I think I have totally confused myself and just wanted to check whether the only if in the statement above even holds. Let us look at some field extension $K/L$ and some $\alpha\in L$. We proved in the lecture that if $\alpha$ is algebraic, then $L=K(\alpha)$ as a simple extension generated by an algebraic element will be algebraic. In particular, $K(\alpha)=K[\alpha]$, where $K(\alpha)$ denotes the smallest field containing both $K$ and $\alpha$ and $K[\alpha]$ is the image of the evaluation homomorphism.
But, the other direction should also hold, shouldn't it. We denote $\phi_\alpha: K[X]\rightarrow L, f\mapsto f(\alpha)$ as the evaluation homomorphism. Suppose now the field $L$ is generated as a $K$-algebra of a single element $\alpha\in L$ (and $\notin K$, otherwise the statement would be pointless). Then, by definition of $K-$algebras, $L=K[\alpha]=$im$(\phi_\alpha)$. Since, in general $K[\alpha]\subset K(\alpha)$ but here $K[\alpha]$ is already a field it holds $K[\alpha]=K(\alpha)$.
Now, comes my question. I think that this already suffices to conclude that $L$ must be algebraic. It suffices to show that $\alpha$ is algebraic. My "proof": Since $K[\alpha]$ is a field there is some $f\in K[X]$ with $\deg(f)\geq1$ such that $\alpha^{-1}=\phi_a(f)$. Since $\phi_\alpha$ is a homomorphism, $\phi_\alpha(X*f-1)=\alpha*\alpha^{-1}-1=0$ but, since $K[X]$ is an integral domain $\deg(fX-1)=\deg(X)+\deg(f)\geq 2$. Which means $\ker(\phi_\alpha)$ is non-trivial and, thus, $\alpha$ must be algebraic.
Thank you very much in advance
