I'm studying algebraic geometry following the Red Book, and a few questions arise in the section about Dimension, that after much try I could not understand.
First question (about theorem 2 of page 41, namely: For a variety $X$, $U\subset X$ open, $g\in \Gamma(U,\underline{o}_X)$, $Z$ an irreducible component of $\{x\in U:g(x)=0\}$. Then if $g\ne 0$, $\dim Z=\dim X-1$). In the proof, the author states that since $Z$ is a maximal irreducible subset of the locus $g=0$, then $Z\cap U_0$ is also maximal irreducible subset of the locus $g\upharpoonright_{ Z\cap U_0}=0$, where $U_0\subset U$ is an open affine subset such that $Z\cap U_0\ne \emptyset$. Why?
Second question. On page 44, corollary 1 (namely: Let $X$ be a variety and $Z$ a maximal closed irreducible set, smaller than $X$ itself. Then $\dim Z=\dim X-1$). How this follows from previous results?
I hope these questions are not too silly.
Thank you very much.
For your first question think on this more general problem (easy and well known): Let $Y$ be a topological space and let $V$ be an open subset of $Y$. Then the map $Z\rightarrow Z\cap U$ gives a bijection between the irreducible components of $Y$ meeting $V$ and the irreducible components of $V$. In your notation, apply this with $Y=\{x\in U_0:g(x)=0\}$ and $V=Y\cap U_0$.
For your second question: take any affine subset $U$ of $X$ meeting $Z$. Then $U\cap Z$ is a maximal irreducible closed subset of $U\cap X$ smaller thatn $U\cap X$. Note that $\dim X=\dim U$ because $X$ is irreducible and for the same reason $\dim Z=\dim Z\cap U$. Then you can assume that $X$ is affine. Then $Z$ is an irreducible component of $V(g)$ for some $g\in \Gamma (X,\mathcal{O}_X)$ (any non zero $g$ in the corresponding minimal prime ideal of $Z$) and then you can use Theorem 2 with $U=X$