My hunch is that $\sqrt{2}\not\in\mathbb{Q}(\sqrt[8]{3})$. For practice, I want to compute the splitting field and its degree of $x^8-3$ over $\mathbb{Q}$. I know the roots are $\sqrt[8]{3},\sqrt[8]{3}\omega,\dots,\sqrt[8]{3}\omega^7$ where $\omega=e^{2\pi i/8}=\cos(\pi/4)+i\sin(\pi/4)$.

So I believe the splitting field is $\mathbb{Q}(\sqrt[8]{3},\omega)=\mathbb{Q}(\sqrt[8]{3},\sqrt{2},i)=:K$.

To find the degree, I write $$ [K:\mathbb{Q}]=[K:\mathbb{Q}(\sqrt[8]{3},\sqrt{2})][\mathbb{Q}(\sqrt[8]{3},\sqrt{2}):\mathbb{Q}(\sqrt[8]{3})][\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}]. $$ The first term on the RHS is $2$, since $i\not\in\mathbb{Q}(\sqrt[8]{3},\sqrt{2})$. The third term on the RHS is $8$. The middle term is at most $2$, and I'm fairly sure it is two, so the degree is $32$, if I've reasoned this correctly.

However, I can't prove what the middle term actually is. I want to show $\sqrt{2}\not\in\mathbb{Q}(\sqrt[8]{3})$, to see that it is $2$. I figured I could also switch the order of the extensions, and instead show $[\mathbb{Q}(\sqrt{2},\sqrt[8]{3}):\mathbb{Q}(\sqrt{2})]=8$ by showing $x^8-3$ is irreducible over $\mathbb{Q}(\sqrt{2})$, but this still seems worse, considering the ways the polynomial could factor.

How can I do this right? Thank you.


Here's one approach to the problem:

If $\sqrt{2} \in \mathbb{Q}(\sqrt[8]{3})$, we have $$8 = [\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}(\sqrt{2})]\cdot [\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2 \cdot [\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}(\sqrt{2})],$$
and so $\sqrt[8]{3}$ must satisfy an irreducible polynomial of degree 4 over $\mathbb{Q}(\sqrt{2})$, call it $g(x)$. Then necessarily, $g(x) \mid x^8-3$ over $\mathbb{Q}(\sqrt{2})$. Thus the four roots of $g(x)$ are of the form $\sqrt[8]{3} \zeta_8^i$ for some $i$. In particular, the constant term of $g(x)$ is of the form $\sqrt[2]{3} \zeta_8^j$. But this constant term is in $\mathbb{Q}(\sqrt{2})$, so it must be real, hence $\zeta_8^j = \pm 1$. But then $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$, but I think you can show this is not true pretty easily.

  • $\begingroup$ Thanks Daniel, this makes perfect sense. $\endgroup$
    – yunone
    Nov 12 '11 at 2:49
  • $\begingroup$ @Dane : Why would $g(x)$ have the constant term of the form $\sqrt 3 \zeta_8^j $ ? $\endgroup$
    – Theorem
    Jan 13 '13 at 18:48
  • $\begingroup$ @Theorem The constant term of $g(x)$ is the product of the roots of $g(x)$, all of which are also roots of $x^8-3$ since $g(x) \mid x^8 -3$. Since these roots are of the form $\sqrt[8]{3} \zeta_8^i$, the product of four of them will be $\sqrt{3} \zeta_8^i$ for some $i$. $\endgroup$ Jan 14 '13 at 2:38

Your hunch can be justified and generalized by using some algebraic number theory, specifically the discriminant $D_K$ of a number field $K$. The discriminant has the following two important properties:

  • If $K \subset L$, then $D_K$ divides $D_L$.
  • The primes that divide $D_K$ are precisely the primes $p$ which ramify in $K$.

It is not hard to see on the one hand that $3$ ramifies in $\mathbb{Q}(\sqrt[8]{3})$ and on the other hand that $2$ is the only prime that ramifies in $\mathbb{Q}(\sqrt{2})$, so the discriminant of the former cannot divide the discriminant of the latter.

  • 7
    $\begingroup$ You have the divisibility backwards, e.g., ${\mathbf Q}$ is in every number field and you don't want to say $D_L$ divides 1. But you can avoid discriminants. The prime 3 is totally ramified in ${\mathbf Q}(\sqrt[8]{3})$, so it is totally ramified in every subfield. Thus if $\sqrt{2}$ were in ${\mathbf Q}(\sqrt[8]{3})$ then 3 would be totally ramified in ${\mathbf Q}(\sqrt{2})$, but it isn't: $X^2-2 \bmod 3$ is irreducible, so $(3)$ is prime in ${\mathbf Z}[\sqrt{2}]$. $\endgroup$
    – KCd
    Nov 12 '11 at 3:14

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