$K(X)$ is not algebraically closed

Question

I'm working through some basic algebra. I'm struggling with the following question:

Suppose $\beta$ is transcendental over $K$. Prove that $K(\beta)$ is not algebraically closed.

I thought that the Tower law might help, and seek a contraction based on the fact that if transcendental, degree is infinite.

Answer 1

Consider the polynomial $$ f(X)=X^2-\beta $$ A root in $K(\beta)$ should be of the form $$ \frac{P(\beta)}{Q(\beta)} $$ where $P$ and $Q$ are polynomials with coefficients in $K$.

Do you remember the proof that $\sqrt{2}$ is irrational?