Associativity of multiplication in a field

You can prove that $(F, \times)$ is associative using the field axioms you cite. Associativity states that for any $a,b,c \in F$, $(ab)c=a(bc)$. If none of $a$, $b$, or $c$ are zero then this equation is in $F^\times$, where it is true. Suppose then that at least one of $a$, $b$, or $c$ are zero. We'll need the following lemma.

Lemma. Let $a \in F$. Then $0a=0$.

Proof. $0a + 0a = (0+0)a =0a$ by distributivity. Then as $(F,+,0)$ is a group, $0a=0$ (subtract $0a$ from both sides).

Using this lemma, we can see that if any of $a,b,c$ were zero then $(ab)c=0$ and $a(bc)=0$. To be rigorous, we can do the 3 cases.

• $a=0$: $a(bc)=0(bc)=0$ and $(0b)c=0c-0$.

• $b=0$: $a(0c)=a0=0$ and $(a0)c=0c=0$.

• $c=0$: $a(b0)=a0=0$ and $(ab)0=0$.

Multiplicative associativity also holds if you include $0$, since $0$ is absorbing, i.e., for all $x\in F$,

$x\cdot 0 = 0$.

Absorption can be proved as follows:

$0=0+0$ and so by multiplying with $x$, $x0 = x0+x0$. Adding $-x0$ on both sides gives $0=x0$.