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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.

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 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.

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.

Multiplication by a nonzero element of an integral domain is injective. If the ring is finite, that map is surjective, so its image contains one and the multiplier has a multiplicative inverse. Hence every nonzero element is invertible and the ring is a field.

For a normal Noetherian ring $A$ that is an integral domain, the integral closure in a finite separable field extension of its fraction field is finite as an $A$-module. An integral field basis by denominator clearing gives a finite lattice. Integral traces place the closure in its finite trace-dual lattice; Noetherianity makes that submodule finitely generated.

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.

A fractional ideal of a number field is a nonzero ${\mathcal{O}}_K$-submodule $I\subset K$ for which $dI\subseteq {\mathcal{O}}_K$ for some nonzero $d\in {\mathcal{O}}_K$.

A generic point is a point whose closure is the whole space. An integral scheme has a unique generic point; on an affine scheme associated to an integral domain, it corresponds to the zero prime ideal. Its local ring is the fraction field of the domain. On every nonempty affine chart of an integral scheme these fields identify with the same function field.

On an integral scheme $X$, every nonempty open $U$ contains its generic point $\eta$. Taking a germ gives an injective ring homomorphism from its regular functions to the function field ${\mathcal{O}}_{X,\eta }$. Injectivity follows on each nonempty affine open subscheme from the injection of an integral domain into its field of fractions, and then from the sheaf gluing axiom. This lets regular functions on different open subsets be compared inside one field.

An ideal $I$ is idempotent when $I^2=I$. Then $I^j=I$ for every $j\ge 1$, so a nonzero idempotent ideal gives a nonzero intersection of all its powers. A semigroup algebra with arbitrarily divisible positive exponents supplies examples in a non-Noetherian integral domain. If $I$ is finitely generated, the determinant trick applied to $I=I^2$ produces $r\in I$ such that $(1+r)I=0$.

For an integral domain $R$, an integer-valued polynomial is a polynomial over its fraction field that maps $R$ into itself. Its coefficients need not lie in $R$: $X(X-1)/2$ maps $\mathbb{Z}$ into $\mathbb{Z}$. These polynomials form a ring containing $R[X]$.

If $A$ is an integral domain with fraction field $K$ and $L/K$ is finite, a $K$-basis of $L$ can be rescaled to consist of integral elements over $A$. For a monic algebraic equation, choose $a\in A$ clearing the coefficient denominators; replacing $u$ by $au$ multiplies the coefficient of degree $r-j$ by $a^j$, putting it in $A$.

An integral ideal is a fractional ideal contained in the ring of integers ${\mathcal{O}}_K$.