Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Noetherian"

 
(expanded)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
−
Let <math>R</math> be a [[ring]] and <math>M</math> a left <math>R</math>-[[module]]. Then we say that <math>M</math> is a '''noetherian module''' if it satisfies the following property, known as the ascending chain condition (ACC):
+
Let <math>R</math> be a [[ring]] and <math>M</math> a left <math>R</math>-[[module]]. Then we say that
 +
<math>M</math> is a '''Noetherian module''' if it satisfies the following
 +
property, known as the [[ascending chain condition]] (ACC):
  
−
* For any ascending chain <math>M_1\subseteq M_2\subseteq M_3\subseteq\cdots</math> of [[submodule]]s of <math>M</math>, there exists an integer <math>n</math> so that <math>M_n=M_{n+1}=N_{n+2}=\cdots</math> (i.e. the chain eventually terminates).
+
:For any ascending chain
 +
<cmath> M_0\subseteq M_1\subseteq M_2\subseteq\cdots </cmath>
 +
:of [[submodule]]s of <math>M</math>, there exists an integer <math>n</math> so that <math>M_n=M_{n+1}=M_{n+2}=\cdots</math> (i.e. the chain eventually stabilizes, or terminates).
  
−
'''Theorem.''' The following conditions are equivalent for a left <math>R</math>-module:
+
We say that a ring <math>R</math> is left (right) Noetherian if it is Noetherian
 +
as a left (right) <math>R</math>-module. If <math>R</math> is both left and right
 +
Noetherian, we call it simply Noetherian.
  
−
* <math>M</math> is noetherian.
+
'''Theorem.''' The following conditions are equivalent for a left
−
* Every submodule <math>N</math> of <math>M</math> is [[finitely generated]] (i.e. can be written as <math>Rm_1+\cdots+Rm_k</math> for some <math>m_1,\ldots,m_k\in N</math>).
+
<math>R</math>-module:
−
* For every collection of submodules of <math>M</math>, there is a [[maximal element]].
+
# <math>M</math> is Noetherian.
 +
# Every submodule <math>N</math> of <math>M</math> is [[finitely generated]] (i.e. can be written as <math>Rm_1+\cdots+Rm_k</math> for some <math>m_1,\ldots,m_k\in N</math>).
 +
# Every collection of submodules of <math>M</math> has a [[maximal element]].
 +
The second condition is also frequently used as the definition for
 +
Noetherian.
  
−
(The second condition is also frequently used as the definition for noetherian.)
+
We also have right Noetherian modules with the appropriate
 +
adjustments.
  
−
We also have right noetherian modules with the appropriate adjustments.
+
''Proof.''  In general, condition 3 is equivalent to [[ACC]].
 +
It thus suffices to prove that condition 2 is equivalent to ACC.
  
−
We say that a ring <math>R</math> is left (right) noetherian if it is noetherian as a left (right) <math>R</math>-module. If <math>R</math> is both left and right noetherian, we call it simply noetherian.
+
Suppose that condition 2 holds.  Let <math>M_0 \subseteq M_1 \subseteq \dotsb</math>
 +
be an ascending chain of submodules of <math>M</math>.  Then
 +
<cmath> \bigcup_{n \ge 0} M_n </cmath>
 +
is a submodule of <math>M</math>, so it must be finitely generated, say
 +
by elements <math>a_1, \dotsc, a_n</math>.  Each of the <math>a_k</math> is contained
 +
in one of <math>M_0, M_1, \dotsc</math>, say in <math>M_{t(k)}</math>.  If we set
 +
<math>N = \max t(k)</math>, then for all <math>n \ge N</math>,
 +
<cmath> \{ a_1, \dotsc, a_n \} \subset M_n , </cmath>
 +
so
 +
<cmath> M_n = M_N = \bigcup_{n\ge 0} M_n . </cmath>
 +
Thus <math>M</math> satisfies ACC.
 +
 
