Timeline for Is a complete homogeneous symmetric polynomial irreducible?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2020 at 13:52 | comment | added | Richard Stanley | For a stronger result (using that $h_a$ is the Schur function $s_a$) see mathoverflow.net/questions/98494/…. | |
| S Aug 29, 2020 at 12:14 | history | suggested | Joshua P. Swanson | CC BY-SA 4.0 |
Cleaned up title and fixed its grammar; I normally wouldn't bother but I want to cite this post in a paper
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| Aug 29, 2020 at 9:29 | review | Suggested edits | |||
| S Aug 29, 2020 at 12:14 | |||||
| May 26, 2012 at 22:23 | vote | accept | Neeraj | ||
| May 26, 2012 at 20:08 | comment | added | Martin Brandenburg | @Will Sawin: Thank you for your explanation! And I also agree with Patricia that answers should not be hidden as comments. | |
| May 26, 2012 at 17:41 | answer | added | François Brunault | timeline score: 17 | |
| May 26, 2012 at 17:35 | comment | added | Patricia Hersh | You're right. You should post this comment above as an actual answer I think, since it's the best answer! | |
| May 26, 2012 at 17:27 | comment | added | Will Sawin | I don't understand why the degree would change. In a polynomial of degree $a$, the constant term has degree $a$. It's $n$ that changes. | |
| May 26, 2012 at 16:55 | answer | added | Will Sawin | timeline score: 1 | |
| May 26, 2012 at 16:51 | comment | added | François Brunault | For the case $n=3$ it would suffice to prove that the projective curve $h_a(x,y,z)=0$ is smooth, in other words that $h_a$ has no common zero with all its partial derivatives in $\mathbf{P}^2(\mathbf{C})$. | |
| May 26, 2012 at 16:43 | answer | added | Igor Rivin | timeline score: 2 | |
| May 26, 2012 at 16:30 | comment | added | Will Sawin | View the polynomial as a monic polynomial in $x_n$. The leading term is $1$, the constant term is the polynomial with $a$ the same and $n$ one less. By induction on $n$, this is irreducible, so each factor must have constant term $1$ or that polynomial. But since the polynomial is homogeneous, the factors are homogeneous, thus degree $0$ or degree $a$, thus no nontrivial factors. | |
| May 26, 2012 at 16:18 | comment | added | Martin Brandenburg | @Will Sawin: What is this induction argument exactly? | |
| May 26, 2012 at 16:15 | answer | added | Denis Serre | timeline score: 2 | |
| May 26, 2012 at 16:14 | comment | added | Denis Serre | You should have a look to McDonald's book about symmetric and Hall polynomials | |
| May 26, 2012 at 15:51 | comment | added | Will Sawin | You can reduce to the case $n=3$ using an induction argument, but I can't immediately see how to solve $n=3$. | |
| May 26, 2012 at 15:19 | answer | added | Patricia Hersh | timeline score: 8 | |
| May 26, 2012 at 14:31 | history | asked | Neeraj | CC BY-SA 3.0 |