Let k ⊆ k(α) be a simple extension, with α transcendental over k. Let E be a subfield of k(α) properly containing k. Prove that k(α) is a finite extension of E.
This is a question from the book "Chapter 0" by Paolo Aluffi. I am confused on where to begin, I have an inkling that it has something to do with containment and how we can write all elements of k(α) as some f(α)/g(α). Any help or hints are appreciated.
Since $E$ is a subfield of $\mathbb{K}(\alpha)$ properly containing $\mathbb{K}$ it exists $\beta\in E\setminus\mathbb{K}$. As you stated since $\mathbb{K}(\alpha)$ is a simple and trascendent extension of $\mathbb{K}$, you should know that $\mathbb{K}(\alpha)$ is isomorphic to the field of rational function over $\mathbb{K}$. So since $\beta\in \mathbb{K}(\alpha)$, you know that it exists $f,g\in \mathbb{K}[x]$, with $g$ non zero such that $$\beta=\frac{f(\alpha)}{g(\alpha)}.$$ Take $F(x)=f(x)-\beta g(x)$, you know that $F(x)\in E[x]$ and $F(\alpha)=0$, so $\alpha$ is algebraic over $E$. So you can conclude that $E(\alpha)=\mathbb{K}(\alpha)$ (why?), so you have done, since $$|\mathbb{K}(\alpha)\,\colon\, E|=|E(\alpha)\,\colon\, E|.$$ Clearly the most important fact is that $E$ contains properly $\mathbb{K}$, since if $E=\mathbb{K}$ the extension would have been simple and trascendental and we cannot say a priori that the extension is finite.
