Nakayama's lemma: Difference between revisions

Revision as of 17:44, 3 March 2008

This article is about the statement of a simple but indispensable lemma in commutative algebra
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Statement

Let $R$ be a commutative unital ring, and $I$ be an ideal contained inside the Jacobson radical of $R$ . Let $M$ be a finitely generated $R$ -module. Then the following are true:

If $I M = M$ then $M = 0$
If $N$ is a submodule of $M$ suc hthat $N + I M = M$ , then $N = M$
If $m_{1}, m_{2}, \dots, m_{n}$ have images in $M / I M$ that generate it as a $R$ -module, then $m_{1}, m_{2}, \dots, m_{n}$ generate $M$ as a $R$ -module

Related facts

The graded Nakayama's lemma is a related fact true for graded rings.

@@ Line 5: / Line 5: @@
 Let <math>R</math> be a [[commutative unital ring]], and <math>I</math> be an [[ideal]] contained inside the [[Jacobson radical]] of <math>R</math>. Let <math>M</math> be a finitely generated <math>R</math>-module. Then the following are true:
-* If <math>IM = M</math> then <math>M = 0</math>
+# If <math>IM = M</math> then <math>M = 0</math>
-* If <math>m_1, m_2, \ldots, m_n</math> have images in <math>M/IM</math> that generate it as a <math>R</math>-module, then <math>m_1, m_2, \ldots, m_n</math> generate <math>M</math> as a <math>R</math>-module
+# If <math>N</math> is a submodule of <math>M</math> suc hthat <math>N + IM = M</math>, then <math>N = M</math>
+# If <math>m_1, m_2, \ldots, m_n</math> have images in <math>M/IM</math> that generate it as a <math>R</math>-module, then <math>m_1, m_2, \ldots, m_n</math> generate <math>M</math> as a <math>R</math>-module
+==Related facts==
+The [[graded Nakayama's lemma]] is a related fact true for [[graded ring]]s.