Greetings fellow Mathematics enthusiasts. I was hoping someone could offer me some advice on proving the following statement about the discriminant of a polynomial with degree $n$.
Let $f(x)=x^n+ax+b$ be irreducible over $\mathbb{Q}$ and $\alpha$ a root of $f$, where $a,b\in \mathbb{Q}, a \neq 0, n \geq 2$. Prove that disc$(\alpha) = (-1)^\frac{n(n-1)}{2}\left( (-1)^{1-n}(n-1)^{n-1}a^n+n^nb^{n-1}\right)$. My professor left us a hint that reads:
Show that $f'(\alpha)= \frac{-((n-1)a \alpha+nb)}{\alpha}$. Now find $N((n-1)a \alpha+nb)$ by noticing that it is a root of $\left( \frac{x-nb}{(n-1)a}\right)^n + a\left( \frac{x-nb}{(n-1)a}\right)+b$. This power-reducing technique was introduced in class to find the discriminant above, but with $n=3$. As for the Norm calculation, we are told to use the constant term of the minimal polynomial.
Here is my work so far:
Given $f'(\alpha)=\frac{-((n-1)a \alpha+nb)}{\alpha}$, let $\beta=(n-1)a\alpha+nb$. So we have $\alpha = \frac{\beta-nb}{(n-1)a}$. Back substituting into the original equation,
$\left( \frac{\beta-nb}{(n-1)a}\right)^n + a\left(\frac{\beta-nb}{(n-1)a}\right) + b \Rightarrow \left(\beta-nb \right)^n + \left((n-1)a\right)^n\frac{\beta-nb}{n-1} + \left((n-1)a\right)^nb=0$. Upon simplifying a bit, we are left with
$\left(\beta-nb\right)^n + \left((n-1)\right)^{n-1}a^n(\beta-nb) + \left((n-1)a\right)^nb=0$.
Using the Binomial Theorem, we expand the first term and obtain
$\left[\beta^n-nb\beta^{n-1} + \ldots + (-nb)^n\right] + \left((n-1)\right)^{n-1}a^n(\beta-nb) + \left((n-1)a\right)^nb=0$.
Could somebody please explain to me why this approach is or is not correct?
Thanks in advance, any suggestions would be greatly appreciated.
Undoubtedly you have seen the formula $$ \operatorname{disc}(\alpha)=(-1)^{n(n-1)/2}\prod_{i=1}^nf'(\alpha_i), $$ where $\alpha_i$, $i=1,2,\ldots,n,$ are the conjugates of $\alpha$. The relations arising from the factorization $$ f(x)=\prod_{i=1}^n(x-\alpha_i)\qquad(*) $$ will also play a role.
The norm calculation that your professor talked about is equivalent to my use of $f(q)$ below. Basically it means that for a rational number $q$ we have $f(q)=N(q-\alpha)$. You may have done related tricks in class, so I cannot tell how familiar you are with this technique.
I first describe how I would do this (this sounds familiar actually - I'm fairly sure I have done this exercise at some point). Then at the bottom I make a few comments about the material you posted. I'm afraid I'm not sure that I will answer your questions.
The relations $$ f'(\alpha_i)=\frac{-n(a\alpha_i+b)+a\alpha_i}{\alpha_i} $$ that hold for all $i$ are the key. We get $$ \prod_{i=1}^nf'(\alpha_i)=\prod_{i=1}^n\frac{a(1-n)\alpha_i-nb}{\alpha_i}.\qquad(**) $$ Here the product of the denominators is $\prod_{i=1}^n\alpha_i=(-1)^nb$, because that product emerges as the constant term of the minimal polynomial $f$ (= the norm of $\alpha$ up to a sign). In the numerator let's write $$ a(1-n)\alpha_i-nb=-a(1-n)\left(\frac{nb}{a(1-n)}-\alpha_i\right) $$ Here the fraction $q=nb/(a(1-n))$ is independent of $i$. Thus the factorization $(*)$ tells us that $$ \begin{aligned} \prod_{i=1}^n(a(1-n)\alpha_i-nb)&=(-1)^n(a(1-n))^n\prod_{i=1}(q-\alpha_i)\\ &=(-1)^na^n(1-n)^nf(q). \end{aligned} $$ Combining this with $(**)$ tells us that $$ \prod_{i=1}^nf'(\alpha_i)=\frac{a^n(1-n)^n}{b}f(q). $$ I'm sure you can take it from here.
You can also use the fact that you mentioned: $$g(x):=\left( \frac{x-nb}{(n-1)a}\right)^n + a\left( \frac{x-nb}{(n-1)a}\right)+b$$ is the minimal polynomial of $(n-1)a\alpha +nb=a(1-n)(q-\alpha)$. You need to first scale $g(x)$ so that it becomes monic. Then you need to expand and find the constant term of that scaled $g$. That can be used much the same way as I used $f(q)$ (that I didn't bother to calculate!). If you pick the terms that do not contain $\beta$ from the left hand side of the equation on the third line from the bottom, you do get this. I'm not sure why you used that $\beta$ though.