If every intermediate ring of a field extension is a field, then the extension is algebraic 
Suppose $E/F$ is an extension of fields.
Prove that if every ring $R$ with $F\subseteq R\subseteq E$ is a field, then $E/F$ is an algebraic extension.

I can show the converse is true by demonstrating that the inverses of elements of $R$ are also in $R$, but I have no idea where to start for this one.
 A: Let $\alpha \in E$ and consider $F[\alpha]$. If this is a field, then $\frac 1 \alpha \in F[\alpha]$, so for some $d \in \Bbb{N}$ and $e_i \in F$, we have $\sum_{i=0}^d e_i\alpha^i=\frac{1}{\alpha}$. Multiply both sides by $a$ and subtract both sides by $1$ and we get:
$$\left(\sum_{i=0}^{d} e_i\alpha^{i+1}\right)-1=0$$
Thus, we have found a polynomial with coefficients in $F$ where $\alpha$ is a root. This means $\alpha$ is algebraic over $F$. However, $\alpha$ was just a generic element in $E$, so this is true for all $\alpha \in E$, so $E/F$ is an algebraic extension.
A: You can prove that if $E/F$ is a transcendental extension, then there exists a ring $R$, with $F\subseteq R\subseteq E$, that is not a field.
Take $a\in E$ that is transcendental over $F$; then $F[a]\cong F[X]$, where $X$ is an indeterminate over $F$. The ring $F[X]$ is not a field.
Alternatively, suppose $a\in E$ and consider the evaluation map $v_a\colon F[X]\to F[a]$. Since $F\subseteq F[a]\subseteq E$, by hypothesis $F[a]$ is a field, so $\ker v_a\ne\{0\}$ because it is a maximal ideal.
A: You can use the following lemma, which is well known:

Lemma: $\alpha$ is algebraic in $K \Leftrightarrow K[\alpha]$ is a field.
Proof: Consider the map
\begin{align*}
\psi:K[X] &\to K[\alpha]
\\ F &\mapsto F(\alpha)
\end{align*}
which is clearly a surjective homomorfism, so that $K[\alpha] \simeq K[X]/\text{ker}(\psi)$. Since $K[\alpha]$ is a domain, $\text{ker}(\psi)$ must be a prime ideal. Because $K[X]$ is a principal ideal domain, we have $\text{ker}(\psi)=(p)$ (possibly $p=0$). If $p=0$, then $K[\alpha]\simeq K[X]/(0)\simeq K[X]$, which is not a field. If $p\neq 0$, then $(p)$ is a maximal ideal (since it is prime), which means $K[\alpha]\simeq K[X]/(p)$ is a field. In other words, $K[\alpha]$ is a field $\Leftrightarrow p\neq 0$. But clearly $p\neq 0  \Leftrightarrow \alpha$ is algebraic in K. $_\blacksquare$
Corollary: $\alpha_1, ..., \alpha_k$ algebraic in $K$ $\Rightarrow$ $K[\alpha_1, ..., \alpha_k]$ is a field.

Back to your question: 
Suppose every ring $R$ between $F$ and $E$ is a field. If we take $\alpha \in E$, then $F[\alpha]$ is obviously a ring with $F\subset F[\alpha]\subset E$ so it must be a field. By the lemma, $\alpha$ is algebraic in $F$. Since $\alpha\in E$ is arbitrary, $E|F$ is algebraic.
Conversely, supose $E|F$ is algebraic and take any ring $R$ with $F\subset R\subset E$. Consider the set
$$\mathcal{U}:=\bigcup_{r_1, ..., r_k\in R \\\,\,\,\,\,\,k\in\mathbb{N}} F[r_1, ..., r_k]$$
We will prove that $\mathcal{U} = R$. Since every $F[r_1, ..., r_k]$ is a subring of $R$, we get $\mathcal{U}\subset R$. Conversely, if $\alpha \in R$, then $\alpha\in F[\alpha]\subset \mathcal{U}$. Now since $E|F$ is algebraic, by the corollary we have $r_1, ..., r_k\in R\subset E \Rightarrow F[r_1, ..., r_k]$ is a field. So for an arbitrary $\alpha\in \mathcal{U}$, we have $\alpha \in F[r_1, ..., r_k]$, for some $r_1, ..., r_k\in R$, $k\in\mathbb{N}$, which is a field, so $\alpha^{-1}\in F[r_1, ..., r_k]\subset \mathcal{U}$, i.e., $\mathcal{U}$ is closed by inversion. That means $R$ is a ring closed by inversion, i.e., $R$ is a field.
