The Algebraic dependence of two rational functions

Question

How to show that the two rational functions $a(t) , b(t) \in \Bbb K(t) $ when $\operatorname{Char}(\Bbb K) \not =2$ are algebraically dependent.

(there is polynomial $P(x,y) \in \Bbb K[x,y]$ such that $P(a(t),b(t))=0$)?

Answer 1

user45765 already gave an answer.

Let us assume that $a$ is not a constant. We may assume $a(t) = \frac{c(t)}{d(t)}$, where $c(t),d(t)\in K[t]$ and $d(t) \neq 0$. Now $f(x)=c(x) - d(x) a(t)\in K(a(t))[x]$ which is not zero (why?), and $f(t) =0$, this shows that $t$ is algebraic over $K(a)$. Therefore $K(t)$ is an algebraic extension of $K(a)$, in particular, $b$ is algebraic over $K(a)$.