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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 has addition and multiplication, with an abelian group under addition and multiplication distributive over addition.

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ is $\mathbb{Z}[\sqrt{2}]$, with integral basis $1,\sqrt{2}$ and field discriminant eight.

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.

A root of a polynomial $f$ is a scalar $a$ satisfying $f(a)=0$.

A topological ring is a ring equipped with a topology making addition, additive inversion and multiplication continuous functions. An adic topology is obtained from powers of an ideal.

For a module $M$ over an integral domain $R$, its torsion submodule isThe domain condition makes this a submodule: products of nonzero annihilators remain nonzero and annihilate sums.

The whole ring $R$, viewed as an ideal. Elements $f_1,\dots ,f_r$ generate the unit ideal exactly when coefficients $g_i\in R$ give ${\sum }_i{g}_i{f}_i=1$.