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Let $R$ be a commutative ring and $R[x]$ the polynomial ring over $R$. Let $f,g \in R[x]$. $D(fg)$ defines the discriminant of $fg$ and $res(fg)$ the respective resultant. I need to prove that the equation

$D(fg)=D(f)D(g)res(f,g)^2$

holds.

So far I fiddled around to no avail. My ideas so far are to use the properties of the definition of the Sylvester-Matrix that is used in $D(f)$ and $res(f)$ and the determinant.
The definitions of $D(f)$ and $res(f,g)$ I'm working with are as follows: $D(f)=(-1)^ \frac{n(n-1)}{2}*res(f,f')$ and $res(f,g)=det(S(f,g))$. $S(f,g)$ is the Sylvester-Matrix of $f$ and $g$, $f'$ denotes the derivative of $f$. Help is much appreciated.

1 Answers 1

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Finally, the answer: We're gonna use two theorems which I am not gonna prove. They are the following:

(1) For f and g in $K[x]$ with $f=\prod_{i=1}^{n} (x-a_i)$ and $g=\prod_{i=1}^{m} (x-b_i)$ it follows that $res(f,g)=\prod_{i=1}^{n}\prod_{j=1}^{m} (a_i-b_i)$

(2) For $f=\prod_{i=1}^{n} (x-a_i) \in K[x]$ it follows that $D(f)=\prod_{1\leq i

Now in the field extension $R'$ we have $f=\prod_{i=1}^{n} (x-a_i)$ and $g=\prod_{i=1}^{m} (x-b_i)$. For $1\leq k \leq n+m$ define $c_k:=a_k$ for $k\leq n$ and $c_k=b_{k-n}$ for $k\geq n+1$. It follows: $D(fg)=\prod_{1\leq k

  • 0
    What field extension are you talking about? – 2017-02-19
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    It seems to me that you proved the equation for only fields. – 2017-02-19
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    The splitting field, to be exact. For the proof to work we briefly "enter" it, to use the definition of the discriminant with the polynomials in linear-factor form but "exit" the splitting field when we "merge" the polynomials back into $D(f)$, $D(g)$, and $res(f,g)^2$ – 2017-02-20