Timeline for Sums of two squares in (certain) integral domains
Current License: CC BY-SA 2.5
16 events
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| Jul 25, 2010 at 0:12 | comment | added | Dror Speiser | The corollary mentioned in edit 3 is about prime elements (like how $4(6+\sqrt{-17}) = (1+\sqrt{-17})^2+(3-\sqrt{-17})^2$), and not about all primes that simply split: $2^n3 = a^2+b^2$ does not have a solution in $\mathbb{Z}[\sqrt{-17}]$ (send $\sqrt{-17}$ to $1$ and look in $\mod{9}$). | |
| Jul 10, 2010 at 19:04 | vote | accept | Pete L. Clark | ||
| Jul 10, 2010 at 19:04 | vote | accept | Pete L. Clark | ||
| Jul 10, 2010 at 19:04 | |||||
| Jul 9, 2010 at 13:34 | comment | added | Franz Lemmermeyer | @Dror: I was slepping in my last edit -( I'll correct as soon as I have something better. | |
| Jul 9, 2010 at 11:29 | comment | added | Dror Speiser | Regarding edit 3: edit 2 says that the class of the prime above two is of order 2, and hence is equal to its own inverse, making the second sentence in edit 3 weird. | |
| Jul 9, 2010 at 10:56 | history | edited | Franz Lemmermeyer | CC BY-SA 2.5 |
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| Jul 9, 2010 at 6:47 | comment | added | Pete L. Clark | @Everyone -- it is a bit distressing to me that my question has many more upvotes than Franz's answer: an expert in algebraic number theory has generously devoted his time to analyzing my (2/3)-baked question, including doing computations that would not be so trivial for me, at least. What's not to upvote? I think that Franz's self-effacing preamble may have dissuaded people, so I have taken the liberty of removing it. | |
| Jul 9, 2010 at 6:42 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jul 9, 2010 at 6:41 | comment | added | Pete L. Clark | @Franz: your response gets more and more useful, thank you! About $\mathbb{Q}(\sqrt{-5})$: note that my original statement had an "up to a unit" in it, but as I mentioned it was premised on a silly MAGMA error. Given that, I have no reason to think the original statement is true (I checked zero cases numerically; sorry!). | |
| Jul 9, 2010 at 6:14 | history | edited | Franz Lemmermeyer | CC BY-SA 2.5 |
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| Jul 8, 2010 at 19:08 | history | edited | Franz Lemmermeyer | CC BY-SA 2.5 |
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| Jul 8, 2010 at 17:11 | comment | added | KConrad | If $R$ is the ring of integers of a number field, the class group of $R[1/2]$ is the quotient of the class group of $R$ by the subgroup generated by ideal classes of primes lying over 2. (More generally, the class group of a localization of $R$ is the quotient of the class group by the subgroup generated by primes that meet the multiplicative set at which you're localizing.) So $R[1/2]$ has class number 1 if and only if the primes lying over 2 generate the class group of $R$. | |
| Jul 8, 2010 at 17:01 | comment | added | Pete L. Clark | @Franz: right, that was my thinking as well. | |
| Jul 8, 2010 at 16:00 | comment | added | Franz Lemmermeyer | I need more time to think this through - off the top of my head, Z_K[1/2] is principal for quadratic number fields with even discriminant if and only if the prime ideal above 2 generates the class group, which it does for your example (in that case, the resulting ring is even norm Euclidean). | |
| Jul 8, 2010 at 13:17 | comment | added | Pete L. Clark | Thanks, Franz. I pretty much knew what you wrote in the first paragraph -- that's why I picked $\mathbb{Q}(\sqrt{-5})$ , because its genus field is obtained by adjoining $−1$ -- but combining this with your second observation is very helpful. Since you are especially knowledgeable about this sort of thing, here's a question: can you find such a domain $R$ as $\mathbb{Z}_K[\frac{1}{2}]$ -- i.e., same as above with $2$ inverted? Off the top of my head, I think that if $K = \mathbb{Q}(\sqrt{-5})$, the resulting ring is a PID. | |
| Jul 8, 2010 at 9:21 | history | answered | Franz Lemmermeyer | CC BY-SA 2.5 |