Proof that a field is an integral domain 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?
 A: 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 $m=1\cdot m=r^{-1}⋅r⋅m=r^{-1}⋅0=0$
A: It looks just fine. I would say $b\cdot a=1$ rather than $ab=1,$ though, to make it all perfectly clear.
To proceed without contradiction, simply remove the assumption that $c\ne 0.$ Then $a\cdot c=0$ implies $b\cdot (a\cdot c)=b\cdot 0,$ which implies (if you've shown/seen that $0$ is multiplicatively absorptive) by associativity that $1\cdot c=0,$ so $c=0$. This shows that if $a\in F$ with $a\ne 0,$ then the only $c\in F$ such that $a\cdot c=0$ is $c=0$. Consequently, it is not possible for the product of two non-$0$ elements of $F$ to be $0,$ so $F$ is an integral domain.
A: This is the right idea, but I'd recommend changing a couple things:


*

*Change the "Let" in "Let $a\cdot c = 0$ to "Suppose that", "Assume", or something to that effect. (You've already defined $a$ and $c$, and so "let" is a bit awkward here.) This is a technicality but definitely helps with the "flow" of your argument.

*Don't argue by contradiction when you don't have to. If you remove the assumption that $c \neq 0$, then what you've shown is that $a\neq 0$ in $F$ and $ac = 0 \implies c = 0$, which (by definition) shows that $F$ is an integral domain.
