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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?

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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.

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

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