Special Case for Bézout Theorem

Question

Over an algebraically closed field one has that by Bézout, two hypersurfaces in projective $n$-space always intersect. Is there an elementary proof for this statement?

Answer 1

Yes, dimension theory does it (assuming $n>1$). One of the most basic results about intersections in projective space is that if $X,Y\subset\Bbb P^n_k$ are irreducible of codimension $r,s$ respectively, then every irreducible component of $X\cap Y$ has codimension at most $r+s$, and if $r+s\leq n$ this intersection is guaranteed to be nonempty. For a proof, see Hartshorne theorem I.7.2.

Alternatively, if you know that the complement of a hypersurface is affine (use the appropriate Veronese embedding to see this) and no affine variety contains a projective variety of positive dimension, you can see that for projective hypersurfaces $X,Y$, $\Bbb P^n_k\setminus X$ cannot contain $Y$ - this means $X\cap Y$ is nonzero, though it doesn't give you the result on dimensions like the previous paragraph.