Is there a definition out there of the notion of conductor of a non-abelian number field (i.e. a finite extension of Q whose Galois group is non-abelian)? If not, is there anyone you know of working on it? The definition for abelian number fields uses class field theory; it comes out of Artin reciprocity (see page 525 of Neukirch's Algebraic number theory).
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$\begingroup$ Sorry, does non-abelian mean "with non-abelian Galois group"? $\endgroup$– Ben Webster ♦Commented Oct 31, 2009 at 16:49
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$\begingroup$ Yeah, sorry. Similarly, abelian means with abelian Galois group. And the Galois group I mean is that over Q. You can also define the conductor of an extension of number fields L/K with abelian Galois group (again using the Artin reciprocity map). $\endgroup$– Rob HarronCommented Oct 31, 2009 at 17:02
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3 Answers
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I think that a good notion of "conductor" isn't going to be intrinsic to the extension K/Q; rather, you might choose some finite-dimensional complex representation rho of Gal(K/Q) and then use the Artin conductor of the resulting Galois representation. When K/Q is abelian, there aren't so many interesting choices of rho.
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I don't know the precise definition of the conductor of a nonabelian Galois extension of Q, but see page 10 of Langlands's expository article "Representation Theory: Its Rise and Its Role in Number Theory" sunsite.ubc.ca/DigitalMathArchive/Langlands/pdf/gibbs-ps.pdf . Presumably the thesis of Joe Buhler referenced therein gives a precise definition, or a reference to one.
answered Oct 31, 2009 at 17:45
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$\begingroup$ Buhler is discussing: given a Galois extension K/Q with Galois group A_5, what is the smallest Artin conductor of a two-dimensional Galois representation factoring through Gal(K/Q) whose image in PGL(2,C) is A_5? Buhler finds it's 800, and occurs for the quintic polynomial Langlands mentions. Langlands (and Buhler 2 or 3 times) refers to this as the "conductor of K", but Buhler generally refers to this as the minimal conductor of a corresponding projective representation. He makes no assertion that this is a definition of the conductor of K. It certainly offers a possibility though. Thanks. $\endgroup$ Commented Oct 31, 2009 at 19:45
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The conductor of an order O \subset K is defined on p. 79 of Neukirch's Algebraic Number Theory (so that for K you define the conductor to be the conductor of its ring of integers). This conductor tells you how to compute how a prime decomposes in K. I think in the abelian case these are the same conductors as in class field theory.
In the non-abelian case, maybe the conductor is related to the zeta function of K (I'm not sure off the top of my head).
answered Oct 31, 2009 at 16:46
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$\begingroup$ The notion of conductor you are talking about is unrelated. It simply measures how far away a ring is from its integral closure. Since the ring of integers is by definition integrally closed, its conductor (in this sense) is the unit ideal. (This conductor tells you where you can't easily determine the decomposition of a prime.) The conductor I'm talking about is defined on page 525 of Neukirch. $\endgroup$ Commented Oct 31, 2009 at 16:59