# Why is the additive identity of a ring always a multiplicative absorbing element?

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.

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

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