## Discriminant of a pretty easy polynomial

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### Discriminant of a pretty easy polynomial

I am trying to find the discriminant of x^m - 1 as an intermediate problem in a bigger proof for an assignment problem. It seems to be, based what I've computed for small values of m, that |D(f)| = m^m, but I'm unable to come up with a direct proof for this without having to use a result like: if f is monic with roots \alpha_1, \ldots, \alpha_n, then:
D(f) = (-1)^{n(n-1)/2} \prod_{i=1}^n f'(\alpha_i)

... but proving this seems like overkill for what seems like such a simple solution.

I don't immediately see how induction would work (but I have not tried very hard). I have proved that the square of the Vandermonde determinant is the discriminant, so that might be of use...

Should I be using results on cyclotomic polynomials? I feel like I am missing something obvious... Can I have a hint please?
HINDYhat

Posts: 19
Joined: Fri Dec 11, 2009 6:54 am UTC

### Re: Discriminant of a pretty easy polynomial

I don't know much about that topic, but your formula looks like it can directly produce the result you want:

f' = m*x^(m-1), you have m such roots which all have |alpha_i|=1. Therefore, the absolute value of your expression is already m^m. You just have to check the phase, where you can use that every complex root has its complex conjugated value as root, too. The only thing left to evaluate is the sign of D(f).
mfb

Posts: 803
Joined: Thu Jan 08, 2009 7:48 pm UTC

### Re: Discriminant of a pretty easy polynomial

Oh yeah, I can use that formula to prove it easily... but I have to prove the formula, and it doesn't look so easy to prove for such a clean result as m^m.

I am not sure, but I think I only need the case when m is prime in my problem, in which case I can use the Vandermonde determinant. Will report back later when I have enough time to try this out.
HINDYhat

Posts: 19
Joined: Fri Dec 11, 2009 6:54 am UTC

### Re: Discriminant of a pretty easy polynomial

I've gotten nowhere... I'm not even sure if this result is necessary to prove what's required.

We have to show that Chebotarev's theorem on the density of primes implies Dirichlet's theorem on arithmetic progressions. A lot of this stuff seems outside the scope of the course, but we only have to show the implication so it's probably not a big deal at all.

So, let f(x) = x^m - 1. The Galois group of f over \unicode{x211a} is
G = \text{Gal}(\unicode{x211a}(\zeta_m)/\unicode{x211a}) \cong (\unicode{x2124}/m\unicode{x2124})^\times

And:
G = \{\sigma_a : \zeta_m \mapsto \zeta_m^a \; | \; (a, m) = 1\}

Since G is abelian, its conjugacy classes are the singleton sets \{\sigma_a\}. Then, by Chebotarev's density theorem, for a such that (a, m) = 1, the set of primes p not dividing the discriminant of f such that \sigma_p is conjugate to \sigma_a, i.e. p \equiv a \; (\text{mod} m) has density 1/|G| = 1/\phi(m).

Everything seems to work except the bolded part...
HINDYhat

Posts: 19
Joined: Fri Dec 11, 2009 6:54 am UTC

### Re: Discriminant of a pretty easy polynomial

There are only finitely many primes that divide the discriminant of f.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

antonfire

Posts: 1768
Joined: Thu Apr 05, 2007 7:31 pm UTC

### Re: Discriminant of a pretty easy polynomial

So, since there are only finitely many primes p' | \Delta(f), roughly speaking these ones do not contribute to the density of primes p \equiv p' \; (\text{mod} m), where
p \equiv a \; (\text{mod} m)
and (a, m) = 1?
HINDYhat

Posts: 19
Joined: Fri Dec 11, 2009 6:54 am UTC