Finding the degree of an algebraic field extension

Question

Let $K(\alpha)/K$ be a field extension of degree 4 such that $\alpha^2$ is not a root of the minimal polynomial of $\alpha$ over $K$. Find the degree of $K(\alpha^2)/K$.

So far I've been able to show two very basic things: such degree divides 4 and it is not 1. Hence, it is 2 or 4. Through examples, it seems to be the case that it is 2, but so far I've been unable to prove it. Any thoughts?

Indeed, if the degree was 1, we would have that $\alpha^2\in K$ and the minimal polynomial of $\alpha$ over $K$ would divide $x^2-\alpha^2$ which, in turn, would imply that the degree of the extension $K(\alpha)/K$ would be 1 or 2, in contradiction with the hypothesis.

I suspect I have to assume that the degree is 4 and conclude that $\alpha^2$ is a root of the minimal polynomial of $\alpha$ over $K$, somehow using that, in this case, $K(\alpha)=K(\alpha^2)$ and, in particular, $\alpha\in K(\alpha^2)$.

Answer 1

• Consider first $a=\sqrt[4]2$. Then the minimal polynomial of $a$ over $\Bbb Q$ is $X^4-2$, and $a^2$ is not a root of it. So $\Bbb Q(a):\Bbb Q$ is of degree $4$, and $\Bbb Q(a^2):\Bbb Q$ of degree two.

• We further consider the algebraic number $a=\sqrt 2+\sqrt 3+\sqrt 6$. Using sage we get its minimal polynomial, and the minimal polynomial of its square:

  sage: (sqrt(2)+sqrt(3)+sqrt(6)).minpoly()
  x^4 - 22*x^2 - 48*x - 23
  sage: ( (sqrt(2)+sqrt(3)+sqrt(6))^2 ).minpoly()
  x^4 - 44*x^3 + 438*x^2 - 1292*x + 529
  

so both algebraic numbers generate a field of degree four over $\Bbb Q$. Because of the obvious inclusion, the two fields are equal.