Prove that the field of quotients of an integral domain $D$ is the smallest field containing $D$. . My Attempt Shown

Question

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..

Answer 1

Since, $F′$ is a field, $F′$ is a sub field of $E$.

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.

Answer 2

Suppose that $\iota:D\to E$ is an injection where $E$ is a field. Then $x\neq 0$ in $D$ gives $\iota x\neq 0$ in $E$; so every nonzero element of $D$ maps to an invertible element. By the universal property of localizations, we get an induced map $\bar \iota:F(D)\to E$. Since $F(D)$ is a field and the map is nonzero, it is injective.

Answer 3

Actually we have: $F=\{(a,b)|a,b \in D,b\ne 0\}$, so for any $(a,b)\in F$, we can define $\phi:F\to E$ \begin{align} \phi((a,b))=ab^{-1},(a,b)\in F \end{align}

We can check that $\phi$ is a homorphism form $F$ to $E$, and Im $\phi$ is a sub-field of $E$,so we complete our proof.