Timeline for Are those $2$ quadratic forms congruent over $\mathbb{Z}[1/q]$
Current License: CC BY-SA 3.0
17 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Dec 12, 2016 at 1:36 | comment | added | thierry stulemeijer | @Noam That's amazing. Thank you very much for your help ! | |
| Dec 12, 2016 at 1:30 | vote | accept | thierry stulemeijer | ||
| Dec 12, 2016 at 1:30 | answer | added | thierry stulemeijer | timeline score: 3 | |
| Dec 12, 2016 at 1:05 | comment | added | Noam D. Elkies | It's much the same thing, using $39$ copies of the formula for multiplying a Hamilton quaternion by say $2(1+i+j)$ or $3+i+j+k$ (and likewise for other $q \equiv 0,3 \bmod 4$, via the four-square theorem). | |
| Dec 12, 2016 at 0:28 | comment | added | thierry stulemeijer | Thank you ! So indeed, the comment of Noam answers my problem for all $q$ congruent to 1 or 2 modulo 4 that are sum of squares (that's one case where the Bruck-Ryser theorem does not say anything). The remaining cases not covered by Bruck-Ryser (or equivalently, when my 2 quadratic forms are in the same genus) are when q is congruent to 0 or 3 modulo 4, e.g. q=12. So, what can be said about the isomorphism over $\mathbf{Z}[1/12]$ of $\sum \limits_{i=1}^{157} x_i^2$ and $x_1^2+\sum \limits_{i=2}^{157}12x_i^2$ ? | |
| Dec 12, 2016 at 0:18 | comment | added | KConrad | @thierrystulemeijer note that $10 = 1^2 + 3^2 = {\rm N}(1+3i)$ and $(1+3i)(x-iy) = (x+3y)+(3x-y)i$. Take the norm of both sides. | |
| Dec 12, 2016 at 0:03 | history | edited | thierry stulemeijer | CC BY-SA 3.0 |
added 1 character in body
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| Dec 11, 2016 at 23:57 | comment | added | thierry stulemeijer | @Noam How did you find that ? Because if there is a technique, there are other values of the parameters that are of great interest to me, as explained in the main question. Can you actually say something about the general question ? And could Magma help at this level of magnitude if necessary ? Anyway, thank you already for this wonderful answer. That's in some sense a bad news, but that was to be expected anyway... | |
| Dec 11, 2016 at 23:47 | comment | added | Noam D. Elkies | If we may invert $10$ then I don't need Magma because I can use $55$ copies of the identity $(x+3y)^2 + (3x-y)^2 = 10(x^2+y^2)$. | |
| Dec 11, 2016 at 23:38 | history | edited | thierry stulemeijer | CC BY-SA 3.0 |
taking comments into account
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| Dec 11, 2016 at 23:31 | comment | added | thierry stulemeijer | @Noam D. Elkies Argh, sure, I got over enthusiastic and forgot that :-) but can we say anything about their isomorphism over $\mathbf{Z}[1/10]$, then? With the help of Magma, maybe ? | |
| Dec 11, 2016 at 23:26 | comment | added | Noam D. Elkies | They can't literally be isomorphic over $\bf Z$ because the determinants don't match. | |
| Dec 11, 2016 at 23:15 | history | edited | thierry stulemeijer | CC BY-SA 3.0 |
Added a related question, added some tags
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| Dec 11, 2016 at 23:15 | history | edited | YCor |
edited tags
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| Dec 11, 2016 at 23:13 | history | edited | thierry stulemeijer | CC BY-SA 3.0 |
Added a related question, added some tags
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| Apr 22, 2016 at 13:47 | history | asked | thierry stulemeijer | CC BY-SA 3.0 |