The proof I have starts of with $\;xy=0\;$ in a field.
Then $x^{-1}$ exists because it is a field.
Then $x^{-1} xy=x^{-1} 0$.
Therefore $y=0$.
But surely if an integral domain can not have any zero divisors, how can we end the proof by saying $y=0$? Surely then y is a zero divisor and hence the field is not an integral domain?
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$\begingroup$Let $A$ be a field.
A field is a commutative division ring. So every non-zero element of $A$ is invertible.
Let $x,y \in A$ be such that $xy = 0$. Assume that $(x \neq 0 \ and \ y \neq 0)$.
As $x \neq 0$, $x$ is invertible, so $x^{-1}$ exists. Look at:
$$xy = 0 \implies x^{-1}(xy) = x^{-1}0 \implies (x^{-1}x)y = 0 \implies ey = 0 \implies y = 0$$
A contradiction.
Therefore $x = 0 \ or \ y = 0$.
This shows that $A$ is an integral domain.
What's invalid in your argument is the assumption $x^{-1}$ exists. This need not be true unless $x \neq 0$. I guess that you already understood this after reading my proof.
$\endgroup$ $\begingroup$You must say before your line about the assertion of $x^{-1}$ existing that $x \neq 0$, or else $x^{-1}$ may not exist. So if one element is nonzero and their product is 0, the other element must be 0.
Otherwise, it's great!
$\endgroup$ $\begingroup$You are missing the point: start with $xy=0$ and $x\neq 0$, because we are in a field $x^{-1}$ exists ( remember $x\neq 0$) and you then prove as you did that $y=0$. In summary whenever $xy=0$ either $x=0$ or $y=0$ and a field is an integral domain
$\endgroup$ $\begingroup$Our aim is to prove xy=0 only when either x=0 or y=0. Let us assume the contradiction, suppose the field F contains zero divisors, then xy=0 for some non zero x and y. As x is non-zero, and F is a field, x^{-1} exists and x^{-1}(xy)=0 which leads to y=0, a contradiction to our assumption that y is non-zero. This contradiction occured as we assume that F contains zero divisors. Hence F does not contain any zero divisors and it will be an integral domain
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