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Observe, trivially, that since quadratic fields correspond to rational integers modulo squares (viz. discriminants), there are (roughly about, but certainly at most) $2^{|S|+1}$ quadratic fields unramified outside $S$.

Consider now cubic fields unramified outside $S$. Presumably, I thought, their number cannot be bounded in terms of $|S|$ alone. Could it? For one thing, if the situation above had a general extension as an exponential bound in $|S|$ on the number of unramified outside $S$ extensions of $\mathbb{Q}$ of a fixed degree $d$, a theorem of Evertse on the number of solutions to $S$-unit equations would immediately yield Brumer and Silverman's conjecture that the logarithm of the number of rational elliptic curves of conductor $N$ is $O(\log{N} / \log{\log{N}})$, in the much stronger form $O(\omega(N))$ that is not even suggested anywhere. [Here is the argument: For $E/\mathbb{Q}$ with conductor $N$, choose a minimal special Weierstrass equation $y^2 = x^3 + Ax + B$ over $\mathbb{Z}[1/6]$, and factor the right-hand side as $(x-\alpha)(x-\beta)(x-\gamma)$. Then $K := \mathbb{Q}(\alpha,\beta,\gamma)$ is a number field of degree at most $6$ and unramified outside $6N$. For each fixed such $K$, noting $T$ the set of places of $K$ lying over $6N$ and $\infty$, we have by minimality of the equation that $\alpha - \beta, \beta - \gamma, \gamma - \alpha \in O_{K,T}^{\times}$, and Evertse's theorem limits $\lambda := (\alpha - \beta) / (\alpha - \gamma)$ to at most $\exp(O(|T|)) = \exp(O(\omega(N)))$ possibilities. But $E$ is isomorphic to $y^2 = x(x-1)(x-\lambda)$ over the field $K(\sqrt{\alpha - \gamma})$ of degree bounded by $12$, and each $K(\sqrt{\alpha - \gamma})$-isomorphism class splits into at most $6^{12}$ isomorphism classes over $\mathbb{Q}$. ]

Hence my question:

Is it possible to demonstrate that there cannot be a bound depending only on $|S|$ and $d$ on the number of unramified outside $S$ degree $d$ extensions $K/\mathbb{Q}$? For instance, as $p$ varies, can there be arbitrarily many cubic fields unramified outside $\{p,\infty\}$?

Added. The hypothesis on exponential boundedness in $|S|$ of the set of degree-$d$ extensions $K/\mathbb{Q}$, which as noted above implies the Brumer-Silveman conjecture and a couple of similar ones (e.g. Schmidt's conjecture that the number of integer solutions of $y^2 = x^3+k$ are bounded by $O_{\epsilon}(k^{\epsilon})$, in fact by $\exp(O(\omega(k)))$), would follow if

(H) the number of degree-$d$ extensions $K/\mathbb{Q}$ with a given discriminant $D$ could be bounded only in terms of $d$.

For $d = 3$ this is similar (but not quite the same as general number fields are not monogenic), to asking for a uniform bound on the number of integer solutions to a minimal Weierstrass equation, which in turn is related to the possible boundedness of ranks of elliptic curves. (Hindry and Silverman have proved that the last statement is implied by the ABC conjecture joint with the boundedness of Mordell-Weil ranks.) The problem of bounding $\mathrm{Cl}(\mathbb{Q}(\sqrt{D}))[3]$, related in Hasse's paper (class field theory) to the problem here, is also of a similar spirit, and in all cases these quantities are bounded on average. Might they be bounded uniformly in the degree $d$?

These are all bold questions, but it seems that nowadays there are people who believe in the boundedness of ranks. If there is any reference mentioning the possible boundedness of degree-$d$ extensions $K/\mathbb{Q}$ of a fixed discriminant, I would very much appreciate to know it! If such a thing could possibly be true, then in view of the above application one could further ask for a quantitative Faltings theorem in the form "there are constants $C(F,g), A(g) < \infty$ such that for any $S \subset \mathrm{spec}(O_F)$, the number of p.p. $g$-dimensional abelian schemes over $O_{F,S}$ does not exceed $C(F,g) \cdot A(g)^{|S|}$."

