Suppose that $K$ is a finite extension of $Q$ and that $\alpha$ is algebraic over $K$. Show that $\alpha$ is algebraic over $Q$.

Question

Suppose that $K$ is a finite extension of $Q$ and that $\alpha$ is algebraic over $K$. Show that $\alpha$ is algebraic over $Q$.

The hint is to use the Tower Law. However, I don't see where I can use the Tower Law.

Answer 1

We have the tower of fields

$\Bbb Q \subset K \subset K(\alpha), \tag 1$

with both

$[K(\alpha):K], [K:\Bbb Q] < \infty; \tag 2$

thus the tower law yields

$[K(\alpha):\Bbb Q] = [K(\alpha):K][K:\Bbb Q] < \infty; \tag 3$

since

$\alpha \in K(\alpha), \tag 4$

it follows then that $1$ and the first $n = [K(\alpha):\Bbb Q]$ powers of $\alpha$,

$1, \alpha, \alpha^2, \ldots, \alpha^n \tag 5$

are linearly dependent over $\Bbb Q$; thus we have some

$q(x) \in \Bbb Q[x], \deg q \le n, \tag 6$

with

$q(\alpha) = 0, \tag 7$

the criterion for $\alpha$ to be algebraic over $\Bbb Q$.