Why is the absence of zero divisors not sufficient for a field of fractions to exist?

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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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.