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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.

Moosh
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  • 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 09 '20 at 18:18
  • yes, we also already know quite a bit about class groups in general. – Moosh Dec 09 '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
  • Related. – Jyrki Lahtonen Dec 11 '20 at 23:04

1 Answers1

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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.

CPCH
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  • @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 02:57
  • @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 04:55
  • @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