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(5): Another possibility? LCPRCP in monoids??? by JLo)
I just developed a proof that LCP implies RCP for finite monoids:
Assume a finite monoid with a left cancellation property, i.e. Ax=Ay implies x=y for all nonzero A and all x and y.
Then x <> y implies Ax <> Ay for all nonzero a.
Just so I can use language, let's say that AB is "A operating on B".
Then non-zero A operating on each element of the monoid is a one-to-one
mapping onto every element of the monoid; otherwise there would exist
some x and y for which x <> y and Ax = Ay. Note that
this is only true because the monoid is finite. Therefore, for
every A <> 0 there exists a unique element A' for which
AA' = 1 (identity element).
We are just about done.
xA = yA (A not zero) implies
(xA)A' = (yA)A' implies
x(AA') = y(AA') implies
x = y
By a similar argument, RCP implies LCP.
Therefore, Richard's statement is true for finite monoids.
re(6): Another possibility? LCPRCP in monoids???
