# How does knowing the norm of an ideal tell us about its factors?

In the ring $$\mathbb{Z}[\sqrt{223}]$$ we have that the ideal $$\mathfrak{p}_3=(3,\sqrt{223}+1)$$ is a prime ideal lying over 3, and that the ideal $$\mathfrak{p}_{11}=(11,\sqrt{223}+5)$$ is a prime ideal lying over 11. If we find an element $$16+\sqrt{223}$$ with norm 33, this is supposed to imply that $$(16+\sqrt{223})=\mathfrak{p}_3\mathfrak{p}_{11}$$ But I am having trouble seeing why this is the case.

• how much do you know about the ideals of rings of integers of number fields? do you know that every ideal in $\mathbb{Q}[\sqrt{223}]$ has a factorization as a product of prime ideals? – Atticus Stonestrom Dec 9 '20 at 18:18
• yes, we also already know quite a bit about class groups in general. – Moosh Dec 9 '20 at 19:13
• An old answer of mine happens to include a proof by brute force calculation of this. Another key is that $\mathfrak{p}_3$ is not principal. Neither is $\mathfrak{p}_{11}$. The conjugate of either is obviously its inverse in the class group. When you find an element with norm $33$ it follows that some product of an ideal of norm $3$ and an ideal of norm $11$ is principal. But this is insufficient information to uniquely identify the pair of prime ideals IMO. – Jyrki Lahtonen Dec 11 '20 at 22:50
• (cont'd) After all, if $\mathfrak{p}_3\mathfrak{p}_{11}$ is principal, so is $\overline{\mathfrak{p}_3}\,\overline{\mathfrak{p}_{11}}$, right? – Jyrki Lahtonen Dec 11 '20 at 22:51
• If you already know that the both $\mathfrak{p}_3$ and $\mathfrak{p_{11}}$ have order three in the class group, then finding an element with norm $33$ shows that they belong to the same subgroup of order three. – Jyrki Lahtonen Dec 11 '20 at 22:56

Let $$K$$ be a number field. There are a couple of facts coming in handy here:

1. Any ideal of $$\mathcal{O}_K$$ has a unique factorization into positive powers of prime ideals.
2. The ideal norm is multiplicative.
3. If an ideal is integral, then its norm is in $$\mathbb{Z}$$.
4. The norm of a prime $$\mathfrak{p}$$ over $$p\in \mathbb{Z}$$ is always a power of $$p$$.
5. The norm of the ideal $$x\mathcal{O}_K$$ is the same as the norm of $$x$$.

Then we just have to put these things together... An ideal of norm $$33$$ has to be the product of some prime lying over $$3$$ and one lying over $$11$$. Now you just have to make sure that these ideals over $$3$$ and over $$11$$ are the right ones.

EDIT: As Jyrki Lahtonen correctly remarked, we are not done here and the computations are not as easy as I hoped:

Be careful here since both $$3$$ and $$11$$ split, so there are $$4$$ ideals with norm $$33$$ and we have to be sure that $$\mathfrak{p}_3\mathfrak{p}_5$$ is the ideal we are looking for and not one of the other three. Let's calculate (by multiplying generators): $$\mathfrak{p}_3\mathfrak{p}_5 = (33, 11\sqrt{223}+11,3\sqrt{223}+15,228+6\sqrt{223})$$

A quick computation yields that $$2\cdot (11\sqrt{223}+11)-7\cdot (3\sqrt{223}+15)+3\cdot 33 = 16+\sqrt{223}$$

Thus $$(16+\sqrt{223})\subset \mathfrak{p}_3\mathfrak{p}_5$$ meaning that $$\mathfrak{p}_3$$ and $$\mathfrak{p}_5$$ divide $$(16+\sqrt{223})$$, so by the first paragraph we have equality.

• @JyrkiLahtonen You know the product of the two ideals of norm 3 is principle, because the product of the two ideals of norm 3 is exactly the ideal (3). – Moosh Dec 12 '20 at 2:57
• @mshoosterman Yes, I know that. The point I wanted to make might be the following. There are 4 ideals of norm $33$, $\mathfrak{p}_3\mathfrak{p}_{11}$, $\mathfrak{p}_3\overline{\mathfrak{p}_{11}}$, $\overline{\mathfrak{p}_3} \mathfrak{p}_{11}$ and $\overline{\mathfrak{p}_3}\overline{\mathfrak{p}_{11}}$. Finding a principal ideal of norm $33$ tells us that at least one of those ideals is principal. Conjugation tells us that at least two of them are principal. So if we find a principal ideal of norm $33$ I don't see how it would immediately follow that it should be the first of these four? – Jyrki Lahtonen Dec 12 '20 at 4:55
• @JyrkiLahtonen Yes, thanks for the remark. I had assumed that the OP was just struggling with the norm argument and I was too lazy. Bad behavoiour on my part. I added another paragraph that should take care of the rest of the argument. Not super aesthetic, but it should work and is rather short and elementary. – CPCH Dec 12 '20 at 17:09
• @mshoosterman If you still need some pointers: Check out the edit. Hope I didn't mess up the computation – CPCH Dec 12 '20 at 17:12
• Nice ending! Of course, it suffices to show that $16+\sqrt{223}$ is in the product ideal. In the other thread I went to great pains proving that it is a generator, but the known norm of the product makes that obvious :-) – Jyrki Lahtonen Dec 12 '20 at 18:50