This is part of an exercise I'm doing for self study. Here, $K = \mathbb Q(\alpha)=\mathbb Q[X]/(X^5-X+1)$, and $L$ is the splitting field.

"Using the fact that any extension of local fields has a unique maximal unramified subextension, prove that for any monic irreducible polynomial $g\in\mathbb Z[X]$, the splitting field of $g$ is unramified at all primes that do not divide $\operatorname{disc} g$. Conclude that $L/\mathbb Q$ is unramified away from primes dividing $\operatorname{disc}\mathcal{O}_K$ and tamely ramified everywhere, and show that every prime dividing $\operatorname{disc}\mathcal{O}_K$ has ramification index 2. Use this to compute $\operatorname{disc}\mathcal{O}_L$."

I have already computed $\operatorname{disc}\mathcal{O}_K = 2869 = 19\times151$. I've used the Dedekind-Kummer theorem to show that the ramified primes $\mathfrak{p}$ dividing 19 and 151 have $e_\mathfrak{p} = 2$, so that $K/\mathbb Q$ is tamely ramified (tamely ramified at all $K_v/\mathbb Q_p$ for $p$ prime and $v|p$).

What I don't understand is how to use the hint to show the primes $p\nmid\operatorname{disc}g$ are unramified in $L$ or how to use this and the other results to compute $\operatorname{disc}\mathcal{O}_L$. Any hints or answers would be very helpful.


2 Answers 2


The fact that $L/\mathbb{Q}$ is unramified away from primes dividing $D=\text{disc } \mathcal{O}_K$ is evident: $L$ is composition of different embeddings of $K$, each such embedding is unramified away from primes dividing $D$, so is their composition $L$.

Now we show that for $p\mid D$, $p$ has ramification index $2$ in $L$. Let $\alpha_i\in L$, $i=1,\cdots,5$ be roots of $f(X) = X^5-X+1$. By factoring $f$ modulo $p$, we see that there are exactly four distinct $\bar{\alpha}_i \in \bar{\mathbb{F}}_p$, say $\bar{\alpha}_1 = \bar{\alpha}_2$ and $\bar{\alpha}_1, \bar{\alpha}_3,\bar{\alpha}_4,\bar{\alpha}_5$ are distinct. Any inertia group above $p$ fixes $\alpha_3,\alpha_4,\alpha_5$, only non-trivial element for inertia group will be the swapping of $\alpha_1$ and $\alpha_2$. Therefore ramification index is $2$.

To compute the discriminant, you can use the discriminant formula for tame ramification. But a more elegant approach is to consider $F = \mathbb{Q}(\sqrt{D})$. Since every $p\mid D$ has ramification $2$ in $L$, $L/F$ is unramified at every finite prime. Note that $[L:F] = 60$, therefore $$|D_{L/\mathbb{Q}}| = |D_{F/\mathbb{Q}}|^{60} = 19^{60} 151^{60}$$

  • $\begingroup$ +1 for the last paragraph. This certainly is a more elegant approach $\endgroup$ Commented Jul 5, 2020 at 6:42
  • $\begingroup$ Very nice solution. It seems that the choice of $F=\mathbb Q(\sqrt{D})$ essentially the choice of maximal unramified subextension when you localize, so I suppose that's what the hint was getting at. You mention a formula for the discriminant in tame ramification, what is this formula (a cursory search through my references hasn't revealed anything)? $\endgroup$
    – Nico
    Commented Jul 5, 2020 at 19:11
  • 1
    $\begingroup$ @Nico I think the formula is $\prod_p \sum_i p^{f_i (e_i-1)}$, where $e_i, f_i$ are ramification indices and inertial degrees for various primes lying above $p$. For any extension $K/\mathbb{Q}$, the discriminant is at least divisible by this number, and exactly equals to when $K/\mathbb{Q}$ is tamely ramified. $\endgroup$
    – pisco
    Commented Jul 6, 2020 at 4:14

Edit: I didn't read the question thoroughly and was going to delete this answer since it does not use the hint supplied. But maybe I will leave it with the note that this does not use the given hint (I am happy to delete if OP wishes).

Consider an irreducible polynomial $g(x) \in \mathbb{Z}[x]$ such that a prime $p$ does not divide $disc(g(x))$. Let $\bar{g}[x] \in \mathbb{F}_p[x]$ be obtained from $g$ by reducing the coefficients. Since $p$ does not divide the discriminant of $g$ we have that $disc(\bar{g}(x)) \neq 0$, in particular $\bar{g}$ has distinct roots in $\mathbb{F}_p$.

Let $\mathfrak{p}$ lie above $p$ in $L$. Now consider the decomposition group $D_{\mathfrak{p}/p}$ and the inertia group $I_{\mathfrak{p}/p}$. We want to show that the inertia group is trivial (since this is the case if and only if $p$ is unramified in $L$.

The group $D_{\mathfrak{p}/p}$ acts on the roots of $g(x)$ faithfully (since these generate the extension of local fields $L_{\mathfrak{p}} / \mathbb{Q}_p$). But notice that the reduction map taking $$\{ \text{roots of } g(x) \} \to \{ \text{roots of } \bar{g}(x) \}$$ is injective (since both polynomials have distinct roots). Thus if $\sigma \in I_{\mathfrak{p}/p}$ (i.e., if $\sigma$ fixes the roots of $\bar{g}(x)$) then $\sigma$ must act trivially on the roots of $g(x)$ by the injectivity noted above. In particular $I_{\mathfrak{p}/p}$ is trivial.

  • $\begingroup$ @Nico, The $\operatorname{Disc}(L/\mathbb{Q})$ part should come from the tame ramification, see this question $\endgroup$ Commented Jul 5, 2020 at 5:14

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