= Right Noetherian domains satisfy the Ore condition
{title2=$aR\cap bR\ne0\quad(a,b\ne0)$}
If two nonzero principal <right ideals> in a <right Noetherian domain> had zero intersection, the sums $bR\oplus abR\oplus\cdots\oplus a^mbR$ 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.
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