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Over the complex numbers, I'm familiar with the fact that the discriminant of a polynomial $f$ and the resultant of $f$ and f' are equal.

Now say you have an arbitrary polynomial $ f(x)=a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0 $ over a fixed field $k$ with algebraic closure $\bar{k}$. Is is still true that \text{disc} f=\text{res} (f,f')? I'm curious because suppose the resultant is defined in terms of the determinant of a matrix $ \left|\begin{matrix} a_m & a_{m-1} & \dots & a_0 & 0 &\dots \\ 0 & a_m & a_{m-1} &\dots & \dots & \dots \\ 0 & 0 & a_m & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix}\right| $ and not as the product of differences of roots. Here I'm taking the definition of discriminant to be the product of differences of roots. I'm sure the formulations are the same, but how can one still make the same conclusion? Thanks.

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    Okay. With this (also very standard) definition, the link I gave above seems to answer your question. Are you satisfied with this?2012-02-15

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