I have this exercise where I am having some problems solving it:
Show that for a field extension $L/K$ the following are equivalent:
(a) $L/K$ is algebraic.
(b) Every ring $R$ with $K ⊆ R ⊆ L$ (subrings) is a field.
For $a \Rightarrow b$
Let $L/K$ be algebraic and let R be a subring of L with $K ⊆ R ⊆ L$. We take a $x \in R\backslash \{ 0 \}$. $x$ is algebraic over $K$. This implies that there exists a polynomial $f \in K[X]$ with $f(x)=a_n x^n+a_{n-1}x^{n-1}+...+a_0=0$
So we have that $x^n+a_{n-1}x^{n-1}+...+a_0+1=1$ and therefore $x(x^{n-1}+a_{n-1}x^{n-2}+...+a_1+\frac{a_0+1}{x})=1$ and therefore $x^{-1}=x^{n-1}+a_{n-1}x^{n-2}+...+a_1+\frac{a_0+1}{x}$. From this it follows that every subring is a field. Am I right?
For $a \Leftarrow b$
Since every subring $R$ is a field, for every $x \in R$ we have that $x^{-1} \in R$. Therefore $x^{-1}=a_nx^n+a_{n-1}x^{n-1}...+a_0$. We have that $x(a_nx^n+a_{n-1}x^{n-1}...+a_0)=1$ and $a_nx^{n+1}+a_{n-1}x^{n}...+(a_0-1)x=0$ so we have fund for every $x \in R$ a polynomial $\in K[X]$ where $x$ is a root. So we have that $L/K$ is algebraic.
I am not sure about my results. Can someone check them?