I came across the following line and was wondering what it meant exactly and how you go about showing it. Let d be a fundamental discriminant. Let P(d) = the divisors of d except for the largest.

The cardinality of P(d) is at most the 2-adic valuation of the h (the class number).

I only have a vague notation of what is meant by the 2-adic valuation, so any clarification on that as well as how to prove the statement would be helpful. Thank you.

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    $\begingroup$ If the discriminant has t prime divisors, then the class number is divisible by 2^{t-2}, and the class number in the strict sense by 2^{t-1}. The 2-adic valuation here is the largest power of 2 dividing the class number. For proofs, see a book on quadratic number fields or google for "genus theory". $\endgroup$ Jul 15, 2010 at 9:10
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    $\begingroup$ It means the highest power of 2 dividing the class number is something like 2^t or 2^(t-1), where t is the number of prime factors of the discriminant. This is related to genus theory and the point is that it predictably tells you information about the quotient group C/C^2, where C is the class group of a quadratic field. $\endgroup$
    – KConrad
    Jul 15, 2010 at 9:13
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    $\begingroup$ Of course you know genus theory provides a lower bound for the class number; the highest power of 2 dividing the class number can be considerably larger if there are elements of large 2-power order. $\endgroup$ Jul 15, 2010 at 9:23

1 Answer 1


This is in the book by Buell I keep mentioning, largely pages 66-73 about genera and the class group. But it is clearer in "Primes of the Form $x^2 + n y^2$" by David A. Cox. Here you want Proposition 3.11 on page 52 and Theorem 3.15 on page 54. Note that Cox concentrates on discriminants that are negative and even, then mentions odd discriminants in various side notes or exercises. For the kind of things you seem to be doing over your past several questions, you ought to have both books.

  • $\begingroup$ Yep I picked it up from the library yesterday and have Cox's book laying around somewhere. I think my main confusion was due to the 2-adic part. Once that was cleared up it made sense and the proof is straight forward. Thanks! $\endgroup$
    – paarshad
    Jul 15, 2010 at 22:44

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