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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, $\alpha\in K(\alpha^2)$.

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  • $\begingroup$ Both $2$ and $4$ can occur. An example of the former is $\alpha=\root4\of2$. In that case $\alpha^2$ is clearly of degree two. An example of the latter is the only positive real zero $\alpha$ of the irreducible quartic $x^4+x-1$ In that case $\alpha^2$ cannot be a root of quadratic for then we would have two linearly independent quartics with $\alpha$ as a zero, and that is absurd. $\endgroup$
    – Jyrki Lahtonen
    Nov 7, 2018 at 22:13
  • $\begingroup$ Mind you, the Galois group of $x^4+x-1$ is full $S_4$. This implies that there are no intermediate fields between $\Bbb{Q}(\alpha)$ and $\Bbb{Q}$ making it rather obvious that $\Bbb{Q}(\alpha)=\Bbb{Q}(\alpha^2)$. Bringing this up because the situation in my example is markedly different from Dan Fulea's (+1) example (where the said field has several quadratic intermediate fields). $\endgroup$
    – Jyrki Lahtonen
    Nov 7, 2018 at 22:19
  • $\begingroup$ What you can say is that for $[K(\alpha):K] = 4$, or $(\alpha^2+c)^2 \in K$ for some $c \in K$ or $K(\alpha) = K(\alpha^2)$. Some degree $4$ extensions don't have sub-quadratic field. $\endgroup$
    – reuns
    Nov 7, 2018 at 22:32

1 Answer 1

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  • 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.

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  • $\begingroup$ Thank you for your answer $\endgroup$
    – Ray Bern
    Nov 8, 2018 at 7:13
  • $\begingroup$ Just curious: how did you come up with this example? $\endgroup$
    – Ray Bern
    Nov 8, 2018 at 7:38
  • $\begingroup$ First it was rather clear for me that $\Bbb Q$ should work as ground field, and that i have to produce an algebraic number having a square that is not in the shape $\frac 12(-b\pm\sqrt\Delta)$ (times rational). So i wanted to insure at least two radicals, "different in their nature". First example was to use $\sqrt {-1}$ and $\sqrt 2$, as a minimal example, but then i was rejecting the $-1$ to keep as far as possible from something that tastes like a cyclotomic root. Of course, $\sqrt 2+\sqrt 3$ is not working, so one has to take also some "other part". For instance $1$ or $\sqrt 6$... $\endgroup$
    – dan_fulea
    Nov 9, 2018 at 20:12

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