Finding discriminant of a monic polynomial.

Question

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.

Thank you very much for your valuable time for reading.

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

Answer 1

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