Here is my attempt at proving this.
Let $F$ be a field and let $a \in F, \ a \neq 0$.
Then $a$ is a unit and hence $\exists \ b \in F$ such that $ab = 1$
Now let $c \in F, \ c \neq 0$
Let $a \cdot c = 0$
Then $b \cdot a \cdot c = b \cdot 0$
$\implies 1 \cdot c = c = 0$
This is a contradiction as $c \neq 0$
Hence $a \cdot c \neq 0$
I.e. F is an integral domain.
Does that look ok? Any ideas on how to show it without using a contradiction?
Best Answer
You could just assume that $a$ is invertible with $ba=1$ and that $a⋅c=0$ for some $c\in F$. Then $$c=1⋅c=ba⋅c=b⋅ac=b⋅0=0$$ so $c$ must be $0$. But the definition of a zero divisor $a$ says that $a⋅c=0$ for some non-zero $c$. In other words, $a$ is not a zero divisor. Since $a$ was an arbitrary field element, this means that $F$ has no zero divisors.
A similar proof shows that an invertible $r$ in a ring $R$ cannot give the product $0$ when multiplied with a non-zero module element $m$, as $m=1\cdot m=r^{-1}⋅r⋅m=r^{-1}⋅0=0$