Exercise 2.3.5

Proof. The polynomial

has two opposite roots iff

Let

$u$ is symmetric, so is a polynomial in ${\sigma }_1,{\sigma }_2,{\sigma }_3,{\sigma }_4$.

We obtain this polynomial with the following Sage instructions

e = SymmetricFunctions(QQ).e()
e1,e2,e3,e4 = e([1]).expand(4),e([2]).expand(4),e([3]).expand(4),e([4]).expand(4)
R.<x0,x1,x2,x3,y1,y2,y3,y4> = PolynomialRing(QQ, order = ’lex’)
J = R.ideal(e1-y1,e2-y2,e3-y3,e4-y4)
G = J.groebner_basis()
u = (x0+x1)*(x0+x2)*(x0+x3)*(x1+x2)*(x1+x3)*(x2+x3)
var(’sigma_1,sigma_2,sigma_3,sigma_4’)
u.reduce(G).subs(y1=sigma_1, y2 = sigma_2,y3=sigma_3,y4=sigma_4)

So

The evaluation ring homomorphism defined by $x_i\mapsto {\alpha }_i,i=1,2,3,4$ verifies

So $({\alpha }_1+{\alpha }_2)({\alpha }_1+{\alpha }_3)({\alpha }_1+{\alpha }_4)({\alpha }_2+{\alpha }_3)({\alpha }_2+{\alpha }_4)({\alpha }_3+{\alpha }_4)=\mathit{bcd}-b^{2}e-d^2$.

Conclusion : $f=x^4+b{x}^3+c{x}^2+\mathit{dx}+e$ is such that one root is the negative of another iff $\mathit{bcd}-b^{2}e-d^2=0$. □