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"...


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: $a + x = x + a = a$?

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

1 Answer 1


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).

  • $\begingroup$ @AlexC Now, the answer is edited. $\endgroup$
    – gete
    Jul 7, 2020 at 20:01
  • $\begingroup$ @AlexC Oh! thanks for the correction. It is corrected now. $\endgroup$
    – gete
    Jul 8, 2020 at 1:51
  • $\begingroup$ 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? $\endgroup$
    – Alex C
    Jul 8, 2020 at 2:35
  • $\begingroup$ @AlexC yes, it is.. $\endgroup$
    – gete
    Jul 8, 2020 at 3:07

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