Timeline for Why do they think this question is not of research level?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 23, 2013 at 19:10 | comment | added | Makoto Kato | I edited the question to make it clearer. | |
| Aug 23, 2013 at 18:29 | comment | added | user9072 | @YemonChoi re chat, click on 'chat' at the top of this page (or on main). If you never did it so far you likely have to login on the page where you arrive. (Same login as here.) Then click on some room, MathOverflow the general one seems the obvious choice. Then just type things in the box at the bottom of the page an click send [or press 'enter'] (works basically like comments). Or create a room clicking on tab 'mine' and 'create room'. | |
| Aug 23, 2013 at 18:25 | comment | added | user9072 | For $p=3$ one simply says let $f(X)= a_0 + a_1 X$ then the product is (up to errors in caluclation) $a_0^2 + a_1^2 -a_0a_1 + (-a_1^2+a_0a_1 + a_1^2X)(1+ X+ X^2)$. And $a_0^2 + a_1^2 -a_0a_1$ is also easily seen to be positive. Is there a general argument along these lines. | |
| Aug 23, 2013 at 18:25 | comment | added | user9072 | @NoahSnyder what I meant by "direct" is using as little 'theory' as possible (or developping it in the process). Say, suppose you want to convince somebody that essentially only knows rational polynomials that the following is true: let $f(X)$ be any rational polynomial (of degree at most $p-2$). Then $\prod_{i=1}^{p-1} f(X^i)= r + g(X)(1+ \dots + X^{p-1})$ for some rational polynomial $g$ and $r$ a positive rational. How do you proceed? | |
| Aug 23, 2013 at 18:13 | comment | added | Yemon Choi | @MichaelZieve I guess this would be something for "chat" but I've never used that functionality, do you know how to do it? | |
| Aug 23, 2013 at 17:50 | comment | added | Michael Zieve | @YemonChoi: could you explain how to formulate this in terms of determinants of circulant matrices? That sounds interesting. | |
| Aug 23, 2013 at 17:48 | comment | added | Makoto Kato | @MichaelZieve I edited the question. | |
| Aug 23, 2013 at 16:15 | comment | added | Yemon Choi | I much prefer Dan Petersen's reformulation of the question, which enables people like me who did take an alg NT course but haven't thought about these things for over 10 years to understand what one is after. (For that matter, it seems to me that you could formulate the question in terms of determinants of circulant matrices with rational entries, and ask for a proof that avoids diagonalizability.) | |
| Aug 23, 2013 at 15:47 | comment | added | Noah Snyder | @DanPeterson: Of course, I know that the statement doesn't require referring to the real numbers. We're not disagreeing about the statement, we're disagreeing about whether the obvious proof "uses the real numbers." In particular, I don't see how your answer "disagrees" with mine. | |
| Aug 23, 2013 at 15:36 | comment | added | Noah Snyder | @quid: What do you mean by "direct"? The proof in the original argument is pretty direct (just group terms in complex conjugate pairs). | |
| Aug 23, 2013 at 12:55 | comment | added | Steven Landsburg | I think quid's "version of the question" is pretty clearly just a restatement of the question as it was intended. | |
| Aug 23, 2013 at 12:39 | comment | added | Michael Zieve | @quid In a comment to his m.SE post, MK said "I'm just asking a proof using only elementary properties of $\mathbb{Q}$". I maintain that Dedekind cuts provide such a proof, and hence answer MK's question on his terms. Your version of the question is more interesting. | |
| Aug 23, 2013 at 12:25 | comment | added | user9072 | @MichaelZieve I agree that questions asking for proofs that seek to avoid one thing or other are always a bit tricky, since it is hard to know what should actually be avoided. But then I rather agree with Dan Petersen that I think what seems asked for here seems relatively tranparent. Namely, there is a product prod f(X^i) one knows this is (the class of) a positive rational, but can one (how can one) see this 'directly'; say, expanding the product, collecting terms, observing certain terms are 0 and only soething rational and positive remains. It seems reasonable to wonder about this. | |
| Aug 23, 2013 at 12:16 | comment | added | Michael Zieve | real numbers via Dedekind cuts, and then the original proof becomes a proof which only mentions rationals. Is that legal? | |
| Aug 23, 2013 at 12:15 | comment | added | Michael Zieve | My point was to get Makoto Kato to think about what ingredients he was allowing to be used in such a proof. For instance, is it legal to extend the total ordering on $\mathbf{Q}$ to a total ordering on the subfield $\mathbf{Q}(\alpha+1/\alpha)$ of your field $K$, by defining the value of $\alpha+1/\alpha$ to be less than a rational number $b$ if and only if $b$ is greater than all rational numbers which are smaller than all positive rational numbers $c$ for which $f(c)>0$, where $f$ is the minimal polynomial of $\alpha+1/\alpha$? Of course I'm just saying that you can describe the relevant | |
| Aug 23, 2013 at 9:00 | history | answered | Dan Petersen | CC BY-SA 3.0 |