Grothendieck's generic freeness lemma: Difference between revisions

Revision as of 22:21, 20 January 2008

Statement

Suppose $R$ is a Noetherian ring and $S$ is a finitely generated $R$ -algebra. Further, suppose $M$ is a free module over $S$ . Then, there exists $0 \neq a \in R$ such that $M [a^{- 1}]$ is free as a module over $R [a^{- 1}]$ . Here $R [a^{- 1}]$ denotes the localization of $R$ at $a$ , and $M [a^{- 1}]$ denotes the localization of $M$ at $a$ .

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	==Statement==		==Statement==

	Suppose <math>R</math> is a [[Noetherian ring]] and <math>S</math> is a finitely generated <math>R</math>-algebra. Further, suppose <math>M</math> is a [[free module]] over <math>S</math>. Then, there exists <math>0 \ne a \in R</math> such that <math>M[a^{-1}]</math> is free at a <math>R[a^{-1}]</math>~~-module~~. Here <math>R[a^{-1}]</math> denotes the localization of <math>R</math> at <math>a</math>, and <math>M[a^{-1}]</math> denotes the localization of <math>M</math> at <math>a</math>.		Suppose <math>R</math> is a [[Noetherian ring]] and <math>S</math> is a finitely generated <math>R</math>-algebra. Further, suppose <math>M</math> is a [[free module]] over <math>S</math>. Then, there exists <math>0 \ne a \in R</math> such that <math>M[a^{-1}]</math> is free as a module over <math>R[a^{-1}]</math>. Here <math>R[a^{-1}]</math> denotes the localization of <math>R</math> at <math>a</math>, and <math>M[a^{-1}]</math> denotes the localization of <math>M</math> at <math>a</math>.