I have a question about an argument in the proof of prop. 8.4 in Bosch's "Algebraic Geometry and Commutative Algebra" (see page 376): Following statement is used:
Let $A$ be a $k$ algebra of finite type and an integal domain. Consider $X=\operatorname{Spec}(A)$ the corresponding affine scheme. Let $x \in X$ arbitrary.
We define $\dim_x X:= \inf\limits _{\substack{U \subset X \text{ open, }\\ x \in U}}\!\! \dim U$
Why does
$$\dim_x X = \dim U$$
hold for every open subset $U \subset X$?
Especially why the inequality $\dim_x X \ge \dim U$?
In the book he argues firstly that every closed point of a open non empty subscheme $U \subset X$ is dense in $X$. But for example if we take $U=X$ and $x$ arbitrary closed point of $X$ then thats of course highly wrong.
Furthermore it should follow from Hilbert's Nullstellensatz. How?
Could anybody explain to me the argument given about to settle $\dim_x X = \dim U$?
If $A$ is a finite type $k$-algebra for any field $k$ and $A$ is furthermore a domain, then if $U$ is any non empty open subset of $\operatorname{Spec} A$, we have $\dim U=\dim A$. This essentially follows from the fact that $\dim A=\operatorname{tr.deg}_k\operatorname{Frac}A$ holds. Since $U$ contains a principal open which is itself the spectrum of a domain of finite type over $k$ that has the same field of fractions as $A$, the equality follows, together with the fact that we always have $\dim U\leq \dim A$ for any open subset $U$.