Ore condition

The Ore condition is a common-multiple condition that allows fractions in a noncommutative ring. For a multiplicative set $S\subseteq R$ containing one and excluding zero, the right Ore condition requires that for every $a\in R$ and $s\in S$ there are $b\in R$ and $u\in S$ with $au=sb$. It lets a numerator pass a denominator while keeping denominators on the right. The left version instead asks for $ua=bs$.

If the denominators are regular elements of a ring, these conditions suffice for the corresponding Ore localization. For example, all nonzero elements of a commutative integral domain satisfy either condition, giving its field of fractions. In a right Noetherian domain, the nonzero elements satisfy the right condition and give a division ring. Without the condition, adjoining formal inverses need not retain the expected single-fraction form.

A multiplicative set $S$ of regular elements of a ring admits a right ring of fractions exactly when it satisfies the right Ore condition. The resulting localization embeds the original ring, inverts $S$, and expresses every element as $a{s}^{-1}$. For denominators that can be zero divisors, an additional denominator reversibility condition is required.

If two nonzero principal right ideals in a right Noetherian domain had zero intersection, the sums $bR\oplus abR\oplus \cdots \oplus a^{m}bR$ would be strictly increasing. Directness follows by splitting the lowest term into $bR$ and the other terms into $aR$, then cancelling $a$ and repeating. This contradicts the ascending chain condition. A nonzero common element $au=bv$ gives the right Ore condition. The left version follows from the opposite ring. Consequently a noncommutative domain Noetherian on both sides has a single division ring of fractions with both left and right fraction presentations.

For a multiplicative set $S$ of regular elements satisfying the left Ore condition, represent a fraction by $(s,a)$. Pairs $(s,a)$ and $(t,b)$ agree when there are $u,v$ with $us=vt\in S$ and $ua=vb$. Addition uses this common denominator. For multiplication choose $v\in S,c\in R$ with $va=ct$, then $(s^{-1}a)(t^{-1}b)=(vs{)}^{-1}cb$. The Ore condition gives common multiples for successive operations; regularity supplies cancellation. In a domain with $S$ all nonzero elements, every nonzero fraction has inverse $a^{-1}s$.

// Destination: associative-algebra.bigb

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 set $S$ containing $1$ and not $0$ is a right denominator set if it satisfies the right Ore condition and right reversibility: $sr=0$, with $s\in S$, implies $rt=0$ for some $t\in S$. Its Ore localization consists of right fractions $r{s}^{-1}$ and has kernel $\{r:rt=0\text{ for some }t\in S\}$. This condition permits zero-divisor denominators; it is automatic for a set of regular elements of a ring. When all denominators are regular, the localization map is injective.

The classical right ring of quotients inverts all regular elements of a ring. Existence requires the right Ore condition for these denominators. The original ring embeds, every regular element becomes a unit, and every fraction has a right denominator. The right Ore condition also permits finitely many fractions to be placed over a common right denominator.

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.

An Ore localization of a noncommutative ring $R$ at a right denominator set $S$ is a ring $R{S}^{-1}$ in which each element has the form $r{s}^{-1}$. Its canonical map sends every member of $S$ to a unit and is universal among ring homomorphisms doing so. The right Ore condition provides common denominators, and denominator reversibility makes fraction equality and multiplication well-defined.

The kernel is $\{r\in R:rs=0\text{ for some }s\in S\}$. Thus the map is injective when every denominator is a regular element of a ring, but it can kill elements when zero divisors are inverted. In a right Noetherian domain, localizing at all nonzero elements produces a division ring, the noncommutative counterpart of a field of fractions. A left Ore fraction construction uses denominators on the other side.

This is the localization of a ring at all its regular elements of a ring, with left fractions. It exists precisely when those denominators satisfy the left Ore condition. If $R$ embeds in a ring $Q$, all regular elements become units in $Q$, and every element of $Q$ is such a fraction, the Ore condition follows by writing $r{c}^{-1}=s^{-1}b$, which gives $sr=bc$.

Let $A$ be a domain satisfying both Ore conditions, with quotient division ring $K$. A matrix over $A$ is regular exactly when it is invertible over $K$: a kernel vector of a singular matrix can have its denominators cleared to give a nonzero annihilator over $A$. An upper triangular matrix is regular exactly when all diagonal entries are nonzero. Each finite list of entries of a matrix over $K$ has a common left denominator $a\in A\setminus \{0\}$. Thus it is a fraction with denominator $a{I}_r$, proving the displayed classical localizations. The triangular localization retains its strictly upper triangular nilpotent ideal; it is not a semisimple matrix ring.

A multiplicative set $S$ in a ring $R$, with $1\in S$ and $0\notin S$, is left Ore when, for every $s\in S$ and $r\in R$, there are $s^{\prime }\in S$ and $r^{\prime }\in R$ such that $s^{\prime }r=r^{\prime }s$.

In a noncommutative domain that is a left Noetherian ring, two nonzero principal left ideals always intersect nontrivially. If $Ra\cap Rb=0$, a relation ${\sum }_{i=0}^m{u}_{i}a{b}^i=0$ first forces $u_{0}a\in Ra\cap Rb$, then $u_0=0$. Right cancellation of $b$ and repetition give all $u_i=0$. Hence ${\sum }_{i\ge 0}Ra{b}^i$ is an infinite direct sum of nonzero left ideals, whose finite partial sums violate the ascending chain condition. A nonzero member of $Rr\cap Rs$ supplies the equality $s^{\prime }r=r^{\prime }s$ with $s^{\prime }\ne 0$, proving the left Ore set condition for all nonzero elements.