Resultant of two polynomials 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!
 A: 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.
A: 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.
A: 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
$$\operatorname{Res}(F,G)=\prod_iG(\alpha_i-\alpha)=\prod_ig(\alpha_i)=\operatorname{Res}(f,g)\ .$$
A: Yes. This is the particular case
$$
\pmatrix{1 & \alpha\\
0 & 1}
$$
of the more general $SL_2$ invariance of the resultant under transformations
$$
x\mapsto \frac{ax+b}{cx+d}
$$
(with clearing of denominators understood).
