Timeline for What is the essence of the constant factor in the standard definitions of the discriminant?
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| Sep 7, 2017 at 9:43 | comment | added | js21 | Moreover the exponent $2m-2$ is optimal since the discriminant of $f_0 x^m + f_1 x^{m-1} + f_m$ is equal to $\pm m^m f_0^{m-1} f_m^{m-1} \pm (m-1)^{m-1} f_{1}^{m} f_m^{m-2}$ which is not divisible by $f_0$. | |
| Sep 7, 2017 at 9:16 | comment | added | js21 | The resultant polynomial $R(f,f')$ is equal to $\pm f_0^{2m-1} \Delta^2$. However $R(f,f')$ is divisible by $f_0$, since the first row of the Sylvester matrix is. Hence $f_0^{-1} R(f,f') =\pm f_0^{2m-2} \Delta^2 $ is a polynomial. | |
| Sep 6, 2017 at 18:18 | comment | added | Mikhail Goltvanitsa | Thank you. And, why the exponent is exactly $2m-2$? | |
| Sep 6, 2017 at 17:37 | history | answered | Robert Israel | CC BY-SA 3.0 |