Timeline for Explicit description of a quaternion algebra with a prescribed set of ramified places
Current License: CC BY-SA 3.0
9 events
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| S Mar 24, 2014 at 2:38 | history | suggested | j0equ1nn | CC BY-SA 3.0 |
The $a$ was missing in the second mention of $ax^2+by^2-abz^2$, so I put that in.
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| Mar 24, 2014 at 2:30 | review | Suggested edits | |||
| S Mar 24, 2014 at 2:38 | |||||
| Jul 10, 2012 at 7:41 | vote | accept | anonymous | ||
| Jul 10, 2012 at 5:48 | comment | added | Noam D. Elkies | $2$ is a quartic residue mod $p \equiv 1 \bmod 4$ iff $p = 64a^2+b^2$. The first example is indeed $73$, with $(a,b) = (1,3)$. Since also $73 = 3^4 - 2^3$ we have $2^3 \equiv 3^4 \bmod 73$, so $2^4 = 2 \cdot 3^4$ and we get the explicit 4th root $2/3 \equiv 25$ of $2 \bmod 73$. | |
| Jul 9, 2012 at 17:47 | history | edited | Will Sawin | CC BY-SA 3.0 |
deleted 33 characters in body
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| Jul 9, 2012 at 17:45 | comment | added | Will Sawin | Actually $41$ doesn't work. $73$ does though, since $2^18=1$ modulo $73$. | |
| Jul 9, 2012 at 17:38 | comment | added | Will Sawin | What you need are four different roots to the equation $x^4-2=0$, so four fourth roots of $2$. The two things you can check with simple congruence conditions are the existence of a square root of $2$ and a square root of $-1$. Then you look at all primes satisfying this congruence condition until you find one where $2$ is a fourth power. You can do this by computing the quartic character $2^{(p-1)/4}$, for instance. If I computed correctly then the first one I found was $41$. | |
| Jul 9, 2012 at 10:51 | comment | added | anonymous | Thanks a lot, this example is very helpful. Do you mind telling me why you chose 41? How do you know it is completely split? I can see that it is split in $\Q(\sqrt{2})$, since $8$ is a quadratic residue mod 41. Is there a similar criterion for extensions of degree 4? | |
| Jul 6, 2012 at 15:55 | history | answered | Will Sawin | CC BY-SA 3.0 |