Is the hierarchy of relative geometric constructibility by straightedge and compass a dense order?

Question

Consider the hierarchy of relative geometric constructibility by straightedge and compass. Namely, given a geometric figure $B$, a set of points in the plane, we define that geometric figure $A$ is constructible from $B$, written as $$A\leq B,$$ if from points in $B$ using straightedge and compass we may construct every point in $A$.

This is a partial preorder on geometric figures, considered as sets of points in the Euclidean plane. The order gives rise to a corresponding strict order notion $A<B$, which holds when from $B$ we may construct $A$ but not conversely. That is, $$A<B\qquad\text{ if and only if }\qquad A\leq B\quad\text{ and }\quad B\not\leq A.$$

Question. Is the hierarchy of relative geometric constructibility a dense order? That is, if $A<B$ for figures in the plane, is there a figure $C$ such that $A<C<B$?

I would also be interested to know whether the answer depends on considering only finite sets of points or infinite sets.

[Update. The main question is now answered by the maximality argument of Pace Nielsen below, but remains open in the case of finite figures. That is, if $A$ and $B$ are finite geometric figures with $A<B$, must there be some figure $C$ strictly between?]

Let me make a few observations. First, relative constructibility is symmetric on line segments. That is, if a line segment $AB$ is constructible from $CD$, then $CD$ is constructible from $AB$. Thus, the relation of relative constructibility on line segments is an equivalence relation, with no occurences of the strict order. The argument is given on my blog post. For this reason, there will be no counterexamples to density using segments only.

The symmetry claim is not true, however, for triangles. For example, from a unit length and $\sqrt[3]{2}$, we can construct a unit isosceles right triangle, but from such a triangle, we cannot construct the length $\sqrt[3]{2}$. This instance is also discussed on my blog post.

Since this case has the feeling of being perhaps a minimal extension, it might be considered as a natural candidate counterexample, and several people (including David Madore and Nikolay Nikolov) have proposed various arguments that make definite progress. But the matter is currently still open.

A constructively closed set $S$ of points in the plane is determined by any two of them and the set of points on the corresponding line $\ell$. The reason is that if point $p\in S$, then the distances between $p$ and two points on $S\cap\ell$ will be a distance witnessed on $S\cap\ell$, and so we can construct $p$ via circles from points on $S\cap\ell$.

Therefore, a constructively closed set of points in the plane is determined by a unit segment and the real field of distances realized by that set. (one can also simply view the points as complex numbers, as determined by a unit segment from the set.) For this reason, the question is equivalent to the following:

Question. If $K< L$ are quadratically closed field extensions of the rationals, must there be a quadratically closed field $F$ strictly between them? $$K< F< L$$ Such a field would correspond to a strictly intermediate set of points between $K$ and $L$ in the hierarchy of relative constructibility.

Regarding the natural candidate counterexample, the question would be: is there a quadratically closed field strictly between the quadratic closure of $\mathbb{Q}$ and the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$?

Answer 1

There are three answers. Throughout let $qcl(F)$ be the quadratic closure of a field $F$ inside $\mathbb{C}$.

Part 1: Yes there is a quadratically closed field strictly between $qcl(\mathbb{Q})$ and $qcl(\mathbb{Q}(2^{1/3}))$. First, find an $S_4$-extension $K/\mathbb{Q}$ containing $\mathbb{Q}(2^{1/3})$. (My friend Darrin Doud tells me that the Galois closure of adjoining a root of $x^6 + 3x^4 + 3x^2 + 3$ will be such an $S_4$ extension.) Note that $K$ is a subfield of $qcl(\mathbb{Q}(2^{1/3}))$, being constructible from $2^{1/3}$ by a sequence of square roots.

Now, $K$ has a unique subfield $\mathbb{Q}(\sqrt{-3})$ quadratic over $\mathbb{Q}$. Then $K/\mathbb{Q}(\sqrt{-3})$ is an $A_4$-extension, and hence there exists a non-Galois subfield $F$ with $[F:\mathbb{Q}(\sqrt{-3})]=4$. Note, in particular, that the Galois closure of $F$ requires a degree $3$ extension. We know that $F$ is not contained in $qcl(\mathbb{Q})$ because $qcl(\mathbb{Q})$ is Galois but contains no cubic extension of $\mathbb{Q}$. Also, $qcl(F)$ is proper in $qcl(\mathbb{Q}(2^{1/3}))$, since $qcl(F)$ is formed only using power-of-2 extensions of $\mathbb{Q}$.

