degree/dimension of field extension if $K=\Bbb Q(\alpha)$ and $F=\Bbb Q$

Question

Say $\alpha\notin\Bbb Q$ but $\alpha^k\in\Bbb Q$ then the minimal polynomial $m_{\alpha,\Bbb Q}$ has degree $k$. Does that mean the dimension of $\Bbb Q(\alpha)/\Bbb Q$ is k? Why?

Answer 1

If the minimum polynomial of $\alpha$ has degree $k$ then $|\Bbb Q(\alpha):\Bbb Q|=k$. But $\alpha\notin Q$ and $\alpha^k\in\Bbb Q$ does not entail $\alpha$'s minimum polynomial having degree $k$. An instructive example is $\alpha^4=-4$.

Answer 2

What you can show is: if $\alpha\in \mathbb{R}$, $\alpha^k\in \mathbb{Q}$, and $\alpha^l\not \in \mathbb{Q}$ for $l<k$, then the degree of $\alpha$ is $k$, that is, the polynomial $X^k - \alpha^k$ is irreducible.

About your other question:

If $f(\alpha)=0$, $f$ monic, and $f$ is of smallest possible degree, then $f$ is irreducible over $\mathbb{Q}$ ( otherwise one of its factors has $\alpha$ as a root).

Let $k$ be the degree of $f$. Let's show that $\{ c_0 + c_1 \alpha + \cdots c_{k-1} \alpha^{k-1}\ | \ c_0, \ldots, c_{k-1}\in \mathbb{Q}\}$ is a field. ( note that the coefficients of each element are uniquely determined). Indeed,let $c_0 + c_1 \alpha + \cdots c_{k-1} \alpha^{k-1}\ne 0$. The polynomial $g=c_0 + c_1 x + \cdots c_{k-1} x^{k-1}$ is not zero, so it is relatively prime to $f$. So there exist ( Euclid algorithm for gcd) $p$, $q$ polynomials such that $p g + q f = 1$ (moreover, we can choose $p$ of degree $<k$) Plug in $x=\alpha$ and get $p(\alpha) g(\alpha)=1$. So we got the inverse of $g(\alpha)$ of the same form.