1

I want to show that a rational prime $p$ is ramified in a cyclotomic field $K = \mathbb{Q}(\zeta_q)$, where $q$ is an odd prime, if and only if it does not divide the discriminant $\Delta_K$.

I have shown that the discriminant is $\Delta_K = (-1)^{(q-1)/2}q^{q-2}$, and so what the question is really asking is to show that $p$ is ramified in $K$ if and only if $p=q$.

My efforts so far: The minimal polynomial of $\zeta_q$ over $\mathbb{Q}$ is $f(X)=X^{q-1}+X^{q-2}+...+X+1$, and $\{1,\zeta_q,...,\zeta_q^{q-2}\}$ is an integral basis for $K$, $f(X)\equiv (X-1)^{q-1}$ mod $q$ (using Binomial coefficients to expand and simplify) and so by the Dedekind-Kummer Theorem we have that $q$ is indeed ramified in $K$.

I now must show that any rational prime $p\neq q$ is not ramified, but have no idea how to do this. When proving an analogous result for quadratic fields, we did this using a case by case approach, but this seems trickier here as $f$ itself is not quite as simple.

Martin
  • 645
  • 1
    "so what the question is really asking is to show that $p$ is ramified in $K$ if and only if $p=q$." Isn't this answered here, using the notation $\ell$ instead of $p$? – Dietrich Burde Apr 01 '20 at 10:52

1 Answers1

3

Well, using Dedekind-Kummer in the same way, what you want to show is that $f$ has no multiple roots mod $p$.

But observe that all roots of $f$ are roots of $x^q - 1 = (x-1)f(x)$ and the latter is clearly separable when $p\neq q$, as its derivative is $qx^{q-1}$, whose only root is $0$ (because, crucially, $q\neq 0$ mod $p$) but $0$ is clearly not a root of $f$.

(using the general fact that if $g$ has a multiple root then it has a root in common with its derivative)