# Computing prime factorization of ideals?

I want to compute the prime factorizations of the ideals $\langle 4\sqrt{-14}\rangle$, $\langle 6\sqrt{-6} \rangle$ and $\langle 4\sqrt{-5} \rangle$ in the ring of algebraic integers of $\mathbb{Q}(\sqrt{-14})$,$\mathbb{Q}(\sqrt{-6})$ and $\mathbb{Q}(\sqrt{-5})$. How can I do that?

Also, is there any way to prove every prime ideal in prime factorization of the ideals above is stable under complex conjugation?

• It is not true that prime ideals are stable under conjugation. – Crostul Apr 25 '16 at 17:48
• The prime ideals in your factorizations are all (completely) ramified and therefore invariant under the Galois group (i.e. ambiguous). – franz lemmermeyer Apr 25 '16 at 19:14
• Hello @franzlemmermeyer, thank you very much for the comment. Is there any way to prove that prime ideals in my factorization are stable under conjugation without actually computing the prime factorization? – grok Apr 25 '16 at 19:51
• I had this problem myself a while ago and I still don't know the answer. It seems a little circular that (I think) we can only compute decomposition and inertia groups of an ideal $\mathfrak{P}$ above $\mathfrak{p}$ when we already know what its prime decomposition is. So, or I'm wrong, or I guessing that it is not a computation tool and it is more like a theorical thing, saying that the Galois group is the invisible hand behind the things that happen in the arithmetic of the ring. Not sure the person asking is asking in this context, though. – Shoutre Apr 25 '16 at 20:08

As an example, take $I = \langle 4\sqrt{-14}\rangle\subset\mathbb Q(\sqrt{-14})$. It's clear that $I = \langle 2\rangle^2\cdot\langle \sqrt{-14}\rangle$. To factorise $I$, it is sufficient to factorise $\langle2\rangle$ and $\langle \sqrt{-14}\rangle$ separately.
To factorise $\langle 2\rangle$ we can use the Kummer-Dedekind theorem: $\mathcal O_{\mathbb Q(\sqrt{-14})}=\mathbb Z[\sqrt{-14}]$, and $$X^2+14\equiv X^2\pmod 2$$ so $\langle 2\rangle = \langle2,\sqrt{-14}\rangle^2$.
To factorise $\langle \sqrt{-14}\rangle$, observe that $\langle \sqrt{-14}\rangle\cap\mathbb Z = 14\mathbb Z$, so $\langle \sqrt{-14}\rangle$ is not prime, and it has norm $14$, so it should factorise as a product of a prime above $2$ and a prime above $7$. Using the Kummer-Dedekind theorem, we see that $7\mathcal O_{\mathbb Q(\sqrt{-14})} = \langle 7,\sqrt{-14}\rangle^2$, and it is simple to check that $\langle\sqrt{-14}\rangle=\langle 2,\sqrt{-14}\rangle\cdot\langle 7,\sqrt{-14}\rangle$.
To illustrate this, consider $J=\langle 4\sqrt{-5}\rangle\subset \mathbb Q(\sqrt{-5})$. Suppose that $\mathfrak p$ is a prime ideal dividing $J$, which is not stable under complex conjugation. Since $J$ is stable under complex conjugation, it follows that $\overline{\mathfrak p}$ also divides $J$, and therefore $\mathfrak p\overline{\mathfrak p} = p\mathcal O_{\mathbb Q(\sqrt{-5})}$ divides $J$, where $p = \mathfrak p\cap\mathbb Z$. In this case $p$ must therefore be $2$. It is simple to check that $2$ is totally ramified in $\mathbb Q(\sqrt{-5})$, which rules out this case.
More generally, if $I$ is an ideal in an imaginary quadratic field that is stable under conjugation, and for every rational prime $p$ such that $p\mathcal O\mid I$, there is a unique prime ideal of $\mathcal O$ lying above $p$ (i.e. $p$ is totally ramified or inert), then the prime ideals dividing $I$ are stable under complex conjugation.