Jump to content

Wikipedia:Reference desk/Archives/Mathematics/2009 July 18

From Wikipedia, the free encyclopedia
Mathematics desk
< July 17 << Jun | July | Aug >> July 19 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


July 18

[edit]

Polynomial whose roots are powers of another polynomial

[edit]

Hello again. This is part of a theorem I am reading on single Hensel lift of a polynomial. Suppose I am given a polynomial in hm in which divides xk-1 as well. By Hensel's lemma I can construct a polynomial h(x) which is equivalent to it mod pm and which divides xk-1 too but in mod pm+1. If we let x to be the root of hm and y=x+pmi the root of h then we have xk=1+pme as hm divides xk-1 in . Also in ; yp=xp and ykp=1. All this is fine. My book now claims (and so does the link) that if I consider another polynomial hm+1 whose roots are the pth powers of the roots of h then these roots coincide mod pm with those of hm. I can't justify this statement. I'll be grateful for any help. Thanks--Shahab (talk) 10:17, 18 July 2009 (UTC)[reply]

Irreducibility of multivariate polynomials

[edit]

(redirect from sci desk) I cannot find anything about that. I can think of two reasons for this: either it's utterly complicated and not well understood or it can be so easily reduced to the question of irreducibility of polynomials of a single variable that no one cares to tell. What is it? 93.132.138.254 (talk) 09:35, 18 July 2009 (UTC) —Preceding unsigned comment added by 83.100.250.79 (talk) [reply]

Start maybe from here : Algebraic geometry.... --pma (talk) 11:47, 18 July 2009 (UTC)[reply]
Ah, your're right! I already have guessed this had something to do with mathematics somehow! 93.132.138.254 (talk) 14:44, 18 July 2009 (UTC)[reply]
My understanding was that Algebraic geometry primarily studied polynomials over algebraically-closed fields. What about polynomials over general rings? Can you provide any particular insight there? Eric. 76.21.115.76 (talk) 21:30, 18 July 2009 (UTC)[reply]
Not sure what kind of insights you are after, but as a starting point, multivariate polynomials over any unique factorization domain form themselves a UFD because of Gauss' lemma. — Emil J. 12:44, 20 July 2009 (UTC)[reply]

How does this work?

[edit]

How does this work. I recently had it emailed to me and it seems to work. Thanks. Computerjoe's talk 22:33, 18 July 2009 (UTC)[reply]

Warning: link plays unsolicited audio. Algebraist 22:34, 18 July 2009 (UTC)[reply]
Let your initial number have first digit x and second y, so the number is 10x+y. Subtracting the sum of digits from this gives 9x. So all the page has to do is label all multiples of 9 with the same option and it wins. Algebraist 22:37, 18 July 2009 (UTC)[reply]
Ah okay. Thanks. Computerjoe's talk 22:46, 18 July 2009 (UTC)[reply]
but in fact it's a bit more subtle. As soon as you get the arithmetic trick, you try to cheat with n≠9x, and it still gets it right... The fact is that it knows the square where you are pointing the mouse on, when you choose it...--pma (talk) 08:19, 19 July 2009 (UTC)[reply]

Ummm... No...
I have not noticed that behavior. --COVIZAPIBETEFOKY (talk) 14:23, 19 July 2009 (UTC)[reply]
oh i really almost thought it read my mind... now i understand! it changes every time. --pma (talk) 16:37, 19 July 2009 (UTC)[reply]