How to find all the ideals of a given norm? I am working on a question:

Find all the ideals of norm $10$ in $\mathcal{O}_K$ where $K=\mathbb{Q}(\sqrt{35})$.

I am given the hint:

Observe that $(2)=(2,\alpha)^2, (5)=(5,\alpha)^2, (\alpha)=(2,\alpha)(5,\alpha)$ where $\alpha=5+\sqrt{35}$.

So I can tell that $(2,\alpha)$ is of norm 2, $(5,\alpha)$ is of norm 5, $(\alpha)$ is of norm 10. I think, to find all ideals of norm 10 I first should find all ideals of norm 2 and 5 since any ideal of norm 10 factorizes into prime ideals of norms 2 and 5.
But in general, how do I find all ideals of norm 2, 5?
 A: You can also approach this the following way:
If $I$ is an ideal of norm $10$, the additive group of $\mathcal O_K/I$ is an abelian group with $10$ elements, hence isomorphic to $\mathbb Z/10\mathbb Z$. We obtain $\mathcal O_K/I \cong \mathbb Z/10\mathbb Z$ as rings, as the ring structure of a cyclic group is uniquely determined by $1 \cdot 1 = 1$.
So ideals of norm $10$ are in bijective correspondence to surjective ring maps $\mathcal O_K \to \mathbb Z/10\mathbb Z$. By the homomorphism theorem such maps correspond to maps $\mathbb Z[X] \to \mathbb Z/10\mathbb Z$, which map $X^2-35$ to zero. By the universal property of the polynomial ring, such maps correspond to elements $a \in \mathbb Z/10\mathbb Z$ with $a^2-35=a^2-5=0$. We get $a=5$ as unique solution. Going reverse in our arguments, this gives us the map $\mathbb Z[\sqrt{35}] \to \mathbb Z/10\mathbb Z, a+b\sqrt{35} \mapsto a+5b$ with kernel $(5+\sqrt{35})$. This is the unique ideal of norm $10$.
After "We get $a=5$ as the unique solution", you could stop and be like "Using abstract arguments, I showed there is a unique such ideal. The hint gives me one such ideal. So I am done".
A: The ideal $(p)$ is the product of all of the prime ideals of norm a power of $p$  (with multiplicity equal to the ramification index).
