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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..

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    $\begingroup$ It does not look correct to me. How do you show that since $F'$ is a field then $F'$ is a sub field of $E$? Maybe you can think intuitively about it like this: if $E$ contains $D$ then since $E$ is a field we have for $d \in D$ ($d \ne 0$) that $d^{-1}$ exists in $E$, but $d^{-1}$ is exactly the kind of element that we added to get $F$, so it seems that elements of $F$ will be in $E$ as well (I'm a bit vague, but trying to give a sense of it) $\endgroup$ – user50948 Jul 18 '14 at 21:23
  • $\begingroup$ Thank you for the comment :-) $\endgroup$ – MathMan Jul 19 '14 at 1:28
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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.

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    $\begingroup$ Thank you for your answer :-) $\endgroup$ – MathMan Jul 19 '14 at 1:16
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    $\begingroup$ Dear @VHP : I asked myself "where should I send elements of $D$? Where should I send inverses of elements in $D$? Did I get everything?" $\endgroup$ – rschwieb Jul 19 '14 at 1:16
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    $\begingroup$ @VHP No problem... hope it helped! $\endgroup$ – rschwieb Jul 19 '14 at 1:16
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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.

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  • $\begingroup$ Judging from the line "Please note that my book hasn't yet introduced polynomials, reducability, divisibility in integral domain or field extensions.", it looks like referring to universal properties and localization is overshooting user's knowledge requirements. Judging from recent posts by this user, it looks like only basics about rings, fields and homomorphisms are at his/her disposal. Regards $\endgroup$ – rschwieb Jul 18 '14 at 21:52
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    $\begingroup$ @rschwieb Answers are not always supposed to go to the OP. Some time later he might come back and understand it. $\endgroup$ – Pedro Tamaroff Jul 18 '14 at 22:05
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    $\begingroup$ That's true, and I'm not faulting any of the content of your answer really. If I had been in your shoes though, I'd have also put a simpler answer in front :) Then whatever stuff happens later is beyond reproach. Regards $\endgroup$ – rschwieb Jul 19 '14 at 1:14
  • $\begingroup$ OK. ${}{}{}{}{}$ $\endgroup$ – Pedro Tamaroff Jul 19 '14 at 1:29

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