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.
On
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.
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.