# upper semi-continuity of dimension of a locally finite type $k$-scheme at closed points

Let $$X$$ be a locally finite type $$k$$-scheme. Let $$p$$ be a closed point. We define the dimension of $$X$$ at $$p$$ to be the largest dimension of an irreducible component of $$X$$ that contains $$p$$, denoted $$\dim_p X$$. Then the function $$\phi:p \mapsto \dim_p X$$ is upper semicontinuous. To see this it is enough to show that for $$n$$ any positive integer $$\phi^{-1}[0, n)$$ is an open set. Indeed, $$\phi^{-1}[0, n)$$ is just the set of closed points of $$X$$ minus those closed points that lie on irreducible components of dimension $$\ge n$$. If there are finitely many such irreducible components their union is a closed set and the inverse image is open. But what if there are infinitely many of them? (Reference: Definition 11.2.I, Vakil November 2017)

• Doesn't locally finite type imply only finitely many irreducible components through any point? – Mohan Oct 2 '19 at 13:11

## 1 Answer

Lemma: For any point $$x$$ in any locally noetherian scheme $$X$$, there exists a neighborhood of $$x$$ which intersects only finitely many irreducible components of $$X$$.

Proof: Take a noetherian affine open neighborhood $$U$$ of $$x$$. As $$\mathcal{O}_X(U)$$ is a noetherian ring, it has finitely many minimal primes, so $$U$$ has finitely many irreducible components. As every irreducible component of $$X$$ intersects $$U$$ in an irreducible component or misses $$U$$ entirely, we have the claim.

To apply this to the situation at hand, $$X$$ is locally noetherian as any scheme locally of finite type over a locally noetherian scheme is again locally noetherian. So there's a neighborhood of $$p\in X$$ which only has finitely many irreducible components, so only finitely many of them can meet $$p$$ and your problem is solved.