One of the field axioms is:

Axiom A

For every element $a$ in field $F$ there an element $-a\in F$ such that $a+(-a)=0_F$, where $0_F$ is the additive identity element of the field.

Is it right to say that the existence of $-0_F$ follow a fortiori from Axiom A? Otherwise, $0_F$ wouldn't have an additive inverse.


2 Answers 2


The additive inverse of $0_F$ is exactly $0_F$.


In other words, using the argument of @eloiPrime below, assume you have any group $(G,+)$ with identity element $0_G$. Then, $0_G + 0_G = 0_G$, hence $0_G$ has an inverse, and it is itself. In any field $(F,+,\cdot)$, the structure $(F,+)$ is a group...

  • $\begingroup$ $0_F=-0_F$ is true, but does that mean $-0_ F$ doesn't exists? $\endgroup$ Jul 3, 2018 at 23:47
  • 2
    $\begingroup$ This follows from the axiom for the existence of an additive identity. Recall that $-0_F$ is what you add to $0_F$ to get $0_F$. Since $0_F+0_F=0_F$, we have $-0_F=0_F$. $\endgroup$
    – user180579
    Jul 3, 2018 at 23:49

Yes. $0_F$ is an element of the field, just a particular instance of $a$, and this axiom guarantees that it too has an additive inverse. Things that are true for all elements of a set are true for any particular element you consider.

(I am assuming that there is a previous axiom says that $0_F$, otherwise I don't know how you could refer to it within the Axiom A.)

BTW, one could imagine a world where inverses are defined (e.g. "The inverse of an element $x$ is an element $y$ such that $x+y=0_f"$.) but we haven't yet given an axiom stating that all elements have inverses. Then we would already know that $0_F$ has an inverse, with out Axiom A, as $0_f + 0_F = 0_F$ from the definition of $0_F$.


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