Can we embed cancellative rings into division rings? Let $(R, +, \cdot)$ be a ring with identity $1$. That is, $(R, +)$ is an abelian group, $\cdot$ is distributive over $+$ and $1x = x1 = x$ for every $x \in R$. Suppose that $R$ is a cancellative ring, ie. $xy = 0$ implies $x = 0$ or $y = 0$ in $R$. 
Can we embed $R$ into a division ring $D$? If not, can we embed $R$ into a ring where every element of $R$ has an inverse? If not, are there some interesting special cases where this possible?
One case where this is true when $R$ is commutative, then we can take $D$ to be the field of fractions. However, I believe the same construction does not work when $R$ is not commutative. The relation $~$ on pairs $(a,b)$ for $a, b \in R$, $b \neq 0$ defined by $(a,b) \sim (c,d)$ if and only if $ad = bc$ is not be an equivalence relation (I haven't checked, but it seems likely).
 A: "Can we embed R into a division ring D? If not, can we embed R into a ring where every element of R has an inverse?"- No. There is a famous result of Malcev.
Mal'tsev, A.I.
On the immersion of an algebraic ring into a field. (English)
Math. Ann. 113, 686-691 (1937)
A: The general question of when one can embed a cancellative ring into a division ring has been studied a lot (and more generally, how far the usual localization techniques carry over to non-commutative rings). A good reference for this is Lam's Lectures on Modules and Rings (chapter 4).
In particular, one of the things to get familiar with is the Ore condition, which is what is needed to get a localization that behaves as one is used to from the commutative case.
A multiplicative subset $S$ of $R$ is said to be a right Ore set if for any $a\in R$ and $s\in S$ we have $aS\cap sR\not\neq \emptyset$ (so given some element $as'$ with $a\in R$ and $s'\in S$ we have some $a'\in R$ such that $as' = sa'$)
If the non-zero elements of $R$ (where $R$ is a cancellative ring) is a right Ore set then we call $R$ a right Ore domain.
One can show that any right Ore domain can be embedded into a division ring $D$ such that the embedding has the usual universal properties of an embedding of a commutative ring into its field of fractions.
