Hi,

I have to proof, that $\displaystyle \mathbb Z[i]$ is a noetherian ring and a faithfully flat $\displaystyle \mathbb Z$-algebra.

Furthermore, I have to proof, that for every ideal $\displaystyle \mathfrak a \neq 0$ in $\displaystyle \mathbb Z[i]$ the ring $\displaystyle \mathbb Z[i]/ \mathfrak a$ is a Artinian ring und every prime ideal in $\displaystyle \mathbb Z[i]$ is a maximal ideal in $\displaystyle \mathbb Z[i]$.

Would you please check my proof? :

$\displaystyle \mathbb Z[i]$ is a Euclidean domain, hence it is principal ideal domain and therefore noetherian, because every principal ideal domain is a noetherian ring.

Or another idea: $\displaystyle \mathbb Z[i]$ is isomorphic to $\displaystyle \mathbb Z \times \mathbb Z$ as $\displaystyle \mathbb Z$-module. The $\displaystyle \mathbb Z$-module $\displaystyle \mathbb Z$ is noetherian (e.g. $\displaystyle \mathbb Z$ is noetherian ring) and therefore $\displaystyle \mathbb Z \times \mathbb Z$ is also a noetherian $\displaystyle \mathbb Z$-module.

faithfully flat $\displaystyle \mathbb Z$-algebra:

Let $\displaystyle M' \to M \to M''$ a exact sequence of $\displaystyle \mathbb Z$-moduls. Then

$\displaystyle M' \otimes (\mathbb Z \times \mathbb Z)\to M \otimes (\mathbb Z \times \mathbb Z) \to M'' \otimes (\mathbb Z \times \mathbb Z) $

$\displaystyle \cong (M' \otimes \mathbb Z) \times (M' \otimes \mathbb Z) \to (M \otimes \mathbb Z) \times (M \otimes \mathbb Z) \to (M'' \otimes \mathbb Z) \times (M'' \otimes \mathbb Z)$

$\displaystyle \cong M' \times M' \to M \times M \to M'' \times M'' $ is exact and we have

$\displaystyle M' \to M \to M''$ is exact $\displaystyle \Longleftrightarrow$ $\displaystyle M' \times M' \to M \times M \to M'' \times M'' $ is exact, this means that $\displaystyle \mathbb Z[i]$ is a faithfully flat $\displaystyle \mathbb Z$-algebra.

Now the other part:

There is a one-to-onecorrespondencebetween the ideals in $\displaystyle \mathbb Z[i]/ \mathfrak a$ and the ideals $\displaystyle \mathfrak b$ in $\displaystyle \mathbb Z[i]$ with $\displaystyle \mathfrak b \supset \mathfrak a $.

$\displaystyle \mathbb Z[i]$ is a principal ideal domain, this means, that for every ideal $\displaystyle \mathfrak b$ in $\displaystyle \mathbb Z[i]$ there exists an element $\displaystyle b \in \mathbb Z[i]$ with $\displaystyle \mathfrak b=(b)$.

Let $\displaystyle \mathfrak a=(a)$.

$\displaystyle (b) \supset (a) \Leftrightarrow b|a $.

There is only a finite number of divisors of a. It follows, that there is only a finite number of ideals $\displaystyle \mathfrak b$ in $\displaystyle \mathbb Z[i]$ with $\displaystyle \mathfrak b \supset \mathfrak a $, thus there is only a finite number of ideals in the ring $\displaystyle \mathbb Z[i]/ \mathfrak a$ and so $\displaystyle \mathbb Z[i]/ \mathfrak a$ is a artinian ring.

It follows: Every prime ideal in $\displaystyle \mathbb Z[i]/ \mathfrak a$ is a maximal ideal in $\displaystyle \mathbb Z[i]/ \mathfrak a$, hence every prime ideal in $\displaystyle \mathbb Z[i] $ is a maximal ideal in $\displaystyle \mathbb Z[i]$.

Are there any mistakes in my proof? I would be glad if someone would check the proof.

Thanks in advance!

Bye,

Lisa