A positive-dimensional projective variety meets every hypersurface

Show that any two curves in $\PP^2$ have a nonempty intersection.
More generally, show that if $Y \subseteq \PP^n$ is a projective variety of dimension $\geq 1$ and $H$ is a hypersurface, then $Y \intersect H \neq \emptyset$.

Hint

Use (Ex. 3.5) and (Ex. 3.1e). See (7.2) for a generalization.

Solution

1. If $Y\subseteq\PP^n$ is projective with $\dim Y\ge1$ and $H\subseteq\PP^n$ is a hypersurface, then $Y\cap H\ne\varnothing$.

Suppose instead that $Y\cap H=\varnothing$. Then \[ Y\subseteq\PP^n\setminus H. \] By P-AGH35COMPLHYP, the hypersurface complement $\PP^n\setminus H$ is affine. Since $Y$ is closed in $\PP^n$, it is also closed in the open subspace $\PP^n\setminus H$. Therefore $Y$ is a closed subvariety of an affine variety and hence is affine.

But $Y$ is projective as well. Part (e) of P-AGH31CONICS shows that a variety which is both affine and projective consists of a single point. That would give $\dim Y=0$, contradicting $\dim Y\ge1$. Thus $Y\cap H\ne\varnothing$, proving (b).

2. Any two curves in $\PP^2$ have a nonempty intersection.

Let $C,D\subseteq\PP^2$ be curves. Each has dimension one. By P-AGH28HYPERSURFACE, a projective variety of dimension $2-1=1$ in $\PP^2$ is a hypersurface. Apply step 1 with $Y=C$ and $H=D$. It gives \[ C\cap D\ne\varnothing, \] which proves (a).

3. Q.E.D.

Step 1 proves (b), and step 2 proves (a).

Source Collections

Hartshorne, Algebraic Geometry, I.3.7

Dependencies

P-AGH28HYPERSURFACEA projective variety has codimension one exactly when it is a hypersurfaceuses
P-AGH31CONICSConics, and varieties that are not isomorphic to one anotheruses
P-AGH35COMPLHYPThe complement of a hypersurface in $\PP^n$ is affineuses

Backlinks

P-AGH45QUADSURFBIRThe quadric surface $xy = zw$ is birational but not isomorphic to $\PP^2$uses