Timeline for Correspondence between binary quadratic representations and proper ideals of quadratic number fields
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2021 at 18:43 | vote | accept | Melanka | ||
| Jan 29, 2021 at 18:43 | |||||
| Jan 29, 2021 at 18:34 | comment | added | Melanka | Sorry, I missed the point about fixing a representative $I$ for each class. Thanks for pointing it out. | |
| Jan 29, 2021 at 18:13 | comment | added | GH from MO | @Melanka: $cI$ and $I$ represent the same ideal class, so both cannot be representatives. I fixed a set of (pairwise inequivalent) representatives $\{Q\}$ for the forms and a set of (pairwise inequivalent) representatives $\{I\}$ for the ideals. I assumed the knowledge of the correspondence $Q\leftrightarrow I$, my proof started from there. You can find the details of this correspondence in the Appendix of gofile.io/d/iqnq4A | |
| Jan 29, 2021 at 15:09 | comment | added | Melanka | Now we do a base change to get back $Q$.But this relies on proving that $(m, \gamma)$ is a basis of $\mathfrak{m}$. Now I need to prove it. | |
| Jan 29, 2021 at 15:08 | comment | added | Melanka | To show existence, suppose $n = m e^2$ where $m > 0$ is properly represented by $Q$ and $e > 0$. Then for some $\sigma \in SL_2(\mathbb{Z})$, $\sigma Q(x, y) = mx^2 + b'xy + c'y^2 = \frac{(mx + \gamma y)(mx + \bar{\gamma y)}}{m}$ by factorising $\sigma Q$ for some $\gamma$ in the ring of integers. So $\sigma Q(x, y) = frac{(emx + e \gamma y)(emx + e \bar{\gamma y)}}{e^2 m} = frac{(emx + e \gamma y)(emx + e \bar{\gamma y)}}{(e \mathfrak{m})(e \bar{\mathfrak{m}})}$ and we set $I = e \mathfrak{m}$ where $\mathfrak{m} \bar{\mathfrak{m}} = (m)$ as ideals in the ring of integers. | |
| Jan 29, 2021 at 15:06 | comment | added | Melanka | Thank you for the answer. But I still do have one question. Why is it that the ideal representative $I$ of $Q$ unique? For example, $cI$ and the basis $(c \omega_1, c \omega_2)$ would also be a valid representative for any positive integer $c$. I think for $I$ to be unique we need to add an extra condition. Given $n > 0$, represented by $Q$ there is a unique such ideal $I$ with norm $n$. Now in this case $I$ is unique if such a $I$ exists. | |
| Jan 29, 2021 at 1:44 | history | edited | GH from MO | CC BY-SA 4.0 |
added 4 characters in body
|
| Jan 29, 2021 at 1:30 | history | edited | GH from MO | CC BY-SA 4.0 |
added 30 characters in body
|
| Jan 29, 2021 at 1:24 | history | edited | GH from MO | CC BY-SA 4.0 |
added 30 characters in body
|
| Jan 28, 2021 at 23:34 | history | answered | GH from MO | CC BY-SA 4.0 |