Given $f ( x ) = x^n + a_{1} x_{n − 1} + ··· + a_{n − 1} x + a_n$, find the discriminant of the these polynomials

Question

Given $f ( x ) = x^n + a_{1} x_{n − 1} + ··· + a_{n − 1} x + a_n$, find the discriminant of the these polynomials

(1) $x^n+a_{n}$

(2) $x^n+a_{n-1}x$

(3) $x^n+a_{n-1}x+a_n$

For (1), a hint says that we should remember the factorization of $x^n-1$ into $n$ linear terms. But I cannot find how to use it.

And (3) is related to (1) and (2), a hint says that it is related to weighted degree.

I am stucked on (1) and do not know what to do with it, maybe the method of using the hint is considering the $n$ th root of $a_n$, but it seems so weird that I am confused.

Could someone give help on these? Thanks in advance!

EDIT: I use Artin's Algebra as the textbook and in this text the discriminant is defined to be $\prod_{i<j}(x_{i}-x_{j})^2$. May I ask for a derivation from this definition?

Answer 1

Hint for (1): use the expression for the discriminant in term of the roots for a monic polynomial with roots $r_1, r_2, \cdots, r_n$:

$$ D = (-1)^{\frac{1}{2}n(n-1)} \prod_{i \ne j}(r_i-r_j) $$

For the polynomial $x^n+a$ the roots are $r_k = (-a)^{1/n}\,\omega^k$ where $(-a)^{1/n}$ is the principal $n^{th}$ root of $-a$ and $\omega$ is a primitive $n^{th}$ root of unity. Then the discriminant can be calculated as:

$$ \begin{align} D & = (-1)^{\frac{1}{2}n(n-1)} \left((-a)^{1/n}\right)^{n(n-1)} \prod_{i \ne j}(\omega^i-\omega^j) \\[5px] &= (-1)^{\frac{1}{2}n(n-1)} (-a)^{n-1} \,\prod_{i=0}^{n-1} \,\prod_{j \ne i} (\omega^i-\omega^j) \\[5px] & = (-1)^{\frac{1}{2}n(n-1)} \,(-1)^{n-1} \,a^{n-1} \prod_{i=0}^{n-1} \left( \omega^i \prod_{k \ne 0} (1-\omega^k)\right) \\[5px] & = (-1)^{\frac{1}{2}n(n-1)} \,(-1)^{n-1} \,a^{n-1} \,\left(\prod_{i=0}^{n-1} \omega^i \right) \left( \prod_{k \ne 0} (1-\omega^k)\right)^n \\[5px] & = (-1)^{\frac{1}{2}n(n-1)} \,(-1)^{n-1} \,a^{n-1} \,(-1)^{n+1} \,n^n \\[5px] & = (-1)^{\frac{1}{2}n(n-1)} \,a^{n-1} \,n^n \end{align} $$

The penultimate step follows from $1 \cdot \omega \cdot\omega^2 \, \cdots \cdot w^{n-1} = -(-1)^n$ by Vieta's relations for $x^n-1$, and $\prod_{k \ne 0} (1-\omega^k) = n$ because $\prod_{k \ne 0} (x-\omega^k) = (x^n-1)/(x-1) = x^{n-1}+x^{n-2}+\cdots+1$.

Answer 2

In general: To determine the discriminant of the polynomial we use resultant of Two Polynomials $\mathcal{R}( f(x),f'(x)) $ where $f(x)=a_0x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n$, then $f'(x)=na_0x^{n-1}+(n-1)a_1x^{n-2}+...+2a_{n-2}x+a_{n-1}$ and $$\Delta=(-1)^{\frac {n(n+1)}{2}} \frac{1}{a_0} \mathcal{R} (f(x),f'(x)) $$

$\mathcal{R}( f(x),f'(x))=\left| \matrix{a_n&a_{n-1}&a_{n-2}& ...&a_0&0&...&0&0 \\0&a_n&a_{n-1}&a_{n-2}&...&a_0&...&0&0\\...&...&...&...&...&...&...&...&...& \\0&0&...&...&...&...&...&a_1&a_0 \\a_{n-1}&2a_{n-2}&...&na_0&0&0&...&0&0\\0&a_{n-1}&2a_{n-2}&...&na_0&0&0&...&0\\...&...&...&...&...&...&...&...&...&\\ 0&0&...&...&...&...&...&(n-1)a_1&na_0 } \right| $

for example: if $n=2$, then $f(x)=a_0x^2+a_1x+a_2$, then $f'(x)=2a_0x+a_1$, so

$\mathcal{R}( f(x),f'(x))=\left| \matrix{a_2&a_1&a_0\\a_1&2a_0&0\\0&a_1&2a_0 } \right|=4a_2a_0^2-a_0a_1^2 $. Therefore $\Delta=(-1)^3\frac{1}{a_0}(4a_2a_0^2-a_0a_1^2)=a_1^2-4a_2a_0$