Finding discriminant of a monic polynomial.

I have now engaged in studying Galois Theory from NPTEL online lecture series which encompasses Finite Fields and Galois Theory. While watching the $$48$$-th lecture on Discriminant of a Polynomial a proposition has been discussed which I failed to understand properly.

Before going to the main proposition let us first define formally the discriminant of a polynomial.

Let $$K$$ be a field. Let $$f_n$$ denote general monic polynomial of degree $$n$$ i.e. it is of the form $$f_n = (X-X_1)(X-X_2) \cdots (X-X_n).$$

Let $$V(X_1,X_2, \cdots, X_n)$$ denote the Vandermonde deteminant in $$X_1,X_2, \cdots X_n.$$ So $$V(X_1,X_2, \cdots , X_n) = \prod\limits_{1 \leq i < j \leq n} (X_j - X_i).$$ Now the discriminant of $$f_n$$ is denoted by $$D(f_n)$$ and it is defined as $$D(f_n):= {V(X_1,X_2, \cdots , X_n)}^2 = \prod\limits_{1 \leq i < j \leq n} {(X_j - X_i)}^2.$$

Now let us take any monic polynomial $$f \in K[X]$$ of degree $$n.$$ Let $$f=X^n + a_1 X^{n-1} + \cdots + a_n.$$ Then by Kronecker's theorem $$\exists$$ a finite field extension $$L|K$$ such that $$f$$ splits completely into linear factors in $$L[X].$$ Let $$x_1,x_2, \cdots , x_n$$ be the zeros of $$f$$ lying in $$L.$$ Then it is clear that $$(-1)^r a_r = S_r (x_1,x_2,\cdots , x_n)$$ for $$r=1,2, \cdots , n$$ where $$S_r$$ is the $$r$$-th elementary symmetric polynomial in $$n$$-variables $$X_1,X_2, \cdots , X_n$$ i.e. $$S_r = \sum\limits_{1 \leq i_1 < i_2 < \cdots < i_r \leq n} X_{i_1} X_{i_2} \cdots X_{i_n}$$ for $$r=1,2, \cdots , n.$$ Now the discriminant of $$f$$ is denoted by $$D(f)$$ and is defined as \begin{align*} D(f) & = D(f_n) (-a_1, \cdots , (-1)^r a_r , \cdots , (-1)^n a_n ) \\ & = D(f_n) (S_1(x_1,x_2, \cdots , x_n), S_2(x_1,x_2, \cdots , x_n), \cdots , S_n (x_1,x_2, \cdots , x_n)). \end{align*}

By Fundamental Theorem of Symmetric Polynomials it is easy to show that $$D(f) \in K.$$ Now let us come back to the main proposition.

$$\textbf {Proposition} :$$ Let $$f(X) \in K[X]$$ be a monic polynomial of degree $$n$$ and $$x_1,x_2, \cdots , x_n \in L$$ be all zeros of $$f$$ in a finite field extension $$L|K.$$ Then $$D(f)= {V(x_1,x_2, \cdots , x_n)}^2 = \prod\limits_{1 \leq i < j \leq n} (x_j - x_i)^2.$$

In the proof of the above proposition the instructor wrote down an equality without giving any proper reasoning behind it. He said that $$D(f_n) (-a_1, \cdots , (-1)^r a_r , \cdots ,(-1)^n a_n ) = D(f_n) (x_1,x_2, \cdots , x_n).$$

But why is it always the case? The thing what he wrote implies $$D(f_n)(x_1,x_2, \cdots , x_n) = D(f_n) (S_1(x_1,x_2, \cdots ,x_n), S_2(x_1,x_2. \cdots , x_n), \cdots , S_n (x_1,x_2, \cdots ,x_n)).$$

But I don't understand why it necessarily holds. For instance let $$K= \Bbb Q$$ and $$L=\Bbb Q (\sqrt 2).$$ Let $$f=X^2-2 \in \Bbb Q[x].$$ Then $$f$$ splits completely into linear factors in $$L[X].$$ The zeros of $$f$$ are $$\pm \sqrt 2 \in L.$$ Let $$x_1 = \sqrt 2$$ and $$x_2 = -\sqrt 2.$$ Then $$S_1(x_1,x_2) = x_1 + x_2 = \sqrt 2 - \sqrt 2 = 0$$ and $$S_2(x_1,x_2) = x_1x_2 = \sqrt 2 (- \sqrt 2) = -2.$$ If the equality holds then we must have $$D(f_2)(\sqrt 2 , - \sqrt 2) = D(f_2) (0,-2).$$ But $$D(f_2) (\sqrt 2, - \sqrt 2) = 8 \neq 4 = D(f_2) (0,-2).$$ So the equality is in general false. So ultimately we get a false proof of the above proposition.

