# Are complex conjugates, of prime ideals, prime ideals?

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.

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

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

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

• Can you prove that $CD \subset \overline{P}$ implies $\overline{C} \, \overline{D} \subset P$ ? What could you deduce from that ? – Joel Cohen Jan 10 at 12:04
• @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. – 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.
• 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 $P$. – Eric Wofsey Jan 20 at 16:19