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Let $Res(f,g)$ be a resultant of two polynomials $f(x)$ and $g(x)$. Is it true that resultant does not change under a linear change of coordinates $x\mapsto x+\alpha$?

Thanks a lot!

3 Answers 3

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Recall the definition of the resultant of two polynomials as the product of differences of their roots. Under the global shift, $x \mapsto x + \alpha$, the differences remain invariant. Hence the resultant remains unchanged as well.

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By the definition given on wikipedia, $\operatorname{Res}(f,g)=\prod_{(x,y)\in\overline k, f(x)=g(y)=0}(x-y).$

Since we are interested in the roots inside $\overline k,$ we factorize $f(x)=a(x-a_1)\cdots(x-a_m)$ and $g(x)=b(x-x_1)\cdots(x-b_n)$ over $\overline k,$ so that

$\operatorname{Res}(f,g)=\prod_{i,j}(a_i-b_j).$

Under the change of variables $x'= x+c,$ we get $f(x')=a(x'-c-a_1)\cdots(x'-c-a_n)=a(x'-(c+a_1))\cdots(x'-(c+a_n))$, and similarly for $g(x')$.

Thus $\operatorname{Res}(f(x'),g(x'))=\prod_{i,j}((c+a_i)-(c+b_j))=\prod_{i,j}(a_i-b_j)$ does not change.

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Yes that's true.

Let's say that $ f(x)=\prod_i(x-\alpha_i) $.

Set $F(x)=f(x+\alpha)$ and $G(x)=g(x+\alpha)$.

Then $F(x)=\prod_i(x-(\alpha_i-\alpha))$.

We have that $\operatorname{Res}(f,g)=\prod_ig(\alpha_i)$ so $\operatorname{Res}(F,G)=\prod_iG(\alpha_i-\alpha)=\prod_ig(\alpha_i)=\operatorname{Res}(f,g)\ .$