(From Artin)Prove the general formula of discriminant of $x^n + px + q$ for $n ≥ 2$

Question

Here is a proof http://www.csie.ntu.edu.tw/~b01902113/artin-sols.pdf and the proof is on page 84. I am searching for help to prove lemma 16.2.4a which is:

I am able to follow the prove before the author conclude that $r(n)=s(n−1)$. I am not able to derive $r(n)=s(n-1)$ from the formula above.

And after this, I cannot follow the argument in this whole part

I cannot understand what $g(n)$ actually is andhow can we get $(-1)^{\frac{n(n-1)}{2}}\prod_{i=1}^{n}\prod_{i \ne j}(\xi_{i}-\xi_{j})$

I know maybe there are just some algebraic manipulations but I am not able to get them so far.Could someone please explain? Thanks in advance!

Answer 1

I cannot understand what $g(n)$ actually is

That's a typo, and it should read $s(n)$ instead of $g(n)\,$. From the previous paragraph:

$$\,D_n (p, q) = s(n−1)\,p^n + s(n)\,q^{n−1}\,$$

Writing it for $\,p=0\,$ and $q=-1\,$ gives:

$$\,D_n (0, -1) = s(n−1) \cdot 0^n + s(n) \cdot (-1)^{n−1}=(-1)^{n-1}\,s(n)\,$$

and how can we get $(-1)^{\frac{n(n-1)}{2}}\prod_{i=1}^{n}\prod_{i \ne j}(\xi_{i}-\xi_{j})$

See this other answer.