Why am I allowed to divide by polynomials even though multiplicative inverse do not exist?

Question

Suppose I have the following equation:

$(x+1)(x-1)p(x) = (x-1)(x^2+x+1)q(x)$

According to the solutions I'm given, this is equivalent to

$(x+1)p(x) = (x^2+x+1)q(x)$

I get that, that one 'cancels out' $(x-1)$ but why am I allowed to do that, when I know that $(x-1)$ has no multiplicative inverse?

Answer 1

The property of an operation that $a*b=a*c$ implies $b=c$ (or in other words, that multiplication by a fixed element on the left is injective) is called (left) cancellativity.

This is certainly true when we have inverses, but it is not necessary. For instance, it is not hard to see that among nonzero elements in a ring (in fact, in a distributive algebra), multiplication is cancellative (on both sides) if there are no zero divisors. This is a simple consequence of the fact that a group homomorphism is injective if and only if its kernel is trivial.

Polynomials over rings without zero divisors don't have zero divisors (proof hint: consider the leading term).

Answer 2

I think that if your solutions actually say that both equations you wrote at the beginning are "equivalent" (which I understand that it means both equations have exactly the same solutions) then that is wrong, as the first equation has the solution $\;x=1\;$ whereas the second doesn't seem to have it unless $\;p(1)=q(1)=0\;$ .

So either cancellation is forbidden in this case, or else one has first to explicitly write down the solution $\;x=1\;$ and then say: "for $\;x\neq 1\;$ we can cancel and..." etc.

Now, if for "equivalent" you understand something else then we must know what it is.