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Edit: Magdiragdag's response clarifies something along the lines of what I was thinking

The canonical proof I'm referring to is the one shown in Artin Algebra (15.5) or the one found in these notes: http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Gao.pdf

Here is the part where I'm encountering trouble:

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These rules seem really restrictive to me, because as far as I can tell, they don't allow for the construction of arbitrary points satisfying certain properties. (see edit: clarification in bold letters)

It is often the case that euclidean construction problems often involve constructing arbitrary points, lines, and circles. (I don't have an example currently at hand but many Euclidea problems can only be done by constructing an arbitrary object)

I thought that the construction of arbitrary points would eventually be addressed in the ongoing proof, but it never happened. All constructions strictly adhered to the rules shown above. And when the proof ended, I felt thoroughly betrayed because the proof completely ignored the case where you can construct arbitrary points as you wish. Isn't this what is naturally done in practice?

My question: is there a way to formalize the concept of "constructing an arbitrary point satisfying some parameters" and integrate it into the canonical proof?

Edit (clarification): the rule set above provides a deterministic existence-consequence "chain reaction" which is how we end up constructing the field extensions, but how can we address the case where we simply add an arbitrary point into the picture? Distinguished from constructed points, arbitrary points needn't always be constructed points.

I want to show that with the added freedom of being allowed to construct arbitrary points, lines, and circles, that trisecting an angle is still deterministically impossible (i.e. there is no general method that will guarantee successful trisection each time the method is repeated).

I have some ideas of what this entails, but I can't combine them into one coherent proof. Here are some facts that can be deduced from the rules of ruler and compass construction above, given (0,0) and (1,0) are constructed points:

  1. given the coordinates of some arbitrary point A, one can construct a point B that is arbitrarily close to A using the rules above

  2. Moral fact: Let S be the set of the points on the x-y space satisfying a set of properties P (e.g. position relative to a line, etc.). If we can deterministically construct an arbitrary point such that we be can certain that it belongs to S, then S must be infinite. (e.g. Let S be the set of points above the constructed line y=x. Then we can construct an arbitrary point such that we are sure it belongs to S).

  3. Another moral fact: The only properties that S can take are boolean: (e.g. above or below the line L, inside or outside the circle C, lies on or lies off line T)

I feel like a proof sketch would look something like this:

  1. We wish to construct an arbitrary point in S satisfying the list of properties P.

  2. We somehow prove that there exists a point in S with an open neighborhood. Thus, S must contain a constructed point from fact 1.

  3. Suppose a method to trisect an angle exists and it involves constructing arbitrary points. Whenever we get to such a step where arbitrary point construction is required, there is always the chance we end up constructing a constructed point. In the case where we end up constructing a constructed point at every step involving arbitrary point construction, the proof becomes equivalent to the original canonical proof provided in Artin and the pdf.

  4. Thus trisecting an angle while allowing arbitrary point construction is still deterministically impossible as was the case in the canonical proof.

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  • $\begingroup$ Is your issuethat only two points are allowed? Because it does not make a difference if those points are at 0,0 and 1,1, or at a,b c,d; they are just two points in a plane. If you allow more than two, the excess points will have some relation to the first two, specializing any proof reliant on them $\endgroup$ – bukwyrm Jul 31 '18 at 14:33
  • $\begingroup$ that isn't my issue. my issue is that constructing an "arbitrary point" is strictly prohibited by the rule set above. (e.g. i make a random point in the first quadrant) $\endgroup$ – user398748 Jul 31 '18 at 14:34
  • $\begingroup$ the rule set above provides a deterministic existence-consequence "chain reaction" which is how we end up constructing the field extensions, but my question is, how can we address the case where we simply add an arbitrary point into the picture? $\endgroup$ – user398748 Jul 31 '18 at 14:36
  • $\begingroup$ if we were to allow more than two points (let's say at the beginning we start off with all the arbitrary points we'll end up using throughout the rest of the construction). if we start off with n points, not all of them need to be in Q, so they'll create some finite extension of Q, whose degree over Q isn't necessarily a power of 2. But the proof that 60 degrees cannot be trisected specifically relies on the fact that x^3-3x-1 has a non power of 2 degree over the original field (in this case Q). How do you guarantee that this holds if we start with a finite extension fields of Q instead of Q? $\endgroup$ – user398748 Jul 31 '18 at 14:53
  • $\begingroup$ @Barycentric_Bash I don't know what you're asking, but adding to the list of allowable constructions essentially increases the number of points that you can build, and you'll get more than the constructible numbers. Of course if you can construct arbitrary real numbers, you can plot anything in $\mathbb R\times \mathbb R$, which is not the case with ordinary straightedge-compass construction. $\endgroup$ – rschwieb Jul 31 '18 at 15:11
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Courant and Robbins in "What Is Mathematics?" chap 3 on geometrical constructions and field extensions handle this very simply. They simply say whenever arbitrary points or radii are chosen, use rational coefficients, so the new item is already in the lowest subfield, Q without any square root extensions.

"In such cases we may choose the element to be rational; i.e. arbitrary points may be chosen with rational coordinates x, y, arbitrary lines ax + by + c = 0 with rational coefficients a, b, c, arbitrary circles with centers having rational coördinates and rational radii. ... if the elements are indeed arbitrary this restriction cannot affect the result of a construction." (p. 127)

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You could add:

  • Given all previously constructed points, lines, and circles, construct a generic point $x$ that has no relation to those previously constructed points, lines, and circles.

(Note, this doesn't allow for the notion of a point close to a line, or inside a circle, or to the left/right of a line. Just a point that you know nothing about.)

Translating this to field extensions:

  • You have an extension field $F$ of ${\mathbb Q}$ that is generated by all previously constructed points. Now change to the field extension $F(X)$ of ${\mathbb Q}$. This is generated by the previously constructed points and the new generic point $X$.

Of course, the field extensions you get this way are not algebraic anymore and the degree argument to show that you cannot trisect an angle doesn't work anymore. I'd guess splitting the ultimate extension in a purely transcendental part and an algebraic part with the Noether Normalization Lemma helps to recover the argument.

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  • $\begingroup$ does "a point you know nothing about" mean that it whether or not X has relationships with the previously constructed objects doesn't matter, or does it mean that there exist NO relationships between X and the previously constructed objects? $\endgroup$ – user398748 Jul 31 '18 at 15:13
  • $\begingroup$ I mean "there is no relationship between $x$ and the previously constructed things". $\endgroup$ – Magdiragdag Jul 31 '18 at 15:16
  • $\begingroup$ why not $F(X,Y)$ ? $\endgroup$ – mercio Jul 31 '18 at 22:39
  • $\begingroup$ @mercio What do you mean? If you wanted to add two generic points, you'd get $F(X,Y)$. $\endgroup$ – Magdiragdag Jul 31 '18 at 22:42
  • $\begingroup$ because a point has two coordinates ? $\endgroup$ – mercio Aug 2 '18 at 5:09

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