Timeline for irreducibility of discriminant
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 27, 2015 at 21:08 | history | edited | Chris Gerig | CC BY-SA 3.0 |
Incorrect subscript
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| Oct 26, 2015 at 10:34 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added 27 characters in body
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| Oct 26, 2015 at 10:28 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added the treatment of the case of characteristic $2$
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| Oct 25, 2015 at 13:53 | comment | added | Geoff Robinson | I didn't suggest it was a flaw, but felt that the exception should be noted. | |
| Oct 25, 2015 at 13:36 | comment | added | Jarek Kuben | By the same argument as in the answer: one can't detect which of the $t$'s are equal using only the $s$'s. | |
| Oct 25, 2015 at 13:30 | comment | added | Fedor Petrov | Why square root of discriminant is irreducible in characteristic 2? | |
| Oct 25, 2015 at 13:17 | comment | added | Jarek Kuben | In characteristic $2$ the polynomial $\prod_{i<j} (t_i-t_j)=\prod_{i<j} (t_i+t_j)$ is symmetric in $t$'s, hence a polynomial in $s$'s (the fundamental theorem of symmetric polynomials holds in all commutative rings), thus the discriminant is a square of an irreducible polynomial. | |
| Oct 25, 2015 at 13:02 | comment | added | Robert Bryant | Oh, also, I just checked: $D$ is reducible in a field of characteristic $2$ when $d = 4$ as well. | |
| Oct 25, 2015 at 13:00 | comment | added | Igor Rivin | This is pretty close to the sort of thing I was thinking of (that is, a more invariant theoretic proof - the factors must be invariants of the alternating group). | |
| Oct 25, 2015 at 12:56 | comment | added | Robert Bryant | Yes, of course, obviously the last step in my argument assumes that the characteristic is not $2$, but, then as Ofir's comment points out, $D$ can, in fact, be reducible in characteristic $2$, so I don't regard that as a flaw, but as a feature. The argument shows exactly where characteristic $2$ is needed. (In fact, $D$ is reducible over characteristic $2$ when $d=2$ and $d=3$, and maybe in all degrees.) | |
| Oct 25, 2015 at 12:51 | comment | added | Geoff Robinson | As Ofir's answer implicitly suggests, this argument doesn't seem to work as it stands in characteristic 2. | |
| Oct 25, 2015 at 11:29 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Cleaned up the argument, corrected typos
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| Oct 25, 2015 at 10:32 | history | answered | Robert Bryant | CC BY-SA 3.0 |