For a very long time I have wanted to derive the class number formula, and recently I was able to wrap my head around the formula in the case of an imaginary quadratic field. Now I want to see if I can wrap my head around the proof for the formula in a general field. The parts of a field I want to understand are:

  • Embeddings (real and complex)
  • Regulator
  • Discriminant of a field

If you were to describe these in the simplest terms possible how would you? (I'm only a junior in high school, so trying to teach myself these concepts is really difficult when the only resources available to me are from colleges/Wikipedia. At the moment, I don't care about how "formal" my understanding is, I just want to be able to understand what these terms "show" about a field. ) Any examples would help a lot too, especially in an imaginary quadratic field as I am most comfortable with that. I would really appreciate any help, thanks.

  • 1
    $\begingroup$ $\Bbb{Q}(i)$ has two complex embeddings $\sigma(a+ib) = a+ib$ and its complex conjugate. For a number field $K$ (assuming it has no real embeddings) the main object is the map $K \to \Bbb{C}^{n/2},\rho(a) = (\sigma_1(a),\ldots,\sigma_{n/2}(a))$ where $\sigma_1,\ldots,\sigma_{n/2}$ are the $n/2$ pairs of complex embeddings. $\rho$ sends $O_K$ to a lattice in $\Bbb{R}^n \cong \Bbb{C}^{n/2}$, the discriminant tells the volume of the fundamental parallelogram of that lattice, the regulator tells the volume of the $\log$-lattice coming from the unit group $O_K^\times$ $\endgroup$ – reuns Jun 27 at 22:44
  • $\begingroup$ the zeta function of $K$ is obtained by summing $(\prod_j |\rho(a)_j|^2)^{-s}$ over $ a \in O_K$ and over $a$ in each ideal class. $\endgroup$ – reuns Jun 27 at 22:45
  • $\begingroup$ @reuns so are embeddings essentially be the same as elements of the galios group of the field? And by the example I'm assuming that a real embedding would be a mapping that only acts on real parts of an element of a field (so conjugation and identity of a+b*sqrt(d) for d>0 in Q(sqrt(d))). $\endgroup$ – uhhhhidk Jun 27 at 23:11
  • $\begingroup$ @reuns if I took the field Q(2^1/4) would the embeddings be the identity function, the real conjugate, and f(2^1/4)=i*2^1/4 and its conjugate? $\endgroup$ – uhhhhidk Jun 28 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.