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Let $R$ be an arbitrary ring. Now, we assume that we don't know whether $R$ has the multiplicative identity or not.

I know that $R$ has no zero divisors if and only if the cancellation law holds. So, suppose $R$ has no zero divisors. Consider for a nonzero element $a\in R$, $ab=a$ for some $b\in R$. Now, I want to apply for the cancellation law, but, if so, we have $b=1$, where $1$ is the multiplicative identity of $R$.

I think it is false because we don't know whether the ring has the unity.

Thus, I'm wondering when the cancellation law holds.

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  • $\begingroup$ For an example consider the even integers, which have no zero divisors, obey the cancellation law for non-zero elements, and have no multiplicative identity. A finite ring which obeys the cancellation law will have an element such that $ab=a$ - just multiply $a$ by all elements in turn. By cancellation, these must be distinct. So every element is included. And this includes $a$. $\endgroup$ – Mark Bennet Mar 1 '15 at 13:16
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If $ab=a$ and $R$ has no zero divizors, then for any $c\in R$ $$ a(bc) = (ab)c = ac, $$ so $bc = c$. So $R$ does have a unity, and it is $b$.

Upd. $cb = c$ holds as well since $(cb)d = c(bd) = cd$.

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    $\begingroup$ It is also true that $cb=c$? I think your conclusion is true only when $R$ is commutative. $\endgroup$ – JeongHobin Mar 1 '15 at 12:12
  • $\begingroup$ Let $d$ be a non-zero element, then $(cb)d = c(bd) = cd$, so $cb = c$ for any $c\in R$. $\endgroup$ – Litho Mar 1 '15 at 12:16
  • $\begingroup$ Ah, then it is true that if $R$ has no zero divisors, then $R$ has the unity? $\endgroup$ – JeongHobin Mar 1 '15 at 12:28
  • $\begingroup$ If $R$ has no zero divizors and $ab=a$ for some $a\neq 0$ and $b$, then yes, it has the unity. $\endgroup$ – Litho Mar 1 '15 at 12:31
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From $ab=a$ you cannot conclude that $b=1$ because the cancellation law says $ab=ac$ implies $b=c$. So, to use the cancellation law in $ab=a$, you have to write $a$ as a product $ac$.

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  • $\begingroup$ Ummm.... it is possible to use the cancellation law when $R$ has no unity? That is, if $R$ has no zero divisors, then does the cancellation law hold regardless of the existence of the unity of $R$? $\endgroup$ – JeongHobin Mar 1 '15 at 12:23
  • $\begingroup$ @JeongHobin, the way I read the cancellation law is that the map $x \mapsto ax$ is injective for all $a$. If one of these maps is not injective, then there are zero divisors. Reciprocally, if there are zero divisors $ab=0$, then both $b$ and $0$ are mapped to $0$ under $x \mapsto ax$. $\endgroup$ – lhf Mar 1 '15 at 12:26
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Cancellation law holds in addition of a ring but in multiplication cancellation law holds if and only if there exist multiplicative inverse of that element which you want to cancel out from both sides

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