Is $\mathbb{Z}[\sqrt{-5}]$ Dedekind?

Question

It is known that $\mathbb{Z}[\sqrt{-5}]$ is a Dedekind ring. I want to prove it but I don't know. Please tell me if you know.

Answer 1

Hint:

Since $-5\not\equiv 1\mod 4$ the integral closure of $\mathbf Z$ in the quadratic field $\mathbf Q\bigl(\sqrt{-5}\bigr)$ is precisely $\mathbf Z\bigl[\sqrt{-5}\bigr]$.

One of the characterisations of Dedekind rings is that they are noetherian integrally closed rings of Krull dimension $1$ (i.e. every non-zero prime ideal is maximal).

Answer 2

$$O=\Bbb{Z}[\sqrt{-5}]=\Bbb{Z}[x]/(x^2+5)$$

• If $p\ne 2,5$ then factorize $x^2+5 \bmod p$, either you'll get that it is irreducible in which case $(p)$ is a maximal ideal of $O$,

  or $x^2+5=(x-a)(x-b)\bmod p$ so that $(p,x-a),(p,x-b)$ are maximal ideals of $O$. In the latter case using that $(p,x-a,x-b)=(1)$ we get that $$(p,x-a)(p,x-b)= (p^2,p(x-a),p(x-b),(x-a)(x-b))$$ $$(p(p,x-a,x-b),(x-a)(x-b))= (p)$$ Thus those two maximal ideals are invertible

(a non-zero ideal $I$ is said invertible if there is another one such that $IJ$ is principal)

• The only maximal ideal above $2$ is $(2,x+1)$ and $(2,x+1)^2=(2)$, and the only maximal ideal above $5$ is $(5,x)$ which satisfies $(5,x)^2=(5)$

Thus every maximal ideal is invertible, which is quite the best characterization of Dedekind domains because (together with that $O_K/I$ is always a finite ring) we easily deduce that every non-zero ideal is invertible as well as the unique factorization in prime ideals from it.

Answer 3

A domain is Dedekind if and only if it is integrally closed, Noetherian, and has Krull dimension $1$. It was already remarked that $R=\Bbb Z[\sqrt{-5}]$ is the ring of integers in $\Bbb Q(\sqrt{-5})$, so it is integrally closed. Since $\Bbb Z[X]$ is Noetherian, so is $R\cong \Bbb Z[X]/(X^2+5)$. Finally its Krull dimension is $1$, compare with this post:

Dimension of $\mathbb Z[\sqrt 5]$