How to calculate the norm of an ideal?

Question

Would someone please help explain how to calculate the norm of an ideal? I can't find a source that explains this clearly. For example, I know that the norm $N_\mathbb{Q(\sqrt10)}(\langle2,\sqrt10\rangle)=2$, but not clear on how to get it myself. Many thanks

Answer 1

In general, given $I=(s_1,\dots,s_n)\lhd\mathcal{O}_K$ in the ring of integers of some number field $K$, we can factor it into primes to try and work out its norm.

You can see the answer here which explains the rational behind it, but the important points are that:

1. If a prime $\mathfrak{p}\mid I$, then $s_1,\dots,s_n\in\mathfrak{p}$. Let $(p)=\mathfrak{p}\cap\mathbb{Z}$. Then by the multiplicativity of the norm, we know that $\text{N}(\mathfrak{p})\mid\text{N}(s_i)$, and $\text{N}(\mathfrak{p})$ is always a prime power. This means that the primes in $\mathcal{O}_K$ dividing $I$ are above the rational primes dividing $\gcd(\text{N}(s_1),\dots,\text{N}(s_n))$: this gives us a finite list to check!
2. If $\mathcal{O}_K=\mathbb{Z}[\alpha]$, then find the possible primes $\mathfrak{p}=(p,g(\alpha))$. Then $\mathfrak{p}\mid I$ if and only if $s_1,\dots,s_n\in(\overline{g})\lhd(\mathbb{Z}/p\mathbb{Z})[\alpha]$.

In your example, $\text{N}(a+b\sqrt{10})=a^2-10b^2$, so $\text{N}(2)=4$, and $\text{N}(\sqrt{10})=-10$. So the only primes dividing $I=(2,\sqrt{10})$ are of norm $2$. In $\mathcal{O}_K=\mathbb{Z}[\sqrt{10}]$, there is in fact only one prime ideal of norm $2$, namely $\mathfrak{p}_2=(2,\sqrt{10})=I$.

---

Alternatively, you can use the definition that $\text{N}(I)=|\mathcal{O}_K/I|$, and this method may well be quicker with simpler examples such as these. It should be quite clear that $$ \mathcal{O}_K/I=\mathbb{Z}[\sqrt{10}]/(2,\sqrt{10})\cong\mathbb{Z}/2\mathbb{Z} $$