My question refers to an argument from D. Mumford's "Red Book of Schemes" (page 211):
By definition: A subvariety $Z \subset X$ has pure dimension $r$ iff every component of $Z$ has dimension $r$.
Futhermore if $X$ is a scheme then the dimension $dim(X)$ equals the supremum of the dimensions of all irreducible components of $X$.
We know by §7 Thm 2 (page 41) that $V_i:= V(t_i)$ are of pure dimension $n-1$
My question is why is the intersection $V_1 \cap V_2 $ pure $(n-2)$-dimensional?
Why it cannot occure that $V_1 \cap V_2 $ has components of lower dimension than $n-2$?
Best Answer
This is nothing but Krull's principal ideal theorem. Pick an affine neighborhood $\operatorname{Spec} R$ of $x'$. Then $R/(t_1)$ is the coordinate ring of $V_1$ in this neighborhood, and $V_1\cap V_2$ is given by $V(t_2)\subset \operatorname{Spec} R/(t_1)$. The irreducible components of $V_1\cap V_2$ are then the minimal primes over $(t_2)\subset R/(t_1)$. By Krull's principal ideal theorem, these are of height at most one, so combining this with the assumption that $V(t_1)$ and $V(t_2)$ don't share any components, the irreducible components of $V(t_2)\subset \operatorname{Spec} R/(t_1)$ are of codimension one, or $V_1\cap V_2$ is of pure codimension two.