Let $\mathbb{K}$ be a field. Show that the elements of $\mathbb{K}(T)$ that are algebraic on $\mathbb{K}$ are the elements of $\mathbb{K}$.
Let $f = \frac{P[T]}{Q[T]}$ an algebraic element of degree $n$ of $\mathbb{K}(X)$. Alors $a_0 + a_1 f + \dots + a_{n-1}f^{n-1} + f^n = 0$ where $a_i \in \mathbb{K}$. Thus $$a_0 Q^n + a_1 Q^{n-1}P + a_2 Q^{n-2}P^2 + \dots + a_{n-1}QP^{n-1} + P^n = 0 .$$
I've read that one can conclude using the fact that $Q$ divides $a_0 Q^n + a_1 Q^{n-1}P + a_2 Q^{n-2}P^2 + \dots + a_{n-1}QP^{n-1}$, but divides not $P^n$. But I don't see why this implies that $n=1$ for that $f \in \mathbb{K}$.
Let us assume that $f = P/Q$ is such that $P$ and $Q$ do not have common divisors. Now $$ P^n = Q \cdot (a_0 Q^{n-1} + \cdots) $$ tells us that every prime divisor of $Q$ is one of $P$. That is only possible if $Q$ does not have any, that is $Q\in K$. So $f$ is a polynomial. Looking at the degree tells you, that $f \in K$.