Timeline for The degree of the cube root of the $j$-invariant
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 6, 2018 at 14:20 | comment | added | Shimrod | Now I see that it is obvious, thank you. | |
| May 6, 2018 at 13:12 | comment | added | Luca Ghidelli | $t>0$ because it is a polynomial of degree 2 with negative discriminant. If you want, you can see it in one variable: t=c^2(x^2-3x+N) with x=d/c (a rational, hence real number). Now, y=x^2-3x+N is a parabola that doesn't cross the x-axis, so it has the graph strictly on the upper half plane. Moreover, if c is nonzero, c^2 is positive as well. So t is strictly positive. | |
| May 6, 2018 at 12:40 | vote | accept | Shimrod | ||
| May 6, 2018 at 12:40 | comment | added | Shimrod | Thank you for your answer. This seems to work. Could you please explain why should $t>0$? | |
| May 6, 2018 at 11:57 | history | answered | Luca Ghidelli | CC BY-SA 4.0 |