Timeline for How to classify a plane complex curve?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2018 at 18:24 | vote | accept | Jianrong Li | ||
| Aug 2, 2018 at 15:46 | answer | added | Xarles | timeline score: 2 | |
| Aug 2, 2018 at 15:30 | history | edited | Jianrong Li | CC BY-SA 4.0 |
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| Aug 2, 2018 at 15:23 | comment | added | Xarles | So, the same for $b_2=p_2$, I guess.... | |
| Aug 2, 2018 at 15:14 | history | edited | Jianrong Li | CC BY-SA 4.0 |
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| Aug 2, 2018 at 15:14 | comment | added | Jianrong Li | @Xarles, thank you very much. Yes, $b_1=p_1$. I will correct it. | |
| Aug 2, 2018 at 15:13 | comment | added | Xarles | Sorry, but $b_1=p_1$ or they are different parameters? Because you did not defined at the beginning. | |
| Aug 2, 2018 at 11:38 | comment | added | Jianrong Li | @Philipp Lampe, thank you very much for your kind help. I will try to do the computations in Sage. | |
| Aug 2, 2018 at 10:38 | comment | added | Philipp Lampe | Macaulay2 or Sage can tell whether a curve is smooth and compute the genus. The answers depend on the chosen constants. For instance, the curve is not smooth for $p_1=b_1=0$ and $p_2=t_2=1$. Note that a curve of genus $2$ is not an elliptic curve since they have genus $1$. | |
| Aug 2, 2018 at 8:30 | history | edited | Jianrong Li | CC BY-SA 4.0 |
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| Aug 2, 2018 at 8:23 | history | edited | Jianrong Li | CC BY-SA 4.0 |
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| Aug 2, 2018 at 8:21 | comment | added | Jianrong Li | @abx, thank you very much. I forgot some terms in the curve. I will add them. | |
| Aug 2, 2018 at 8:21 | comment | added | Jianrong Li | @PhilippLampe, thank you very much. I will correct the post. | |
| Aug 2, 2018 at 8:08 | comment | added | Philipp Lampe | I cannot follow the construction of the Newton polygon. Where does the term $(0,2)$ come from? Why is there no term $(1,1)$ (coming from the term on the right hand side)? The line $\{\,(\lambda,1)\mid \lambda\in\mathbb{R}\,\}$ seems to be a boundary component of the Newton polygon, so why does $(2,1)$ lie in the interior? Moreover, the answer depends on the choice of the constants since points in the Newton polygon vanish when some of the constants are zero. | |
| Aug 2, 2018 at 7:59 | comment | added | abx | It is certainly not smooth, not even irreducible: it contains the line $c_2=0$. | |
| Aug 2, 2018 at 7:49 | history | asked | Jianrong Li | CC BY-SA 4.0 |