements.

Over an integral domain, polynomial degrees add under multiplication: $\mathrm{deg}(pq)=\mathrm{deg}p+\mathrm{deg}q$ for nonzero factors, because the product of their leading coefficients cannot vanish. Under addition, cancellation may lower the degree: $(X^2+1)+(-X^2+X)=X+1$. This explains why degree is reliable for counting factors and bounding polynomial roots, but cannot always be inferred from the degrees of summands alone.

The ring $k\times k$, for a field $k$, has two prime ideals, $0\times k$ and $k\times 0$. Their localizations at a prime ideal are both isomorphic to $k$. Nevertheless $(1,0)(0,1)=0$ is a product of nonzero elements, so $A$ is not an integral domain. Nonzero orthogonal idempotents separate the two components of its spectrum of a commutative ring.

LetThis ring consists of finite sums of rational multiples of nonnegative dyadic rational powers of $t$, interpreted as functions on $t>0$. Every pair of elements lies in one polynomial ring $\mathbb{Q}[u_n]$, so $R$ is an integral domain. The constant term defines a surjective ring homomorphism $R\to \mathbb{Q}$. Its kernel of a ring homomorphism $\mathfrak{m}$ is nonzero and is a maximal ideal, by the maximal ideal quotient criterion.

Every $f\in \mathfrak{m}$ has the form $u_{n}g(u_n)$ for some $n$ and some polynomial $g$. The identityexpresses it as a product of two elements of $\mathfrak{m}$. Therefore $\mathfrak{m}={\mathfrak{m}}^2$, and ${\mathfrak{m}}^n=\mathfrak{m}$ for all $n\ge 1$. This shows that strict descent of powers of an ideal can fail in an integral domain, even for a nonzero maximal ideal. By the determinant trick, such an ideal cannot be finitely generated, so this ring is not a Noetherian ring.

Apply clearing denominators relative to an independent module subset to find $0\ne a$ with $aM$ lying in a finite free module. Multiplication by $a$ is injective on a torsion-free module, giving the required embedding. A submodule need not itself be free over a general integral domain.

On ${\mathbb{P}}_k^n$ with $n>0$, let $\mathcal{F}=i_{\ast }k$ at a rational point. It has one-dimensional global sections, but its dual of a sheaf is zero: a local map from the residue field into the local polynomial ring must have image annihilated by a nonzero parameter, which is impossible in an integral domain. Thus $H^n({\mathcal{F}}^{\vee }(-n-1))=0$. Serre duality for arbitrary coherent sheaves instead uses an Ext functor; replacing this by the ordinary dual can lose torsion information.