 +
On the other hand, suppose that condition 2 does not hold, that
 +
there exists some submodule <math>M'</math> of <math>M</math> that is not finitely
 +
generated.  Thus we can recursively define a sequence of elements
 +
<math>(a_n)_{n=0}^{\infty}</math> such that <math>a_n</math> is not in the submodule
 +
generated by <math>a_0, \dotsc, a_{n-1}</math>.  Then the sequence
 +
<cmath> (a_0) \subset (a_0, a_1) \subset (a_0, a_1, a_2) \subset \dotsb </cmath>
 +
is an ascending chain that does not stabilize.  <math>\blacksquare</math>
 +
 
 +
''Note:  The notation <math>(a,b,c \dotsc)</math> denotes the module
 +
generated by <math>a,b,c, \dotsc</math>.''
 +
 
 +
[[Hilbert's Basis Theorem]] guarantees that if <math>R</math> is a Noetherian
 +
ring, then <math>R[x_1, \dotsc, x_n]</math> is also a Noetherian ring,
 +
for finite <math>n</math>.  It is not a Noetherian <math>R</math>-module.
 +
 
 +
 
 +
== See also ==
 +
 
 +
* [[Artinian]]
 +
* [[Hilbert's Basis Theorem]]
 +
 
 +
 
 +
[[Category:Ring theory]]
 +
[[Category:Commutative algebra]]

Latest revision as of 21:18, 10 April 2009

Let $R$ be a ring and $M$ a left $R$-module. Then we say that $M$ is a Noetherian module if it satisfies the following property, known as the ascending chain condition (ACC):

For any ascending chain

\[M_0\subseteq M_1\subseteq M_2\subseteq\cdots\]

of submodules of $M$, there exists an integer $n$ so that $M_n=M_{n+1}=M_{n+2}=\cdots$ (i.e. the chain eventually stabilizes, or terminates).

We say that a ring $R$ is left (right) Noetherian if it is Noetherian as a left (right) $R$-module. If $R$ is both left and right Noetherian, we call it simply Noetherian.

Theorem. The following conditions are equivalent for a left $R$-module:

  1. $M$ is Noetherian.
  2. Every submodule $N$ of $M$ is finitely generated (i.e. can be written as $Rm_1+\cdots+Rm_k$ for some $m_1,\ldots,m_k\in N$).
  3. Every collection of submodules of $M$ has a maximal element.

The second condition is also frequently used as the definition for Noetherian.

We also have right Noetherian modules with the appropriate adjustments.

Proof. In general, condition 3 is equivalent to ACC. It thus suffices to prove that condition 2 is equivalent to ACC.

Suppose that condition 2 holds. Let $M_0 \subseteq M_1 \subseteq \dotsb$ be an ascending chain of submodules of $M$. Then \[\bigcup_{n \ge 0} M_n\] is a submodule of $M$, so it must be finitely generated, say by elements $a_1, \dotsc, a_n$. Each of the $a_k$ is contained in one of $M_0, M_1, \dotsc$, say in $M_{t(k)}$. If we set $N = \max t(k)$, then for all $n \ge N$, \[\{ a_1, \dotsc, a_n \} \subset M_n ,\] so \[M_n = M_N = \bigcup_{n\ge 0} M_n .\] Thus $M$ satisfies ACC.

On the other hand, suppose that condition 2 does not hold, that there exists some submodule $M'$ of $M$ that is not finitely generated. Thus we can recursively define a sequence of elements $(a_n)_{n=0}^{\infty}$ such that $a_n$ is not in the submodule generated by $a_0, \dotsc, a_{n-1}$. Then the sequence \[(a_0) \subset (a_0, a_1) \subset (a_0, a_1, a_2) \subset \dotsb\] is an ascending chain that does not stabilize. $\blacksquare$

Note: The notation $(a,b,c \dotsc)$ denotes the module generated by $a,b,c, \dotsc$.

Hilbert's Basis Theorem guarantees that if $R$ is a Noetherian ring, then $R[x_1, \dotsc, x_n]$ is also a Noetherian ring, for finite $n$. It is not a Noetherian $R$-module.


See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Noetherian&oldid=31213"