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.

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