# If $(p)$ is a proper subset of a proper ideal $I$, then is $I$ prime?

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 $$P$$.

• 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. – Wojowu Jan 20 at 9:29
• @Wojowu I changed to $(p)=P \overline P \subset I \subset R$ – Ekhin Taylor R. Wilson Jan 20 at 10:02
• I see, now it makes more sense. – Wojowu Jan 20 at 10:04

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.
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 $$I=\overline{P}$$.