Let $R$ be the ring of algebraic integers of a quadratic imaginary number field $\mathbb Q[\sqrt{d}]$ for a negative square-free integer $d$.

If $P$ is a prime ideal of $R$, then is $\overline P$ the ideal consisting of the complex conjugates of elements of $P$, also a prime ideal?

  • Our definition of a prime ideal $P$ is that $P$ is nonzero and if the product $CD$ of two ideals $C$ and $D$ is a subset of $P$, then $C$ or $D$ is a subset of $P$.

  • I know that in $R$, an ideal $C$ is a subset of an ideal $D$ if and only if $D$ divides $C$, that is $C=DE$ for another ideal $E$.

  • I know this has to do with $M \overline M = (m)$: the product of a nonzero ideal $M$ of $R$ and its conjugate $\overline M$ is a principal ideal generated by some positive integer $m$.

  • I started by assuming $CD \subseteq \overline P$ and then I did a lot of conjugate multiplications and trying to see whether or not $\frac 1 e G$ is an ideal for some positive integer $e$ and ideal $G$ in order to try to get something that looks like $EG \subseteq P$ but to no avail.

Thanks in advance!

Here's my answer:

$CD \subseteq \overline P \implies CD =\overline P G \implies \overline{CD} = P \overline G \implies \overline{CD} = \overline{C} \ \overline{D} \subseteq P \implies \overline{C} \subseteq P \ $ or $\ \overline{D} \subseteq P$

Suppose $\overline{C} \subseteq P$. Then

$\overline{C} = P H \implies C = \overline{P H} = \overline{P} \ \overline{H} \implies C \subseteq \overline P$

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    $\begingroup$ Can you prove that $CD \subset \overline{P}$ implies $\overline{C} \, \overline{D} \subset P$ ? What could you deduce from that ? $\endgroup$ – Joel Cohen Jan 10 at 12:04
  • $\begingroup$ @JoelCohen If that's true because $A \subset \overline B$ implies $\overline A \subset B$, then the answer to my question is obviously yes. Does $A \subset \overline B$ imply $\overline A \subset B$? If that's true, then I shall try to prove it myself and ask a new question if I cannot. $\endgroup$ – Ekhin Taylor R. Wilson Jan 10 at 12:15

Yes: complex conjugation is a ring isomorphism $f:R\to R$, so it preserves any property defined only using the ring structure. Explicitly, if $P$ is a prime ideal, then $CD=f(P)$ is equivalent to $f^{-1}(C)f^{-1}(D)=P$ (just apply $f^{-1}$ to all the elements). Since $P$ is prime, this means $f^{-1}(C)\subseteq P$ or $f^{-1}(D)\subseteq P$, so then applying $f$ to everything we see $C\subseteq f(P)$ or $D\subseteq f(P)$. That is, $f(P)$ is prime.

  • $\begingroup$ Is my answer in my latest review also correct? $\endgroup$ – Ekhin Taylor R. Wilson Jan 20 at 9:04
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    $\begingroup$ I don't know why you claim that $CD \subseteq \overline P \implies \overline{CD} \subseteq P$ is wrong. Every element of $\overline{CD}$ is the conjugate of some element of $CD$, and therefore the conjugate of some element of $\overline{P}$, and therefore the conjugate of the conjugate of some element of $P$, and therefore an element of $P$. $\endgroup$ – Eric Wofsey Jan 20 at 16:19

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