Noetherian Ring

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• Sep 7th 2010, 02:38 AM
lisa
Noetherian Ring
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-one correspondence between 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
• Sep 7th 2010, 06:50 AM
NonCommAlg
Quote:

Originally Posted by lisa
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? :

Ok!

Quote:

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

Quote:

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.
you've only proved that your ring is Noetherian $\displaystyle \mathbb{Z}$-module. you haven't proved that it's a Noetherian "ring" yet.

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

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

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

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

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

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

Quote:

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.
well, $\displaystyle \mathbb{Z}[i]$ is a free $\displaystyle \mathbb{Z}$-module and every free module is faithfully flat.

Quote:

Now the other part:
There is a one-to-one correspondence between the ideals in $\displaystyle \mathbb Z[i]/ \mathfrak a$ 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.
correct.

Quote:

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

Quote:

hence every prime ideal in $\displaystyle \mathbb Z[i]$ is a maximal ideal in $\displaystyle \mathbb Z[i]$.
why?
• Sep 7th 2010, 08:40 AM
lisa
Hi,

thank you for your answer.

Quote:

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

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

What is $\displaystyle S$? 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 $\displaystyle 0\to M' \to M \to M'' \to 0$ the modules $\displaystyle M'$ and $\displaystyle M''$ are noetherian if and only if $\displaystyle M$ is noetherian. So I would proof it with induction over $\displaystyle n$.
If $\displaystyle 0\to M_2 \to M_1\oplus M_2 \to M_1 \to 0$ is exact, and $\displaystyle M_1, \ M_2$ are noetherian, it follow that $\displaystyle M_1 \oplus M_2$ is noetherian.

Is $\displaystyle 0\to M_n \to M_1\oplus ... \oplus M_n \to M_1\oplus ... \oplus M_{n-1} \to 0$ an exact sequence and $\displaystyle M_n$ and $\displaystyle M_1\oplus ... \oplus M_{n-1}$ are noetherian, then $\displaystyle M_1\oplus ... \oplus M_n$ 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 $\displaystyle \mathbb Z[i]/( a)$ has the form $\displaystyle (b)+(a)$ with $\displaystyle (b)$ is a ideal in $\displaystyle \mathbb Z[i]$. If the ideal $\displaystyle (b) \subset \mathbb Z[i]$ is a prime ideal but not a maximal ideal, then it exists an ideal $\displaystyle (c) \subset \mathbb Z[i]$ with $\displaystyle (b) \subset (c) \subset \mathbb Z[i]$. Then $\displaystyle (b)+(a) \subset \mathbb Z[i]/ (a)$ is prime but not maximal, because $\displaystyle (c)+(a) \subset \mathbb Z[i]/ (a)$ is "bigger". This is a contradiction, because $\displaystyle (b)+(a) \subset \mathbb Z[i]/ (a)$ should be a maximal ideal.

Bye,
Lisa
• Sep 7th 2010, 10:08 AM
NonCommAlg
first take a look at my previous post again. i added a couple of things which are in red.

regarding your proofs:

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 $\displaystyle a$ in your last proof? here's how you should do it:

suppose that $\displaystyle P$ is a non-zero prime ideal of $\displaystyle R=\mathbb{Z}[i].$ choose $\displaystyle 0 \neq a \in P$ and let $\displaystyle I=\langle a \rangle.$ then $\displaystyle P/I$ is a prime ideal of $\displaystyle R/I$ and thus it is maximal. it's now clear that $\displaystyle P$ is maximal in $\displaystyle R.$
• Sep 7th 2010, 10:56 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 $\displaystyle \mathbb Z[i]$ is clear now.

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

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

Buy,
Lisa
• Sep 7th 2010, 11:05 AM
NonCommAlg
Quote:

Originally Posted by lisa
Hello.

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

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

Quote:

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