# Rings Trapped Between Fields [duplicate]

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.

• Does this answer your question? Intermediate ring between a field and an algebraic extension. Note that finite implies algebraic. – user208649 Jun 30 '20 at 5:18
• It also seems your question is a duplicate of the other reference you just edited in. – user208649 Jun 30 '20 at 5:20
• 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? – Gae. S. Jun 30 '20 at 6:05

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 $$x$$.
• 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. – user208649 Jun 30 '20 at 5:22
• 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. – user208649 Jun 30 '20 at 5:33
• @TokenToucan I think it is necessarily the case that $c_0\neq0$ if $n$ is minimal, which was my intent. – alex.jordan Jun 30 '20 at 7:48