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

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.

• math.stackexchange.com/questions/3314752/… – Viktor Vaughn Aug 6 '19 at 23:24
• Oh, I don't know anything about ring norms, but I will read that one. At least I should be able to understand the isomorphism. – P.Luis Aug 6 '19 at 23:46
• How is it that two users came to ask the same question within a few hours of each other? Is there something you're not telling us, P.L? – Gerry Myerson Aug 7 '19 at 0:24
• Well, thats all. I haven't read the rules in a long time and I think that making homework questions is forbidden. But after toying with a diophantic equation solver trying to come out with a solution I got really curious of this. Moreover, most people I have talked about this problem with think it is false. Sorry if this was against the rules. I should read them to refresh it, this is my honest answer. – P.Luis Aug 7 '19 at 0:57

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.

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.

• It's a easy result to deduce? Where can I read about that? It sounds very useful but I didn't know that. I should have taken the linear algebra approach like Galois groups does and use matrix representations of my rings. I still don't see why the determinant is that one. Sorry, this is my first course with ring theory and it barely started a 6 weeks ago. All I know about rings before I had learned it reading Dummit and Foote on my free time. It's a little embarrasing to me, but I hadn't have ring theory on my undergrad. – P.Luis Aug 8 '19 at 0:13