# Ramification in a splitting field

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.

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 $$|D_{L/\mathbb{Q}}| = |D_{F/\mathbb{Q}}|^{60} = 19^{60} 151^{60}$$

• +1 for the last paragraph. This certainly is a more elegant approach – Mummy the turkey Jul 5 at 6:42
• 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)? – Nico Jul 5 at 19:11
• @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. – pisco Jul 6 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.

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