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.

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

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 $\operatorname{Frac}(R)\subseteq k$.

  • $\begingroup$ 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

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