How many integral ideals $\mathfrak{a}$ are there with the given norm $\mathfrak{N}(\mathfrak{a})=n$? It is Exercise 1.6.1 in Jürgen Neukirch's number theory textbook(P38).
I think the number $n$ here means $[K:\mathbb{Q}]$, the degree of field extention for algebraic number field K.
I can image how simple this question is, but I was still floored by it.
Any help would be appreciated.
 A: No, $[K:\Bbb{Q}]$ has nothing to do with it. You could rephrase the exercise as

Given a fixed positive integer $n$, how many ideals are there with norm $n$?

Hint: You know that $\mathfrak{N}$ is multiplicative, i.e. that $\mathfrak{N}(\mathfrak{a}\mathfrak{b}) = \mathfrak{N}(\mathfrak{a})\mathfrak{N}(\mathfrak{b})$. Furthermore, you know that if $\mathfrak{p}$ is a prime ideal of $\mathcal{O}_K$ such that $\mathfrak{p} \cap \Bbb{Z} = p$, then $\mathfrak{N}(\mathfrak{p}) = p^f$, where $f = [\mathcal{O}_K/\mathfrak{p}:\Bbb{F}_p]$ is the relative degree of $\mathfrak{p}$.
A: It’s a question of how the various rational primes split in $K$, and has relatively little arithmetic content, but looks to me more like a combinatorial problem. Let’s just look at $\Bbb Q(i)$, the Gaussian numbers, and its ring of integers $\Bbb Z[i]$.
The norm of an ideal $\mathfrak a$ of the ring of integers $R$ is the cardinality of $R/\mathfrak a$. The prime ideals of $\Bbb Z[i]$ are, as you probably know, $(1+i)$ above $2$, with norm $2$; the rational primes $p$ with $p\equiv3\pmod4$ and they are still Gaussian primes, they have norm $p^2$; and above each prime $p=a^2+b^2\equiv1\pmod4$, there are two primes, $(a\pm bi)$. Each of these has norm $p$.
So, in $\Bbb Z[i]$, there are no ideals of norm $15$ because there’s no prime of norm $3$. On the other hand, there are $2$ primes of norm $45$, namely $(3(2+i))$ and $(3(2-i))$. I think you can see where the combinatorial questions come in, especially in extensions of higher degree, where there will be rational primes that split completely into $[K:\Bbb Q]$ primes upstairs.
A: In order to end this question, I demonstrate an example of cyclotomic field $K=\mathbb{Q}(\zeta_5)$, which is not so simple as a PID (equal as UFD in Dedekind domains) (Since it's well-known that $\mathbb{Q}(\sqrt{5})\subset\mathbb{Q}(\zeta_5)$, and $2^2=(\sqrt{5}+1)(\sqrt{5}-1)$ in $\mathbb{Q}(\sqrt{5})$). 
Here consider $n=[K:\mathbb{Q}]=4$, the property of $\mathfrak{N}(\mathfrak{a})$ mentioned above implies that:
prime idea $\mathfrak{p}$ lying over $\mathfrak{a}$ $\Rightarrow$ $\mathfrak{p}\cap\mathbb{Z}=(2)$, and $\mathfrak{N}(\mathfrak{p})$ equals $2$ or $4$.
which is impossible
(since we can calculate the degree of $\mathfrak{p}$ by formula $2^f\equiv1(5)$ $\Rightarrow f=4$).
so the number is zero in the case. You can get similar result in $\mathbb{Q}(\zeta_7)$ by the same method. Here $\zeta_n=e^{2\pi i/n}$.
