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For a field $K\subset \mathbb{Q}(\zeta_p)$ $~$($\zeta_p$ a primitive pth root of unity, p a prime), it seems to be the case that the discriminant of $K$ is $p^{[K:\mathbb{Q}]-1}$ (according to Sage). How can I prove that/ is the proof implied by something written down somewhere?

I have a feeling this is somewhat deep so I ask it here; if it's insufficiently deep I'm happy to close it and ask elsewhere.

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    $\begingroup$ The Führerdiskriminantenproduktformel tells you that is it the product of conductors of characters, but all but the trivial character must have conductor p. $\endgroup$
    – Chris Wuthrich
    Commented Oct 18, 2019 at 12:31

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The Führerdiskriminantenproduktformel tells you that is it the product of conductors of characters, but all but the trivial character must have conductor $p$.

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This would also work for non-abelian, and even non-Galois extensions: if $L/\mathbb{Q}$ is totally ramified and tamely ramified at $p$ and $K$ is an intermediate field, then $K$ is also totally tamely ramified at $p$, so the exponent of $p$ in the discriminant of $K$ is $[K:\mathbb{Q}]-1$ (Serre, Corps Locaux, Proposition 13).

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    $\begingroup$ This is (IMO) a better answer. First, the different (and discriminant) of a totally tamely ramified extension is certainly a simpler notion than understanding the conductor of an abelian extension. Second, it is more general. Of course, both answers are correct. $\endgroup$
    – user145307
    Commented Oct 18, 2019 at 21:29
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    $\begingroup$ @ElectricPenguin theres a very elementary definition of conductor in this special case: the smallest n s.t. $\chi$ factors through (Z/pZ)*-->(Z/nZ)* $\endgroup$
    – dessin d'enfant terrible
    Commented Oct 19, 2019 at 22:52
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    $\begingroup$ @dessind'efantterrible To split hairs, I would say that this is a computation of the conductor in this case rather than a definition. The distinction is that this if you actually want to prove the Führerdiskriminantenproduktformel (in the abelian case) for a local extension $L/K$, you really need to grapple with $N_{L/K}(L^*)$ in some way (say via basic local class field theory), whereas one could compute (with a full proof) the different of a totally tamely ramified extension $L/K$ in a single MO comment. $\endgroup$
    – user145307
    Commented Oct 20, 2019 at 0:33
  • $\begingroup$ @GHfromMO The extension $\mathbf{Q}_p(p^{1/e})$ with $e > 1$ prime to $p$ (for example) is certainly totally tamely ramified and is never generated by a root of unity. Perhaps you are thinking about the fact that an abelian extension of $\mathbf{Q}_p$ is cyclotomic (local Kronecker-Weber)? $\endgroup$
    – user145307
    Commented Oct 20, 2019 at 0:44

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