How do I manage to overcome the mistake in the lecture to prove the above proposition? Any suggestion regarding this will be highly appreciated.

Source $$:$$ https://youtu.be/PPI_3yVTHzQ?list=PLOzRYVm0a65dsCb_gMYe3R-ZGs53jjw02&t=1219

What I observed is that the actual problem lies in the definition of discriminant of a monic polynomial. Below is a way to prove the desired proposition by redefining the discriminant of a monic polynomial properly in the following way $$:$$

Let us first state the following theorem due to Jacobi without proof (the proof is very simple thoough!)

Theorem $$:$$ Let $$V = V(X_1,X_2, \cdots , X_n) = \prod\limits_{1 \leq i < j \leq n} (X_j - X_i) \in K[X_1,X_2, \cdots , X_n),$$ the Vandermonde's determinant in $$n$$ unknowns $$X_1,X_2, \cdots , X_n.$$ Then for any $$\sigma \in S_n$$ $$\sigma (V) = \text{sgn} (\sigma)\ V$$ where $$\text {sgn} (\sigma)$$ is defined as follows $$:$$

$$\text {sgn} (\sigma) = \left\{ \begin{array}{ll} 1 & \quad \text {if}\ \sigma\ \text {is even} \\ -1 & \quad \text {if}\ \sigma\ \text{is odd} \end{array} \right.$$

With the help of the above theorem it is easy to see that $$D(f_n),$$ the discriminant of the general monic polynomial of degree $$n,$$ is fixed by every permutation $$\sigma \in S_n.$$ Because $$D(f_n) = V^2 = \prod\limits_{1 \leq i < j \leq n} (X_j - X_i)^2 \in K[X_1,X_2, \cdots , X_n].$$ So for any $$\sigma \in S_n$$ when it extends to an automorphism of $$K(X_1,X_2, \cdots ,X_n)$$ defined by $$X_i \mapsto X_{\sigma(i)}$$ for all $$i=1,2,\cdots , n$$ and leaving all elements of $$K$$ fixed then we have $$\sigma (D(f_n)) = \sigma (V^2) = {\sigma (V)}^2 = V^2,$$ because for any permuatation $$\sigma \in S_n$$ we have $${\text {sgn}(\sigma)}^2 = 1.$$ This shows that $$D(f_n)$$ is a symmetric polynomial in $$X_1,X_2, \cdots , X_n.$$ So by Fundamental theorem of Symmetric Polynomials (also known as Newton's theorem) it follows that $$\exists$$ $$D \in K[X_1,X_2, \cdots , X_n]$$ such that $$D(f_n) = D(S_1,S_2, \cdots , S_n)$$ where $$S_i$$ is the $$i$$-th elementary symmetric polynomial in $$X_1,X_2, \cdots , X_n.$$ Now let $$f = X^n + a_1 X^{n-1} + \cdots + a_n \in K[X]$$ be a monic polynomial. Let us denote discriminant of $$f$$ by $$\text {Disc} (f)$$ (for avoiding confusion with $$D$$ I already defined). Then $$\text {Disc} (f)$$ is defined as follows $$:$$ $$\text {Disc} (f) : = D(-a_1, \cdots , (-1)^i a_i, \cdots , (-1)^na_n).$$

With the help of the revised definition of Discriminant of a Monic Polynomial it is now very easy to prove the desired proposition.

Let $$x_1,x_2, \cdots , x_n$$ be the zeros of $$f$$ lying in some finite field extension $$L|K.$$ Then we first note that $$S_r (x_1,x_2, \cdots , x_n) = (-1)^r a_r$$ for $$r=1,2, \cdots , n.$$ Then we have \begin{align*} \prod\limits_{1 \leq i < j \leq n} (x_j - x_i)^2 & = D(f_n) (x_1,x_2, \cdots , x_n)\\ & = D(S_1(x_1,x_2, \cdots , x_n), S_2(x_1,x_2, \cdots , x_n), \cdots , S_n(x_1,x_2, \cdots , x_n))\\ & = D(-a_1, \cdots , (-1)^i a_i , \cdots , (-1)^na_n)\\ & = \text {Disc} (f). \end{align*}

So we have $$\text {Disc} (f) = \prod\limits_{1 \leq i < j \leq n} (x_j - x_i)^2 = {V(x_1,x_2, \cdots , x_n)}^2,$$ as required.

This completes the proof of the proposition.

QED

• Yeah sorry for the typo. Fixed it. Can you please check now @ancientmathematician? – math maniac. Dec 5 '19 at 8:31