Part 2: The order is not dense. Just use Zorn's lemma (or transfinite induction, if you want to avoid the axiom of choice) to construct a quadratically closed subfield of $qcl(\mathbb{Q}(2^{1/3}))$ that is maximal with respect to not containing $2^{1/3}$.

Part 3: If by "geometrical figure" you mean a finite set of points, then I don't know the answer to the first question anymore. (Edited to add: I've now also answered this question in the negative. I've put it as a separate answer.)

Answer 2

There are a lot of computations in this answer, so I hope there is no mistake. Answer to the small last question: there is a quadratically closed field strictly between the quadratic closure of $\mathbb{Q}$ and the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$. Let $K$ be the quadratic closure of $\mathbb{Q}$, and $L$ be the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$. The roots of the polynomial $P(x) := x^4 + 36x + 54$ are not in $K$, but they are in $L$, and $P$ is irreducible because it is irreducible modulo $5$, hence any root $\alpha$ of it is of degree $4$, and the quadratic closure of $\mathbb{Q}(\alpha)$ is strictly between $K$ and $L$. (I used https://www.alpertron.com.ar/POLFACT.HTM for the calculations modulo 5.)

First, why are the roots of $P$ not in $K$? It is because mod $7$, it decomposes in $(x+6)(x^3+x^2+x+2)$, which are irreducible, and the discriminant of this polynomial is $-5038848$, which is not divisible by $7$, therefore by Dedekind's theorem there is a cycle of order $3$ in the Galois group of $P$ (Again, calculation modulo 7 with https://www.alpertron.com.ar/POLFACT.HTM, and calculation of the discriminant with https://planetcalc.com/8188/.)

Finally, it is in the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$, because we can look at the explicit formula for the roots of a quartic:

In our case, $a = b = 0$, therefore $e = \sqrt{(27c^2)^2-4(12d)^3} = 11664$, and $f = \sqrt[3]{27c^2 + e} = \sqrt[3]{2^6 \cdot 3^6} = 2^2 \cdot 3^2$. Except that, only square roots and $\sqrt[3]{2}$ appear in the formula, thus the roots are in $L$.

Answer 3

I've figured out how to do the finite graph case, in the negative. The ordering of quadratically closed fields is not dense in that case either. Throughout let $qcl(F)$ denote the quadratic closure of $F$.

Let $\alpha$ be such that $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$ and there is no intermediate field. (For instance, take any $S_4$ or $A_4$ extension of $\mathbb{Q}$, and take a fixed field for any index $4$ subgroup.) We know then that $qcl(\mathbb{Q}(\alpha))$ properly contains $qcl(\mathbb{Q})$. We will show there is no intermediate quadratically closed field.

Let $\beta\in qcl(\mathbb{Q}(\alpha))$. Let $K$ be the Galois closure of $\mathbb{Q}(\alpha,\beta)$ over $\mathbb{Q}(\alpha)$. Since $\beta$ can be captured by a sequence of quadratic extensions, we know that $K/\mathbb{Q}(\alpha)$ is also created by a sequence of quadratic extensions, and in particular $[K:\mathbb{Q}]$ is a power of $2$.

Assume $qcl(\mathbb{Q}(\beta))\subsetneq qcl(\mathbb{Q}(\alpha))$. Then $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\beta)]=4$, since $\alpha$ is not constructible from $\beta$. Therefore, $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]=[\mathbb{Q}(\beta):\mathbb{Q}]$. In particular, if $f$ is the minimal polynomial of $\beta$ over $\mathbb{Q}$, it remains irreducible over $\mathbb{Q}(\alpha)$. Therefore, all the roots of $f$ lie in $K$. Hence, the Galois group of $f$ (i.e., the Galois closure of $\mathbb{Q}(\beta)$ over $\mathbb{Q}$) is a $2$-group. Therefore, $\beta\in qcl(\mathbb{Q})$.