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I'm interested in the class number formula derived from the Dedekind zeta function, but I have no idea how to derive it. From what I've read, turning the Dedekind zeta function into a normal Dirichlet series and analyzing the coefficients can lead to the formula. I know that for a field K (I'm interested in imaginary quadratic fields, so assume K=$mathbbQ(sqrt-n), n notequiv 3 mod(4)$), $zeta _K(s)=sum_ain mathcalO_Kfrac1N(a)^s$. If my intuition is correct this can be rewritten as $zeta _K(s)= sum_m=1^inftyfracc_mm^s$ where $c_m=| a : N(a)=m, ain mathcalO_K |$, which basically means the number of ideals that have norm $m$ in $mathcalO_K$. Now my problem is approximating $c_m$. Heres what I think:

So if we look at $C_m=sum_n=1^mc_n$, this equates to finding the amount of lattice points of $mathcalO_K$ inside a circle of radius $sqrtm$. Since a circles area grows at the same rate as a square, the ratio of the area of the circle and the "fundamental region" should provide a decent estimate, especially for large m $fracpi msqrt$. However, in the context of this problem, since $(a)=(ua) ; uinmathcalO_K^ times$ we are "overcounting" these points so if $w=mathcalO_K^ times$ the original estimate should be fixed to $fracpi mwsqrt$. So what (I think at least) I have is $sum_n=1^mc_n$ is reasonably approximated by $fracpi mwsqrt$. From here I have no idea how to continue.

Now I know I'm missing the most fundamental part of the entire thing ($h$) but I have no idea how the ideal class group would play into this, to me it makes no sense. I think this approach should work but I just have no idea how to continue. If someone could explain how the ideal class group affects counting these points inside a circle or how to finish the proof of the class number formula, I'd really appreciate it. I apologize if something is being grossly misunderstood, as a junior in high school using only online resources trying to understand problems like this is terribly difficult. Thanks for any help.

For $K=BbbQ(sqrt-d)$ imaginary quadratic field the class number formula follows from

$$zeta_K(s) = sum_c in C_K sum_I subset O_K,Isim c N(I)^-s=frac1O_K^times sum_c in C_K sum_a in I_c^-1, a ne 0 (fracN(a I_c)N(I_cI_c^-1))^-s$$
where $C_K$ is the ideal class group, $I subset O_K $are the ideals of $O_K$, $I_c$ is an ideal in the class of $c$, so $I_cI_c^-1$ is a principal ideal.

Then

$$|O_K^times|(2pi)^-sGamma(s) zeta_K(s) = int_0^infty x^s-1(f(x)-|C_K|)dx$$ where

$$f(x) = sum_c in C_K sum_a in I_c, a ne 0 e^^2 x/N(I_c)= sum_c in C_K sum_n,m e^^2$$

and $I_c = N(I_c)(u_c BbbZ+v_c BbbZ)$ is a lattice in $O_K$ and $u_c BbbZ+v_c BbbZ$ is a lattice in $K$ whose area of fundamental parallelogram doesn't depend on $c$ thus is of area $D/4$, $D=Disc O_K$.

From there it suffices to show the asymptotic of $sum_n,m e^^2$ as $x to 0$

depends only on the area of the fundamental parallelogram of $u_c BbbZ+v_c BbbZ$ so that $sum_n,m e^^2sim x^-1D^-1/2$, a claim which follows from the asymptotic $sum_n e^-pi n^2 x sim x^-1/2$.

On the other hand that $f(x) sim |C_K|D^-1/2 x^-1$ implies $|O_K|^times(2pi)^-sGamma(s) zeta_K(s) sim fracC_Ks-1$ as $s to 1$ ie. from $zeta_K(s) = zeta(s) L(s,chi)$ where $chi(p) = (fracdp) = (fracpD)$ then $|C_K|D^-1/2 = |O_K^times|(2pi)^-1L(1,chi)$.