Let $k$ be a field. $$k(t) = \{ f(t)/g(t) \mid f(x), g(x) \in k[x], g(x)\neq0 \}$$ be an extension over $k$. I wanna prove that, for any $m(t)$ in $k(t)\setminus k$, $m(t)$ is transcendental in $k$, i.e. If there exist an $f(x)$ in $k[x]$ so that $$f(m(t)) = 0$$ holds in $k(t)$, then $f$ must be 0 polynomial, namely $$f(x) = 0.$$
I think $k(t)$ over $k$ is a very important example for field extension. Write $$m(t) = g(t)/h(t).$$ I think it ok to add that $$gcd(g, h) = 1,$$ and $h$ is monic. Then I thought it difficult to conclude from $$f(g(t)/h(t)) = 0$$ that $$f = 0.$$ So I am wondering if there is some good approach. Thank you!
Suppose $f(m(t))=0$ where $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. Then, \begin{align} 0&=h(t)^nf(m(t))=h(t)^nf(g(t)/h(t))\\ 0&=h(t)^n\left(a_n\frac{g(t)^n}{h(t)^n}+a_{n-1}\frac{g(t)^{n-1}}{h(t)^{n-1}}+\cdots+a_1\frac{g(t)}{h(t)}+a_0\right)\\ 0&=a_ng(t)^n+a_{n-1}g(t)^{n-1}h(t)+\cdots+a_1g(t)h(t)^{n-1}+a_0h(t)^n \end{align} Therefore, $t$ is algebraic.