A ring is an algebraic system that supports unlimited addition, subtraction, and multiplication, with all the familiar laws (such as the distributive laws a(x+y)=ax+ay and (x+y)b=xb+yb) holding except that there may possibly be a,b pairs for which ab=ba does not hold. The ordinary integers are an example of a ring (where, however, ab=ba does always hold).

A ring has the left-cancellation property if ax=ay implies x=y for all nonzero a and all x and y, and has the right-cancellation property if xb=yb implies x=y for all nonzero b and all x and y.

Your mission should you choose to accept it: Prove that a ring has the left-cancellation property if and only if it has the right-cancellation property.

(In reply to re: Counter example for Monoids (not valid) by Steve Herman)

You had me scared for a moment, Steve! Now I looked at your calculation and I think associativity is OK. This is how the calculation should read:

((a,b)*(0,1))*(1,c) = (a,b+1)*(1,c) = (a+1,c)

(a,b)*((0,1)*(1,c)) = (a,b)*(1,c) = (a+1,c)

I think you have misread the rule "(a,b)*(x,y)=(a+x,y) for x unequal 0" as "(a,b)*(x,y)=(a+x,x) for x unequal 0". What do you think?


BTW, the way I came up with this monoid was to consider the free monoid over a two letter alphabet {r,t}. This is simply the set of all strings with letters r and t where concatenation is the monoid operation. Then you consider all strings in the monoid equal if the y contain the same number of r's and the string to the right of the right most r is identical (Mathematically speaking "you divide with an equivalence relation") What you end up is a monoid of strings r^a*t^b with integers a and b and the multiplication operation as defined above. That makes the multiplication rule a little easier to remember. Still, quite useless monoid this is, I admit...

Posted by JLo on 2006-08-07 15:23:54