Timeline for Quadratic refinements of a bilinear form on finite abelian groups
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 4 at 16:30 | comment | added | Sean Eberhard | $q_i$ is an element of $\mathbb R/\mathbb Z$, not $C_i$. In $\mathbb R/\mathbb Z$ there are always exactly two solutions to $2x=y$. If $y$ has finite odd order $k$ then one of the solutions has order $k$ and the other has order $2k$. | |
| Aug 4 at 13:58 | comment | added | Andrea Antinucci | I am not sure I understand. If $r_i$ is odd $2$ is invertible in $C_i$ and any $q_i$ such that $2q_i=\chi(g_i,g_i)$ satisfies $q_i=2^{-1}\chi(g_i,g_i)$. But then since the order of $\chi(g_i,g_i)$ divides $r_i$ I get $r_iq_i=2^{-1}r_i\chi(g_i,g_i)=0$. Here it seems to me I am not assuming $q_i$ to have odd order. Actually on a cyclic group of odd order I think there is no concept of even or odd. | |
| Aug 4 at 13:06 | comment | added | Sean Eberhard | @AndreaAntinucci For $q$ to be well-defined the expression on the right must depend only on $n_i$ modulo $r_i$. If you replace $n_i$ with $n_i + r_i$ the difference is $2n_i r_i q_i + r_i^2 q_i + \sum_{j \ne i} \chi(g_i, g_j) r_i n_j = r_i^2 q_i$, and this must be zero. | |
| Aug 4 at 12:53 | comment | added | Andrea Antinucci | Why do we need $q_i$ to have odd order when $r_i$ is odd? | |
| Aug 4 at 12:53 | vote | accept | Andrea Antinucci | ||
| Aug 4 at 11:34 | history | answered | Sean Eberhard | CC BY-SA 4.0 |