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A real number is algebraically constructible exactly when it lies in a finite tower of real quadratic adjunctions

Statement

A real number $x$ is algebraically constructible if and only if there is a tower

$Q = K_{0} \subseteq K_{1} \subseteq \dots \subseteq K_{r} \subseteq R$

with $x \in K_{r}$ and, for every $i$ , an element $a_{i} \in K_{i - 1}$ with $a_{i} > 0$ such that

$K_{i} = K_{i - 1} (\sqrt{a_{i}}) and [K_{i} : K_{i - 1}] = 2.$

The tower may have length zero.

Facts & Assumptions

Given: The algebraically constructible field $C \subseteq R$ .

[L1]

The field $C$ is the smallest real subfield containing $Q$ and closed under positive square roots (Algebraically constructible real numbers as the smallest real subfield closed under positive square roots).

[L2]

An element algebraic over a field generates a finite simple extension, with degree equal to its minimal-polynomial degree (An element is algebraic over $F$ if and only if its simple extension $F (a) / F$ is finite).

[L3]

Every positive real element has a unique positive square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L4]

The minimal polynomial of an algebraic element divides every polynomial over the base field vanishing at it (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

Proof

technique · direct

1.1

Let $T$ be the union of all terminal fields of finite towers obtained from $Q$ by adjoining positive square roots, omitting any adjunction whose square root is already present. The empty tower puts $Q$ in $T$ .

givenL3

1.2

Conversely, induction along any such tower shows $K_{i} \subseteq C$ : the base $Q$ lies in $C$ , and square-root closure puts each generator and hence its generated field in $C$ . Thus $T \subseteq C$ .

2.1

If $x, y \in T$ , append the generators of their two towers to one another. Any redundant adjunction is omitted; every remaining step has degree $2$ , because its generator $\sqrt{a}$ is a root of $t^{2} - a$ over the base, so by [L4] its minimal polynomial divides $t^{2} - a$ and has degree at most $2$ , while degree $1$ would put $\sqrt{a}$ in the base, which the omission of redundant adjunctions excludes; [L2] then gives $[K_{i - 1} (\sqrt{a}) : K_{i - 1}] = 2$ . Thus $x - y$ , $x y$ , and $x^{- 1}$ for $x \neq 0$ lie in a common terminal field, so $T$ is a subfield.

step 1.1L2L4

3.1

If $0 < a \in T$ , append $\sqrt{a}$ to a tower containing $a$ , or do nothing if it is already present. Hence $T$ is closed under positive square roots. By minimality in [L1], $C \subseteq T$ .

step 1.1step 2.1L1L3

4.1

Therefore $T = C$ . Membership in $T$ is exactly the existence of a displayed finite quadratic tower, including the zero-length case.

step 3.1step 1.2∎

Depends on

Used by

An algebraically constructible real algebraic number has degree over ℚ equal to a power of two Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 52 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

A. W. Knapp, Basic Algebra, 2nd ed., Chapter IX, Section 5 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, Theorem 1.37 through consequence 1.41 (standard reference, not scraped)