# 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?

• $[K(\alpha):K] = [K(\alpha):K(\beta)][K(\beta):K]$ – Max Apr 1 '18 at 21:38
• If $K(\beta)/K$ and $K(\alpha)/K(\beta)$ were both algebraic, then $K(\alpha)/K$ would be too. – anomaly Apr 1 '18 at 21:42

## 1 Answer

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 $K$,

$[K(\beta):K] = \infty. \tag{12}$