0

We know that an additive identity in a ring is always a multiplicative annihilator. But this doesn't need always be true in case of a semiring. Consider that $e$ is the additive identity of a semiring $S$, then for any $a\in S$, we see that $a.e=a.(e+e)=a.e+a.e \implies a.e=e \iff -a.e \in S.$ Looking for a better suggestion. Thanks

gete
  • 1,352
  • you don't have cancellation so you cannot infer $a\cdot e + a \cdot e= a \cdot e \implies a \cdot e =e$ – baharampuri May 12 '18 at 10:27
  • Yeah i meant $a\cdot e + a\cdot e =a\cdot e \implies a \cdot e=e$ will hold only if inverse exists in $S$ or cancellation law holds in it and as a consequence of this argument, we can conclude that additive identity is not multiplicative annihilator in a general semiring. But $e$ being additive identity is multiplicative annihilator only if semiring is cancellative. – gete May 12 '18 at 10:55

1 Answers1

1

But this doesn't need always be true in case of a semiring

I’m not sure what definition you’re using, but semirings are usually defined to require that the additive identity is absorbing.

If you’d like an example of something slightly less than a semiring which has a non absorbing zero, see Examples for almost-semirings without absorbing zero.

rschwieb
  • 153,510
  • Dear @rschieb, do you mean that additive identity if exists is multiplicatively absorbing too? Is converse true? i.e, if an element of a semiring is multiplicatively absorbing then will it always be an additive identity? – gete May 12 '18 at 14:52
  • @gete In the standard definition of a semiring, additive identity exists and is absorbing. The converse is probably usually false. For example, you can make any semigroup into a semiring using zero multiplication. – rschwieb May 12 '18 at 15:13