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Let $(R, m)$ be a local ring and $M$ a finitely generated projective $R$-module.

Let $n$ be the number of elements in a minimal generating set of $M$. Then by Nakayama's lemma $M/m M$ any minimal generating set of $M/ m M$ as an $R$-module must have at least $n$ elements.

It seems that this implies that $M/mM$ has dimension $n$ as a $R/mR$ module. Why is this?

Bernard
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Yuugi
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  • The way I think about it, so that I can be checked too. First of all, M is not necessarily projective. Nakayama’s Lemma proves that you can lift a generating system in the quotient to a generating system of M. This gives the fact: consider a minimal generating set, pass it to the quotient, (it generates the vector space), refine it to obtain a basis of the vector space and lift it again by Nakayama’s Lemma. By minimality, you have the same system of generators, hence that must be the dimension of the basis. Works? – ZenoCozeno Apr 01 '18 at 07:18
  • But why does it necessarily generate the vector space? – Yuugi Apr 01 '18 at 17:55
  • Nakayama only tells us that it generates the quotient as an $R$-module, not as an $R/mR$-module. – Yuugi Apr 01 '18 at 18:10

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