# If an element has prime norm in the ring of quadratic integers, then it is a prime element

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!

• 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 $(I,b) \supset I \implies O/(I,b) = (O/I)/(b) \implies N((I,b))\ |\ N(I)$. – reuns Aug 6 '19 at 3:18
• Thank you @reuns for your answer. I'm not familiar with ideal norms and field norms. Is there another approach for this problem? – Albert Aug 6 '19 at 4:05
• 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 $\Bbb{Q}+\sqrt{d} \Bbb{Q}$ – reuns Aug 6 '19 at 4:19

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 $$\mathcal{O}$$.