## Bounds on the degrees of some algebraic elements

Let $\alpha$ be an algebraic element of degree $n$ over $\mathbb{Q}$. Show that $\alpha^2$ is algebraic over degree at most $n$ over $\mathbb{Q}$, and that $\sqrt{\alpha}$ is algebraic of degree at most $2n$ over $\mathbb{Q}$. In each case, show by an example that both “<” and “=” are possible.

By Theorem 4.9 in TAN, since $\alpha^2 \in \mathbb{Q}(\alpha)$ (clearly) the degree of $\alpha^2$ is at most $n$. Now suppose the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $p(x)$; then certainly $\sqrt{\alpha}$ is a root of $p(x) \circ x^2 \in \mathbb{Q}[x]$, which has degree $2n$. Since the minimal polynomial of $\sqrt{\alpha}$ divides $p(x) \circ x^2$, the degree of $\sqrt{\alpha}$ is at most $2n$.

Let $\alpha = 2$. Since $\alpha \in \mathbb{Q}$, $\alpha$ is algebraic over $\mathbb{Q}$ of degree 1. $\alpha^2 = 4$ is also algebraic of degree 1, so that $\mathsf{deg}_\mathbb{Q} \alpha = \mathsf{deg}_\mathbb{Q} \alpha^2$. However, $\sqrt{\alpha} = \sqrt{2}$ has degree 2, so that $\mathsf{deg}_\mathbb{Q} \alpha < \mathsf{deg}_\mathbb{Q} \sqrt{\alpha}$.

On the other hand, if $\beta = 4$, then $\sqrt{\beta} = 2$, and $\mathsf{deg}_\mathbb{Q} \beta = \mathsf{deg}_\mathbb{Q} \sqrt{\beta}$.

Now consider $\gamma = i$; this element is algebraic of degree 2, while $\gamma^2$ has degree 1. So $\mathsf{deg}_\mathbb{Q} \gamma > \mathsf{deg}_\mathbb{Q} \gamma^2$.