Let $D$ be an integral domain and let $F$ be the field of quotients of $D$. Show that if $E$ is any field that contains $D$, then, $E$ contains a sub field that is ring isomorphic to $F$. Hence, the field of quotients of an integral domain is the smallest field containing $D$.
Attempt: Let $S$ be a sub ring of the field of quotient $F$ such that $S \approx D$
We need to show that $F \approx $ a sub field of $E$.
Let $F'$ be the field of quotients of $F$.
If $K$ be a field, then, the field of quotients of $K$ is ring isomorphic to $K$ is a result.
Hence, We know that $F' \approx F$ . Hence, $F'$ must contain $D$.
Since, $F'$ is a field, $\implies F'$ is a sub field of $E$.
Hence Proved that there exists a sub field of $E$ which is isomorphic to $F$.
Is my attempt correct?
Please note that my book hasn't yet introduced polynomials, reducability, divisibility in integral domain or field extensions.
Thank you for your help..
Best Answer
This line is pretty much assuming what you are currently are trying to prove. You will have to actually make reference to how $D$ lies in $E$ to prove the statement.
Why not just try to make a map from $F$ into $E$? Since you already have $D\subseteq E$, it's natural to just say $\phi(a)= a\in E$ for all $a\in D$.
What about other elements of $F$? For each $b\neq 0$ in $D$, there is an element $b^{-1}\in E$ which is $b$'s inverse, and an element $b^{-1}\in F$ which plays the same role in $F$. Naturally you'd want $\phi(b^{-1})=b^{-1}\in E$, where the first $b^{-1}$ is in $F$ and the latter is in $E$.
To map the remaining elements of $F$, we would require $\phi(ab^{-1})=\phi(a)\phi(b^{-1})=ab^{-1}$ (again the first $b^{-1}$ is the one in $F$ and the last one is in $E$.)
Verify this gives a well-defined injective ring homomorphism from $F$ into $E$. The image of $\phi$ is the copy of $F$ that you seek.