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I am looking for number fields $K$ which satisfy the following properties:

  1. $[K:\mathbb{Q}]=5$.
  2. The Galois closure of $K$ has Galois group $S_5$.
  3. For each prime $p$ which ramifies in $K$, there exists a prime ideal $\mathfrak{p}$ of $K$ of inertia degree and ramification index $1$ above $p$, i.e., we have $$(p) = \mathfrak{p} \cdot \prod_{i=1}^r \mathfrak{p}_i^{e_i},$$ with $N(\mathfrak{p}) = p$.

It is well-known that one can construct infinitely many fields satisfying conditions 1 and 2 using Hilbert's irreducibility theorem. It is the last condition 3 which is the most important one, and says something like the ramification of $K$ is very mild.

I have been able to find such fields by looking at databases of number fields. For example, one such field is given by $$t^5-t^4-5t^3+5t^2+2t-1 =0.$$ This has discriminant equal to $101833$, which is prime. One checks that we have the factorization $$(101833) = \mathfrak{p}_1^2 \mathfrak{p}_2\mathfrak{p}_3\mathfrak{p}_4,$$ where each ideal has norm $101833$.

Do there exist infinitely many such fields?

I'm possibly willing to weaken condition 2 to ask instead that the Galois group is a solvable (but still non-abelian) subgroup of $S_5$, if it helps. However, in my application I cannot remove condition 1.

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3 Answers 3

up vote 5 down vote accepted

Since you are okay with other non-abelian Galois groups, let $F$ be a quadratic extension of $\mathbb{Q}$ with class number divisible by $5$. (There are infinitely many, see this paper and references therein). By class field theory, $F$ has an unramified $\mathbb{Z}/5$ extension $L$. I claim that $L/\mathbb{Q}$ is a $D_5$ extension, and I claim that the degree $5$ subextensions corresponding to the order $2$ subgroups of $D_5$ have the desired property.

Verification that $L/\mathbb{Q}$ is a $D_5$ extension Let $H$ be the class group of $F$. Let $A/L$ be the unramified extension corresponding by class field theory to $H/5H$. Then $A/\mathbb{Q}$ is Galois and we have a short exact sequence $0 \to H/5H \to \mathrm{Gal}(A/\mathbb{Q}) \to \mathbb{Z}/2 \to 0$. We know that $\mathrm{Gal}(L/\mathbb{Q})$ acts by negation on $H$, and $GCD(|H/5H|,2)=1$, so the short exact sequence is semidirect, and $\mathrm{Gal}(A/\mathbb{Q}) \cong \mathbb{Z}/2 \ltimes H / 5H$. In particular, if we take a particular $\mathbb{Z}/5$ quotient of $H/5 H$, we get a $D_5$ extension of $\mathbb{Q}$.

Verification of condition (3) Let $p$ be a prime which ramifies in $K$ (and hence in $L$) and choose a prime $\pi$ of $L$ lying above $p$. Since $L/F$ is unramified, the ramification degree of $\pi/p$ must be $2$. Let $I_{\pi}$ be the inertia group and $D_{\pi}$ the decomposition group. Now, $I_{\pi}$ is $2$, so it is conjugate to $\{\pm 1 \} \ltimes \{ 0 \}$ in $D_5$. Since $I_{\pi}$ is normal in $D_{\pi}$, we must have $D_{\pi} = I_{\pi}$.

But then we compute that $p$ factors in $K$ as $\mathfrak{p}_1^2 \mathfrak{p}_2^2 \mathfrak{p}_3$, and $\mathfrak{p}_3$ is the desired prime.

The problem in $S_5$ is that we can have $I_{\pi}$ be the group generated by $(12)$ and $D_{\pi}$ the group generated by $(12)$ and $(345)$.

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Thanks very much for your detailed answers, I very much appreciate it. I think that these $D_5$-extensions might suffice for my purposes, but I need to go away and think about it some more. In any case, I will accept your answer. –  Daniel Loughran Jul 22 at 15:58

I'm not sure what is and isn't written down, but Manjul Bhargava's work on counting quintic fields can be used to get an exact asymptotic for the number of quintic extensions of Q with discriminant a SQUAREFREE integer between 0 and X. In fact, the number of such will be asymptotic to a constant multiple of X. Squarefree discriminant implies your conditions 2 and 3.

This may be massive overkill, for all I know!

Update: Sorry, I didn't pay careful enough attention to DL's condition! No, squarefree doesn't imply condition 3. That's a stronger local condition. But my guess will be that the method Bhargava uses will still give you a positive proportion, though I'm less certain. (Namely: there is still an Euler factor at p given by the total mass of all etale quintic extensions of Q_p satisfying your conditions.)

