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Jul 21, 2014 at 23:53 comment added David E Speyer Agreed. I was confused by Bhargava talking about $K$ etale over $\mathbb{Q}_p$. I thought he was abusing language and meant $\mathcal{O}_K$ etale over $\mathbb{Z}_p$ (at which point, other things didn't make sense) but the statement is perfectly sensible taken literally, and then Jeremy Rouse is right that this isn't enough.
Jul 21, 2014 at 20:42 comment added Jeremy Rouse I don't think Theorem 1.3 from Bhargava's paper is enough, given that for sufficiently large primes, the "admissible" local conditions must include all of those that are consistent with the discriminant being squarefree.
Jul 21, 2014 at 19:55 comment added David E Speyer I haven't digested Bhargava's terminology well enough to tell, but does Theorem 1.3 in arxiv.org/abs/1402.0031 do the job?
Jul 21, 2014 at 19:25 history edited JSE CC BY-SA 3.0
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Jul 21, 2014 at 17:03 comment added Daniel Loughran @David Speyer: Thanks for the reference. I will have a look at this paper, it might turn out to be useful for me
Jul 21, 2014 at 16:55 comment added Daniel Loughran @JSE: I agree with David Speyer, square-free discriminant does not imply my condition (3). Explicitly, consider the number field $K$ defined by the polynomial $x^5 -5x^3 + 4x -1$. This has discriminant $38569$, which is a prime, and moreover its splitting field has Galois group $S_5$. However this prime splits as $\mathfrak{p}^2\mathfrak{q}$, where $\mathfrak{p}$ has inertia degree $1$ and $\mathfrak{q}$ has inertia degree $3$.
Jul 21, 2014 at 16:10 comment added David E Speyer I don't follow why square free implies (3). If $p$ factors as $\mathfrak{p}^2 \mathfrak{q}$, where $\mathcal{O}_K/\mathfrak{q} \cong \mathbb{F}_{p^3}$, I think the discriminant is squarefree. That said, you might want to look at Kedlaya arxiv.org/abs/1103.5728 for some statements about sieving for square free discriminant.
Jul 21, 2014 at 15:42 history answered JSE CC BY-SA 3.0