0
$\begingroup$

There's a theorem that states that cancellation laws hold in a ring $R$ if and only if $R$ has no zero divisors. Note that Integral Domains have no zero divisors. However, from my understanding in group theory, cancellation law happens by multiplying the (multiplicative) inverse on both sides, i.e. $$a^{-1}\cdot ab=a^{-1}\cdot ac\implies b=c.$$ Equivalently, $$ba\cdot a^{-1}=ca\cdot a^{-1}\implies b=c.$$ Going back to rings, this feels counterintuitive as this is only possible if all elements have a multiplicative inverse. But take note that Integral Domains are not necessarily division rings. So how does cancellation exactly work in rings?

Our professor said we should not multiply the inverse on both sides since we're not guaranteed that a multiplicative inverse exists for all elements of a ring $R$. So for the rest of the discussion, she only canceled terms without explicitly stating why it happens.

$\endgroup$
  • 7
    $\begingroup$ Cancellation means you can cancel, it doesn't mean that it HAD to happen by operating by inverses. $\endgroup$ – Randall May 21 '18 at 16:32
3
$\begingroup$

Suppose $ab=ac$. This means that

$$a(b-c)=0$$

but if the ring is an integral domain, either $a=0$ or $b-c=0$. If $a\neq0$, this implies that $b=c$.

$\endgroup$
1
$\begingroup$

Consider the example of the polynomial ring ${\Bbb R}[x]$. Since it is an integral domain, it holds for a non-zero polynomial $p$ that $$ pq_1=pq_2\quad\Rightarrow\quad q_1=q_2, $$ however, an inverse of the polynomial $p$ does not exist.

$\endgroup$
0
$\begingroup$

I would like to illustrate that with an example: take for example the integers $modn$ with n composite, then if $n=6$ for example you can see that $2x=0mod6$ does not imply $x=mod6$ but it could also mean $x=3$. This is because you have zero divisors in that ring.

$\endgroup$
0
$\begingroup$

For variety....

Every integral domain $D$ has a field of fractions $F$, and $D$ is a subring of $F$. (The general procedure for constructing rings of fractions is called localization)

If $ab = ac$ holds for $a,b,c \in D$, then in $F$ we have $\frac{ab}{1} = \frac{ac}{1}$. If $a \neq 0$, then $\frac{1}{a} \in F$, and multiplying through and applying the fact that $\frac{xy}{x} = \frac{y}{1}$ gives an equation $\frac{b}{1} = \frac{c}{1}$. Since $D$ is a subring of $F$, this implies $b=c$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.