Timeline for A special integral polynomial
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Apr 16, 2010 at 17:12 | vote | accept | Roberto Svaldi | ||
| Mar 13, 2010 at 23:31 | history | edited | Pete L. Clark |
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| Mar 13, 2010 at 21:49 | answer | added | Robin Chapman | timeline score: 9 | |
| Mar 13, 2010 at 18:02 | comment | added | Pete L. Clark | No, the condition of having no real roots is not generic. Rather, it defines a nonempty open set (in the analytic topology) of the space of all degree $2n$ polynomials: e.g. for quadratic polynomials the condition is just $b^2-4ac < 0$. Thus if you endow this space with some reasonable measure, the locus you want will have positive, but not full, measure. In contrast the locus of the set where the Galois group is $S_{2n}$ will have full measure, so morally there's your existence proof. But I didn't immediately see how to make this rigorous, so I did something totally different below. | |
| Mar 13, 2010 at 17:57 | answer | added | Pete L. Clark | timeline score: 6 | |
| Mar 13, 2010 at 17:51 | comment | added | Douglas Zare | In some sense a generic polynomial with no real roots should have a Galois group of S_2n, but I don't know a sense in which a generic polynomial doesn't have any real roots. | |
| Mar 13, 2010 at 17:43 | comment | added | Roberto Svaldi | Is reasonable to think that the generic monic integral polynomial will have that form? I do not have a precise meaning here for the word generic, but maybe it might be given a number theoretic (and scheme-theoretic on Spec Z?) sense. | |
| Mar 13, 2010 at 16:59 | history | asked | Roberto Svaldi | CC BY-SA 2.5 |