# How to determine the ideal in a ring of integers of a number field corresponds to a given norm.

I want to determine the prime ideal decomposition of $$I=(19-11\sqrt{-5})\subseteq \mathcal O_K$$

Where $$K=\mathbb Q(\sqrt{-5})$$ and $$\mathcal O_K=\mathbb Z[\sqrt{-5}]$$.

Since the ideal $$I$$ seems very odd, I may try the norm of $$I$$ as $$19^2+5\cdot 11^2=2\cdot 3\cdot 7\cdot 23$$, since $$I$$ is principal.

Although I can calculate the behaviour of $$2,3,7,23$$ and calculate their ideal above in $$\mathcal O_K$$ whether they ramify, inert, split, I am not sure that these induced primes will correspond the decomposition of the given original ideal $$I$$ above.

So how can we calculate the prime decomposition of $$I$$ above, is this norm idea appropriote?

• See this duplicate, for $I=(19+11\sqrt{-5})$, which works very similar. Dec 23, 2022 at 19:03
• $69=8^2+5\cdot1^2,$ $6=1^2+5\cdot 1^2,$ etc. So $23$ come from one of the ideals $\langle 23,8\pm\sqrt{-5}\rangle.$ $3$ comes for some ideal $\langle 3,1\pm \sqrt{-5}\rangle,$ etc. Dec 23, 2022 at 19:04
• Not duplicate because of the title and my last comment, I want to see if possible the connection between norm, not just the decomposition. Dec 23, 2022 at 19:26

You say "the ideal $$I$$ seems very odd", but in fact it is very even, since its norm is a multiple of $$2$$. :)

Anyway, the norm of an ideal is a very important numerical value for figuring out how an ideal factor factors.

Theorem. If $$I$$ is a nonzero ideal in $$\mathcal O_K$$ and its norm is divisible by a prime $$p$$, then some prime ideal in $$\mathcal O_K$$ dividing $$p$$ is a factor of $$I$$.

Proof. The ring $$\mathcal O_K/I$$ has size equal to the norm of $$I$$ (do you know that important combinatorial interpretation of the ideal norm?), so we are told $$\mathcal O_K/I$$ has size divisible by $$p$$. Viewing $$\mathcal O_K/I$$ merely as an additive group, since its order is divisible by $$p$$, Cauchy's theorem from group theory tells us that there is an $$\alpha \bmod I$$ with additive order $$p$$, so $$\alpha \not\equiv 0 \bmod I, \ \ \ p\alpha \equiv 0 \bmod I.$$ That is equivalent to $$I \nmid (\alpha)$$ and $$I \mid (p\alpha)$$. Since $$(p\alpha) = (p)(\alpha)$$, $$I \nmid (\alpha), \ \ \ I \mid (p)(\alpha).$$ If $$I$$ is not divisible by any prime ideal dividing $$(p)$$, then $$I$$ and $$(p)$$ are relatively prime ideals, so from $$I \mid (p)(\alpha)$$ we get $$I \mid (\alpha)$$, but that contradicts the property $$I \nmid (\alpha)$$. Thus $$I$$ must be divisible by some prime ideal factor of $$(p)$$. QED

Thus for your ideal, with norm $$2 \cdot 3 \cdot 7 \cdot 23$$, it must factor as a product of prime ideals with norms $$2, 3, 7$$, and $$23$$. A prime ideal has prime-power norm, so when a prime factor of the norm has multiplicity $$1$$, there must be a single prime ideal factor with that prime norm. Things would be more subtle if an ideal has norm divisible by a higher power of a prime, say $$3^2$$. Then the ideal might be divisible by two prime ideals of norm $$3$$ or a single prime of norm $$9$$.

Example. In $$\mathbf Z[\sqrt{-5}]$$, let $$I = (7+\sqrt{-5})$$. Then $$I$$ has norm $$49 + 5 = 54 = 2 \cdot 3^3$$, so the prime ideal factors of $$I$$ are among the prime ideal factors of $$2$$ and $$3$$.

In $$\mathbf Z[\sqrt{-5}]$$, $$(2) = \mathfrak p^2$$ where $$\mathfrak = (2,1+\sqrt{-5})$$ and $$(3) = \mathfrak q\mathfrak q'$$, where $$\mathfrak q = (3,1+\sqrt{-5})$$ and $$\mathfrak q' = (3,1-\sqrt{-5})$$. So $$I$$ must be divisible by $$\mathfrak p$$ (the only ideal of norm $$2$$). It can't be divisible by both $$\mathfrak q$$ and $$\mathfrak q'$$, by contradction: if $$(7+\sqrt{-5})$$ were divisible by $$\mathfrak q$$ and $$\mathfrak q'$$ then it would be divisible by their product $$(3)$$, and having $$(3) \mid (7+\sqrt{-5})$$ as principal ideals in $$\mathbf Z[\sqrt{-5}]$$ forces $$3 \mid (7+\sqrt{-5})$$ as elements in $$\mathbf Z[\sqrt{-5}]$$, but $$7 + \sqrt{-5} \not= 3(m+n\sqrt{-5})$$ for integers $$m$$ and $$n$$ since $$7$$ is not a multiple of $$3$$ (and $$1$$ isn't either). Thus the only way to explain $$I$$ having norm divisible by $$3^3$$ is to have $$I$$ be divisible by $$\mathfrak q^3$$ or $$\mathfrak q'^3$$, so $$I = \mathfrak p\mathfrak q^3 \ \ {\sf or } \ \ I = \mathfrak p\mathfrak q'^3.$$ Which one is it? We just need to figure out if $$\mathfrak q \mid I$$ or $$\mathfrak q' \mid I$$, or equivalently, $$7+\sqrt{-5} \stackrel{?}{\equiv} 0 \bmod \mathfrak q \ \ {\sf or } \ \ 7+\sqrt{-5} \stackrel{?}{\equiv} 0 \bmod \mathfrak q'.$$ Since $$\mathfrak q = (3,1+\sqrt{-5})$$, it's quite easy to write $$7+\sqrt{-5}$$ in terms of the generators of $$\mathfrak q$$: $$7 + \sqrt{-5} = 3 \cdot 2 + 1 + \sqrt{-5} \in \mathfrak q$$, so $$\mathfrak q \mid (7+\sqrt{-5})$$. Thus $$(7+\sqrt{-5}) = \mathfrak p\mathfrak q^3 = (2,1+\sqrt{-5})(3,1+\sqrt{-5})^3.$$

• So as you have shown, if we were given a norm which is not square free let say $18=2\cdot 3^2$, then corresponding ideals should come from decomposition of $(2)$ and $(3)$, in $\mathcal O_K$ right? See the following question which is a follow-up to your last sentence : "Things would be more subtle if an ideal has norm divisible by a higher power of a prime, say 32. Then the ideal might be divisible by two prime ideals of norm 3 or a single prime of norm 9."math.stackexchange.com/questions/4604522/… Dec 23, 2022 at 19:33
• @MichealBrainHurts you really should make that other question self-contained. Your second line, "This requires the follow-up question because the norm is now not squarefree," will make no sense to anyone who reads that page without seeing your earlier question above: they will have no idea what "This requires..." is supposed to mean. Rewrite it in a self-contained way and then I will post a reply to it.
– KCd
Dec 23, 2022 at 20:21