# The union of passing irred. components minus others is open

2019-11-19

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

## Statement

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.