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theorem on constructible numbers∗
Wkbj79†
2013-03-21 23:16:39
Theorem 1. Let F be the field of constructible numbers and α ∈ F. Then there
exists a nonnegative integer k such that [Q(α) : Q] = 2k .
Before proving this theorem, some preliminaries must be addressed.
First of all, within this entry, the following nonconventional definition will
be used:
Let S be a subset of C that contains a nonzero complex number and α ∈ C.
Then α is immediately constructible from S if any of the following hold:
• α = a + b for some a, b ∈ S;
• α = a − b for some a, b ∈ S;
• α = ab for some a, b ∈ S;
• α = a/b for some a, b ∈ S with b 6= 0;
p
iθ
• α = |z|e 2 for some z ∈ S with z 6= 0 and θ = arg(z) with 0 ≤ θ < 2π.
The following lemmas are clear from this definition:
Lemma 1. Let S be a subset of C that contains a nonzero complex number
and α ∈ C. Then α is constructible from S if and only if there exists a finite
sequence α1 , . . . , αn ∈ C such that α1 is immediately constructible from S, α2 is
immediately constructible from S∪{α1 }, . . . , and α is immediately constructible
from S ∪ {α1 , . . . , αn }.
Lemma 2. Let F be a subfield of C and α ∈ C. If α is immediately constructible
from F , then either [F (α) : F ] = 1 or [F (α) : F ] = 2.
Now to prove the theorem.
∗ hTheoremOnConstructibleNumbersi
created: h2013-03-21i by: hWkbj79i version:
h39614i Privacy setting: h1i hTheoremi h12D15i
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Proof. By the first lemma, there exists a finite sequence α1 , . . . , αn ∈ C such
that α1 is immediately constructible from Q, α2 is immediately constructible
from Q ∪ {α1 }, . . . , and α is immediately constructible from Q ∪ {α1 , . . . , αn }.
Thus, α2 is immediately constructible from Q(α1 ), . . . , and α is immediately constructible from Q(α1 , . . . , αn ). By the second lemma, [Q(α1 ) : Q] is
equal to either 1 or 2, [Q(α1 , α2 ) : Q(α1 )] is equal to either 1 or 2, . . . , and
[Q(α1 , . . . , αn , α) : Q(α1 , . . . , αn )] is equal to either 1 or 2. Therefore, there
exists a nonnegative integer m such that [Q(α1 , . . . , αn , α) : Q] = 2m . Since
Q ⊆ Q(α) ⊆ Q(α1 , . . . , αn , α), it follows that there exists a nonnegative integer
k such that [Q(α) : Q] = 2k .
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