Dimension of local ring at closed point on integral scheme of finite type over an algebraically closed field.

Question

Dimension of local ring at closed point of an integral scheme of finite type over algebraically closed field is equal to dimension of scheme. Can someone help me with the proof?

Edit: Can someone explain me why $\dim(X)=\dim(U)$?

Answer 1

Question: "Dimension of local ring at closed point of an integral scheme of finite type over algebraically closed field is equal to dimension of scheme. Can someone help me with the proof?"

Answer: The local ring of an integral scheme $X$ of finite type over an algebaically closed field is catenary (see Matsumura's book). If $X:=Spec(A)$ it follows for any maximal ideal $\mathfrak{m} \subset A$ that $\dim(X):=\dim(A)=\dim(A_{\mathfrak{m}})$. Since $X$ is integral, you may choose an open affine subscheme $U:=Spec(A) \subseteq X$ and

$$\dim(X)=\dim(U)=\dim(A_{\mathfrak{m}})=\dim(\mathcal{O}_{X,\mathfrak{m}}).$$

The equality $\dim(X)=\dim(U)$ is HH.ex.II.3.20. There is an "elementary" version in HH.I.1.10.