Wikipedia:Reference desk/Archives/Mathematics/2021 September 9
| Mathematics desk | ||
|---|---|---|
| < September 8 | << Aug | September | Oct >> | Current desk > |
| Welcome to the Wikipedia Mathematics Reference Desk Archives |
|---|
| The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
September 9
[edit]Field extensions of the rationals by closure under arithmetic antidifferentiation
[edit]Certain integers, such as 2 and 3, have no arithmetic antiderivative.(there is no rational number whose arithmetic derivative is 2 or 3, for example). What is the structure of the minimal differentially closed field under the arithmetic derivative containing the rationals? It seems hard to describe, but I think the Galois group of such an extension is a boolean group since, e.g. if we adjoin a formal symbol x to the rationals with arithmetic derivative 2, then the only nontrivial member of the Galois group is sending x to -x. This is very rough and handwavy as I don't know of a good rigorous formulation akin to using quotients of polynomial rings to study algebraic field extensions..--Jasper Deng (talk) 06:05, 9 September 2021 (UTC)
- Note: In the article on differentially closed fields, we have "Shelah also showed that the prime differentially closed field of characteristic 0 (the differential closure of the rationals) is not minimal; this was a rather surprising result, as it is not what one would expect by analogy with algebraically closed fields. " but this statement is unclear because the derivation is not specified.--Jasper Deng (talk) 06:08, 9 September 2021 (UTC)
- Actually, I cannot determine the Galois group in analogy to the algebraic case and in fact I think this is a transcendental extension, because it would seem that powers of the adjoined arithmetic anti derivatives are all linearly independent over the rationals. This makes the question all the more interesting. I still think the Galois group’s elements of finite order should be involuntary though. @RDBury and Lambiam: for thoughts.—Jasper Deng (talk) 18:03, 9 September 2021 (UTC)
- Thanks for having such confidence in my expertise, but I'm afraid this one is over my head. I'm not even sure what it would look like if you just added a single antiderivative of 2. For example, if you're given D(x) = 2, how do you compute D(x+1)? Our article claims that the derivative over a polynomial ring is the same as the regular derivative, but the derivative of 2 as a polynomial is 0 while its arithmetic derivative is 1. So I'm finding the topic very confusing and so far I've been following the maxim "Better to remain silent and be thought a fool than to speak and remove all doubt." --RDBury (talk) 22:45, 9 September 2021 (UTC)
- @RDBury: I think it works if you insist that D(x+a) = D(x) + D(a) for all rationals a. Heuristically, D(x2 - 1) = 4x = D((x+1)(x-1)) = (x+1)D(x-1) + (x-1)D(x+1).—Jasper Deng (talk) 04:17, 10 September 2021 (UTC)
- If D(x+a) = D(x) + D(a), then D(x+1) = D(x) + D(1) = D(x), so D(n) = 0 for all naturals. --Lambiam 05:39, 10 September 2021 (UTC)
- @Lambiam: No, this only applies to the sum of "unlike species"; x here is just a formal symbol denoting an antiderivative of 2 and not a natural number. Also, this extension has to be limited to monomials (i.e. naively trying to distribute over will fail). We then have to appeal to the fundamental theorem of algebra to obtain derivatives (though this would pose the problem of extending it to all algebraic numbers instead of just roots).--Jasper Deng (talk) 07:00, 10 September 2021 (UTC)
- Is the sum rule forced, or is it a choice? If u is a number that is not in the range of D, and given a closure, let u✽ denote the adjoined element such that D(u✽) = u. How to handle D(2✽ + 3✽) – are these summands unlike species? And what about D(2✽ + 3✽✽)? Is it possible to falsify the conjecture that any set of adjoined elements is independent (not just linearly but also algebraically)? Aren't also u✽✽, u✽✽✽, and so on, all independent? --Lambiam 07:38, 10 September 2021 (UTC)
- @Lambiam: I think it's a choice. It seems like the most natural way to extend the arithmetic derivative to the field generated by x. One attempt to more formally define "species" would be that the (n + 1)st antiderivative of a particular number is a different species from the nth (and below) antiderivatives if it is not contained by the field generated by the nth and below antiderivatives. The idea would seem to be akin to Liouvillian functions.--Jasper Deng (talk) 04:38, 11 September 2021 (UTC)
- Is the sum rule forced, or is it a choice? If u is a number that is not in the range of D, and given a closure, let u✽ denote the adjoined element such that D(u✽) = u. How to handle D(2✽ + 3✽) – are these summands unlike species? And what about D(2✽ + 3✽✽)? Is it possible to falsify the conjecture that any set of adjoined elements is independent (not just linearly but also algebraically)? Aren't also u✽✽, u✽✽✽, and so on, all independent? --Lambiam 07:38, 10 September 2021 (UTC)
- @Lambiam: No, this only applies to the sum of "unlike species"; x here is just a formal symbol denoting an antiderivative of 2 and not a natural number. Also, this extension has to be limited to monomials (i.e. naively trying to distribute over will fail). We then have to appeal to the fundamental theorem of algebra to obtain derivatives (though this would pose the problem of extending it to all algebraic numbers instead of just roots).--Jasper Deng (talk) 07:00, 10 September 2021 (UTC)
- If D(x+a) = D(x) + D(a), then D(x+1) = D(x) + D(1) = D(x), so D(n) = 0 for all naturals. --Lambiam 05:39, 10 September 2021 (UTC)
- @RDBury: I think it works if you insist that D(x+a) = D(x) + D(a) for all rationals a. Heuristically, D(x2 - 1) = 4x = D((x+1)(x-1)) = (x+1)D(x-1) + (x-1)D(x+1).—Jasper Deng (talk) 04:17, 10 September 2021 (UTC)
- Thanks for having such confidence in my expertise, but I'm afraid this one is over my head. I'm not even sure what it would look like if you just added a single antiderivative of 2. For example, if you're given D(x) = 2, how do you compute D(x+1)? Our article claims that the derivative over a polynomial ring is the same as the regular derivative, but the derivative of 2 as a polynomial is 0 while its arithmetic derivative is 1. So I'm finding the topic very confusing and so far I've been following the maxim "Better to remain silent and be thought a fool than to speak and remove all doubt." --RDBury (talk) 22:45, 9 September 2021 (UTC)
- Actually, I cannot determine the Galois group in analogy to the algebraic case and in fact I think this is a transcendental extension, because it would seem that powers of the adjoined arithmetic anti derivatives are all linearly independent over the rationals. This makes the question all the more interesting. I still think the Galois group’s elements of finite order should be involuntary though. @RDBury and Lambiam: for thoughts.—Jasper Deng (talk) 18:03, 9 September 2021 (UTC)
