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  "body": "\n## Statement\n\nLet $z \\in \\mathbb{C}$ be a [[constructible number|0FMN2MN7TM]].\nThen $z$ is algebraic over $\\mathbb{Q}$, and $[\\mathbb{Q}(z) : \\mathbb{Q}] = 2^r$ for some integer $r \\ge 0$.\n\n## Remarks\n\nThe proof: there is a tower $\\mathbb{Q} = F_0 \\subseteq F_1 \\subseteq \\dots \\subseteq F_n$ with $z \\in F_n$ and $[F_j : F_{j-1}] = 2$ for each $j$. By the [[tower law|0A7KJ68XHT]], $[F_n : \\mathbb{Q}] = 2^n$, and $[\\mathbb{Q}(z) : \\mathbb{Q}]$ divides it.\nSo the [[minimal polynomial|0GK1WB9R98]] of $z$ over $\\mathbb{Q}$ has degree a power of two.\nThe converse is false: some algebraic numbers whose degree is a power of two are not constructible.\nA sufficient condition: the splitting field of the minimal polynomial of $z$ has degree a power of two over $\\mathbb{Q}$.\n\n## Explanations\n\n### L1 human\n\nEach construction step at most doubles the degree of the field.\nSo a number whose minimal polynomial has degree $3$, like $\\sqrt[3]{2}$, can never be constructed.\nThis simple test settles several problems that were open for two thousand years.\n",
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