Ore theorem

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