1. ## First part of radicals question

Let f (x) = x^4 + px^2 + qx + r ∈ Q[x] be an irreducible quartic. Let α1 , . . . , α4 be the roots of f in a splitting field L over Q. Define elements β1 , . . . , β3 ∈ L by

β1 = (α1 + α2 )(α3 + α4 ),
β2 = (α1 + α3 )(α2 + α4 ),
β3 = (α1 + α4 )(α2 + α3 ).

(a) Show that β1 , β2 , β3 are the roots of the cubic (the so-called resolvent of f )
g(x) = x^3 − 2px^2+(p^2 − 4r)x + q^2 ∈ Q[x].

Any help appreciated. (The hint in the question says to use the theory of symmetric functions...)

2. The theorem of symmetric polynomials says:
$\displaystyle f(x) = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4) = x^4+\sum_{k=1}^{4}(-1)^ks_kx^{4-k} = x^4+px^2+qx+r$
$\displaystyle s_1 = \alpha_1+\alpha_2+\alpha_3+\alpha_4$
$\displaystyle s_2 = \alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_1\alpha_4 +\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_ 4$
$\displaystyle s_3 = \alpha_1\alpha_2\alpha_3+\alpha_1\alpha_2\alpha_4+ \alpha_1\alpha_3\alpha_4+\alpha_2\alpha_3\alpha_4$
$\displaystyle s_4 = \alpha_1\alpha_2\alpha_3\alpha_4$

Observe that for f(x) we have: $\displaystyle s_1 = 0, s_2 = p, s_3= -q, s_4=r$

The same way we have that:
$\displaystyle (x-\beta_1)(x-\beta_2)(x-\beta_3) = x^3-(\beta_1+\beta_2+\beta_3)x^2+(\beta_1\beta_2+\beta _1\beta_3+\beta_2\beta_3)x-\beta_1\beta_2\beta_2$

Since $\displaystyle g(x) = x^3-2px^2+(p^2-4r)x+q^2 = x^3-2s_2x^2+(s_2^2-4s_4)x+s_3^2$

You want to show that:

$\displaystyle \beta_1+\beta_2+\beta_3 = 2p (=2s_2)$
$\displaystyle \beta_1\beta_2+\beta_1\beta_3+\beta_2\beta_3 = p^2-4r (= s_2^2-4s_4)$
$\displaystyle \beta_1\beta_2\beta_2 = -q^2 (= -s_3^2)$