This question was originally asked in MSE about a year ago. Nobody has answered it. So I posted it here. I wonder why they think it's not of research level.
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5$\begingroup$ Since nobody has answered the question for a year in MSE, I think it's not obvious that it's not of research level. $\endgroup$– Makoto KatoCommented Aug 22, 2013 at 0:37
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5$\begingroup$ Makoto Kato: if you would sincerely like to hear why people think your question is not research level, then you should ask this in a different way. The way you have asked it is presumptuous and aggressive, and does not invite reasoned discourse. $\endgroup$– Michael ZieveCommented Aug 22, 2013 at 20:15
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$\begingroup$ @MichaelZieve I deleted the FLT part. I hope you agree with my edit. $\endgroup$– Makoto KatoCommented Aug 22, 2013 at 20:23
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5$\begingroup$ Now the question is much better. $\endgroup$– Michael ZieveCommented Aug 22, 2013 at 20:31
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2$\begingroup$ I just added a comment to the question sketching an answer. This is perhaps a tricky exercise in algebra, which requires knowing about real fields, but is not a research question. $\endgroup$– Felipe VolochCommented Aug 23, 2013 at 13:34
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$\begingroup$ @FelipeVoloch I edited the question. $\endgroup$– Makoto KatoCommented Aug 23, 2013 at 17:43
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3 Answers
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It is for days I want to write this answer. Now that I've come to write it, your question is not there anymore. Thus please consider my answer as a general impression of another "newcomer" about how MO works. Generally speaking:
People in MO land do not like "naked" questions.
That means, more often than not, questions like "solve this" or "I couldn't solve this, could you?" and so on have little chance to get enough attention. People often like to see the surrounding story as well as the question per se. That includes your personal attempt to solve the problem, the relation of the problem to other things, the reason that you are interested in the problem and so on. As you can see, the comments above, all coming from more experienced people than me, somehow try to clarify the surrounding story of your question. And I am sure those people are more qualified than me to tell the story of the importance of the surrounding stories in general.
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$\begingroup$ Dear Amir, if you want to write an answer to the question, it is still left on math.SE: math.stackexchange.com/questions/173775 . I don't think the question has yet received a complete answer. $\endgroup$ Commented Aug 26, 2013 at 9:12
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$\begingroup$ @DanPetersen Dear Dan. Thanks for the link. $\endgroup$ Commented Aug 26, 2013 at 10:21
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I respectfully disagree with Noah Snyder; I think the question makes perfect sense as stated. I would phrase it as follows: "Here is a proof that the norm map $K = \mathbf Q[\alpha]/(1+\alpha+\ldots+\alpha^{p-1}) \to \mathbf Q$ takes only positive values. The proof uses that $K$ embeds in the complex numbers. Is there a different proof that avoids this fact?"
Certainly one can make sense of the statement $N(K) \subseteq \mathbf Q^+$ without mentioning the real numbers. I don't understand what Michael Zieve is getting at in his comments. For what it's worth I don't know how to answer Makoto Kato's question.
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3$\begingroup$ 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 $\endgroup$ Commented Aug 23, 2013 at 12:15
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2$\begingroup$ real numbers via Dedekind cuts, and then the original proof becomes a proof which only mentions rationals. Is that legal? $\endgroup$ Commented Aug 23, 2013 at 12:16
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1$\begingroup$ @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. $\endgroup$– user9072Commented Aug 23, 2013 at 12:25
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$\begingroup$ @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. $\endgroup$ Commented Aug 23, 2013 at 12:39
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1$\begingroup$ I think quid's "version of the question" is pretty clearly just a restatement of the question as it was intended. $\endgroup$ Commented Aug 23, 2013 at 12:55
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$\begingroup$ @quid: What do you mean by "direct"? The proof in the original argument is pretty direct (just group terms in complex conjugate pairs). $\endgroup$ Commented Aug 23, 2013 at 15:36
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$\begingroup$ @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. $\endgroup$ Commented Aug 23, 2013 at 15:47
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1$\begingroup$ 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.) $\endgroup$ Commented Aug 23, 2013 at 16:15
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$\begingroup$ @MichaelZieve I edited the question. $\endgroup$ Commented Aug 23, 2013 at 17:48
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$\begingroup$ @YemonChoi: could you explain how to formulate this in terms of determinants of circulant matrices? That sounds interesting. $\endgroup$ Commented Aug 23, 2013 at 17:50
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$\begingroup$ @MichaelZieve I guess this would be something for "chat" but I've never used that functionality, do you know how to do it? $\endgroup$ Commented Aug 23, 2013 at 18:13
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$\begingroup$ @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? $\endgroup$– user9072Commented Aug 23, 2013 at 18:25
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$\begingroup$ 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. $\endgroup$– user9072Commented Aug 23, 2013 at 18:25
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$\begingroup$ @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'. $\endgroup$– user9072Commented Aug 23, 2013 at 18:29
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$\begingroup$ I edited the question to make it clearer. $\endgroup$ Commented Aug 23, 2013 at 19:10
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As Hurkyl explained on m.SE this question doesn't make sense. You already know how to prove this result, and the proof doesn't use "properties of the real numbers." All it uses is basic facts about positivity, and you already need positivity to state the result.
Even if this question made sense (which it doesn't), it still wouldn't be "research level" since the proof is at the undergraduate level.
