$\alpha$ is transcendental and there exists some $\beta$ such that $f(\beta) =\alpha$. Show that $\beta$ is transcendental. I am starting to study field theory and I encountered this question :
Suppose that $L:K$ is an extension, that $\alpha$ is an element of L which is
transcendental over K, and that $f$ is a non-constant element of
$K[x]$. Show that $f(\alpha)$ is transcendental over $K$. Show that, if $\beta$ is an
element of L which satisfies $f(\beta) =\alpha$, then $\beta$ is transcendental over
$K$.
I have already shown that $f(\alpha)$ is transcendental over $K$ but I'm having trouble showing that $\beta$ is transcendental. Any help would be appreciated.
 A: If $\beta$ would be algebraic over $K$, then $\alpha = f(\beta) \in K(\beta)$ in contradiction with $\alpha$ transcendental over $K$.
Note: remember that any element of a simple extension $K(\gamma)$ of finite degree is algebraic over $K$.
A: Assume $f(\alpha)$ is algebraic over $K$; then there is some
$g(x) \in K[x] \tag 1$
such that
$(g \circ f)(\alpha) = g(f(\alpha)) = 0; \tag 2$
that is, $\alpha$ is a root of the polynomial
$(g \circ f)(x) \in K[x];  \tag 3$
this implies $\alpha$ is algebraic over $K$. Thus, by contraposition, the assumption that $\alpha$ is transcendental over $K$ forces $f(\alpha)$ to be transcendental as well.
Now if
$\beta \in L \tag 4$
is algebraic over $K$, with
$f(\beta) = \alpha, \tag 5$
then
$\alpha \in K(\beta), \tag 6$
and
$K(\alpha) \subseteq K(\beta); \tag 7$
furthermore, $\beta$ algebraic over $K$ implies that
$[K(\beta):K] < \infty, \tag 8$
so in light of (7)
$[K(\alpha):K] < \infty, \tag 9$
as well, which in turn implies $\alpha$ algebraic over $K$.  But $\alpha$ is by hypothesis transcendental over $K$ in contradiction to (9); thus we see that $\beta$ is also transcendental over $K$.
