Let $\mathcal{O}$ be a quadratic integer ring, that is $\mathcal{O}=\mathbb{Z}[\lambda_d]$ where $$ \lambda_d = \begin{cases} \sqrt{d} & \text{ if } d\equiv 2,3 \; (\text{mod }4),\\ \frac{1+\sqrt{d}}{2} & \text{ if } d\equiv 1 \; (\text{mod }4), \end{cases} $$ where $d\neq 0$ is a square-free integer.

Let $\alpha=a+b\lambda_d\in \mathcal{O}$, with $a,b\in\mathbb{Z}$. It is well known that if $N(\alpha)=\alpha\overline{\alpha}$ is a prime in $\mathbb{Z}$, then $\alpha$ is irreducible in $\mathcal{O}$. My question is: is it true that if $N(\alpha)$ is a prime in $\mathbb{Z}$ then $\alpha$ is a prime in $\mathcal{O}$?

I believe that the answer is no. A counterexample must be given in a quadratic integer ring that is not a factorial ring (unique factorization domain), but I could'nt find such a counterexample.

Any help will be appreciated. Thank you in advance!

  • 2
    $\begingroup$ The field norm is also the ideal norm $|N_{K/Q}(\alpha)| = N((\alpha)) = \# O/(\alpha)$. If $N(I)=p$ is prime then $O/I$ is of characteristic $p$ and $O/I\cong \Bbb{Z/pZ}$. Equivalently $I$ is a maximal ideal since $(I,b) \supset I \implies O/(I,b) = (O/I)/(b) \implies N((I,b))\ |\ N(I)$. $\endgroup$ – reuns Aug 6 '19 at 3:18
  • $\begingroup$ Thank you @reuns for your answer. I'm not familiar with ideal norms and field norms. Is there another approach for this problem? $\endgroup$ – Albert Aug 6 '19 at 4:05
  • 1
    $\begingroup$ Not really. Take a group isomorphism $f : O \to \Bbb{Z}^2$, multiplication by $\alpha$ becomes a matrix $A \in M_2(\Bbb{Z})$, then $\# O/(\alpha) =\# \Bbb{Z}^2/A \Bbb{Z}^2= |\det(A)|$. The field norm is $N_{K/Q}(\alpha)= \det(A)$, its value doesn't depend on $f$, on the chosen basis of $O$ or $K$, you'll find it by looking at the matrix of the multiplication by $\alpha$ on $\Bbb{Q}+\sqrt{d} \Bbb{Q}$ $\endgroup$ – reuns Aug 6 '19 at 4:19

I'll give an "elementary" answer to this question based on the comments in the question:

Let $\alpha\in \mathcal{O}$ with $N(\alpha)=p$ a prime in $\mathbb{Z}$. Write $\mathcal{O}=\mathbb{Z}\oplus \lambda_d\mathbb{Z}$ and consider the abelian group homomorphism $\varphi:\mathbb{Z}\oplus \lambda_d\mathbb{Z}\to \mathbb{Z}\oplus \lambda_d\mathbb{Z}$ given by $\varphi(x)=\alpha x$. In the base $\{1,\lambda_d\}$ we can represent $\phi$ by a matrix $[\varphi]$, and a simple computation (considering $d\equiv 1$ (mod $4$) and $d\equiv 2,3$ (mod $4$) separately) gives $$ \det([\varphi]) = N(\alpha). $$ By using the Smith normal form, there are matrices $P,Q\in GL(n,\mathbb{Z})$ such that $\det(P)=\det(Q)=1$ and $$ [\varphi] = P\begin{pmatrix} d_1 & 0 \\ 0 & d_2 \end{pmatrix}Q, $$ and then $$ \varphi(\mathbb{Z}\oplus \lambda_d\mathbb{Z}) \cong (\mathbb{Z}/d_1\mathbb{Z})\oplus (\mathbb{Z}/d_2\mathbb{Z}), $$ so we have that $|\varphi(\mathbb{Z}\oplus \lambda_d\mathbb{Z})|=d_1d_2=\det([\varphi])=N(\alpha)=p$. But $\varphi(\mathbb{Z}\oplus \lambda_d\mathbb{Z})=(\alpha)$ is the principal ideal generated by $\alpha$ in $\mathcal{O}$, so, by the first isomorphism theorem, $|\mathcal{O}/(\alpha)|=p$. Then we can conlude (see for example this answer) that $\mathcal{O}/(\alpha)$ is isomorphic to the field $\mathbb{F}_p$, so $(\alpha)$ is a prime ideal and $\alpha$ is a prime elment in $\mathcal{O}$.


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