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I've isolated a property of rings (integral domains, associative, unitary, non necessarily commutative) that is useful to me :

$$xR\cap yR\neq\{0\}\quad\text{ whenever $x$ and $y$ are non zero.}$$

Question. Does this property has a name, or falls into a known and labelled class of rings ?

Commutative integral domains and division rings obviously have this property.

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These rings are called right uniform rings.

Generally, a right ideal $I$ is called (right) uniform if all nonzero right subideals $J,K\subseteq I$ have nonzero intersection: $J\cap K \neq 0$.

Note: Your condition is equivalent to the condition that any two nonzero right ideals have nonzero intersection.

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  • $\begingroup$ I think not. And this because right uniform rings are not supposed to be integral domains (unless I am mistaken). Please have a look at the MO background (integral domains, associative, unitary, non necessarily commutative). I think the answer is rather the subclass of right uniform rings called right Ore domains. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Feb 29, 2016 at 17:47
  • $\begingroup$ Uniform rings form the largest class of rings with the desired property. If the OP is just interested in rings without zero-divisors, I agree that right Ore domain is the appropriate name. $\endgroup$
    – Todd Leason
    Commented Feb 29, 2016 at 18:53
  • $\begingroup$ @ToddLearson OK, one can interpret the question like that (i.e. focused on the last property). My answer is more contextual $\endgroup$
    – Duchamp Gérard H. E.
    Commented Feb 29, 2016 at 19:04
  • $\begingroup$ I think we can keep both answers, since the title uses the general wording "On rings ..." which might also be of interest for others. However, if the OP may want to change the title into "On domains ...", I would be happy to delete my answer. $\endgroup$
    – Todd Leason
    Commented Feb 29, 2016 at 19:18
  • $\begingroup$ Let's wait and see what the OP has really in mind, I can delete my answer too if needed. Thank you for this high quality interaction. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Feb 29, 2016 at 20:31
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As Andreas Thom mentioned (he could have put his comment as an answer), this is the Ore condition in the case when the ring is a domain and the multiplicative set is $R\setminus \{0\}$, then, in your case, called a right Ore domain. This allows you to build a right (possibly skew) field of fractions.

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