# Basic field axiom proof involving the additive inverse axiom

I have recently started out learning about fields and the basic field axioms, so I have tried proving a couple of questions related to that. However, when I encountered the question:

Let $$F$$ be a field, and $$a ∈ F$$. Then, prove that $$a · 0$$ = $$0$$.

I wasn't sure if such an argument (which I mentioned down below) is allowed without further justification.

Basically, to prove this claim, I have used one of the basic axioms which states that "For any $$x ∈ F$$, there is a $$w ∈ F$$ such that $$x + w = 0$$" (introducing the notion of additive inverse). Therefore, I claimed that $$a⋅0 = (a⋅0) + (-a⋅0) \quad \textrm{(where we have x = (a \cdot 0) \quad \textrm{and} w = -(a \cdot 0))}$$ Thus, $$a⋅0 = (a⋅0) + (-a⋅0) = 0 \quad \textrm{(by the additive inverse basic axiom)}$$ So, my question is - would this be considered an acceptable argument?

• How did you claim that $a\cdot0=(a\cdot0)+(-a\cdot0)$? Jan 28, 2020 at 2:54
• I'm trying to only use the basic field axioms to prove this claim, so I tried applying the additive inverse axiom to prove this claim is true, which I'm not sure if I did correctly.. Jan 28, 2020 at 2:56

Firstly, you cannot prove an axiom. An axiom is something you take to be true. Therefore the existence of additive inverse in a field is a known fact. However, it seems difficult to use this fact directly to prove your claim. Instead, we can use the distributivity of multiplication over addition and the fact that $$0$$ is the additive identity of the field, and then use the additive inverse of $$a \cdot 0$$ to prove that $$a \cdot 0 = 0$$. The proof goes as follows:-
$$a \cdot 0 = a \cdot \left( 0 + 0 \right) = a \cdot 0 + a \cdot 0$$
Since $$F$$ is a field, we know that $$a \cdot 0 \in F$$ and from the above equality, we also know that it is idempotent with respect to the addition operation. This is only possible if it is the additive identity, i.e., $$0$$. Hence, $$a \cdot 0 = 0$$.
• Idempotent means when an element is operated with itself it results in the same element. If I say $a$ is idempotent, it means $a+a=a$. It is easy to check that the only element with this property is the additive identity $0$ of the field. Jan 28, 2020 at 3:33