1

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?

MyGanton
  • 155
  • 1
    @DietrichBurde This has nothing to do with $x-1$ being irreducible. In any integral domain (UFD or not) you can cancel any non zero factor. Proving a polynomial ring is a domain is also much easier than proving it is a UFD. – Mark Dec 03 '23 at 12:45
  • It depends whether the original equation is a polynomial identity or whether we want to determine the real solutions of the original equation. I assume the first case and then , the above comments apply. In the second case , we can lose the solution $x=1$ , if we divide by $x-1$. – Peter Dec 03 '23 at 12:58
  • 1
    Simply because the ring of polynomials is not a group with respect to multiplication, so that division by a polynomial is not a multiplication by the inverse of the divisor. – uniquesolution Dec 03 '23 at 13:05
  • Yes, sorry, of course integral domain, not UFD. I wanted to say, for the ring $\Bbb Z$, $2n=2m$ also implies $m=n$, although $2$ is not invertible in $\Bbb Z$. Then it is perhaps more obvious what is happening. – Dietrich Burde Dec 03 '23 at 13:42
  • You did not say what $x$ is. If $x$ is a real number such that $x-1$ is not zero we have commutative and associative multiplication. And we have unique inverses (no zero divisors is equivalent) and the distributive property. And then it makes sense to divide by it. Otherwise you get the trivial $0=0$. And division by $0$ is not allowed. This is also true if $x$ is a complex number. Or for any case with commutative and associative multiplication and the distributive property and unique inverses (equivalent to no zero divisors ). If it holds in general, it also holds mod "something ". Clear ? – mick Aug 10 '24 at 19:32

2 Answers2

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).

tomasz
  • 37,896
0

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.

DonAntonio
  • 214,715
  • You seem to be misunderstanding the question. The variable there is $p$, while $x$ is not even a variable, but a monomial. The two are equivalent equations about $p$ for just about any reasonable context. – tomasz Dec 04 '23 at 05:50
  • @tomasz What you say seems to be your own opinion since the OP doesn't mention anything of the like, and in almost any more or less reasonable context, without further explanation, it is the case $;x;$ is the variable and $;p,q;$ are functions, or just polynomials, in $;x;$ . And for the value $;x=1;$ , what this is. the first equation is fulfilled whereas the second one is not. – DonAntonio Dec 04 '23 at 15:32
  • Given that OP's solution sheet says that the two are equivalent (and the fact that he's using abstract algebraic terminology), I think the interpretation that $p$ is the variable and not $x$ is much more likely. And certainly likely enough not to warrant outright dismissal of the equivalence of the two equations, since they absolutely are equivalent in a very reasonable context. – tomasz Dec 04 '23 at 16:02
  • @tomasz I cannot see, in the OP's notation, how could it be possible that $;p;$ is the (free, to be sure) variable. That'd be outrightly unusual and weird, and unless another interpretation is given from that book or at least the OP explains further, I'd say that book's solution to that problem is incorrect...unless it has a particular meaning to "equivalent equation" that I am not aware of. – DonAntonio Dec 04 '23 at 17:41