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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.

A propositional geometric theory has primitive propositions and sequents between formulas built with finite conjunction and arbitrary set-indexed disjunction. Start with the free frame on the primitives, identify conjunction with meet and disjunction with join, and impose each sequent as an inequality. The resulting quotient frame is $\mathcal{O}(X_T)$. In the sheaf topos $\mathrm{Sh}(X_T)$, interpret each primitive as the subterminal sheaf corresponding to its generating open. Operations on these subterminal sheaves are precisely the frame operations, so every imposed sequent is valid. This constructs the canonical model.

For a commutative ring $R$, denote the generator attached to $r$ by $U_r$. The presentation impliesThe product equality uses the multiplication implication in one direction and both factor implications in the other; the latter follow by commuting $r,s$. A ring homomorphism $\phi :R\to S$ sends these relations to the corresponding ones in $S$. It therefore gives a frame homomorphism $U_r\mapsto U_{\phi (r)}$ and hence a map $X_S\to X_R$ of locales.

If $ar+bs=1$, thenso the two opens cover. Conversely let $S=R/(r,s)$ and pull back a putative cover along $X_S\to X_R$. Both generators become $U_0$, so $1=0$ in $\mathcal{O}(X_S)$. The allowed degeneracy criterion gives $0=1$ in $S$, equivalently $1\in (r,s)$. ThusThis necessity argument never selects a prime ideal.

Finite intersections of the generating opens are again generators, so they form a basis. Assign $R[r^{-1}]$ to $U_r$. The localization of a ring carries its ordinary addition and multiplication of fractions; equality of $a/r^n$ and $b/r^m$ means that some power of $r$ annihilates the cross-multiplied difference. On $U_r\cap U_s=U_{rs}$, restriction is further localization.

We also justify restriction for any inclusion of basic opens. Adding the condition $U_t=1$ to the presentation of $X_R$ gives the open $U_t$ and the presentation of $X_{R[t^{-1}]}$: the generator for $a/t^n$ corresponds to $U_a\wedge U_t$. This is well-defined since equality of fractions, after multiplication by a power of the now-invertible $t$, makes the two membership propositions equivalent. More explicitly, $U_t=1$ implies $U_{t^{k}z}=U_z$. If $t^{k}z=0$, this gives $U_z=0$. If $x-y$ is killed by a power of $t$, then $U_{x-y}=U_{y-x}=0$ (since $U_{-1}=1$), and the additive inequalities for $x=y+(x-y)$ and $y=x+(y-x)$ give $U_x=U_y$. A zero element in the localized ring has false membership, multiplication gives intersections, and the additive relation follows by clearing denominators. The reverse assignment on $a\in R$ proves that these maps are inverse. Thus $X_{R[t^{-1}]}\cong U_t$. If $U_t\subseteq U_r$, then in this localized theory $U_r=1$. Apply the quotient argument above, with the second generator $0$, to conclude $1\in (r)$ in $R[t^{-1}]$. Therefore $r$ is invertible there, and the universal property of localization gives the restriction $R[r^{-1}]\to R[t^{-1}]$. These restrictions compose uniquely.

It remains to glue, and the allowed basis criterion reduces this to a whole-locale cover $U_r\vee U_s=1$. Let $\alpha \in R[r^{-1}]$ and $\beta \in R[s^{-1}]$ have equal restrictions to $R[(rs{)}^{-1}]$. Choose a common initial exponent $n$, writing $\alpha =a/r^n$, $\beta =b/s^n$. Equality on the overlap supplies $k\ge 0$ withSet $N=n+k$, $A=r^{k}a$ and $B=s^{k}b$. Thenwith the last equality in $R$ itself. This clearing step is necessary when the commutative ring has zero divisors.

Since $(r,s)=R$, also $(r^N,s^N)=R$. Choose $\lambda r+\mu s=1$. For $N\ge 1$, expand $(\lambda r+\mu s{)}^{2N-1}=1$; each term contains either $r^N$ or $s^N$, producing $C,D\in R$ with $C{r}^N+D{s}^N=1$. For $N=0$ choose $C=1,D=0$. DefineIn $R[r^{-1}]$,and similarly in $R[s^{-1}]$,Thus $t$ glues the matching pair. If $t^{\prime }$ is another gluing, some $r^p$ and $s^q$ annihilate $t-t^{\prime }$. The same binomial argument gives $(r^p,s^q)=R$, so $t-t^{\prime }=0$. This proves uniqueness without cancellation or a domain assumption.

