# Understanding steps of calculating discriminants.

Here is the question I am asking about some steps in its keen answer:

For $n =3,$ write $\Delta^2$ as an element of $A = \mathbb{Q}[e_{1}, e_{2}, e_{3}.]$(manually)

The answer is given below:

What a strange hint. There's a much better way to do it, and I have no idea why the author didn't set $$a_3 = 1$$ from the start.

I learned this argument from David Speyer. Computing the discriminant as a polynomial in the $$e_i$$ is equivalent to computing the discriminant of the polynomial $$f(x) = x^3 - e_1 x^2 + e_2 x - e_3$$. Up to a translation, which does not affect the discriminant, we can remove the quadratic term from this polynomial: if we substitute $$x = y + \frac{e_1}{3}$$, then we get a new polynomial (brace yourself)

\begin{align} g(y) &= \left( y^3 + e_1 y^2 + \frac{e_1^2}{3} y + \frac{e_1^3}{27} \right) - e_1 \left( y^2 + \frac{2e_1}{3} y + \frac{e_1^2}{9} \right) + e_2 \left( y + \frac{e_1}{3} \right) - e_3 \\ &= y^3 + \left( - \frac{e_1^2}{3} + e_2 \right) y + \left( -\frac{2e_1^3}{27} + \frac{e_1 e_2}{3} - e_3 \right). \end{align}

Set $$p = - \frac{e_1^2}{3} + e_2, q = -\frac{2e_1^3}{27} + \frac{e_1 e_2}{3} - e_3$$, so that this new polynomial can be written $$g(y) = y^3 + py + q$$. Now it suffices to compute the discriminant of this new polynomial as a function of $$p$$ and $$q$$, then plug in these expressions for $$p$$ and $$q$$ in terms of the $$e_i$$.

The point of doing this is that we've now removed most of the terms from the discriminant: $$p$$ has degree $$2$$, $$q$$ has degree $$3$$, and $$\Delta^2$$ has degree $$6$$, which implies that it must be some linear combination $$ap^3 + bq^2$$ of $$p^3$$ and $$q^2$$. So we only have two coefficients to compute. We can actually compute these coefficients by computing the discriminant of two specific polynomials:

• Set $$p = -1, q = 0$$. Then $$g(y) = y^3 - y = y(y + 1)(y - 1)$$ has roots $$0, \pm 1$$, so its discriminant is $$\left( (0 - 1)(1 - (-1))( (-1) - 0) \right)^2 = 4$$, which gives $$a = -4$$.
• Set $$p = 0, q = -1$$. Then $$g(y) = y^3 - 1 = (y - 1)(y - \omega)(y - \omega^2)$$ has roots $$1, \omega, \omega^2$$ where $$\omega$$ is a primitive third root of unity (satisfying $$1 + \omega + \omega^2 = 0$$), so its discriminant is $$\left( (1 - \omega)(\omega - \omega^2)(\omega^2 - 1) \right)^2 = (1 - \omega)^6 = (- 3 \omega)^3 = -27$$, which gives $$b = -27$$.

So we get $$\boxed{ \Delta^2 = -4p^3 - 27q^2 }$$ (already a useful formula in practice), and substituting $$e_1, e_2, e_3$$ in gives the full cubic discriminant (of more dubious value; I have never used it).

My questions are:

In this part of the solution: " * Set $$p = -1, q = 0$$. Then $$g(y) = y^3 - y = y(y + 1)(y - 1)$$ has roots $$0, \pm 1$$, so its discriminant is $$\left( (0 - 1)(1 - (-1))( (-1) - 0) \right)^2 = 4$$, which gives $$a = -4$$.

