Let $K$ be a field and $K(X)$ the field of rational functions with one variable over $K$. If $T(X) \in K(X)$ show that the field extension $K(X)/K(T)$ is finite.
The definition of the field of rational function was:
$$K(X):=\{\frac{f}{g} | f,g \in K[X], g \neq 0\}$$
The field operations for $K(X)$ are:
$$\frac{f_1}{g_1} + \frac{f_1}{g_1} = \frac{f_1 g_2 + f_2 g_1}{g_1 g_2}$$ $$\frac{f_1}{g_1}\cdot\frac{f_1}{g_1} = \frac{f_1 f_2}{g_1 g_2}$$
I have the feeling it has something to do with separability. There is a subfield of $K(T)$ which is isomorph to $\mathbb{Q}$, therefore the charateristic of $K(T)$ is $0$, which means it is a perfect field. Then $K(X)/K(T)$ is seperable.
But I have no idea how to connect this to $[K(X):K(T)]<\infty$. Maybe I have a completely wrong approach. Any help would be appreciated and I'm sorry if this is a duplicate.
Best Answer
Let $f,g\in K[X]$ be such that $T=\frac{f}{g}$, and consider the polynomial $$h:=Tg(Y)-f(Y)\in (K(T))[Y].$$ Note that $h=0$ if and only if $f=g$, in which case $T\in K$ is constant and the extension $K(X)/K(T)$ is in fact not finite.
If $h\neq0$ then the fact that $h(X)=0$ implies that $X$ is algebraic over $K(T)$, and hence that $K(X)$ is a finite extension of $K(T)$.