Timeline for Expressing field inclusions by polynomial equalities on coefficients
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 16, 2009 at 11:14 | vote | accept | Ewan Delanoy | ||
| Dec 8, 2009 at 11:56 | comment | added | Ewan Delanoy | Your proof is complete now, Thorny. BTW, here is a way to see that reducibility is not a nowhere dense condition : The Pell equation x^2-8y^2=1 has integer solutions with y arbitrarily large. Then the polynomials X^2+(1/y^2)X-2 are all reducible, and converge to X^2-2. | |
| Dec 8, 2009 at 10:53 | history | edited | Thorny | CC BY-SA 2.5 |
erroneous argument
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| Dec 8, 2009 at 10:52 | comment | added | Thorny | Sorry, it isn't. However, the set of irreducible $x^2+ux+v$ is dense because it is equivalent to $v-(u/2)^2$ not being square in $\mathbb{Q}[\sqrt{2}]$. This latter condition is satisfied by all numbers of the form $\frac{1}{n^2}(r+s\sqrt{2})$ with $r$ even and $s$ odd, say. | |
| Dec 8, 2009 at 10:21 | comment | added | Ewan Delanoy | Why is reducibility a nowhere dense condition ? | |
| Dec 8, 2009 at 9:33 | history | answered | Thorny | CC BY-SA 2.5 |