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I'm currently reading Datskovsky and Wright's "Density of discriminants of cubic extensions" and came across the following (paraphrased) statement. Let $K$ be a number field, and $K'$ a cyclic cubic extension of $K$. Let $\dagger$ denote the conductor of the extension, thought of as an ideal inside the ring of integers of $K$, and write $$ \dagger=\prod{\mathfrak{p}_i}^{n_i}. $$ The goal is to characterize what $\dagger$ may look like. Clearly if $\mathfrak{p}_i$ does not lie above $3$ it must ramify tamely at worst, and hence in this case $n_i\leq 1$ by the relationship between conductors and ramification. Let $\mathfrak{q_1},\dots,\mathfrak{q_l}$ denote the primes above $3$.

The statement I do not know how to justify is that there exists a $c$ depending solely on $K$ (in particular, not on $K'$ or $\mathfrak{q}$) such that the contribution of $\mathfrak{q}_i$ to $\dagger$ is at most $\mathfrak{q}_i^c$.

The reference given for this is unfortunately in German; it is Hasse's "Vorlesungen über Klassenkörpertheorie." Since I cannot read German, would anyone know of an English reference to a statement like the one above, or possibly know the idea behind it? In my head I'm imagining that unbounded ramification must in some sense drive the degree of the extension up, but to be honest my intuition for these class field theoretic results is not great. Thanks.

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  • $\begingroup$ How can an extension of a number field be wildly ramified? Or do you mean a residual extension $K'/\mathfrak{q}_i$ of $K/\mathfrak{p}_i$? $\endgroup$
    – Servaes
    Mar 11, 2019 at 7:05
  • $\begingroup$ @Servaes This is the language used in the paper; I guess what they mean is that when you complete at $\mathfrak{p}$ you have wild ramification? $\endgroup$
    – TomGrubb
    Mar 11, 2019 at 7:14
  • $\begingroup$ I guess they mean that $e(\mathfrak{q}_i/\mathfrak{p}_i)\in\mathfrak{p}_i$, never mind my earlier comment. $\endgroup$
    – Servaes
    Mar 11, 2019 at 7:24
  • $\begingroup$ Doesn't this follow from Kummer-Dedekind by the way? $\endgroup$
    – Servaes
    Mar 11, 2019 at 7:34
  • $\begingroup$ @Servaes I'm not sure it does, since (if I'm thinking of the same theorem as you) there are assumptions of either monogenecity or being prime to the conductor $\endgroup$
    – TomGrubb
    Mar 11, 2019 at 7:44

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