Basic field axiom proof involving the additive inverse axiom

Question

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?

Answer 1

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, $a \cdot 0 = 0$.