Added later. Perhaps I should add that an analog for Riemann surfaces of the main statement is obviously true: the number of degree-$d$ coverings of $\widehat{\mathbb{C}}$ branched over $S$ is bounded only in terms of $d$ and $|S|$, and the dependence on $|S|$ is exponential. A little thought reveals that the stronger statement (H) also holds for Riemann surfaces.

From the point of view of Galois theory, the "reason" we know this in the complex function field case is that the fundamental group of $\widehat{\mathbb{C}} - S$ only depends on $|S|$. The analog of this group in the number field situation is the Galois group $G_S$ of $\mathbb{Q}^{(S)}/\mathbb{Q}$, the maximal extension of $\mathbb{Q}$ with ramification limited to $S$. The situation here is much more subtle of course: consider $G_S^{\mathbb{ab}}$ and class field theory. And yet, granting the principle that "all primes are equal" in algebraic number theory, would it be completely misguided to think that the abstract group $G_S$ "essentially" - up to some sort of finite discrepancy, say - ought to depend on $S$ only through the cardinality $|S|$?

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    See Hasse, http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002371952&IDDOC=85276 – Franz Lemmermeyer Feb 02 '15 at 20:50
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    In other terms, since certainly (for a fixed $d$) the set of possible discriminants of such a field is bounded exponentially in $|S|$, the question concerns the optimal type of bound one can expect on the number of degree-$d$ extensions $K/\mathbb{Q}$ of a fixed discriminant $D$. By Davenport-Heilbronn type conjectures, one expects at least the number of such fields to be bounded on average. – Vesselin Dimitrov Feb 02 '15 at 21:43
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    "Silverman has proved that the last statement is implied by the ABC conjecture joint with the boundedness of Mordell-Weil ranks." This is actually a joint result with Marc Hindry. – Joe Silverman Feb 03 '15 at 05:21
  • @VesselinDimitrov: I probably misunderstand your question, but ${\mathbb Q}(\sqrt[3]{p})/{\mathbb Q}$ are unramified outside $S={3,p,\infty}$ for all $p$. Isn't that a counterexample for $|S|=3? – Tim Dokchitser Feb 03 '15 at 16:11
  • @TimDokchitser: This gives only one extension for each $p$. The question is whether, e.g., there is a bound $M < \infty$ such that for every $p$ there are at most $M$ cubic extensions unramified outside ${3,p,\infty}$. – Vesselin Dimitrov Feb 03 '15 at 17:12
  • @VesselinDimitrov Ah, ok, thank you, I see now. – Tim Dokchitser Feb 03 '15 at 17:37
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    On a slightly related note, a cubic extension obviously has Galois group (of its Galois closure) $A_3$ or $S_3$, so in some sense your question about cubic fields is a much stronger version of the question: are there infinitely many number fields with Galois group $S_n$ or $A_n$ unramified outside $S$? (Where $n$ is not fixed.)

    My understanding is that even this question is open. Here is a paper where the authors provide some evidence that there might be infinitely many for certain $S$. http://arxiv.org/abs/1401.7379

    – Bobby Grizzard Feb 03 '15 at 18:35
  • Actually, I think I misunderstood your question when I made my comment, and I'm not exactly sure why you would go the other direction on your original question: you are talking about varying $S$, and I was talking about a fixed $S$ (but varying degree). – Bobby Grizzard Feb 03 '15 at 20:18
  • @BobbyGrizzard: I was thinking about the analogy with complex function fields, but then I was notified by Michael Stoll in the linked question that Cohen-Lenstra heuristics actually predict unboundedness. Apparently the situation in arithmetic is very different. – Vesselin Dimitrov Feb 03 '15 at 21:03
  • @VesselinDimitrov When passing between number fields and complex function fields one has to remember to pass through function fields over finite fields first. A prime $p$ of $\mathbb Z$ corresponds to an irreducible polynomial of degree at most $\log p$. When you pass to the complex fields the degree becomes more significant than the irreducibility for questions of this type. – Will Sawin Nov 26 '16 at 23:27

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