# Additive zero of a semiring

The Encyclopedia of Mathematics states that a semiring is
"A non-empty set S with two associative binary operations + and $$\cdot$$, satisfying the distributive laws"...

https://encyclopediaofmath.org/wiki/Semi-ring

Thus, we may expect the following four elements in a semiring:

• Additive zero $$a$$: $$a + x = x + a=a$$ for all $$x$$;
• Additive identity $$b$$: $$b + x = x + b = x$$ for all $$x$$;
• Multiplicative zero $$c$$: $$c\cdot x = x \cdot c = c$$ for all $$x$$;
• Multiplicative identity $$d$$: $$d\cdot x = x \cdot d = x$$ for all $$x$$.

But the Encyclopedia of Mathematics says:
"An additive zero in a semiring $$S$$ is an element $$a$$ such that $$a + x = x + a = x$$ for all $$x$$".

What is the name of the "true" additive zero of a semiring in this case?
Is it possible to add it into any semiring?

For example:

• $$a + x = x + a = a$$,
• $$a \cdot x = x \cdot a = 0$$ (the multiplicative zero)

for any $$x$$.

Is there an example of a semiring in which $$0 \ne a \cdot x \ne a$$ for some $$x$$,
where $$a$$ is the "true" additive zero: $$a + x = x + a = a$$?

• Multiplicative unity $1$ of a semiring $S$ is also called as additive zero.
– gete
Jul 7, 2020 at 17:14
• @gete The question is how we call the "true" additive zero. Jul 7, 2020 at 18:23

Note that the definition of the notion of semiring is not yet settled. In a semiring $$(S, +, \cdot)$$, some authors consider that $$(S, +)$$ and $$(S, \cdot)$$ are semigroups, while others consider $$(S, +)$$ as a commutative monoid with identity $$0$$ (say). When $$(S, \cdot)$$ is a monoid with unity $$1$$ (say), the structure $$(S, +, \cdot)$$ is also called hemiring.

Edit: Here, $$1$$ is additive zero if and only if $$x+1=1=1+x$$ for all $$x\in S$$, and $$0$$ is multiplicative zero as $$0\cdot x=0=x\cdot 0$$. To your last question, the answer is yes.

For example in case of distributive lattice $$(L, \vee, \wedge, 0, 1)$$, where $$0$$ and $$1$$ are bottom and top elements, respectively of $$L$$, $$1$$ is additive zero. While if we take $$S=\Bbb N_{0}$$, the set of non- negative integers endowed with the operations $$(+)$$ as usual addition and $$(\cdot)$$ as usual multiplication on $$S$$, then $$1$$ will not be additive zero (though it is multiplicative unity).

• @AlexC Now, the answer is edited.
– gete
Jul 7, 2020 at 20:01
• @AlexC Oh! thanks for the correction. It is corrected now.
– gete
Jul 8, 2020 at 1:51
• In case of the distributive lattice: $0$ is a multiplicative zero and an additive identity, and $1$ is an additive zero and a multiplicative identity (or vice versa if we swap the addition and multiplication). Correct? Jul 8, 2020 at 2:35
• @AlexC yes, it is..
– gete
Jul 8, 2020 at 3:07