$\alpha$ is transcendental and there exists some $\beta$ such that $f(\beta) =\alpha$. Show that $\beta$ is transcendental.

Question

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.

Answer 1

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 $K$.

Answer 2

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 $K$.