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Following a book "Elementary Algebraic Geometry" by K. Hulek, I would like to ask how to use Nakayama's Lemma to prove the following statement:

Let $C$ be a smooth curve. Then dim$_k m_P / m_P^2=$ dim$C=1$. If $t \in m_P$ and its residue class $\bar{t} \in m_P / m_P^2$ spans the $k$-vector space $m_P / m_P^2$, then $t$ generates $m_P$.

I can not come up with the proof. Unless $m_P$ is always (or in this case) a finitely generated $\mathcal{O}_{C,P}$-module.

Thank you.

Kristina
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    I think this is Prop 2.8 in Atiyah-Macdonald. – hwong557 Nov 20 '16 at 21:39
  • Exactly! I tried to use this proposition, but it assumes that M (in the Prop 2.8) is a finitely generated A-module, doesn't it? – Kristina Nov 20 '16 at 21:44
  • The local ring is noetherian isnt it? – hwong557 Nov 20 '16 at 21:47
  • Apparently there is a counterexample: http://math.stackexchange.com/questions/128838/example-of-a-non-noetherian-complete-local-ring – Kristina Nov 20 '16 at 22:04
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    You're working on a curve so those counterexamples don't work. The local ring of an affine curve is always noetherian. You can calculate the local ring explicitly as $A_P$ where $A$ is the coordinate ring and $P$ is the maximal ideal corresponding to a point. If your curve is projective, zoom in to an affine patch. – hwong557 Nov 20 '16 at 22:10
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    Oh and $A_P$ is noetherian because $A$ is the image of a polynomial ring over a field, and noetherian passes to localization. – hwong557 Nov 20 '16 at 22:11
  • I see, now it's fully clear. Thank you! – Kristina Nov 20 '16 at 23:03

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