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For $(x, y) \in \mathbb{R}^2$ to be a constructible point, we must necessarily have $[\mathbb{Q}(x, y):\mathbb{Q}] = 2^r.$

According to a book I'm reading on Galois theory, questions about sufficiency are much more difficult. There is a nice exercise in the book to show that $\mathbb{Q} = F_0 \subseteq F_1 \subseteq \cdots \subseteq F_r = \mathbb{Q}(x, y)$ with $[F_{j + 1} : F_j] = 2$ is a sufficient condition for $(x, y) \in \mathbb{R}^2$ to be a constructible point.

But is just $[\mathbb{Q}(x, y):\mathbb{Q}] = 2^r$ a sufficient condition for $(x, y) \in \mathbb{R}^2$ to be a constructible point?

Oscar
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  • There is just a question then about requiring a Galois extension. Because if you have a Galois extension, it seems you can create the subnormal series (with degree 2 extensions) and then the exercise in the book (see above) would show that the point is constructible. What happens if $[\mathbb{Q}(x, y) : \mathbb{Q}] = 2^r$ but $\mathbb{Q}(x, y) : \mathbb{Q}$ is not Galois? Do we still have a sufficient condition? – Oscar Oct 24 '20 at 02:49
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    @Oscar: no, a necessary and sufficient condition is that the Galois closure has degree a power of $2$. There are number fields of degree a power of $2$ whose Galois closure doesn't have a power of $2$: https://math.stackexchange.com/questions/1126788/converse-of-the-theorem-lf-a-real-number-c-is-constructible-then-c-is-alge?rq=1 – Qiaochu Yuan Oct 24 '20 at 03:23
  • @QiaochuYuan Thanks, that link does answer my question with degree 4 counterexamples. The condition that I asked about is not sufficient. – Oscar Oct 24 '20 at 03:49

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