• Set $$p = 0, q = -1$$. Then $$g(y) = y^3 - 1 = (y - 1)(y - \omega)(y - \omega^2)$$ has roots $$1, \omega, \omega^2$$ where $$\omega$$ is a primitive third root of unity (satisfying $$1 + \omega + \omega^2 = 0$$), so its discriminant is $$\left( (1 - \omega)(\omega - \omega^2)(\omega^2 - 1) \right)^2 = (1 - \omega)^6 = (- 3 \omega)^3 = -27$$, which gives $$b = -27$$. " 1 -in the first paragraph why is the discriminant looks like that $$\left( (0 - 1)(1 - (-1))( (-1) - 0) \right)^2 = 4$$,, what are the first terms $$0,1,(-1)$$ in the brackets representing and similarly what are the second terms $$0,1,(-1)$$ representing? should not $$a = -2$$ instead?

2-In the second paragraph, why we are factorizing $$y^3 - 1$$ in terms of roots of unity, why we needed that? why this root of unity satisfies $$1 + \omega + \omega^2 = 0$$?

3-In the second paragraph, How in generally we calculate the discriminant of roots of unity?

Could anyone help me answer those questions please?

I think it is always easier to calculate the discriminant of a trinomial by noting that $$\Delta^2$$ is equal to the Vandermonde Matrix times its transpose.

If the roots are $$\alpha_1,\alpha_2,\dots,\alpha_n$$ then we have $$\Delta^2= \begin{bmatrix} 1 & 1 &\dots &1\\ \alpha_1&\alpha_2&\dots&\alpha_n\\ \vdots &\vdots &\ddots&\vdots\\ \alpha_1^{n-1}&\alpha_2^{n-1}&\dots&\alpha_n^{n-1}\\ \end{bmatrix} \begin{bmatrix} 1 & 1 &\dots &1\\ \alpha_1&\alpha_2&\dots&\alpha_n\\ \vdots &\vdots &\ddots&\vdots\\ \alpha_1^{n-1}&\alpha_2^{n-1}&\dots&\alpha_n^{n-1}\\ \end{bmatrix}^{T}$$ which is equal to $$\begin{bmatrix} s_0 &s_1 &\dots&s_{n-1}\\ s_1& s_2 &\dots &s_{n}\\ \vdots &\vdots&\ddots&\vdots\\ s_{n-1}&s_{n}&\dots&s_{2n-2} \end{bmatrix}$$ where the $$s_k:=\sum_{i=1}^{n}\alpha_{i}^{k}$$ are the sums of the $$k$$-th powers of the roots. Hence $$s_0=\sum_{i=1}^{n}1=n$$, $$s_1=\sum_{i=1}^{n}\alpha_i=e_1$$ (in your notation), and so on.

In the case of $$X^{3}-1$$ the roots are $$1,\omega,\omega^2$$, where $$\omega:=e^{\frac{2\pi i}{3}}$$. As the coefficient of $$X^2$$ is $$0$$ we have that the sum of the roots is $$0$$.

Hence we see that $$s_0=3$$, $$s_1=0$$, $$s_2=1+\omega^2+\omega^4=1+\omega^2+\omega=0$$; and then recursively $$s_3=3$$, $$s_4=0$$ and so $$\Delta^2= \begin{bmatrix} 3 &0 & 0\\ 0& 0 & 3\\ 0 & 3& 0\\ \end{bmatrix}=-27.$$

We could as easily have done the cubic $$X^3+pX+q$$ this way: find $$s_1$$ and $$s_2$$ by hand the rest follow recursively as $$X^{k+3}+pX^{k+1}+qX^{k}=0$$ at the roots.

It is almost as easy to work out the discriminant of $$X^n-1$$: I leave it for you to check.

• In the case of $X^{3}-1$ why the roots are $1,\omega,\omega^2$?
– user777833
Oct 15, 2020 at 18:05
• Does $k$ ranges from 0 to $n-1$?
– user777833
Oct 15, 2020 at 18:06
• I do not see why "$s_0=3$, $s_1=0$, $s_2=0$" .... could you please elaborate that?
– user777833
Oct 15, 2020 at 18:08
• Can you please explain to me what is the way used in the link I mentioned?
– user777833
Oct 15, 2020 at 18:08
• I have annotated the answer. I think you should ask the person who answered your question last time about their solution. Oct 16, 2020 at 9:11