Discriminant of $f(x)=x^3+ax+b$

Question

Suppose we have the polynomial $f(x)=x^3+ax+b$, with roots $\alpha, \beta, \gamma$ in $\mathbb{C}$, and let $\Delta = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$.

Is there any quick way of show that the discriminant $\Delta^2$ is $-4a^3-27b^2$?

It seems as though we need to exploit certain properties of symmetric functions. Is there some way of writing out $\Delta$ in terms of elementary symmetric polynomials perhaps?

Answer 1

$\Delta$ is not expressible in terms of elementary symmetric polynomials (because it is not invariant by all permutations of the roots, and so $-4a^3-27b^2$ is not a square).

The standard way is to develop $\Delta^2$ and write that in terms of elementary symmetric polynomials.

I guess you can take a shortcut by noticing that $\Delta^2$ is homogeneous of degree $6$ in the roots, so it has to be a linear combination of $(\alpha\beta\gamma)^2$ and $(\alpha\beta+\alpha\gamma+\beta\gamma)^3$ (the other products involving $\alpha+\beta+\gamma$ vanish because this is $0$ in our case)

So you only have to figure out the two coefficients. Then by picking some custom polynomials for which you know the roots, like $X(X-1)(X+1)$ and $(X-1)(X-2)(X+3)$, the discriminant is simple to compute, $a$ and $b$ are too, so you should be able to deduce the right coefficients from there.

Answer 2

As the others have said, you need to consider $Δ^2$. Then by sorting the factors you can assemble them to form the product (assuming for the moment that none of the roots is zero, i.e., $b\ne 0$) \begin{align} -Δ^2&=Res_x(f(x),f'(x)) =f'(α)f'(β)f'(γ) =(3α^2+a)(3β^2+a)(3γ^2+a)\\[.5em] &=-(2a+\tfrac{3b}α)(2a+\tfrac{3b}β)(2a+\tfrac{3b}γ) =\frac1b(2aα+3b)(2aβ+3b)(2aγ+3b) \end{align} which you can then combine using $$ 0=(α+β+γ)^2=α^2+β^2+γ^2+2a $$

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One can see the initial identity in terms of formal derivatives of polynomials or ordinary derivatives of polynomial functions with complex coefficients. If extending the argument to a splitting field for a polynomial over a general domain, the formal derivative only can be used. \begin{align} f'(x)&=3x^2+a =\frac{d}{dx}(x-α)(x-β)(x-γ)\\ &=(x-α)(x-β)+(x-β)(x-γ)+(x-α)(x-γ). \end{align} Then $$ f'(α)=3α^2+a=(α-β)(α-γ) $$ etc. so that the product of the three derivatives, after three sign changes, is $Δ^2$, \begin{align} f'(α)f'(β)f'(γ) &=(α-β)(α-γ)\,(β-α)(β-γ)\,(γ-α)(γ-β)\\ &=(-1)^3\,(α-β)^2(β-γ)^2(γ-α)^2 \end{align}