Suppose that $f(u)$ is algebraic over field $F$, $f \in F[x]$, $f \notin F$. Show $u$ is algebraic over $F$. I was just hoping that somebody could check my proof, and perhaps show me an alternate method of proof. Thanks!
Since $f(u)$ is algebraic over $F$, we have that $F \subseteq F(f(u))$ is a finite field extension:
$$[F(f(u)):F]< \infty$$
Now, we have that
$$f(x) - f(u)$$
Is a polynomial living in $F(f(u))[x]$ that has $u$ as a root. Thus $F(f(u)) \subseteq F(u)$ is a finite extension and so $[F(u):F(f(u))] < \infty$
Then, by the tower rule we have that:
$$[F(u):F]=[F(u):F(f(u))][F(f(u)):F] < \infty$$
Since all finite extensions are algebraic, this shows that $u$ is algebraic over $F$.