Noetherian Ring

• Sep 7th 2010, 03:38 AM
lisa
Noetherian Ring
Hi,

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

Would you please check my proof? :

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

faithfully flat $\mathbb Z$-algebra:

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

$M' \otimes (\mathbb Z \times \mathbb Z)\to M \otimes (\mathbb Z \times \mathbb Z) \to M'' \otimes (\mathbb Z \times \mathbb Z)$
$\cong (M' \otimes \mathbb Z) \times (M' \otimes \mathbb Z) \to (M \otimes \mathbb Z) \times (M \otimes \mathbb Z) \to (M'' \otimes \mathbb Z) \times (M'' \otimes \mathbb Z)$
$\cong M' \times M' \to M \times M \to M'' \times M''$ is exact and we have

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

Now the other part:
There is a one-to-one correspondence between the ideals in $\mathbb Z[i]/ \mathfrak a$ and the ideals $\mathfrak b$ in $\mathbb Z[i]$ with $\mathfrak b \supset \mathfrak a$.
$\mathbb Z[i]$ is a principal ideal domain, this means, that for every ideal $\mathfrak b$ in $\mathbb Z[i]$ there exists an element $b \in \mathbb Z[i]$ with $\mathfrak b=(b)$.

Let $\mathfrak a=(a)$.

$(b) \supset (a) \Leftrightarrow b|a$.
There is only a finite number of divisors of a. It follows, that there is only a finite number of ideals $\mathfrak b$ in $\mathbb Z[i]$ with $\mathfrak b \supset \mathfrak a$, thus there is only a finite number of ideals in the ring $\mathbb Z[i]/ \mathfrak a$ and so $\mathbb Z[i]/ \mathfrak a$ is a artinian ring.
It follows: Every prime ideal in $\mathbb Z[i]/ \mathfrak a$ is a maximal ideal in $\mathbb Z[i]/ \mathfrak a$, hence every prime ideal in $\mathbb Z[i]$ is a maximal ideal in $\mathbb Z[i]$.

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

Bye,
Lisa
• Sep 7th 2010, 07:50 AM
NonCommAlg
Quote:

Originally Posted by lisa
Hi,

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

Would you please check my proof? :

Ok!

Quote:

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

Quote:

Or another idea: $\mathbb Z[i]$ is isomorphic to $\mathbb Z \times \mathbb Z$ as $\mathbb Z$-module. The $\mathbb Z$-module $\mathbb Z$ is noetherian (e.g. $\mathbb Z$ is noetherian ring) and therefore $\mathbb Z \times \mathbb Z$ is also a noetherian $\mathbb Z$-module.
you've only proved that your ring is Noetherian $\mathbb{Z}$-module. you haven't proved that it's a Noetherian "ring" yet.

there are at least three other ways to prove that $R=\mathbb{Z}[i]$ is Noetherian:

1) $\mathbb{Z}$ is a Noetherian subring of $R$ and $R$ is finitely generated over $S.$ thus $R$ is a Noetherian ring. why?

(i meant
$S=\mathbb{Z}.$ this is true in general, i.e. if $R$ is a ring and $S$ is a Noetherian subring of $R$ and $R$ is finitely generated $S$-module, then $R$ is also a Noetherian ring.)

2) the direct product of finitely many Noetherian rings is always a Noetherian ring. why?

(never mind this part! $\mathbb{Z}[i]$ and $\mathbb{Z} \times \mathbb{Z}$ are not isomorphic "as rings".)

3) $R \cong \mathbb{Z}[x]/ \langle x^2 + 1 \rangle$ and $\mathbb{Z}[x]$ is Noetherian by the Hilbert bases theorem.

Quote:

faithfully flat $\mathbb Z$-algebra:

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

$M' \otimes (\mathbb Z \times \mathbb Z)\to M \otimes (\mathbb Z \times \mathbb Z) \to M'' \otimes (\mathbb Z \times \mathbb Z)$
$\cong (M' \otimes \mathbb Z) \times (M' \otimes \mathbb Z) \to (M \otimes \mathbb Z) \times (M \otimes \mathbb Z) \to (M'' \otimes \mathbb Z) \times (M'' \otimes \mathbb Z)$
$\cong M' \times M' \to M \times M \to M'' \times M''$ is exact and we have

$M' \to M \to M''$ is exact $\Longleftrightarrow$ $M' \times M' \to M \times M \to M'' \times M''$ is exact, this means that $\mathbb Z[i]$ is a faithfully flat $\mathbb Z$-algebra.
well, $\mathbb{Z}[i]$ is a free $\mathbb{Z}$-module and every free module is faithfully flat.

