Let $A$ be an integral domain. I want to prove that the field of fractions $ Q(A) := (A \setminus \{0\} \times A )/ \sim$ , with $ (a,b) \sim (c,d) \Leftrightarrow ad - cb = 0 $ , is actually a field.
We name the elements in $ Q(A) $, $\frac {a}{b} $ , meaning the equivalence class of $(a,b)$.
I know that $ Q(A) $ is a ring with the sum $ \frac {a}{b} + \frac {c}{d} = \frac {ad+cb}{bd} $, and the product $ \frac {a}{b} \cdot \frac {c}{d} = \frac {ac}{bd} $.
I have tried seeing if every element $ \frac {a}{b} \in Q(A) $ has an inverse , by doing this :
$ \frac {a}{b} \cdot \frac {x}{y} = \frac {1}{1}$ , then $\frac {ax}{by} = \frac {1}{1} $ , so I have $ (ax, by ) \sim (1,1) $ , but I do not know what to do next.
Hint:
$$ (ax,by)\sim(1,1)\iff ax=by. $$ There is an obvious solution .