I've recently begun to read Skew Fields: The General Theory of Division Rings by Paul Cohn.

On page 9 he writes,

Let us now pass to the non-commutative case. The absence of zero-divisors is still necessary for a field of fractions to exist, but not sufficient. The first counter-example was found by Malcev [37], who writes down a semigroup whose semigroup ring over $\mathbb{Z}$ is an integral domain but cannot be embedded in a field. Malcev expressed his example as a cancellation semigroup not embeddable in a group, and it promped him to ask for a ring $R$ whose set $R^\times$ of nonzero elements can be embedded in a group, but which cannot itself be embedded in a field.

The cited paper [37] is On the immersion of an algebraic ring in a skew field, Math. Ann 113 (1937), 686-91. (EDIT by M.S: doi: 10.1007/BF01571659, GDZ.)

I've had no luck finding this freely available online, nor at the library. Does anyone have reference to this paper, or at least the part where Malcev demonstrates these two parts of his counter-example? I would greatly appreciate seeing it. Thanks.

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    $\begingroup$ Mathematische Annalen are freely available at GDZ (I've edited link to your post.) You would have found that link if you have checked Wikipedia article. This is my general advice - if a journal is available freely online and there is a Wikipedia article about this journal, you will most probably find a link there. (Anyway, getting the paper directly from Springer would be better, since you would have an OCRed version.) $\endgroup$ – Martin Sleziak Jul 1 '12 at 6:25
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    $\begingroup$ BTW I was not sure whether I should post the link to the paper as a comment or as an answer - the reason is that title of your post is somewhat different from the body. The title suggest that you would like to see an explanation, but from the body it seems that you just want to find the paper, if it is available online. $\endgroup$ – Martin Sleziak Jul 1 '12 at 6:27
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    $\begingroup$ Thanks Martin, the link was quite helpful. I've used your advice on finding papers a handful of times already since you've posted these comments! $\endgroup$ – Yong Pan Jul 3 '12 at 6:43
  • $\begingroup$ See also here: Is there any way to read articles without subscription?. $\endgroup$ – Martin Sleziak Jul 3 '12 at 6:55

You can find in most good noncommutative algebra books a discussion of how the normal "field of fractions" construction fails in some noncommutative domains. Your keywords to look for are "Ore condition" and "right Ore domain". If you haven't had luck downloading Cohn's example, there will be many under these keywords.

In short, the right Ore condition is necessary and sufficient for the normal field of fractions definition to work. Without it, it may be impossible to add or multiply fractions, because there will problems finding common denominators.

Here's what I mean: If you carry out the normal equivalence relation in an effort to create a right division ring of fractions, you use $(x,y)\sim(w,z)$ if there exists nonzero $s$ and $t$ such that $ys=zt\neq0$ and $xs=wt$. This allows you to bring things to common denominators, but notice you are only allowed to introduce things on the right.

Suppose you want to add $(a,b)+(c,d)$ where $b,d$ are nonzero. You would like to define this as $(stuff,bd)$. You can form $(ad,bd)\sim (a,b)$ but you are unable to form $(bc,bd)$ because you cannot introduce $b$ on the left.

If you have access through googlebooks or otherwise, Lam's Lectures on Modules and Rings recounts Mal'cev's example of a domain which is not embeddable into a division ring on page 290.

  • $\begingroup$ I think equivalence relation is rather : $(x,y) \sim (w,z)$ iff $\exists t,s \neq 0$ st $yt=zs$ and $xt=ws$. $\endgroup$ – user10676 Jul 2 '12 at 15:26
  • $\begingroup$ @user10676 Yes thanks... fixed. I don't know how I screwed that up here... I'm working on that very definition in another post simultaneously! Must be the 100+ degree weather... $\endgroup$ – rschwieb Jul 2 '12 at 16:30
  • $\begingroup$ Thanks for the reference, it's quite illuminating for me. $\endgroup$ – Yong Pan Jul 3 '12 at 6:43

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