In problems concerned with finding the units in a ring, my textbook seems to always ignore the additive identity as a possibility. In combination with learning the definition of a field (a ring in which every nonzero element is a unit) and the fact that in every ring I've encountered so far, the additive identity is a multiplicative absorbing element, this led me to the suspicion that maybe this is always the case.

The Wikipedia page on additive identities confirms this and proves it by stating:

$s\cdot0 = s\cdot \left(0 + 0\right) = s\cdot0 + s\cdot0 \Rightarrow s\cdot 0 = 0$ (by cancellation)

However, my textbook also shows that rings do not satisfy the cancellation law of multiplication in general, so I guess this 'proof' is not sufficient then. Is there a way to prove it without assuming the multiplicative cancellation property?

  • 3
    $\begingroup$ This uses the cancellation law of addition which all rings satisfy (because they are abelian groups under addition). $\endgroup$ – Karl Kronenfeld Jun 15 '13 at 11:46
  • 2
    $\begingroup$ the proof you state uses additive cancelation, not multiplicative cancelation. $\endgroup$ – Ittay Weiss Jun 15 '13 at 11:47

In the last step of the proof we use cancellation of addition, not multiplication. We are effectively adding the additive inverse of $s\cdot 0$ to each side.

  • $\begingroup$ Oh, wow. I really don't know how I managed to overlook that. Thanks! $\endgroup$ – Jasper Driessens Jun 15 '13 at 12:13
  • $\begingroup$ No worries. We all do it from time to time. $\endgroup$ – john Jun 15 '13 at 12:21

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.