3

While doing Galoistheory I came across this question:

Find $\alpha \in \mathbb{C}$ with $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 2^k$ but $\alpha$ inconstructible.

I know $\alpha$ is inconstructible if the Galoisgroup of its minimal polynomial is not of order a power of $2$.

I believe $\zeta_6 \cdot \sqrt{2}$ works, because its minimal polynomial is $f = X^4 + 2X^2 + 4$ and its splitting field $L$ I believe is $\mathbb{Q}(\zeta_6, \sqrt{2})$ with $[L : \mathbb{Q}] = 2\cdot [\mathbb{Q}(\zeta_6): \mathbb{Q}(\sqrt{2})]$ and that makes that degree $6$? Because $[\mathbb{Q}(\zeta_6): \mathbb{Q}(\sqrt{2})] = 3$ with minimal polynomial $X^3 + 1 = 0$.

However, I often make mistakes while doing Galoistheory, so am I doing this correct?

Krijn
  • 1,945
  • 16
  • 32
  • 4
    But $\zeta_6\cdot\sqrt{2}$ manifestly is constructible with a straight edge and a compass. – Olivier Bégassat Jan 20 '15 at 22:16
  • Yes, $\zeta_6$ is constructible, as is $\sqrt{2}$, so their product has to be. Also, that polynomial is clearly solvable with square roots. First solve $Y^2+2Y+4$, then take the square root of $Y$ to get a root. – Thomas Andrews Jan 20 '15 at 22:19
  • Aha yes, I was to stuck in my maths to see that obvious conclusion. But the question of course still remains. – Krijn Jan 20 '15 at 22:21
  • Since there is a solution to the general quartic equation with square roots, you'll need to go to degree $8$, and find one that cannot be solved by radicals. – Thomas Andrews Jan 20 '15 at 22:24
  • @ThomasAndrews: I take a look at the "general solution" and it involves cubic roots. So I think a carefully chosen one might work. – Bombyx mori Jan 20 '15 at 22:39
  • If you can find an irreducible quartic polynomial with Galois group $G = A_4$ or $G = S_4$, then any root of that polynomial will do the trick. Consider the Galois correspondence between subgroups of $G$ and subfields of the Galois closure. $\mathbb{Q}(\alpha)$ corresponds to a subgroup $H$ of $G$ with index four. If $\alpha$ were constructible, there would have to be some subgroup intermediate between $H$ and $G$. But there isn't one. – user208259 Jan 20 '15 at 22:57

1 Answers1

1

After having spoken to my professor about this problem he proposed the following solution:

Pick $f(x) = x^4 + bx + c$ such that $f$ is irreducible and its Galois-group is either $S_4$ or $A_4$ (just as user208259 mentioned in the comments) and then just pick $\alpha$ as $f(\alpha) = 0$. Trying to write out this $\alpha$ is useless as it is often just horrible. To establish the Galois-group you could use the Resolvent Cubic to make that easier.

$f(x) = x^4 + 3x + 1$ should do the trick.

Krijn
  • 1,945
  • 16
  • 32
  • 2
    This works (+1). The reason is that $[\Bbb{Q}(\alpha):\Bbb{Q}]=4$, but there is no intermediate quadratic field that constructibility requires. As suspected, this is better explained using Galois correspondence to translate it into the language of groups. See this question for a bit of discussion and generalizations. – Jyrki Lahtonen Jan 22 '15 at 12:40