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I have a couple of question regarding this proof of the following theorem:

A regular local ring is a UFD.

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  1. "Since $R$ is regular, $P$ has a FFR" -- is it obvious or it's some deep theorem?
  2. Why does $R[x^{-1}]_{\mathfrak n}$ isomorphic to $R_{R\cap \mathfrak n}$ and why does the dimension decrease here?
  3. Could you please enlarge on the last but one sentence viz "Looking at localizations, we see that $C_{\mathfrak n}$ is zero for all $\mathfrak n$". First, why does this hold, and second, how does this prove the projectivity of $Q$?
user557
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1 Answers1

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1) A regular local ring has finite global dimension. Hence all modules have finite projective dimension, i.e. they have a finite projective resolution. Furthermore, on a local ring, finitely generated projective modules are free.

2) This is because $R[x^{-1}]_\mathfrak n=R_{\mathfrak n\cap R}[x^{-1}]_\mathfrak n$, and $R_{\mathfrak n\cap R}[x^{-1}]=R_{\mathfrak n\cap R}$ since, as $x\notin \mathfrak n\cap R$, $x$ is a unit in $R_{\mathfrak n\cap R}$.

The (Krull) dimension is no decreasing by localisation, and indeed it decreases here because $\mathfrak mR_{\mathfrak n\cap R}=R_{\mathfrak n\cap R}$.

3) $C_\mathfrak n$ is $0$ because $Q$ has finite presentation, so $$\DeclareMathOperator{\Hom}{Hom} \bigl(\Hom_{R[x^{-1}]}(Q,M)\bigr)_\mathfrak n\simeq\Hom_{R[x^{-1}]_\mathfrak n}(Q_\mathfrak n,M_\mathfrak n)\simeq M_\mathfrak n$$ since $Q_\mathfrak n$ is free of rank $1$. Thus $C_\mathfrak n$ is isomorphic to the cokernel of $M_\mathfrak n\longrightarrow N_\mathfrak n$, and localisation is exact…

This proves $C=0$, in other words the functor $\Hom_{R[x^{-1}]}(Q,\ \cdot)$ is right exact, which is the definition of $Q$ being a $R[x^{-1}]$ projective module.

user26857
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Bernard
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