Why does inconstructibility of $\sqrt[3]{2}$ imply impossibility of cube doubling? [closed]

In this question "constructing" and "doubling" is meant in the compass-and-straightedge sense.

On my desk I have five Basic Algebra texts treating constructability in the plane $\mathbb{C}$ or $\mathbb{R}^2$ as an application of basic field theory. After appropriate definitions of the possible construction steps, four of these, namely Hornfeck, Jacobson, Lorenzen, and Meyberg, prove that $\sqrt[3]{2}$ is inconstructible starting from $\{0,1\}$ or $\{(0,0),(1,0)\}$, respectively, but then conclude without further justification that the duplication of the cube is impossible.

For a while I believed this last step to be obvious. But now, having to teach this for the first time the day after tomorrow, I have doubts: Being given a cube, say in $\mathbb{R}^3$, should mean being given its eight corners, and then I could use these to do constructions in space, using lines through two points and circles around one point and through two points. Or, to put it differently, I could take any three noncollinear points given or already constructed and do plane constructions in the plane spanned by these. Restricting the constructions initially to one particular coordinate plane containing one face of the cube appears unjustified to me.

My specific questions are:

How does one treat this problem honestly and elegantly, with a minimum of coordinate computations?

Is the problem I see perhaps the reason why the fifth of my books, by M. Artin, does not mention cube doubling?

• I don't know how the proof you talk about works, but for the proof I know doing compass constructions in different planes won't change anything: At each step of the construction, either the field generated by the coordinate of the marked points is unchanged either it became a degree 2 extension of the previous one. SO if you start with point with rational coordinate you can only get point with coordinate in fields of degree 2^n, which excluded 2^(1/3) – Simon Henry Feb 8 '15 at 21:23
• @Pietro Majer: Starting just in one coordinate plane, as the books do, does not allow you to get out of it. – Lutz Mattner Feb 8 '15 at 21:29
• Think how much easier it would be to construct a regular pentagon if one only had to make five equilateral triangles. – Aaron Meyerowitz Feb 8 '15 at 22:12
• I would have thought a "compass and straight edge construction in $\mathbb R^3$" meant that you could draw a straight line between any two known points and you could draw a sphere with center any known point and passing through any other known point. Then you may intersect such drawings, and you "know" any isolated point of intersection. – Theo Johnson-Freyd Feb 8 '15 at 22:41
• Because there is no such phrase as "field extension theory." You can say "using field extensions" if you want, but "field extension theory" does not exist, just like "number field theory" and "Galois group theory" do not exist. The spelling "constructable" looks as awkward to me as "invisable" does. They ought to end in "ible." – KConrad Feb 9 '15 at 13:48

3 Answers

Disclaimer: The following perhaps isn't an answer to your question as stated, so my apologies if this answer is useless to you. However, you're asking for how to treat this problem "honestly", and I think that adding the right kind of historical perspective falls under the heading of honesty.

Anyway, I think it is important to observe here that the ancient Greeks themselves did not limit their solutions to plane constructions. As can be read in Sir Thomas L. Heath's A History of Greek Mathematics, Vol. 1, pp. 246-9, Archytas proposed a solution to the problem where he intersected three surfaces of revolution in Euclidean $3$-space (a cone, a cylinder, and a torus) to obtain a point whose coordinates generate the field extension $\mathbb{Q}(\sqrt[3]{2})$.

It is somewhat misleading, I think, to keep referring to these problems as "the three famous unsolved problems of Greek mathematics", because the Greeks in fact solved them many times over, Archytas' solution being only one out of a multitude. Moreover, they even recognized that the solution could not be achieved by plane methods, in a way: Pappus has it that the Greeks classified construction problems as "plane", "solid", and "linear", according to the methods with which the problem could be solved. Of course, they never tried to make this very precise, let alone tried to prove it, but then they weren't trying to do the impossible either.

• That is definitely helpful, thanks! – Lutz Mattner Feb 9 '15 at 7:28

You can define the problem however you want in your classroom. If you think that three dimensional operations should be used, make a list of which three dimensional operations you think are allowed and work out which field extensions they will give rise to. If there are all quadratic, then the result is still true.

The one time I taught this, I decided modern students had no reason to care about straight edge and compass, and just talked about a pocket calculator with $+$, $-$, $\times$, $\div$ and $\sqrt{\ }$ keys. (Of course, soon, no one will care about this either.)

• This resonates with my comment above. On the other hand, I think straight edge and compass are cool tools even in the modern world, and every student should care about them! – GH from MO Feb 8 '15 at 21:42
• @GHfromMO: I wholeheartedly agree with you that they are cool tools: euclidthegame.com :) – José Figueroa-O'Farrill Feb 8 '15 at 22:31
• At some point, following a course in basic Galois theory, I bought a compass and ruler and started to draw things I could prove are constructible – Lior Bary-Soroker Feb 9 '15 at 6:59

I think it is straightforward to prove by induction that starting from the eight vertices $(\pm 1, \pm 1, \pm 1)$ of a cube in $\mathbb{R}^3$, the constructible points in the OP's extended sense all lie in $K^3$, where $K$ is the union of all Galois extensions of $2$-power degree over $\mathbb{Q}$. It follows that the possible distances determined by the so constructed points also lie in $K$, whence $\sqrt[3]{2}$ is not among them.

In short, the cube cannot be doubled even in the OP's extended sense. On the other hand, it will be hard to verify what the oracle of Delos had in mind.

• Well, but this does not fit well to the story of the oracle of Delos asking for the doubling of a certain physically given cube. – Lutz Mattner Feb 8 '15 at 21:34
• @LutzMattner: A physically given cube determines three distances between their vertices: each of these is constructible from the edge length, so the problem told by the oracle of Delos is equivalent to the one I described. Note that a compass is designed for plane drawings. – GH from MO Feb 8 '15 at 21:37
• I dont' think so. From points just in the $x$-$y$-plane you can not construct any point not belonging to it. – Lutz Mattner Feb 8 '15 at 21:46
• @LutzMattner: See my added section. – GH from MO Feb 8 '15 at 21:58
• Wait, shouldn't $K$ be the union of all Galois extensions of $2$-power degree over $\mathbb{Q}$? I don't think the set of all algebraic numbers of $2$-power degree forms a field (e.g. any two degree-$4$ subextensions of an $S_4$-extension give rise to a degree-$12$ compositum...). – RP_ Feb 8 '15 at 23:43