Show that $α^2$ is algebraic over $K$

Question

Let $K ⊂ L$ be an extension. Let $α ∈ L$ algebraic over $K$. Show that $α^2$ is algebraic over $K$.

Let $n$ be the degree of the minimal polynomial $p(x)$ of $α$ over $L$ and $m$ be the degree of the minimal polynomial $q(x)$ of $α^2$ over $L$.

Since $α^2 ∈ L(α)$, we have $L(α^2) ⊂ L(α)$, then $m\leq n$.

I have trouble understanding the last statement.

How do I know that $α^2 ∈ L(α)$ and why is $L(α^2) ⊂ L(α)$

Answer 1

Since

$\alpha \in L \tag 1$

is algebraic over $K$,

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

and since

$\alpha^2 \in K(\alpha), \tag 3$

$K(\alpha^2) \subset K(\alpha), \tag 4$

and thus

$[K(\alpha^2): K] \le [K(\alpha): K] < \infty, \tag 5$

which in turn implies $\alpha^2$ is algebraic over $K$.

Note Bene, 8 September 2020 10:20 AM PST: The above demonstration is easily extended to show that $\alpha^n$ is algebraic over $K$ for all $n \in \Bbb N$; one merely need observe that (3) may be replaced with

$\alpha^n \in K(\alpha), \tag 6$

whence

$K(\alpha^n) \subset K(\alpha), \tag 7$

from which

$[K(\alpha^n): K] \le [K(\alpha): K] < \infty, \tag 8$

that is, $\alpha^n$ is algebraic over $K$. End of Note.