Noetherian: Difference between revisions

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:

$M$ is Noetherian.
Every submodule $N$ of $M$ is finitely generated (i.e. can be written as $Rm_1+\cdots+Rm_k$ for some $m_1,\ldots,m_k\in N$ ).
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.

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-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.
+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]]
-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.
 [[Category:Ring theory]]
+[[Category:Commutative algebra]]
-{{stub}}

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Noetherian: Difference between revisions

Latest revision as of 21:18, 10 April 2009

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