Field of fractions

Question

Consider $A$, an integral domain, and $K(A)$, its field of fractions. If $B$ is an other integral domain such that $A\subset B \subset K(A)$, I need to prove that $K(A)$ is isomorphic to $K(B)$.

Thanks in advance!

Jad

Answer 1

$B\subset K(A)$ leads to $K(B)\subset K(K(A))=K(A)$. $A\subset B$ leads to $K(A)\subset K(B)$. So you end up with $K(A)=K(B)$.

addendum meeting Bills comment:

There is an extension $\psi:K\left(B\right)\rightarrow K\left(A\right)$ of inclusion $B\hookrightarrow K\left(A\right)$.

So if $F:=\psi\left(K\left(B\right)\right)$ then $\hat{\psi}:K\left(B\right)\rightarrow F$ defined by $x\mapsto\psi\left(x\right)$ is an isomorphism and $A\subset B\subset F\subset K\left(A\right)$.

There is an extension $\phi:K\left(A\right)\rightarrow F$ of inclusion $A\hookrightarrow F$.

Then composite $K\left(A\right)\stackrel{\phi}{\rightarrow}F\hookrightarrow K\left(A\right)$ is an extension of inclusion $A\hookrightarrow K\left(A\right)$.

That means that this composite must equalize the identity on $K\left(A\right)$ which is the unique extension of that inclusion.

This leads to $F=K\left(A\right)$, hence $\hat{\psi}$ is an isomorphism $K\left(B\right)\rightarrow K\left(A\right)$.

Answer 2

You need two facts about integral domains from which the result follows immediately. First, if $A \subset B$ then $K(A) \subset K(B)$ and second $K(K(A)) = K(A)$. Try proving each of these first.

Answer 3

Somewhat more elementarily:

$K(A)$ contains quotients of elements of $A$. $ A \subset B $, hence the quotients of elements of B include those quotients; $ K(A) \subseteq K(B)$.

This provides us with a very natural and well behaved mapping from $K(A)$ to $K(B)$: $$ \varphi \left( \frac{c}{d} \right) := \frac{c}{d} $$

$\varphi$ is trivially an injective homomorphism, so all that needs to be demonstrated is that $\varphi$ is onto.

Take $ \frac{c}{d} \in K(B) $. By definition of $K(B)$, $c$, $d \in B$ and $ d \not = 0 $.

$B \subset K(A) \implies c, d \in K(A)$

K(A) is a field and $d \not = 0$, hence $\frac{1}{d} \in K(A)$

By closure, $\frac{c}{d} \in K(A)$, giving: $ \varphi \left( \frac{c}{d} \right) := \frac{c}{d} $; $\varphi$ is onto.