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I have seen the phrase "minimal number of generators of an ideal" (in a Noetherian local ring) several times. I am unable to see how this is a well defined. Explicitly, how do we show, if $x_1,...,x_m$ and $y_1,...,y_n$ are minimal generating sets of an ideal in a Noetherian local ring, then $m=n$.

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1 Answer 1

HINT $\ $ Over a local ring there is a nicely behaved notion of minimal set of generators by way of Nakayama's lemma. See section 20.1, The uniqueness of free resolutions, in Eisenbud: Commutative Algebra: with a View Toward Algebraic Geometry.

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Thanks for the response. I don't have access to the book. This is what I gather from your hint. For any ideal $I$ in a noetherian local ring $(R,m)$, $I/mI$ is a finite dimensional vector space over $R/m$. Then any set of generators of $I$ is minimal iff their images modulo $mI$ is a basis of $I/mI$ as a $R/m$-vector space. So all such must have the same size. –  Yan Etor Mar 9 '11 at 3:10

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