Timeline for An application of Maschke's theorem
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 8, 2015 at 19:29 | comment | added | KConrad | @DavidHill, the ring $\mathbf Z[\sqrt{2}]$ has a division algorithm, so it is a unique factorization domain: all integral domains with a division algorithm have unique factorization. Don't try to find more counterexamples to unique factorization in $\mathbf Z[\sqrt{2}]$; they do not exist and if you think you found one then you're making an error. | |
| Apr 8, 2015 at 19:27 | comment | added | KConrad | @DavidHill, that is not a counterexample! The numbers $7$ and $5\sqrt{2}\pm 1$ are not prime. What you wrote is like saying $\mathbf Z$ lacks unique factorization because $4\cdot 9 = 6 \cdot 6$. Each factorization can be broken down further and rearranged to get the other one: $4 \cdot 9 = 2 \cdot 2 \cdot 3 \cdot 3 = 6 \cdot 6$. In ${\mathbf Z}[\sqrt{2}]$, $7 = (3+\sqrt{2})(3-\sqrt{2})$ and $5\sqrt{2}-1 = (3-\sqrt{2})^2(\sqrt{2}+1)$ while $5\sqrt{2}+1 = (3+\sqrt{2})^2(\sqrt{2}-1)$. The factors $3\pm \sqrt{2}$ are prime while the factors $\sqrt{2}\pm 1$ are units (their product is $1$). | |
| Apr 8, 2015 at 1:46 | comment | added | KConrad | You write that you "ignored the fact that $\mathbf Z[\sqrt{2}]$ is not a unique factorization domain," but in fact it is a unique factorization domain! | |
| Apr 8, 2015 at 0:38 | answer | added | Angel del Rio | timeline score: 0 | |
| Apr 28, 2014 at 13:17 | history | edited | David Hill | CC BY-SA 3.0 |
added 752 characters in body
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| Apr 28, 2014 at 13:01 | vote | accept | David Hill | ||
| Apr 27, 2014 at 14:53 | answer | added | Geoff Robinson | timeline score: 3 | |
| Apr 26, 2014 at 20:31 | comment | added | Geoff Robinson | Are you sure that you mean $d = \frac{q-1}{p}$ here? Don't you want to choose $d$ so that $d^{p} \equiv 1$ (mod $q$)? | |
| Apr 25, 2014 at 17:34 | comment | added | David E Speyer | Your $d=2$ proof is broken: $1^2+7^2=2 \times 5^2$; $7^2+17^2 = 2 \times 13^2$. In general, if $a^2+b^2=p$, then $(a^2-2ab-b^2)^2 + (a^2+2ab-b^2)^2=2 p^2$. (Found by playing with Gaussian integers.) | |
| Apr 25, 2014 at 15:59 | answer | added | Ben Webster♦ | timeline score: 9 | |
| Apr 25, 2014 at 15:37 | history | edited | Yemon Choi |
Added NT tag since it seems relevant
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| Apr 25, 2014 at 15:36 | comment | added | Nick Gill | Ha! Yes, you're right. sorry, i commented in haste. | |
| Apr 25, 2014 at 15:18 | comment | added | David Hill | @NickGill: I think you are misreading the condition. | |
| Apr 25, 2014 at 15:13 | comment | added | Benjamin Steinberg | The dimensions of the simple modules divide the order of the group. Since $pq-p<q^2$, they must all have dimension p. I believe you can also get the simple modules by inducing certain $1$-dim characters of $C_q$ and using Mackey's criterion. | |
| Apr 25, 2014 at 15:09 | comment | added | Nick Gill | $3^2+19^2+31^2=11^3$, which suggests that you won't be able to prove this via number theory alone. | |
| Apr 25, 2014 at 14:55 | history | asked | David Hill | CC BY-SA 3.0 |