if $\alpha$ is trascendental over K then $f(\alpha)$ is also trascendental over K?

Question

Consider the field extension $(L:K)$ and $\alpha \in L$ trascendental over K. Let $f(x) \in K[x]$ a not-constant polynomial, then $f(\alpha)$ is also trascendental over $K$.
I was thinking of supposing that $f(\alpha)$ is algebraic, then $\exists g(x)\in K[x]$ s.t. $g(f(\alpha))=0$, then , can i compose to say $(g\circ f)(\alpha)=0\equiv h(\alpha)=0$ meeting a contradiction?

Answer 1

That is exactly what you can do. You just have to show that $h$ is not constant, and that it is a polynomial, and you're done.

Explicitly, you've shown $f(\alpha)$ algebraic $\implies$ $\alpha$ algebraic. No contradictions or anything here. Just the contrapositive of the statement you were originally asked to solve. Since contrapositives are equivalent, you're done.