Let $k \subset F$ be a field extension, and let $\alpha \in F$. Prove that the field $k(\alpha)$ consists of all the elements of $F$ which may be written as rational functions in $\alpha$, with coefficients in $k$.
I think that if $\alpha$ is transcendental, then $k(\alpha) \cong k(t)$ via $\alpha \leftrightarrow t$.
If $\alpha$ is algebraic, and $p(x)$ is it minimal polynomial, then $k(\alpha) \cong k[x]/(p(x))$. However, I don't see how to write elements of $k(\alpha)$ as rational functions in $\alpha$.
The isomorphism $k(\alpha) \cong k[x]/(p(x))$ comes from $k[x] \to k(\alpha)$ sending $x \mapsto \alpha$. This factors through $k[x]/p(x)$ giving the isomorphism.
So, it seems every element of $k(\alpha)$ can be written as $f(\alpha)$, which is technically a rational function in $\alpha$.
However, I believe the point of the exercise is to show that an element of $k(\alpha)$ is $f(\alpha)/g(\alpha)$.
Is there more to this exercise?
Hint: don't worry about whether $\alpha$ is algebraic over $k$. $k(\alpha)$ is the intersection of all subfields of $F$ that contain $k$ and $\alpha$. Now prove (1) that any subfield of $F$ that contains $k$ and $\alpha$ also contains any (well-defined) rational function $f(\alpha)/g(\alpha)$ with all coefficients in $k$ and (2) that the set of all such well-defined rational functions of $\alpha$ is a subfield of $F$. ("Well-defined" here just means that $g(\alpha) \neq 0$.)