Quote:

Now the other part:
There is a one-to-one correspondence between the ideals in $\mathbb Z[i]/ \mathfrak a$ and the ideals $\mathfrak b$ in $\mathbb Z[i]$ with $\mathfrak b \supset \mathfrak a$.
$\mathbb Z[i]$ is a principal ideal domain, this means, that for every ideal $\mathfrak b$ in $\mathbb Z[i]$ there exists an element $b \in \mathbb Z[i]$ with $\mathfrak b=(b)$.

Let $\mathfrak a=(a)$.

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

Quote:

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

Quote:

hence every prime ideal in $\mathbb Z[i]$ is a maximal ideal in $\mathbb Z[i]$.
why?
• Sep 7th 2010, 09:40 AM
lisa
Hi,

Quote:

Originally Posted by NonCommAlg
there are at least three other ways to prove that $R=\mathbb{Z}[i]$ is Noetherian:

1) $\mathbb{Z}$ is a Noetherian subring of $R$ and $R$ is finitely generated over $S.$ thus $R$ is a Noetherian ring. why?

What is $S$? I think, it's because every finitely generated algebra over a noetherian ring is noetherian.

Quote:

Originally Posted by NonCommAlg
2) the direct product of finitely many Noetherian rings is always a Noetherian ring. why?

I know that for every exact sequence of modules $0\to M' \to M \to M'' \to 0$ the modules $M'$ and $M''$ are noetherian if and only if $M$ is noetherian. So I would proof it with induction over $n$.
If $0\to M_2 \to M_1\oplus M_2 \to M_1 \to 0$ is exact, and $M_1, \ M_2$ are noetherian, it follow that $M_1 \oplus M_2$ is noetherian.

Is $0\to M_n \to M_1\oplus ... \oplus M_n \to M_1\oplus ... \oplus M_{n-1} \to 0$ an exact sequence and $M_n$ and $M_1\oplus ... \oplus M_{n-1}$ are noetherian, then $M_1\oplus ... \oplus M_n$ is noetherian.

Quote:

Originally Posted by NonCommAlg
Quote:

Originally Posted by lisa

why?

I am not sure whether my idea is good or not:
Every ideal in $\mathbb Z[i]/( a)$ has the form $(b)+(a)$ with $(b)$ is a ideal in $\mathbb Z[i]$. If the ideal $(b) \subset \mathbb Z[i]$ is a prime ideal but not a maximal ideal, then it exists an ideal $(c) \subset \mathbb Z[i]$ with $(b) \subset (c) \subset \mathbb Z[i]$. Then $(b)+(a) \subset \mathbb Z[i]/ (a)$ is prime but not maximal, because $(c)+(a) \subset \mathbb Z[i]/ (a)$ is "bigger". This is a contradiction, because $(b)+(a) \subset \mathbb Z[i]/ (a)$ should be a maximal ideal.

Bye,
Lisa
• Sep 7th 2010, 11:08 AM
NonCommAlg
first take a look at my previous post again. i added a couple of things which are in red.

1) you didn't prove that the direct product of finitely many Noetherian rings is a Noetherian ring. your proof works for modules only.

2) what is $a$ in your last proof? here's how you should do it:

suppose that $P$ is a non-zero prime ideal of $R=\mathbb{Z}[i].$ choose $0 \neq a \in P$ and let $I=\langle a \rangle.$ then $P/I$ is a prime ideal of $R/I$ and thus it is maximal. it's now clear that $P$ is maximal in $R.$
• Sep 7th 2010, 11:56 AM
lisa
Hello.

Ok, to proof that the direct product of finitely many Noetherian rings is a Noetherian ring, I still must think about...

The thing with the maximal ideals in $\mathbb Z[i]$ is clear now.

But may I ask a question? Why are $\mathbb Z[i]$ and $\mathbb Z \times \mathbb Z$ not isomorphic as rings?

But they are isomorphic as $\mathbb Z$-modules, aren't they?

Lisa
• Sep 7th 2010, 12:05 PM
NonCommAlg
Quote:

Originally Posted by lisa
Hello.

But may I ask a question? Why are $\mathbb Z[i]$ and $\mathbb Z \times \mathbb Z$ not isomorphic as rings?

because the equation $x^2+1=0$ has a solution in $\mathbb{Z}[i]$ but it has no solution in $\mathbb{Z} \times \mathbb{Z}.$

Quote:

But they are isomorphic as $\mathbb Z$-modules, aren't they?
yes, they are.