2
$\begingroup$

The axioms of a semiring include:

$$0·a = 0 = a·0$$

Is there a name for an algebraic structure which satisfies all the axioms of a semiring except for this one?

(if it helps, the structure I have in mind happens to satisfy $0·a·0=0$)

$\endgroup$
2
  • 2
    $\begingroup$ Many authors do not include this axiom when defining a semiring. When this is satisfied usually it is called a semiring with zero, when there is a multiplicative identity it is called division semiring. Look at how math.chapman.edu/cgi-bin/structures.pl?HomePage defines it for example $\endgroup$ Jan 5, 2010 at 8:53
  • $\begingroup$ @Gjergji: Thanks for this comment. I didn't know that and always wondered about the definition which can be found at Wikipedia and of course also other sources. However, I think that Adam assumed the additive structure to be a commutative monoid. And there it is quite natural to impose the condition $0a=0=a0$ (together with distributivity, it just says that left and right multiplication are monoid endomorphisms). The question is probably about the setting where we have have a zero for the addition, but don't require it to be absorbing. $\endgroup$ Dec 15, 2011 at 21:48

1 Answer 1

1
$\begingroup$

If you were looking for a field with convenient terminology for its structures, I should warn you that the field of semirings is pretty bad :)

In

"Graphs, dioids and semirings: new models and algorithms" by Michel Gondran, Michel Minoux

they use "presemiring" to mean a set with two associative binary opearations, + and X where + is commutative and X distributes over + on both sides.

To make a presemiring a semiring, they require identity elements for both operations, and require that the additive identity 0 is absorbing, as you describe.

I wouldn't put too much stock in the one book though. Really there are so many authors throwing around so many terms about these things it's probably useless to find out which is the most common.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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