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Let $R$ be the ring of algebraic integers of a quadratic imaginary number field $\mathbb Q[\sqrt{d}]$ for a negative square-free integer $d$. For a prime integer $p$, $(p)$ is a prime ideal or is the product $P \overline P$ of some prime ideal $P$ and $\overline P$, the ideal consisting of the complex conjugates of elements of $P$. Why does this mean if $(p)$ is a proper subset of a proper ideal $I$ of $R$, then $I$ is prime?

  • If $(p)$ is a prime ideal, then $(p)$ is a maximal ideal so $(p)=I$.

  • I don't know how to say $(p)=P \overline P \subset I \subset R$ implies $I$ is 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 $P$.

Thanks in advance!

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    $\begingroup$ I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $\mathbb Z[i]$. Then $(1+i)^2\in I$ but $1+i\not\in I$, so $I$ is not prime. $\endgroup$ – Wojowu Jan 20 at 9:29
  • $\begingroup$ @Wojowu I changed to $(p)=P \overline P \subset I \subset R$ $\endgroup$ – Ekhin Taylor R. Wilson Jan 20 at 10:02
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    $\begingroup$ I see, now it makes more sense. $\endgroup$ – Wojowu Jan 20 at 10:04
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Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.

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I will focus on the case you still don't solve.

Suppose that $I$ is a proper ideal such that $P\overline P\subsetneq I$ (notice that the inclusion should be strict, otherwise $P\overline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $P\overline P\subsetneq M$ implies either $P\subsetneq M$ or $\overline{P}\subsetneq M$. WLOG we have $P\subsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).

A more elucidative proof would be using the properties of the ideal factorization in number fields. If $P\overline{P}\subsetneq I$ then we have that $I$ properly divides $P\overline{P}$ and hence either $I=P$ or $I=\overline{P}$.

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