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A principal ideal ring is a ring whose left ideals and right ideals each have one generator. For a commutative ring, this says that every ideal is a principal ideal. The examples here use commutative rings. Unlike a principal ideal domain, it may have zero divisors: $\mathbb{Z}/4\mathbb{Z}$ has this property, with ideals $(0)$, $(2)$, and $(1)$. A finite direct product of rings that are principal ideal rings is again a principal ideal ring. Indeed, the coordinate idempotent elements show that every ideal is a product of coordinate ideals; a tuple of their generators generates the product.

For a two-sided ideal $I$ of a ring $R$, the quotient $R/I$ has addition and multiplication induced from $R$.

Let $R$ be a commutative unital ring and $f(T)=a_{1}T+a_2{T}^2+\cdots$ with $a_1$ a unit. Set $g(T)={\sum }_{n\ge 1}{b}_n{T}^n$. The coefficient of $T$ in $f(g(T))$ gives $b_1=a_1^{-1}$. At every degree $n>1$, the coefficient is $a_1{b}_n$ plus an expression involving only $b_1,\dots ,b_{n-1}$; choose $b_n$ to cancel that expression. This gives a unique right compositional inverse of a formal power series.

Similarly, the coefficient of $T^n$ in $h(f(T))$ is $a_1^n{h}_n$ plus previously fixed terms, so the same recursion gives a left inverse $h$. Composition of zero-constant-term formal power series is associative, since every coefficient involves finitely many terms. Therefore $h=h\circ (f\circ g)=(h\circ f)\circ g=g$, proving $f\circ g=g\circ f=T$.

A regular element of a ring is a two-sided non-zero-divisor. The two cancellation conditions must both hold in a noncommutative ring. Products of regular elements remain regular.

Parametrize the affine cuspidal cubic $x^2=y^3$ by $x=t^3,y=t^2$. Removing the zero set of $g=x-y$ removes exactly the parameter values zero and one. Its regular-function ring is the localization $k[t^2,t^3{]}_g$. To recover the normalization coordinate inside this ring, use $g^2=y(y^2-2x+y)$: inverting $g$ makes $y$ a unit, so $t=x/y$ lies in the localization. As $g=t^2(t-1)$ is a unit, so are $t$ and $t-1$, proving the displayed equality. This calculation is valid in every characteristic. At the smooth point $(1,1)$ of the original curve the local ring is instead $k[t{]}_{(t-1)}$: $y$ is a unit there, and every polynomial in $t$ nonzero at one is invertible. That point is absent from the displayed open subset, so its local ring is computed in the original curve.

The right annihilator of a subset $X$ of a ring is the right ideal of elements killed on the left by every member of $X$. Its sidedness matters for a noncommutative ring.

A ring is right Artinian if its right ideals satisfy the descending chain condition, equivalently its right regular module is an Artinian module. Left and right chain conditions should be distinguished for a general ring.

A right module over a unital ring $R$ is an abelian group $M$ with an additive scalar action on the right satisfying $(mr)s=m(rs)$, $m(r+s)=mr+ms$ and $m1=m$. The order of multiplication matters for a noncommutative ring. Equivalently it is a left module over the opposite ring. Row vectors acted on by matrices on the right give a basic example.

A multiplicative subset $S\subseteq A$ with $1\in S$ and $0\notin S$ of a ring $A$ satisfies the right Ore condition if for every $a\in A$ and $s\in S$ there are $u\in S$ and $b\in A$ such that $au=sb$. For a noncommutative domain, taking $S=A\setminus \{0\}$ gives a ring of right fractions $a{s}^{-1}$; every nonzero fraction is invertible. The usual additional denominator reversibility condition is automatic when the elements of $S$ are nonzero in a noncommutative domain.

A multiplicative subset $S$ of a ring is a right Ore set if $1\in S$, $0\notin S$ and it satisfies the right Ore condition. In a noncommutative domain, this condition lets nonzero denominators be used in Ore localization.

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.

A property of rings is a ring isomorphism invariant when any two rings related by a ring isomorphism either both have the property or both lack it. Being an integral domain is such a property: a ring isomorphism preserves nonzero elements, products, and zero.