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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. –  David Speyer Jul 21 at 16:10
@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 $3$. –  Daniel Loughran Jul 21 at 16:55
@David Speyer: Thanks for the reference. I will have a look at this paper, it might turn out to be useful for me –  Daniel Loughran Jul 21 at 17:03
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? –  David Speyer Jul 21 at 19:55
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. –  Jeremy Rouse Jul 21 at 20:42

The point of this answer, building on a comment by "so-called friend Don", is to point out that behavior has density $0$ among all quintic extensions.

My guides here are the introductory portions of The geometric sieve and the density of squarefree values of invariant polynomials and Mass Formulae for Extensions of Local Fields and Conjectures on the Density of Number Field Discriminants, both by Bhargava.

Let $k$ be a field. Recall that an etale $k$-algebra is a direct sum of finitely many finite separable extension fields of $k$. (We'll be in characteristic zero, so you can ignore the adjective separable.) We'll write $\mathrm{Aut}(K/k)$ for the automorphism group of $K$ preserving $k$; this is a finite group. Let $K$ be a degree $p$-adic extension of $\mathbb{Q}_p$ of degree $n$. As I understand the philosophy of these papers, the probability of that the $p$-adic completion of a degree $n$ number field will be isomorphic to $K$ is supposed to be $$\frac{p-1}{p} \frac{1}{\mathrm{Disc}_p(K)} \frac{1}{\# \mathrm{Aut}(K/\mathbb{Q}_p)}.$$

Moreover, these probabilities for distinct primes are supposed to be independent.

For $K$ an etale $\mathbb{Q}_p$ algebra, we write say that $K$ has symbol $(f_1^{e_1}, f_2^{e_2}, \ldots, f_r^{e_r})$ if $K \cong \bigoplus_{i=1}^r K_i$ where $K_i/\mathbb{Q}_p$ is a field extension of ramification degree $e_i$ and residue field extension of degree $f_i$. A lemma in the Mass Formula paper (Prop 2.1) allows us to group together any place where we sum over the set of all etale $\mathbb{Q}_p$-algebras with a given symbol.

So, our desired Euler factor is:

The unramified extensions These are symbols where all the $e_i$ are $1$; namely $(1,1,1,1,1)$, $(2,1,1,1)$, $(3,1,1)$, $(4,1)$, $(5)$. We compute: $$\frac{p-1}{p} \left( \frac{1}{120} + \frac{1}{12} + \frac{1}{8} + \frac{1}{6} + \frac{1}{4} + \frac{1}{5} \right) = \frac{p-1}{p}.$$ It is not a coincidence that the sum came out to $1$; see equation (2.4) in the Mass formula paper.

The other symbols It is also okay to have symbol $(1^2, 1,1,1)$, $(1^2, 2,1)$, $(1^2, 1^2, 1)$ or $(2^2, 1)$. Each of these symbols corresponds to several possible etale $\mathbb{Q}_p$-algebras (for example, for $p$ an odd prime, there are $2$ ramified quadratic extensions of $\mathbb{Q}_p$, so $(1^2, 1,1,1)$ describes two options), but Prop 2.1 in the Mass formula paper means we don't have to think about this. I compute that the respective terms are

$$\frac{p-1}{p} \left( \frac{1}{p} \left( \frac{1}{6} + \frac{1}{2} \right) + \frac{1}{p^2} \left( \frac{1}{2} + \frac{1}{2} \right) \right)$$

Putting everything together, our Euler factor is $$\frac{p-1}{p} \left( 1+\frac{2}{3p} + \frac{1}{p^2} \right) = 1-\frac{1}{3 p} + \frac{1}{3 p^2} - \frac{1}{p^3}.$$

The point is that $$\prod_p \left( 1-\frac{1}{3 p} + \frac{1}{3 p^2} - \frac{1}{p^3} \right) =0.$$ Now, Bhargava's paper doesn't directly allow us to us the prudct in this manner because his best theorem, Theorem 1.3 in the sieve paper, only applies when, for all sufficiently large $p$, we include all the terms with discriminant $1$ or $p$. However, let $p_N$ be the probability that the ramification is as you wish for all $p < N$. Then Bhargava's Theorem 1.3 does show that $p_N$ exists and equals $\prod_{p<N} \left( 1-\frac{1}{3 p} + \frac{1}{3 p^2} - \frac{1}{p^3} \right)$. So this product is an upper bound for the probability that a quintic behaves as desired for all primes. Sending $N \to \infty$, we see that the proportion of quintics with the desired behavior is $0$.

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Doesn't the hypothesis in Theorem 1.3 only require what you said for all large enough p? So why can't you (or can you?) run your argument up to an arbitrarily large finite height to get an arbitrarily small upper density, and hence a density of 0? –  so-called friend Don Jul 22 at 19:44
@so-calledfriendDon Excellent point! You are right! Strange that Bhargava doesn't point that out; this argument shows that Bhargava's formula is always an upper bound. –  David Speyer Jul 22 at 19:54
I have now edited in Don's point. –  David Speyer Jul 22 at 20:00

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