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1$\begingroup$ The question makes sense. Please read my last comment to Yemon Choi in the thread. No, I don't know the answer to my question. If you know it, could you answer it in MSE? $\endgroup$ Commented Aug 22, 2013 at 23:30
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2$\begingroup$ @MakotoKato: Noah is saying the question doesn't make sense because you didn't define what it means for a rational number to be positive. Of course, everyone knows what it means for a real number to be positive, but the point of your question is to avoid mentioning real numbers. So you need to say what definition you have in mind for positive rationals. Also, Noah didn't say that you know the answer to your question, he said that you know a proof of the result in question. $\endgroup$ Commented Aug 23, 2013 at 1:21
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2$\begingroup$ @MichaelZieve
[@MakotoKato: Noah is saying the question doesn't make sense because you didn't define what it means for a rational number to be positive. Of course, everyone knows what it means for a real number to be positive, but the point of your question is to avoid mentioning real numbers. So you need to say what definition you have in mind for positive rationals.]Let $r = a/b$ be a rational number where $a$ and $b$ are rational integers. If $a \gt 0$ and $b\gt 0$, we say $r$ is positive. Regards, $\endgroup$ Commented Aug 23, 2013 at 3:01 -
1$\begingroup$ And how do you define positivity of an integer? $\endgroup$ Commented Aug 23, 2013 at 3:15
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5$\begingroup$ @MichaelZieve
[And how do you define positivity of an integer?]Natural numbers $1, 2, 3,\dots$ are exactly (strictly) positive integers. $\endgroup$ Commented Aug 23, 2013 at 3:21 -
3$\begingroup$ @quid: of course you can define positivity in many ways. I was trying to get MK to give a definition, since doing so would indicate what type of proof he sought. Let me add that I don't understand the upvotes for MK's definition that an integer is positive if it is a positive integer. $\endgroup$ Commented Aug 23, 2013 at 12:25
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1$\begingroup$ @MichaelZieve I only saw your other comment after having written mine. Regarding MK's definition I think it depends how one starts; as you know, a usual construction of the integers is as equivalence classes of pairs of natural numbers. And than one imbeds the natural numbers. Then the natural number so to say 'existed' before the integers and the definition is not circular. (Or one could also say as usual 2 is shorthand for 1 + 1 and so on, then it also makes sense the positive one are thos that can be written as sums of 1, the mult identity of the domain 'integers') $\endgroup$– user9072Commented Aug 23, 2013 at 12:30
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5$\begingroup$ Oh, for heaven's sakes. Of course the question makes sense, because there is just one way to make the rationals into an ordered field. This fact is independent of whether or not the OP knows the formal details (and I think it would be more productive not to badger the OP over these, since he/she might not know them, but to help him/her by incorporating them into an answer). I am glad Michael Zieve gave a productive answer at MSE. $\endgroup$ Commented Aug 23, 2013 at 13:49
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1$\begingroup$ But @ToddTrimble, Michael Zieve's answer at m.SE is saying exactly the same thing as my answer here: the original argument doesn't "use the real numbers." $\endgroup$ Commented Aug 23, 2013 at 15:45
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2$\begingroup$ Noah, the original argument of Kato's did refer to complex conjugation (and hence the real numbers as fixed field). Michael's answer fleshed out what the OP needed, which involves precisely the fact that the fixed field of the automorphism that sends $\zeta$ to $\zeta^{-1}$ is formally real; for that one needs a total ordering on the fixed field. Your answer says nothing about any of that, and IMO was not helpful. $\endgroup$ Commented Aug 23, 2013 at 16:17
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1$\begingroup$ MK has now left a comment on the MO question, saying that as far as he is concerned " A real closure of Q is isomorphic to the field of real algebraic numbers. Hence the proof using it is basically the same as stated in the question." So would those who think it is "obvious" what the OP meant in his question, but who think that Felipe Voloch's argument answers the question, please enlighten me as to what is going on? $\endgroup$ Commented Aug 23, 2013 at 16:22
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1$\begingroup$ @YemonChoi It's not obvious to me why Felipe's argument wouldn't count as "purely algebraic", but it's possible that MK doesn't think it's pure enough since the theory of real closed fields uses Zorn's lemma at places, whereas this result ought to be explicable purely in terms of finite-dimensional extensions of $\mathbb{Q}$. $\endgroup$ Commented Aug 23, 2013 at 16:32
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3$\begingroup$ @ToddTrimble OK, in that case I agree with your reading (on both counts). My point was: if MK is dissatisfied with the "obvious" interpretation of his question, then maybe, just maybe, it wasn't that well posed in the first place. (I like my circulant formulation better, but then I'm far from unbiased.) I also find the attitude displayed by "no one wanted to answer my question on MSE, so it must be research-level in the sense of MO", rather trying - see an earlier version of this meta.MO question which invoked FLT $\endgroup$ Commented Aug 23, 2013 at 16:36
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5$\begingroup$ @YemonChoi Agreed, definitely not research level, and definitely a better fit for MSE than MO (all the more so if less than perfectly posed). $\endgroup$ Commented Aug 23, 2013 at 19:14
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3$\begingroup$ @MakotoKato Well, if you already understand all this, I now join the rest in wondering what the problem is. Did Michael Zieve answer your query satisfactorily at MSE? $\endgroup$ Commented Aug 23, 2013 at 19:17