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?

  • 1
    $\begingroup$ 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 $N((n)) = n^2 = N(II') \implies II' = (n)$. $\endgroup$ – reuns Feb 3 at 9:32
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ In this context Dedekind domain = ideals have unique factorization in prime ideals $\endgroup$ – reuns Apr 9 at 21:12
  • $\begingroup$ they are equivalent. $\endgroup$ – Ignorant Mathematician Apr 9 at 21:16
  • $\begingroup$ 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. $\endgroup$ – reuns Apr 9 at 21:19

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