1

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.

Martin Sleziak
  • 53,687
Yong Pan
  • 740
  • 3
  • 14

1 Answers1

0

Malcev states before that in a word $ \dotsc m n p \dotsc$ we cannot replace $mn$ and $np$ by another word. This means that the relations don't overlap. Regardless how you change a word by the relations, the classes of the letters $\{a,b,c,d\}$ and $\{x,y,u,v\}$ stay the same at each position. So if we want to change $\alpha \beta$ into $\gamma \delta$, we have to change $\alpha$ into $\gamma$ and $\beta$ into $\delta$ separatedly (this is the case that the letters $m,m',n,n'$ don't appear; probably one should add this), or at one step we come across the two words and make something like $(\mu m)(n \nu) \sim (\mu m')(n' \nu)$. Of course this is not really formal. It can be made precise using induction on the length of the deduction.

Martin Brandenburg
  • 163,620