The exactness of the unit-ideal localization Čech complex proves the sheaf axiom for the required basic covers, and hence the assignment with its restrictions defines a sheaf of rings $\tilde{R}$ with $\tilde{R}(U_r)=R[r^{-1}]$. On the empty basic open, localization is the zero ring, whose underlying set has one section, as required. The construction and proof remain valid for the zero ring.

Use the standard convention that an integral domain has $1\ne 0$. Since $(n)=(-n)$, the sign of a nonzero $n$ makes no difference. For $|n|=p$ prime, every nonzero residue has an inverse by Bézout's identity, so the quotient ring is a field and hence an integral domain. If $|n|=ab$ is composite with $1<a,b<|n|$, the nonzero classes of $a$ and $b$ multiply to zero, giving zero divisors. For $n=\pm 1$, the quotient is the zero ring, excluded by the convention above. Finally, $n=0$ gives $\mathbb{Z}/(0)\cong \mathbb{Z}$, an integral domain. ThusIf a convention restricts this notation to positive moduli, the answer is simply the positive primes.

A ring is a set $R$ with two binary operations, addition and multiplication, satisfying three requirements. Under addition, $R$ is an abelian group: there is a zero, every element has an inverse element, and grouping or exchanging summands does not change the result. Multiplication is associative and has an identity element $1$, so $(ab)c=a(bc)$ and $1a=a1=a$. Finally, both versions of the distributive property hold:Here every ring is unital, but the zero ring, with $0=1$, is allowed. Commutativity of multiplication is an additional condition defining a commutative ring. These conventions agree with ^[ref].

The integers form a ring with their usual operations. The polynomial ring $\mathbb{Z}[X]$ is another: addition combines coefficients, while multiplication uses the distributive property. For example,The same rules justify symbolic calculations in both rings; this is why one definition can organize apparently different kinds of arithmetic. In every ring, $a0=0a=0$, and $a(-b)=-(ab)$. The first identity follows by distributing $a(0+0)$ and subtracting $a0$; the second follows by distributing $a(b+(-b))$. Thus familiar sign rules are consequences of the structure.

Order still matters in some rings. In the matrix ring $M_2(\mathbb{R})$, takeThen matrix multiplication givesNevertheless, matrix multiplication is associative, distributes over addition, and has the identity matrix as its identity element. This is a valid ring. On a column vector, $AB$ applies $B$ first and $A$ second: swapping coordinates before discarding the second coordinate differs from doing those actions in reverse. The failure of commutativity records a meaningful difference in function composition.

For a positive integer modulus $n$, a residue class $[a{]}_n$ is a unit when it has a modular multiplicative inverse: some integer $u$ satisfies $au\equiv 1(\mathrm{mod}n)$. Modulo twelve, the units are exactlyIn particular, five is its own modular inverse, since $5\cdot 5=25\equiv 1(\mathrm{mod}12)$. Multiplying all twelve residue classes by five gives a permutation; multiplying the outputs by five recovers the inputs. By contrast, multiplying by six gives only zero and six, with six inputs for each output. It cannot be reversed.

In general, the exact criterion isFor the forward direction, $au-kn=1$ implies every common divisor of $a,n$ divides one. Conversely, if their greatest common divisor is one, Bézout's identity supplies $ua+vn=1$, so $u$ is a modular inverse. The extended Euclidean algorithm computes these coefficients. The units form the unit group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ under multiplication. ^[ref]

Multiplicative cancellation while keeping the same integer modulus is valid for a unit modulo n: from $ca\equiv cb(\mathrm{mod}n)$, multiply by a modular inverse of $c$ to get $a\equiv b(\mathrm{mod}n)$. A nonunit cannot be cancelled as a general rule. For example, $6\cdot 0\equiv 6\cdot 2(\mathrm{mod}12)$, yet $0\not\equiv 2(\mathrm{mod}12)$. Thus division by six cannot undo this multiplication; solving a linear congruence instead keeps track of all possible classes.

This distinction marks the boundary between commutative rings and fields. Modulo a prime number $p>1$, every nonzero residue class is a unit, so $\mathbb{Z}/p\mathbb{Z}$ is a finite field. Modulo twelve it is a commutative ring with zero divisors, and is not a field. For $n=1$, the sole class satisfies $[0{]}_1=[1{]}_1$ and is a unit; this one-element zero ring is not a field. ^[ref]