The union of passing irred. components minus others is open


This is a simple proof that I think is quite interesting to share.


Let \(X\) be a locally Nœtherian scheme and let \(x\in X\). Then we claim that the union of the irreducible components which pass through \(x\) minus the irreducible components which don't pass through \(x\) is open.

A proof

The property of being open is local, so we can assume that \(X\) is affine. Suppose \(X = \text{Spec}(R)\). Then in fact our set \(Y\) is nothing but \(D(\prod_i p_i)\), where \(\left\{p_i \right\}\) is the set of minimal primes in \(R\) not contained in the prime ideal \(p\) corresponding to the given point \(x\).

Since \(X\) is locally Nœtherian, we may assume that \(R\) is a Nœtherian ring, so that there are only a finite number of minimal primes. So for a prime ideal to contain one of the \(p_i\) it is equivalent with containing \(\prod_i p_i\). Since \(Y\) is exactly the set of prime ideals not containing any of the primes \(p_i\), it is clear that \(Y=D(\prod_i p_i)\); consequently \(Y\) is open.

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Author: JSDurand


Date: 2019-11-19 Mar 00:44:00 CST

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