# 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?

• Notice that all ideals of prime norm is a prime ideal. Thus it amounts to the same to finding the prime ideal decompositions of $2\mathbb Z$ and $5\mathbb Z$ in the given extension. – awllower Mar 4 '13 at 12:46
• Why is finding a prime ideal of norm $2$ the same as factorizing $(2)$? – Spook Mar 4 '13 at 12:48
• If the prime ideal is of norm $2$, then it divides $2$. But you already know that $2$ is totally ramified, so that means all prime ideals dividing $2$ must be of norm $2$.In shorter terms, look at the unique factorisation of ideals, and notice that having norm $p$ means dividing $p$. – awllower Mar 4 '13 at 12:51
• In which case, the only ideal of norm $2$ is $(2,\alpha)$? – Spook Mar 4 '13 at 12:56
• An ideal f norm $2$ is prime, and it divides $2$, so that it must be $(2,\alpha)$. – awllower Mar 4 '13 at 12:57

## 2 Answers

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".

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).

• Can you give me a proof or this? I know that the ideal $(p)$ is the product of prime ideal with norm a power of $p$, but how i prove that they are all? – Yecabel May 20 '18 at 14:00
• @Yecabel If $\mathfrak{p}$ is a nonzero prime ideal, then $\mathcal{O}_K/\mathfrak{p}$ is a finite field. Suppose that the order of this field is $p^r$. Then $p+\mathfrak{p}=0+\mathfrak{p}=\mathfrak{p}\in \mathcal{O}_K/\mathfrak{p}$ and $p\in \mathfrak{p}$ and $\langle p\rangle\subseteq \mathfrak{p}$. Write $\langle p\rangle$ as the prime ideals decomposition $\mathfrak{p}_1^{r_1}\mathfrak{p}_2^{r_2}\cdots \mathfrak{p}_s^{r_s}$. – bfhaha May 31 '18 at 9:30
• Then $\mathfrak{p}_1^{r_1}\mathfrak{p}_2^{r_2}\cdots \mathfrak{p}_s^{r_s}=\langle p\rangle \subseteq \mathfrak{p}$ and $\mathfrak{p}_i\subseteq \mathfrak{p}$ for some $i\in \{1, 2, ..., s\}$ because $\mathfrak{p}$ is a prime ideal. In the Dedekind domain $\mathcal{O}_K$, $\mathfrak{p}_i$ is maximal and the prime ideal $\mathfrak{p}\neq \mathcal{O}_K$ (by the definition of a prime ideal). It follows that $\mathfrak{p}=\mathfrak{p}_i$. That is, every prime ideal $\mathfrak{p}$ must appear in the prime ideals decomposition of some $\langle p\rangle$. – bfhaha May 31 '18 at 9:30