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It can be shown that given $F \subset R \subset K$, where $K/F$ is an algebraic extension and $R$ is an integral domain, that $R$ is then an intermediate field. However, is there an explicit counter example for this statement when $K$ is not algebraic over $F$.

I was considering $\mathbb{Q}(\pi)/ \mathbb{Q}$ which is a transcendental extension because there does not exist a polynomial with rational coefficients for which $\pi$ is a root. However how can one show that there does not exist intermediary fields. Is it recommended that I got about in showing that for $R$ where $\mathbb{Q} \subset R \subset \mathbb{Q}(\pi)$, $R$ does not contain multiplicative inverses? Or should I focus on the infinite degree of $[\mathbb{Q}(\pi):\mathbb{Q}]$. Any and all help would be appreciated!

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  • $\begingroup$ Inside $\mathbb{Q}(\pi)$ you certainly find $\mathbb{Q}(\pi^2)$. Are you thinking to $R=\mathbb{Q}[\pi]$, perhaps? $\endgroup$ – egreg Feb 3 '15 at 23:26
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If $K$ is transcendental over $F$, there are infinitely many subfields $F\subset L\subset K$.

Indeed, if $\alpha\in K$ is transcendental over $F$, there is the infinite chain of subfields $$ K\supset F(\alpha)\supset F(\alpha^2)\supset \dots\supset F(\alpha^{2^n}) \supset \dotsb $$ It's sufficient to prove that $F(\alpha^2)\ne F(\alpha)$. Otherwise $$ \alpha=\frac{f(\alpha^2)}{g(\alpha^2)} $$ for some polynomials $f$ and $g$ in $F[X]$, and $g(\alpha^2)\alpha-f(\alpha^2)=0$ would mean $\alpha$ is algebraic over $F$.

What's true is that there exists a ring $R$, with $F\subset R\subset K$ and $R$ is not a field. Consider $R=F[\alpha]$: then $\alpha$ is not invertible here, because otherwise it would be algebraic.

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