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.
Best Answer
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.