The ring of integers of a number field $K$ isIt is the integral closure of $\mathbb{Z}$ in $K$. Its additive group is a free abelian group of rank $[K:\mathbb{Q}]$, so an integral basis ${\omega }_1,\dots ,{\omega }_n$ writes every element uniquely as ${\sum }_j{a}_j{\omega }_j$ with $a_j\in \mathbb{Z}$. Its field of fractions is $K$.^[ref]

For $K=\mathbb{Q}$, this gives ${\mathcal{O}}_K=\mathbb{Z}$; for $K=\mathbb{Q}(\sqrt{2})$, it gives $\mathbb{Z}[\sqrt{2}]$. The Gaussian integers give the corresponding example inside $\mathbb{Q}(i)$. The ring of integers is determined by the field, and need not equal $\mathbb{Z}[\theta ]$ for a chosen algebraic integer generator: in $\mathbb{Q}(\sqrt{5})$, the larger ring $\mathbb{Z}[(1+\sqrt{5})/2]$ contains $\mathbb{Z}[\sqrt{5}]$ with index two.

This is the setting for divisibility, global units, and prime ideal factorization. A ring of integers is a Dedekind domain, so every nonzero proper ideal factors uniquely into nonzero prime ideals, even when unique factorization of elements fails.^[ref]

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ iswith integral basis $1,\sqrt{2}$ and field discriminant eight. Multiplication is ordinary polynomial multiplication followed by $(\sqrt{2}{)}^2=2$:For example, $(1+\sqrt{2}{)}^2=3+2\sqrt{2}$ and $(1+\sqrt{2}{)}^3=7+5\sqrt{2}$. The integer coefficients stay integral, as required for a ring.

Under the two field embeddings, the element $a+b\sqrt{2}$ becomes the point $(a+b\sqrt{2},a-b\sqrt{2})$. The integral basis vectors become $(1,1)$ and $(\sqrt{2},-\sqrt{2})$, giving a Minkowski embedding of a real quadratic field. Their determinant has absolute value $2\sqrt{2}=\sqrt{8}$, consistent with the field discriminant.

For a group acting on a ring by ring automorphisms, the elements fixed by every group element form a subring containing $1$. For finite $G$ acting on a commutative ring, each $r$ satisfies the monic orbit equation ${\prod }_{g\in G}(T-g(r))$, whose coefficients are fixed. Thus $R$ is integral over its invariant subring. This uses an orbit product, not division by the group order.

For a possibly reducible algebraic variety $X$, take pairs $(U,f)$ with $U$ a dense open subset and $f$ a regular function on $U$. Identify pairs agreeing on a common dense open subset. Addition and multiplication on intersections make these classes a ring. For an affine variety with reduced coordinate ring $A$, this ring is its total quotient ring. For an irreducible variety, it is the function field. An isomorphism of dense open subsets induces an isomorphism of these rings.

A rng has an abelian group under addition, an associative multiplication, and the distributive property; a multiplicative identity element is not required. Every ring under this wiki's unital convention is a rng. A nonunital ring may happen to have an identity: the terminology removes an axiom rather than imposing the absence of an identity. For example, the even integers with their usual operations have no multiplicative identity.

Up to an operation-preserving bijection, there are two rngs with two elements. Write the elements as $0,u$. The additive abelian group must satisfy $u+u=0$. The distributive property forces all products involving zero to vanish, leaving only the choice $u^2=0$ or $u^2=u$. The first gives a zero-multiplication rng; the second gives the prime field ${\mathbb{F}}_2$, with identity $u$. Both satisfy the axioms, and they are not isomorphic because an isomorphism must send the unique nonzero element to the unique nonzero element and preserve its square. In particular, there is just one unital ring with two elements.

A root of a polynomial $f$ is a scalar $a$ satisfying $f(a)=0$ in the chosen field or ring. For $p(X)=(X+1)(X-2{)}^2$ over the real numbers, the distinct roots of a polynomial are $-1$ and $2$. The second has multiplicity of a root two, so there are three roots when counted with multiplicity.

The domain matters. $X^2+1$ has no real polynomial root, but has the two complex number roots $i,-i$. The fundamental theorem of algebra guarantees a polynomial root in the complex numbers for every nonconstant complex polynomial, and complete factorization counts its degree of a polynomial many roots with multiplicity of a root. It does not promise that those roots are real or that they have a simple expression.

Fix an $R$-bimodule $M$. A unital ground-ring-linear split extension with square-zero kernel $M$ can be identified with $R\oplus M$, with multiplication $(a,u)(b,v)=(ab,av+ub+f(a,b))$. Associativity is exactly $\delta f=0$. Changing a unital linear section by $g:R\to M$ changes $f$ by $\delta g$. Thus equivalence classes of these split extensions are $H{H}^2(R,M)$. Over a field every such vector-space extension splits; over a general ground ring a splitting hypothesis is essential for this ordinary cochain classification.