Krull's intersection theorem
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regular ring
Let \(R\) be a local Noetherian ring and let \(n\) be the minimal number of generators of its maximal ideal \({\mathfrak{m}}_R\). By Krull’s intersection theorem, \begin{align*}n\geq \operatorname{krulldim}(R),\end{align*} so define \(R\) to be regular iff this is an equality. Note that a minimal set of generators is a regular system.
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Noetherian
For \(X = {\mathbf{C}}\) equipped with the sheaf \({\mathcal{F}}= C^\infty({-}; {\mathbf{R}})\), the stalk \({\mathcal{F}}_0\) is the ring of germs of smooth functions at the origin. Take \(f(x) \coloneqq e^{-{1\over x^2}}\) and patch \(f(0) = 0\) to get a smooth-nonanaltic function, contradicting Krull’s intersection theorem.
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Commutative Algebra: Main Results
What is Krull’s intersection theorem?
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Artin-Rees theorem
Used to prove the Krull’s intersection theorem: this is a separable topology iff \(1+I\) contains no zero divisors,which holds e.g. if \(I \subseteq {J ({R}) }\) (the Jacobson radical).