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Notation/Setup. Let $p$ be a prime greater than 2 and $d$ be a square free integer such that $\gcd(d,p) = 1$.

Question. Suppose we pick $d\equiv 3\mod 4$, in which case $\mathbf{Q}(\sqrt{d})$ is ramified at 2 over $\mathbf{Q}$. Moreover, we know that the norm of $\mathfrak{q}|2$ in $\mathbf{Q}(\sqrt{d})$ is equal to 2. From this, how could one conclude that $\mathbf{Q}(\sqrt{d})$ has no cyclic extension of degree $p$ in which $\mathfrak{q}|2$ ramifies?

Notes. I believe that this should follow from some global class field theory, however, I cannot immediate see that. Also the question does provide some help, however, the cyclic extension of degree $p$ need not be of the form $\mathbf{Q}(\sqrt{d},\sqrt[p]{a})$ for some $a\in \mathbf{Q}(\sqrt{d})$.

This question comes from a tersely written paper I am reading, and I would like to get some more insight into why this is true.

Any references/answers/suggestions are welcome.

Edit. Here is some empirical evidence (Magma code) supporting this claim for $p = 3$:

Q<k> := PolynomialRing(Integers()); A:= k^2 + k + 7; P<t> := PolynomialRing(Rationals()); F := [t^2 + 1,t^2 - 2,t^2 +2,t^2 - 6, t^2 + 6, t^2 - 10, t^2 + 10, t^2 - 14, t^2 + 14]; //Ramified quadratic extensions at 2,3 for i in [1..#F] do L := NumberField(F[i]); D1 := Factorization(2*Integers(L))[1,1]; //Prime above 2 K<x> := PolynomialRing(L); for i in [1..10] do E := Evaluate(A,i); V := x^3 - E*x + E; K := NumberField(V); //An infinite family over Q of number fields with Z/3 Galois group if #GaloisGroup(V) eq 3 then IsRamified(D1,Integers(K)); end if; end for; "********************************************"; end for;

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    $\begingroup$ uh.. is it necessary to introduce $K$ ? seems like you can ask the question without talking about it so I'm a bit confused. $\endgroup$ – mercio Dec 11 '15 at 4:28
  • $\begingroup$ @mercio I do understand that it is a bit confusing. The paper actually uses the ramification (or lack there) of for $\mathbf{Q}(\sqrt{d})$ to conclude information about $K$. For the sake of the question, it is not necessary, so I'll make the edit $\endgroup$ – Jackson Morrow Dec 11 '15 at 12:45
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    $\begingroup$ My first guess would be that it has something to do with the setup forcing the ramification groups to have an impossible structure. But, I never really understood that theory inside out, and I desperately need a nap anyway. Cool question, though! $\endgroup$ – Jyrki Lahtonen Dec 11 '15 at 13:05
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This is an immediate consequence of local class field theory. Namely, if a number field field $E$ has a cyclic extension of degree $p$ ramified at $\mathfrak{q}$, then certainly the completion $K = E_{\mathfrak{q}}$ admits a cyclic extension of degree $p$ ramified at $\mathfrak{q}$. Since $p$ is odd and $\mathfrak{q}$ has residue characteristic two, this extension is tamely ramified. Yet, by local class field theory, the tamely ramified cyclic extensions of a local field $K$ all have degree dividing $|k^{\times}|$, where $k$ is the residue field. In this case, $k = \mathbf{F}_2$, and so there are no tamely ramified (and not unramified) extensions.

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  • $\begingroup$ Thank you! A very nice consequence of local class field theory indeed. $\endgroup$ – Jackson Morrow Dec 22 '15 at 4:49
  • $\begingroup$ @JacksonMorrow Do either of you have a reference for "Yet, by local class field theory, the tamely ramified cyclic extensions of a local field $K$ all have degree dividing $|k^* |$" $\endgroup$ – Ravi May 9 '16 at 22:57
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Here is an argument which could complete the answer given by @Epargyreus , but does NOT use CFT. With your hypotheses, the completion $K$ of your quadratic field $\mathbf Q(\sqrt d)$ at the prime $\mathfrak q$ is a quadratic extension of $\mathbf Q_{2}$, hence has residual characteristic $2$. If $\mathbf Q(\sqrt d)$ admitted a cyclic extension of odd prime degree $p$ in which $\mathfrak q$ is ramified (hence necessarily totally ramified), then the completion at $\mathfrak q$ would be a cyclic extension $L/K$ of degree $p$ which is tamely ramified since $p \neq 2$.

Such an extension $L/K$ can be characterized by the theory of ramification groups, see e.g. Cassels-Fröhlich, chapter $1$, §8. Let $G=Gal(L/K)$. Since $L/K$ is totally ramified, $G = G_0$, the inertia subgroup. If $G_1$ denotes the $2$-Sylow subgroup of $G_0$ (recall that the residual characteristic is $2$), the fixed field of $G_1$ is the maximal tamely ramified subextension $L_1$ of $L/K$. Because $p \neq 2$, $G_1$ is trivial, a contradiction.

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