1. ## Discriminant

Consider $\displaystyle f(x)\in F[x]$. Define its discriminant to be $\displaystyle D=\prod_{i<j}(\alpha_i-\alpha_j)^2$, where $\displaystyle \alpha_i$ are the roots of $\displaystyle f(x)$ in its splitting field $\displaystyle K$.

I'm having trouble seeing why $\displaystyle D\in F$. I know it has to do with symmetric functions, but for some reason I can't quite see it. I think I'm missing something small here.

Thanks!

2. Originally Posted by chiph588@
Consider $\displaystyle f(x)\in F[x]$. Define its discriminant to be $\displaystyle D=\prod_{i<j}(\alpha_i-\alpha_j)^2$, where $\displaystyle \alpha_i$ are the roots of $\displaystyle f(x)$ in its splitting field $\displaystyle K$.

I'm having trouble seeing why $\displaystyle D\in F$. I know it has to do with symmetric functions, but for some reason I can't quite see it. I think I'm missing something small here.

Thanks!

Hmmm...it's not hard but it's not easy, either: if you know that the resultant of f(x) is in F and that the discriminant of f(x) is the resultant times some power of the leading coefficient of f(x) times (-1) to some power then we're done, otherwise we can try:

1) The discriminat is a symmetric function on the roots of f(x) (this is trivial)

2) If s_1,...,s_n are the symmetric pol's on the roots of f(x), then any symmetric pol. on these roots is in F[s_1,...,s_n] = the ring of pol's on s_1,...,s_n.

Of course, the problem is 2 and its proof isn't immediate (and I can't remember it right now, but some book must have it). I think the trick was about partial ordering the pol's in n variables in some way.

Try to work on this, and I'll try to make some time to consult some books.

Tonio

As you said, $\displaystyle D$ is symmetric in the roots of $\displaystyle f(x)$, so $\displaystyle D$ is fixed by all automorphisms of the Galois group of $\displaystyle f(x)$.

This implies $\displaystyle D \in F$ since $\displaystyle F$ is the fixed field of $\displaystyle \text{Gal}(K/F)$.

(We know $\displaystyle F$ is the fixed field of $\displaystyle \text{Gal}(K/F)$ since $\displaystyle [K:F]=|\text{Gal}(K/F)|$, so by the Fundamental Theorem of Galois Theory we get what we want.)

4. Originally Posted by chiph588@

As you said, $\displaystyle D$ is symmetric in the roots of $\displaystyle f(x)$, so $\displaystyle D$ is fixed by all automorphisms of the Galois group of $\displaystyle f(x)$.

This implies $\displaystyle D \in F$ since $\displaystyle F$ is the fixed field of $\displaystyle \text{Gal}(K/F)$.

(We know $\displaystyle F$ is the fixed field of $\displaystyle \text{Gal}(K/F)$ since $\displaystyle [K:F]=|\text{Gal}(K/F)|$, so by the Fundamental Theorem of Galois Theory we get what we want.)

Excellent! I didn't even remember to mention Galois Theory because I tried to make it easier, but if you can deal with that then it's way
easier and short...and elegant, imo.

Tonio

5. Originally Posted by tonio
Excellent! I didn't even remember to mention Galois Theory because I tried to make it easier, but if you can deal with that then it's way
easier and short...and elegant, imo.

Tonio
Yeah, the 'simple' thing I was forgetting was that $\displaystyle D$ was symmetric with respect to $\displaystyle \alpha_i$.