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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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    For any nonzero polynomial $f$ over a field $k$, one of the most common *definitions* of $\operatorname{disc}(f)$ is as $\operatorname{res}(f,f')$. Evidently this is not *your* definition of $\operatorname{disc}(f)$, or you would not be asking the question...So, what is your definition of $\operatorname{disc}(f)$? The one with the differences of the roots?2012-02-15
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    By the way, googling -- discriminant of a polynomial, resultant -- the second hit is this link http://www.win.tue.nl/~aeb/2WF02/resultant.pdf, which explains why the resultant of two polynomials as defined by a determinant is equal to the expression involving the differences of the roots (up to a constant involving the leading coefficients).2012-02-15
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    @PeteL.Clark I've added what definition of discriminant I want to use.2012-02-15
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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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