Timeline for Is the hierarchy of relative geometric constructibility by straightedge and compass a dense order?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 1, 2021 at 0:54 | answer | added | Pace Nielsen | timeline score: 6 | |
| Nov 30, 2021 at 13:37 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
deleted 5 characters in body
|
| Nov 30, 2021 at 13:36 | comment | added | Joel David Hamkins | Yes, I should have said pre-order--now edited. My definition of the strict order is the one appropriate for pre-orders. | |
| Nov 30, 2021 at 13:34 | comment | added | user3840170 | Isn’t ≤ actually a preorder rather than a partial order? The point c = ½a + ½b is compass-and-straightedge constructible from { a, b }, but for a ≠ b, { a, b } and { a, b, c } are hardly equal sets, which means antisymmetry is not satisfied. (Also, { a, b, c } ≤ { a, b } looks a bit funny.) I am sure this can be addressed by constructing a suitable quotient, so this is pretty nitpicky, but still… | |
| Nov 29, 2021 at 23:07 | comment | added | Pace Nielsen | ...of $F_0$, and hence $Gal(K/F_0)$ is $A_4$. Then the argument in my post should apply. | |
| Nov 29, 2021 at 23:06 | comment | added | Pace Nielsen | @QuinnLesquimau First, to make computations slightly cleaner, we can replace the base field $\mathbb{Q}(\alpha^3)$ with $F_0=\mathbb{Q}(\omega,\alpha^3)$, where $\omega$ is a cube-root of unity, since this is a quadratic extension. Now, consider the field $K=F_0(\alpha,t_1=\sqrt{1+\alpha+\alpha^2},t_2=\sqrt{1+\omega \alpha+\omega^2 \alpha^2})$. This is a Galois extension of $F_0$, since the only (seemingly) missing Galois conjugate is $t_3=\sqrt{1+\omega^2\alpha + \omega \alpha^2}=(\alpha^3-1)/(t_1t_2)$. It should have order $12$ over $F_0$, and it should not contain any quadratic extension | |
| Nov 29, 2021 at 16:25 | comment | added | Tourbon Kitsch | @JoelDavidHamkins Exactly. However, the interesting case of finite subsets with algebraic coordinates would still be open. | |
| Nov 29, 2021 at 16:18 | comment | added | Joel David Hamkins | @QuinnLesquimau Good idea! If you are right, it would settle the density question for finite figures negatively. | |
| Nov 29, 2021 at 16:14 | comment | added | Tourbon Kitsch | The next natural candidate for a counterexample would be the quadratic closure of $\mathbb{Q}(\alpha^3)$ and the quadratic closure of $\mathbb{Q}(\alpha)$, with $\alpha$ a transcendental number. I thought that it was obvious that there is nothing between them, because it would mean that there is some element $\beta$ of degree $2^n$ which is expressible with $\alpha$ and $\alpha^3$ and square roots, but not with $\alpha^3$ alone, and then using the minimal polynomial of $\beta$, we would have a contradiction that $\alpha$ is transcendental, but I'm unable to conclude rigorously. | |
| Nov 29, 2021 at 9:57 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 263 characters in body
|
| Nov 29, 2021 at 9:49 | vote | accept | Joel David Hamkins | ||
| Nov 28, 2021 at 22:16 | comment | added | Gerry Myerson | How about $F^{\sqrt{}}$? | |
| Nov 28, 2021 at 19:28 | answer | added | Pace Nielsen | timeline score: 19 | |
| Nov 28, 2021 at 18:43 | history | became hot network question | |||
| Nov 28, 2021 at 18:30 | answer | added | Tourbon Kitsch | timeline score: 10 | |
| Nov 28, 2021 at 17:27 | comment | added | Akiva Weinberger | Incidentally, is there a notation for "the quadratic closure of"? | |
| Nov 28, 2021 at 10:58 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 224 characters in body
|
| Nov 28, 2021 at 10:51 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 67 characters in body
|
| Nov 28, 2021 at 10:42 | history | asked | Joel David Hamkins | CC BY-SA 4.0 |