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

Solution 1:

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.