Timeline for How to make a product of polynomials irreducible?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| May 22, 2017 at 23:39 | history | reopened |
Neil Strickland R.P. Stefan Kohl♦ Michael Albanese Chris Godsil |
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| May 22, 2017 at 20:08 | review | Reopen votes | |||
| May 22, 2017 at 23:39 | |||||
| May 22, 2017 at 19:53 | history | edited | Eva | CC BY-SA 3.0 |
specified problem, adding context
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| May 22, 2017 at 17:47 | comment | added | R.P. | I agree that this is a likely reading. Voting to reopen -- although this doesn't seem a very hard question, I think it is probably too difficult to be likely to get an answer at math.SE... | |
| May 22, 2017 at 17:38 | review | Reopen votes | |||
| May 22, 2017 at 18:10 | |||||
| May 22, 2017 at 17:10 | comment | added | Neil Strickland | I presume that the OP means that $p(0,0)=q(0,0)=0$ and she seeks an irreducible $r(x,y)$ such that $\mathbb{R}[\![x,y]\!]/(r)\simeq\mathbb{R}[\![x,y]\!]/(pq)$ as $\mathbb{R}$-algebras, or something like that. For example you could have $p=x-y$ and $q=x+y$ and $r=pq-y^3=x^2-y^2-y^3$: this is the nodal cubic curve, which is globally irreducible, but reducible in a formal neighbourhood of the origin. | |
| May 22, 2017 at 16:48 | history | edited | Stefan Kohl♦ |
edited tags
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| May 22, 2017 at 14:15 | history | closed |
Max Alekseyev coudy Michael Albanese Chris Godsil Ira Gessel |
Needs details or clarity | |
| May 22, 2017 at 14:05 | comment | added | Dirk | What do you mean by "going through $[0,0]$"? As these are polynomials in two variables, the graph of the corresponding function would be in 3D rather than 2D... Also, what domain are you considering? a field, a ring, the reels,...? And why do you assume that your given candidate is irreducible? I think it will not be in many cases, even in cases where $p$ and $q$ are both irreducible. All in all you might want to add quite some background on your question. | |
| May 22, 2017 at 13:06 | comment | added | Michael Albanese | What do you mean by "having same qualities as $p(x, y)\cdot q(x, y)$"? | |
| May 22, 2017 at 11:42 | review | Close votes | |||
| May 22, 2017 at 14:17 | |||||
| May 22, 2017 at 9:16 | review | First posts | |||
| May 22, 2017 at 9:17 | |||||
| May 22, 2017 at 9:13 | history | asked | Eva | CC BY-SA 3.0 |