# Prove that the ring $\mathcal O_K$ of algebraic integers of $K = \Bbb Q (\sqrt d)$ is a Dedekind domain.

Prove that the ring $$\mathcal O_K$$ of algebraic integers of $$K = \Bbb Q (\sqrt d)$$ ($$d$$ is a square free integer) is a Dedekind domain.

I have taken an ideal $$I \subseteq \mathcal O_K$$. Consider the set $$I' = \{a - b \sqrt {d} : a + b \sqrt d \in I \}$$. Then it is easy to show that $$I'$$ is an ideal of $$\mathcal O_K$$. Our instructor has left as an exercise to prove that $$II' = (n)$$ for some $$n \in \Bbb Z$$ i.e. $$II'$$ is principal. So for any ideal $$I$$ there exists $$(0) \neq I' \subseteq \mathcal O_K$$ such that $$II'$$ is a principal ideal. This will prove that $$\mathcal O_K$$ is a Dedekind domain. But how can I show that $$II'$$ is principal?

• The easiest case is $(I,I') = (1)$ (for example $I$ prime ideal $\ne I'$) which means $O_K/II' \cong O_K/I \times O_K/I'$ so $n=\# O_K/I=N(I) = N(I')$ is zero in the product of rings thus $n \in II'$ and $N((n)) = n^2 = N(II') \implies II' = (n)$. – reuns Feb 3 at 9:32
• It depends from what results. If it is from scratch then let $S = \{p, (p)$ is not a prime ideal, there is only one prime ideal above $p\}$ and $R = O_K[S^{-1}]$ then for every $p \not \in S$, or $(p)$ is a prime ideal in $R$, or $\mathfrak{p},\mathfrak{p}'$ are two different prime-maximal ideals above $p$, so from my previous comment $\mathfrak{p}\mathfrak{p}' = N(\mathfrak{p}) = (p)$, therefore any prime ideal is inversible, any ideal is a product of prime ideals and hence $I I' = (n)$ for some $n$. The remaining ideals of $O_K$ (ramified primes $\in S$) are harder to treat. – reuns Feb 16 at 18:48

Here's one way to see it. All you gotta show it's a Noetherian integrally closed domain of dimension $$1$$.

First of all it is an integrally closed domain by definition (integral closure of $$\mathbb Z$$ in $$\mathbb Q(\sqrt d)$$ ).

Next we observe $$\mathcal O_K |_{\mathbb Z}$$ is an integral extension and hence we have $$\dim \mathcal O_K=\dim \mathbb Z =1$$.

Thus all we have to show is it is a Noetherian ring. This is slightly tricky. Say $$\alpha +\beta \sqrt d \in \mathcal O_K$$ with $$\alpha, \beta \in \mathbb Q$$. Then the minimal polynomial is $$X^2-2\alpha X +\alpha^2-d\beta^2$$. Thus we have $$2\alpha , \ \alpha^2-d\beta^2 \in \mathbb Z$$

The upshot is $$\alpha \in \frac{1}{2} \mathbb Z , \ \beta \in \frac{1}{2}\mathbb Z$$. So we have $$\mathcal O_k\subset \mathbb Z$$-$$\mathrm{span}(\frac{1}{2}, \frac{\sqrt d}{2})$$ and hence $$\mathcal O_k$$ is a finitely generated $$\mathbb Z$$-module and hence a Noetherian ring.

Thus $$\mathcal O_K$$ is a Dedekind Domain.

• In this context Dedekind domain = ideals have unique factorization in prime ideals – reuns Apr 9 at 21:12
• they are equivalent. – Ignorant Mathematician Apr 9 at 21:16
• once you have proven a lot of theorems... in this context the point is to prove the minimal subset of those in order to have the unique factorization in prime ideals. In my comments I have shown when inverting the ramified primes we can show we have a Dedekind domain in just a few lines. The problem is those few ramified primes for which we need more work. – reuns Apr 9 at 21:19