If every ring $R$ such that $K \subset R \subset L$ is a field, then $L \supset K$ is an algebraic extension.

Question

I did manage to prove the other implication ($L \supset K$ algebraic $\Rightarrow$ every commutative ring $R$ such that $K \subset R \subset L$ is a field).

I'm stuck on the begining and I can't seem to find any result that helps me on this implication. The only one close to it is:

If $\alpha \in L \supset K$ and if $\Psi: K[x] \rightarrow L$ is defined by $\Psi(f(x)) = f(\alpha)$, then $\Psi$ is a homomorphism such that:

(iii) if $\alpha \in L$ is algebraic over K and $p(x) = \mathrm{irr}(\alpha,K)$, then $N(\Psi) = K[x] \cdot p(x)$ is a maximal ideal of $K[x]$.

I don't think it helps me that much, because of the order of implication.

Can anyone give me a hint on this one?

Answer 1

Let $0\ne\alpha\in L$. Then by assumption $K[\alpha]$ is a field, and so $\alpha^{-1}\in K[\alpha]$. So we can write:

$\alpha^{-1}=c_0+c_1\alpha+...+c_n\alpha^n$

For some $c_0, c_1,..., c_n\in K$. Can you show $\alpha$ is algebraic from here?