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?

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

1 Answer 1


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 $$ (7+\sqrt{-5}) = \mathfrak p\mathfrak q^3 = (2,1+\sqrt{-5})(3,1+\sqrt{-5})^3. $$

  • $\begingroup$ 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/… $\endgroup$ Dec 23, 2022 at 19:33
  • $\begingroup$ @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. $\endgroup$
    – KCd
    Dec 23, 2022 at 20:21

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