Ring of integers of quadratic field extension, every element with prime norm is a prime

Question

Give a prime number $q$, let be $\omega = \sqrt{q}$ is $q \equiv 2,3 ( \: \mod \: 4)$ and $\omega = \dfrac{1+\sqrt{q}}{2}$ if $q \equiv 1 ( \: \mod \: 4)$. Let $R= \mathbb{Z}[\omega]$. (I think that's the ring of integers of the quadratic field extension $\mathbb{Q}[\sqrt{p}]$).

If $\alpha \in R$ is such that $N(\alpha)=p$ with $p$ a prime integer, where $N$ is the complex norm restricted to $R$, then $\alpha$ is a prime.

I know how to prove that's irreducible, which is easy. If $\alpha = \beta \gamma$ then $N(\alpha) = N( \beta) N(\gamma)$ and as $N(\alpha)=p$ is a prime, then $N(\gamma)=1$ or $N(\beta)=1$. So $\gamma$ or $\beta$ is a unit.

But now I don't know how to proced. I know that $\mathbb{Z} \cap M = p \mathbb{Z}$ if $\alpha \in I$ and $I$ is an ideal, because $p= \alpha \overline{\alpha} \in I$ because $I$ is an ideal, and $p \mathbb{Z}$. I guess that the only ideals $I$ of $R$ such that $I \cap \mathbb{Z} = p \mathbb{Z}$ are $(\alpha)$ and $(-\alpha)$.

But I don't know how to prove it. Well, I am not even 100% sure this is true.

Answer 1

If $N(\alpha)=p$ then $(p) \subset (\alpha)$ as $\alpha \overline{\alpha} = p \subset ( p )$. Also, every element on $(p)$ can be written as a product $\beta p$, and $ N(\beta p)=N(\beta)N(p) = N( \beta ) p^{2}$, so $\alpha \notin (p)$ and the inclusion $(p) \subset (\alpha)$ is proper.

I will use the third isomorphism theorem for groups.

\begin{equation*} \dfrac{\mathbb{Z}[w]}{(\alpha)} \cong \dfrac{\dfrac{\mathbb{Z}[w]}{(p)}}{ \dfrac{(\alpha)}{(p)} } \end{equation*} As abelian groups: $$ \dfrac{\mathbb{Z}[w]}{(p)} = \dfrac{ \mathbb{Z} \oplus \mathbb{Z} \omega } {p \mathbb{Z} \oplus p \mathbb{Z} \omega } \cong \dfrac{\mathbb{Z}}{ p \mathbb{Z} } \oplus \dfrac{\mathbb{Z} \omega}{ p \mathbb{Z} \omega } \cong \dfrac{ \mathbb{Z} }{ p \mathbb{Z} } \oplus \dfrac{ \mathbb{Z} }{ p \mathbb{Z} } $$ So $\dfrac{\mathbb{Z}[w]}{(p)}$ is an abelian group of order $p^{2}$. Also $(\alpha) \subset (p)$ is a proper contention, that is $(\alpha) \neq (p)$. So $\dfrac{\alpha}{(p)}$ has order larger than $1$. As

\begin{equation*} 1 < \left| \dfrac{\mathbb{Z}[w]}{(\alpha)} \right| \cong \dfrac{\left| \dfrac{\mathbb{Z}[w]}{(p)} \right| }{ \left|\dfrac{(\alpha)}{(p)} \right|} < \left| \dfrac{\mathbb{Z}[w]}{(p)} \right| = p^{2} \end{equation*} (The $1<$ is because $\dfrac{\mathbb{Z}[w]}{(\alpha)}$ isn't the trivial group)

As $\left| \dfrac{\mathbb{Z}[w]}{(\alpha)} \right|$ divides $\left| \dfrac{\mathbb{Z}[w]}{(p)} \right|=p^{2}$, under these restriction it can only be $\left| \dfrac{\mathbb{Z}[w]}{(\alpha)} \right| = p$ as the other divisors of $p^{2}$ are $1$ and $p^{2}$ and we already ruled out those possilibities. Thus the abelian group structure of $\dfrac{\mathbb{Z}[w]}{(\alpha)}$ is $\mathbb{Z} / p \mathbb{Z}$. The only ring with unity of $p$ elements is $\mathbb{F}_{p}$, and as out quotient has unity, it must be isomorphic to $\mathbb{F}_{p}$. Thus our ring structure is $\dfrac{\mathbb{Z}[w]}{(\alpha)} \cong \mathbb{F}_{p}$, and as it's a field, $(\alpha)$ is maximal, thus a prime ideal.

Therefore, $p$ is prime.

PD: My professor meant irreducibles when he wrote the exercise. But this one was a very nice exercise and it motivated me to study a lot, so I liked it.

Answer 2

Let $f : Z[w] \to Z^2$ be an isomorphism of abelian group. The multiplication by $\alpha$ in $Z[w]$ becomes the multiplication by a matrix $A\in M_2(Z)$ in $Z^2$.

$Q[\sqrt{q}]=Q+\sqrt{q}Q$ is a $Q$-vector space isomorphic to $Q^2$ and the multiplication by $\alpha=a+ b \sqrt{q}$ in $Q+\sqrt{q}Q$ becomes the multiplication by the matrix $B = \pmatrix{a & b q \\ b & a}$ in $Q^2$.

The field norm of $\alpha$ is defined as $N_{K/Q}(\alpha)=\det(B) = a^2-b^2 q$. We also have $\det(B) = \det(A)$ since $B = M A M^{-1}$.

Finally $$\# Z[w] / \alpha Z[w] = \# f^{-1}(Z[w])/f^{-1}( \alpha Z[w]) = \# Z^2/ A Z^2 = |\det(A)| = |N_{K/Q}(\alpha)|$$

Thus if $|N_{K/Q}(\alpha)|=p$ is prime then $Z[w]/(\alpha) \cong Z/ p Z$ is the field with $p$ elements and $(\alpha)$ is a maximal ideal.

The exact same argument holds in larger number fields.