Say I have a variety $X$ with irreducible components $C_1,…,C_n$ and an open subset $U\subset X$. the irreducible components of $U$ are just $U\cap C_i$. Fix $i$ and set $C=C_i$. My question is: is the dimension of $C$ the same as the dimension of $C\cap U$? If so how do you prove that?
Let $A_1\subset A_2 \subset … \subset A_k$ be a chain of irreducible closed (in $U\cap C$) subsets of $U\cap C$ with $k= \dim(U\cap C)$. Then if you close the $A_i$ in $X$, you get a maximal chain of irreducible closed subsets of $C$, and thus $\dim(C\cap U)\leq \dim(C)$. so it would suffice to prove that every maximal chain of irreducible closed subsets of $C$ (an irreducible Noetherian topological space) has the same length. That might be just a standard fact that I couldn’t find by googling.
Best Answer
The claim in your first paragraph is correct assuming that you require $C\cap U\neq\emptyset$. You may deduce it from the fact that the dimension of an integral variety is the transcendence degree of it's function field, which can be calculated from any nonempty open subset. Details of this are available in most algebraic geometry books: for instance, Vakil's FOAG has it in section 11.2, Hartshorne develops it in exercise II.3.20, etc.
The claim in your second paragraph is unfortunately not correct. If $X$ is a topological space, we call it catenary iff for every pair of irreducible closed subsets $T\subset T'$, there exists a maximal chain of irreducible closed subsets $$T=T_0\subset T_1\subsetneq\cdots\subsetneq T_n=T'$$ and every chain has the same length. There are irreducible noetherian topological spaces which are not catenary: for instance, the spectrum of any non-catenary noetherian domain suffices. Such a thing may be observed in the article A non-catenary, normal, local domain by Heitmann.