Nontrivial projective hypersurface sections

Given: The objects and hypotheses in the statement.

If $X\subseteq {\mathbf{P}}_k^N$ is a nonempty projective algebraic set, then $\dim C(X)=\dim X+1$. Over $X_i=X\cap D_{+}(T_i)$, the locus $C(X)\cap D(T_i)$ is isomorphic to $X_i\times {\mathbf{G}}_m$. If $X$ is irreducible, so is $C(X)$. Work over a fixed algebraically closed field $k$, with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points. (A nonempty projective cone raises dimension by one).

Let $X$ be irreducible affine and $0\mathrm{\ne }f\in k[X]$ be a nonunit. Then $V_X(f)$ is nonempty and every irreducible component has dimension $\dim X-1$, hence codimension one. Work over a fixed algebraically closed field $k$, with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points. (A nontrivial principal section has pure codimension one).

If $U$ is a nonempty open of an irreducible classical variety $X$, then $\dim U=\dim X$. Every proper closed subvariety $Z⊊X$ has $\dim Z<\dim X$. Work over a fixed algebraically closed field $k$, with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points. (Nonempty opens preserve irreducible dimension).