Silverman, p. 104:
Show that the polynomials $$f=x^4−b_4x^2−2b_6x−b8 \qquad \text{and}\qquad g=4x^3+b_2x^2+2b_4x+b_6$$ appearing in the duplication formula (III.2.3d) are relatively prime if and only if the discriminant of the associated Weierstrass equation is nonzero.
I tried to do this in the most straightforward way with computer algebra, and I am surprised that it didn't work. Probably, there is a nice way to do it by hands, but I don't understand how my method gave me the wrong answer. I used SAGE to compute the resultant of $f$ and $g$ with respect to $x$, and I got:
SAGE: resultant$(f,g,x)=$ $-3*b_2^2*b_4^2*b_6^2 + 4*b_2^3*b_6^3 + 4*b_2^2*b_4^3*b_8 - 6*b_2^3*b_4*b_6*b_8 + b_2^4*b_8^2 + 108*b_4^3*b_6^2 - 162*b_2*b_4*b_6^3 - 144*b_4^4*b_8 + 240*b_2*b_4^2*b_6*b_8 + 6*b_2^2*b_6^2*b_8 - 48*b_2^2*b_4*b_8^2 + 729*b_6^4 - 1296*b_4*b_6^2*b_8 + 384*b_4^2*b_8^2 + 192*b_2*b_6*b_8^2 - 256*b_8^3$
My hope was that the discriminant, which is given by the formula below would be a factor of resultant$(f,g,x)$, but apparently not. I used the gcd and the factor function, and I got:
SAGE: gcd(resultant$(f,g,x)$, $D$)
$1$
SAGE: factor(resultant$(f,g,x)$)
resultant$(f,g,x)$