# Proving that $\mathbb{Z}[i]$ is a noetherian ring

Claim: the ring $\mathbb{Z}[i]$ is a noetherian ring

My proof

1) $\mathbb{Z}[i]$ is a finitely generated $\mathbb{Z}$-module.

2) $\mathbb{Z}$ is a noetherian ring.

3) Every finitely generated module over a noetherian ring is a noetherian module, hence $\mathbb{Z}[i]$ is a noetherian $\mathbb{Z}$-module.

4) By definition of noetherian module, every $\mathbb{Z}$-submodule of $\mathbb{Z}[i]$ is finitely generated as a $\mathbb{Z}$-module

5) an ideal $\mathfrak{i}$ of $\mathbb{Z}[i]$ is in particular a $\mathbb{Z}$-submodule of $\mathbb{Z}[i]$

6) $\mathfrak{i}=\mathbb{Z}x_1+\ldots +\mathbb{Z}x_n$

7) since $\mathfrak{i}$ is finitely generated as a $\mathbb{Z}$-module, it is also finitely generated as an ideal

Do you think my proof works?

• Wouldn't it be much easier to prove directly that the Gaussian integers are an euclidean domain and thus even a PID? – DonAntonio Jun 4 '13 at 9:21
• @DonAntonio you are surely right, but here i'm just wondering if there is some mathematical absurdity in my proof – bateman Jun 4 '13 at 9:24
• The first point is interesting (and true, don't worry), but: how do you know $\,\Bbb Z[i]\,$ is a f.g. abelian group = f.g. $\,\Bbb Z$-module) ? – DonAntonio Jun 4 '13 at 9:26
• @DonAntonio I show that $1,i$ is a $\mathbb{Z}$-basis – bateman Jun 4 '13 at 9:29
• I think you can apply Hilbert-basis theorem, with the fact that quotients of the noetherian domains are noetherian. But I think that this is essentially your approach. – awllower Jun 4 '13 at 9:34

By Hilbert's Theorem $\mathbb{Z}[X]$ is noetherian. Hence $\mathbb{Z}[i]$ is also noetherian as a factor-ring of $\mathbb{Z}[X]$.