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Some Background and Motivation: In this question, it is shown that an integral domain $D$ such that $F \subset D \subset E$, $E$ and $F$ fields with $[E:F]$ finite, is itself a field. However, a significantly more general result holds and seems worthy, of independent address; hence,

Let

$F \subset E \tag 1$

be fields with

$[E:F] < \infty; \tag 2$

if $R$ is a ring such that

$F \subset R \subset E, \tag 3$

show that $R$ is in fact a field.

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    $\begingroup$ Does this answer your question? Intermediate ring between a field and an algebraic extension. Note that finite implies algebraic. $\endgroup$ – TokenToucan Jun 30 at 5:18
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    $\begingroup$ It also seems your question is a duplicate of the other reference you just edited in. $\endgroup$ – TokenToucan Jun 30 at 5:20
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    $\begingroup$ Isn't the second theorem exactly the same as the first one, except for the additional observation that subrings of integral domains are integral domains? $\endgroup$ – Gae. S. Jun 30 at 6:05
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If $x\in R\setminus F$, then for some minimal $n\in\mathbb{N}$, $x$ is a root of a polynomial $\sum_{i=0}^nc_ix^i$ over $F$ of degree $n$. Otherwise, $\{1,x,x^2,\ldots\}$ is a basis of an infinite dimensional vector space over $F$, but $[E:F]$ is finite. And note that $c_0\neq0$. Otherwise $x$ would be a zero-divisor, and this all happens within field $E$.

So for any $x\in R\setminus F$, you have $x\cdot\overbrace{\left(-\frac{1}{c_0}\sum_{i=1}^{n}c_ix^{n-1}\right)}^{\in R}=1$. So $R$ contains the inverse of $x$.

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  • $\begingroup$ It is not the case that eventually $x^n \in F$, rather that $x^n$ is eventually a linear combination of smaller powers. Your reasoning would imply, for instance, that all finite extensions are given by radicals which is not true. $\endgroup$ – TokenToucan Jun 30 at 5:22
  • $\begingroup$ @TokenToucan Thanks. Rattling things off too quickly. $\endgroup$ – alex.jordan Jun 30 at 5:27
  • $\begingroup$ Also it isn't necessarily the case that $c_0 \neq 0$ but if it does you can always factor out $x$'s and cancel them (since one is in a domain) to get a linear combination which has a nonzero constant term. $\endgroup$ – TokenToucan Jun 30 at 5:33
  • $\begingroup$ @TokenToucan I think it is necessarily the case that $c_0\neq0$ if $n$ is minimal, which was my intent. $\endgroup$ – alex.jordan Jun 30 at 7:48

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