1
$\begingroup$

If $R$ is an integral domain, show that there is no subfield $k$ of Frac($R$) containing $R$.

I think the way to prove this is by contradiction. So, let $R$ be an integral domain, and let $k$ be a subfield of Frac($R$) such that $R$$\subseteq \!\,$$k$. I think we need to assume that two elements are in $k$ and then show that at least one of these elements is not in $R$. So, to show that two elements are not in $R$, we need the product of these two elements to equal zero.

Let $h$,$j$ $\in \!\ k$. I am honestly not sure what to do from here.

$\endgroup$
  • 3
    $\begingroup$ No proper subfield. $\endgroup$ – drhab Mar 25 '16 at 15:50
4
$\begingroup$

Just use the fact that $\operatorname{Frac}(R)$ has all field operations applicable to elements of $R$ and that $k$ is closed under the field operations. Since $a,b\in R$ with $b\ne 0$ implies $a+b, ab, a/b\in k$, this is exactly the elements of $\operatorname{Frac}(R)$, so $\operatorname{Frac}(R)\subseteq k$.

$\endgroup$
  • $\begingroup$ Why does the conclusion that $Frac(R)$$\subseteq \!\,$$k$ satisfy the proof? A field is a subfield of itself. Doesn't the issue revolve around showing that R cannot be contained in a subfield of Frac($R$)? $\endgroup$ – Jason Smith Mar 25 '16 at 15:51
  • $\begingroup$ @JasonSmith The proof shows that if $k$ is any field containing $R$, it contains $Frac(R)$, that means a proper subfield cannot contain $R$. $\endgroup$ – Adam Hughes Mar 25 '16 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.