Number of ideals of norm $100$ of the Kleinian integers The question to begin with is: how many ideals of norm $100$ does the ring of integers of $K=\mathbb{Q}(\sqrt{-7})$ have?
I know the following stuff for certain: $\Delta_K=-7$ and $\mathscr{O}_K=\mathbb{Z}[\frac{1+\sqrt{-7}}{2}]=\mathbb{Z}[x]/(x^2-x+2)$ (the ring of Kleinian integers).
What I want to know is what is the formula for the norm? The norm in $\mathbb{Z}[\sqrt{d}]$ is $N(a+b\sqrt{d})=a^2+db^2$, and for the Einsteinian integers we have something like $N(a+b\alpha)=a^2-ab+b^2$, where $\alpha=\frac{1+\sqrt{-3}}{2}$. I suppose that the Kleinian integers must have such a nice formula too. Secondly, that formula will only help me in finding the number of principal ideals in $\mathscr{O}_K$ of norm $100$, and Sage tells me that the class number is $1$, so those will be all ideals of norm $100$, right? Is there an alternative for this computation in stead of computing the class group of $K$?
Lastly, this way seems to require computing all ideals explicitly and then counting them. Isn't there something more smart to do to count ideals of a certain norm in a ring?
Added: The norm can be computed by taking the determinant of the matrix $M_x:\mathscr{O}_K\to\mathscr{O}_K,b\mapsto xb$ for $x\in\mathscr{O}_K$. However, I don't know what such a matrix must be in this case. For the norm in a simpler ring $\mathbb{Z}[\sqrt{d}]$, we just have the matrix $[[x,dy],[y,x]]$ with determinant $x^2−dy^2$. But honestly, I don't know where this matrix comes from...
More added: I already found what the norm should be. We take $\overline{\frac{1+\sqrt{d}}{2}}=\frac{1-\sqrt{d}}{2}$ and then the norm in $\mathscr{O}_{\mathbb{Q}(\sqrt{d})}=a^2-db^2$ when $d=2,3\;(4)$ and $\mathscr{O}_{\mathbb{Q}(\sqrt{d})}=a^2+ab+\frac{1-d}{4}b^2$ when $d=1\;(4)$; found on page 230 of Dummit. However, still don't know how to find that matrix.
More added: $(2)$ is definitely not prime in $\mathscr{O}_K$, since it is the product $\alpha\cdot\overline{\alpha}$ with $\alpha=\frac{1+\sqrt{-7}}{2}$.  We have $2\mathscr{O}_K=(2,\alpha)(2,\alpha-1)$ and $5\mathscr{O}_K=(5,\alpha^2-\alpha+2)$ by Kummer-Dedekind. How to determine the ideals of norm $100$?
 A: Many of us out here could answer the question for you, but that would deprive you of all the fun.
I think you want to use the fact that every ideal of $\mathscr O_K$ is uniquely expressible as product of prime (maximal) ideals of that ring; so all you need to know is how to factor $(2)=2\mathscr O_K$ and $(5)$. To see whether these ideals are already prime in your ring, you can probably make use of the norm form for your ring, analogous to $a^2-ab+b^2$ for $\Bbb Z[\frac{-1+\sqrt{-3}}2]$. You must work that out for yourself. It’s not hard.
P.S.: Please don’t use $\alpha$ and $a$ in the same formula: it’s hard on the eyes of us oldsters.
A: As you said, the norm of $x$ is the determinant of the multiplication-by-$x$ map $M_x:\mathcal{O}_K \to \mathcal{O}_K$, $z\mapsto zx$.
Indeed this map is a linear map, and a $\mathbb{Z}$-basis of $\mathcal{O}_K$ is given by $1, \frac{1+\sqrt{-7}}{2}$.
The matrix of $M_x$ is determined by the images of the basis elements.
If $x = a+b\frac{1+\sqrt{-7}}{2}$ then $1x = a+ b\big(\tfrac{1+\sqrt{-7}}{2}\big)$ and:
$$\big(\tfrac{1+\sqrt{-7}}{2}\big)x = \big(\tfrac{1+\sqrt{-7}}{2}\big)a+\big(\tfrac{1+\sqrt{-7}}{2}\big)^2b=\big(\tfrac{1+\sqrt{-7}}{2}\big)a+\big(\tfrac{1+\sqrt{-7}}{2}\big)b - 2b = \big(\tfrac{1+\sqrt{-7}}{2}\big)(a+b) - 2b.$$
So the matrix of $M_x$ with respect to the basis $1, \frac{1+\sqrt{-7}}{2}$ is $\begin{pmatrix}a & -2b\\ b & a+b\end{pmatrix}$.
The norm of $x$ is the determinant $\det(M_x) = a(a+b) + 2b^2 = a^2+ab+2b^2$.

You know that $\alpha = \frac{1+\sqrt{-7}}{2}$ and $1-\alpha$ satisfy $x^2 - x + 2 = 0$.
So $2 =\alpha - \alpha^2 = (1-\alpha) - (1-\alpha)^2$ implies that $(2,\alpha) = (\alpha)$ and $(2,1-\alpha) = (1-\alpha)$ are principal.
Also $(5,\alpha^2-\alpha+2) = (5,0) = (5)$ is principal.

If an ideal $I$ has norm $100$ then $100 \in I$; hence $(100) \subset I$.
Since $\mathcal{O}_K$ is a Dedekind domain, "contains" = "divides", i.e. $I \mid (100)$.
We know $(100) = (5)^2(\alpha)^2(\alpha-1)^2$ is the factorization into primes of $\mathcal{O}_K$.
So $I = (5)^{e_1}(\alpha)^{e_2}(\alpha-1)^{e_3}$ with $0 \leqslant e_i \leqslant 2$.
The norm of $(5)$ is $5^2$ and the norms of $(\alpha)$ and $(1-\alpha)$ are $2$. The norm is multiplicative.
So if $I$ has norm $100$ then $e_1 = 1$ and $e_2+e_3=2$.
That is $I=(5)(\alpha^2)=(5)(\alpha-2)$ or $I=(5)(1-\alpha)^2=(5)(-\alpha-1)$ or $I = (5)(\alpha)(1-\alpha) = (5)(2) = (10)$.
To get the elements of norm 100 you can multiply these (generators) by units of $\mathcal{O}_K$.
See also How many elements in a number field of a given norm?
