Why is $\beta$ transcendental over $K$? 
Let $\alpha$ transcendental over $K$ and $\displaystyle \beta = \frac{f(\alpha)}{g(\alpha)} \not\in K$ with $f(x),g(x)$ not zero and $\gcd(f,g)=1$. Then $\beta$ is transcendental over $K$.

In the proof of this proposition, the author says:

"...$\alpha$ is algebraic over $K(\beta)$. Therefore, $\beta$ is transcendental over $K$".

Before that, the author just proved that $\alpha$ is algebraic over $K(\beta)$. But I don't understanding this implication. Someone can explain to me?
 A: We are given that $\alpha$ is transcendental over $K$; thus
$[K(\alpha):K] = \infty. \tag 0$
I am assuming that
$f(x), g(x) \in K[x]. \tag 1$
Consider the polynomial
$F(x) = f(x) - \beta g(x); \tag 2$
by (1), since $K \subset K(\beta)$, we have
$F(x) \in K(\beta)[x]; \tag 3$
furthermore, 
$F(\alpha) = f(\alpha) - \beta g(\alpha) = 0, \tag 4$
since
$\beta = \dfrac{f(\alpha)}{g(\alpha)}; \tag 5$
it follows from (2)-(4) that $\alpha$ is algebraic over $K(\beta)$; hence
$[K(\beta)(\alpha):K(\beta)] < \infty; \tag 6$
by (5),  
$\beta \in K(\alpha), \; K(\beta) \subset K(\alpha), \tag 7$
whence
$K(\alpha) \subset K(\alpha)(\beta) = K(\beta)(\alpha) \subset K(\alpha)(\alpha) = K(\alpha); \tag 8$
(6) thus yields
$[K(\alpha):K(\beta)] = [K(\beta)(\alpha):K(\beta)] < \infty; \tag 9$
if now $\beta$ is algebraic over $K$, then
$[K(\beta):K] < \infty, \tag{10}$
and we have
$[K(\alpha):K] = [K(\alpha):K(\beta)][K(\beta):K] < \infty, \tag{11}$
affirming that $\alpha$ is algebraic over $K$, and contradicting (0); therefore $\beta$ must be transcendental over $K$,
$[K(\beta):K] = \infty. \tag{12}$
