1
$\begingroup$

I'm reading the paper Immersion of an algebraic ring into a skew field by Malcev. Doi: 10.1007/BF01571659, GDZ.

On the third page of the paper, he writes that

If $\alpha\beta\sim\gamma\delta$ and the words $\alpha$ and $\gamma$ have the same length then $$ \alpha=\mu m,\quad \beta=n\nu,\quad \gamma\sim\mu m',\quad\delta\sim n'\nu,\quad mn\sim m'n' $$ where $m,n,m',n'$ are each one of the letters $a,b,c,d,x,y,u,v$.

How is this property easily seen? Malcev starts be starting with the semigroup of all words generated by eight letters $a,b,c,d,x,y,u,v$, where the operation is concatenation. He defines the pairs of two-letter words as "corresponding" $(ax,by)$, $(cx,dy)$ and $(au,bv)$ and says two words $\alpha$ and $\beta$ are equivalent if one can be obtained from the other by changing $ax$ to $by$, $cx$ to $dy$, etc. or vice versa.

He also proves that there are never any overlap ambiguities of what two letters could be replaced, but I don't see why the above quoted property follows so easily. Thank you.

  • 0
    The link is broken.2012-07-06
  • 0
    I've copied the links from the [other post](http://math.stackexchange.com/questions/165194/why-is-the-absence-of-zero-divisors-not-sufficient-for-a-field-of-fractions-to-e), I guess it is the same paper.2012-07-06

1 Answers 1