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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).

spin
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2 Answers2

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"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)

Boris Novikov
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  • -1 for several reasons: this is a reference only answer, particularly because google does not turn up a usable link. The paper (I think) only explicitly claims to answer one of the questions and not the other. And from what I have gleaned incidentally, it doesn't answer the OP's question because it is critical to his approach that we consider rings that do not contain unity, as it involves semigroups that are counterexamples to a theorem that holds true for cancellation monoids. –  Mar 21 '13 at 16:18
  • @Hurkyl Which theorem for cancellation monoids were you referring to? – Jose Brox Oct 11 '18 at 16:11
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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.

Tobias Kildetoft
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  • +1 One thing to add is that there are other domains which embed into division rings without the "density" property that makes Ore embeddings so nice. (I like how Lam put it: the Good, Bad and Ugly :) ) – rschwieb Mar 21 '13 at 14:54
  • Very interesting is Paul Cohn's historical article on noncommutative localization mentioned in this related post. – Math Gems Mar 21 '13 at 14:55