Let $A$ be a domain. Suppose that for all $a,b\in A-\{0\}$, $Aa \cap Ab \neq \{0\}$ and $aA \cap bA \neq \{0\}$. How to show that the field of fractions of $A$ is $\{ab^{-1}: a \in A, b \in A-\{0\}\}$?
The condition $Aa \cap Ab \neq \{0\}$ is equivalent to the condition: there are some $a', b' \in A$ such that $ab^{-1}=b'^{-1}a'$.
The condition $aA \cap bA \neq \{0\}$ is equivalent to the condition: there are some $a', b' \in A$ such that $b^{-1}a=a'b'^{-1}$.
Therefore $$ (a_1 b_1^{-1})(a_2 b_2^{-1}) = b_1'^{-1}a_1'a_2b_2^{-1} = (a_1'a_2)'b_1''^{-1}b_2^{-1} \in \{ab^{-1}: a \in A, b \in A-\{0\}\}. $$ We have \begin{align} & ab^{-1}+cd^{-1} \\ & = b'^{-1}a'+cd^{-1} \\ & = b'^{-1}(a'd+b'c)d^{-1} \\ &= (a'd+b'c)'b''^{-1}d^{-1} \in \{ab^{-1}: a \in A, b \in A-\{0\}\}. \end{align} Can we conclude that the field of fractions of $A$ is $\{ab^{-1}: a \in A, b \in A-\{0\}\}$? I think that we need both conditions $Aa \cap Ab \neq \{0\}$ and $aA \cap bA \neq \{0\}$. Is it correct? Thank you very much.
Edit: here $A$ can be non-commutative.