# Counting number of ideals in quadratic number field

Let $$K$$ be a quadratic number field and $$R$$ be its number ring, and if $$a(n)$$ denotes number of ideals of norm $$n$$, if $$n$$ is a prime number, then number of ideals of norm $$n$$ is $$1+(d|n)$$, where $$d$$ is the discriminant of $$K$$ and $$(d|n)$$ is Legendre-Jacobi-Kronecker symbol. I wantn to find the number of ideals of norm $$p^m$$ where $$p$$ is a prime and $$m$$ is natural number greater than $$1$$.

To avoid misunderstanding, I first reprove your assertion in the case $$n=$$ a prime $$p$$ (excluding the particular case $$p=2$$). A rational odd prime $$p$$ decomposes in a quadratic field $$K$$ of discriminant $$D$$ in 3 possible manners : (i) $$p \mid D, p$$ is totally ramified, i.e. $$(p)=P^2, N(P)=p$$ ; (ii) $$p \nmid D$$, and $$(\frac Dp)=1$$ iff $$p$$ splits, i.e. $$(p)=P.P', N(P)=N(P')=p$$ ; (iii) $$p \nmid D$$, and $$(\frac D p)=-1$$ iff $$p$$ is inert, i.e. $$(p)=P^2, N(P)=p^2$$ . Considering these 3 cases, as well as the uniqueness of prime factorization in a Dedekind ring, one gets immediately that the number $$a(p)$$ of ideals of norm $$p$$ is $$1$$ if $$p \mid D$$, resp. $$1+(\frac Dp)$$ if $$p \nmid D$$ .
The calculation of $$a(p^m)$$ proceeds exactly in the same way because of the uniqueness of prime decomposition : if you have a prime decomposition of $$(p)$$, then you get a prime decomposition of $$(p^m)$$, and uniqueness tells you that this is the only possible decomposition of $$(p^m)$$. Examining the 3 cases and recalling that the norm of ideals is a multiplicative function, you get that $$a(p^m)=a(p)$$ in the first 2 cases since $$N(P^m)=p^m$$. In the 3-rd case, you have $$(p^{2k})=P^{4k}, N(P^k)=p^{2k}$$ and $$a(p^{2k})=1$$, whereas $$a(p^{2k+1})=0$$ obviously.
• You should mention you are using quadratic reciprocity to go from $(\frac{\delta}{p})$ to $(\frac{p}{D})$. It is condensed in $a(n) = (\frac{.}{D}) \ast 1 = \prod_{p^k \| n} (\sum_{m=0}^k (\frac{p^m}{D}))$. – reuns Apr 17 at 18:05