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I got two questions regarding the proof:

  1. Why we can just assume $A_\mathfrak{m}$ as $A$ if we pick a sufficiently small neighborhood of $x$? (the 7-th line)

  2. Why's the last isomorphism true?

Thank you very much in advance for answering my question.

Ivan So
  • 797
  • There is no assumption that $A_\mathfrak{m}=A$. What do you mean? 2. Do you know the definition of the Koszul complex? If you do, you should be able to plug in and calculate this in a straightforwards manner. If not, you probably should look up what the last few terms of the Koszul complex are.
    • – KReiser Nov 12 '19 at 07:46

  • But the why can we assume the Koszul complex to be the free resolution of the quotient ring?

  • I understand what Koszul complex is. But what involved in the proof was the right derived functor of \mathscr{Hom}. I am new to the subject and I hope to have a more explicit picture.

    • – Ivan So Nov 12 '19 at 09:06
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    This is a consequence of generic freeness. See for instance this. Stalk-locally at $\mathfrak{m}$, the Koszul complex is free, so then there's some neighborhood $D(f)$ where the Koszul complex is free. Use $D(f)\subset U$ instead of $U$ and you're set. – KReiser Nov 12 '19 at 09:11