TITLE: Exercise 3.7 Hartshorne
QUESTION [8 upvotes]: Problem. Show that any two curves in $\mathbb{P}^2$ have a nonempty intersection.
This seems to follow immediately from the Projective Dimension Theorem, but I was wondering if anyone could provide a more 'elementary' proof?
Thanks

REPLY [3 votes]: I'm going to post this as an answer since it received positive reviews in the comments. This is probably the intended solution (also, by the way the OP responded to my comments, not the one he was thinking), but is certainly not as nice as Georges's or Brenin's answer.
We want to take two projective curves $V_+(f)$ and $V_+(g)$ (where $f,g\in k[x,y,z]$ are homogenous) in $\mathbb{P}^2_k$ and show they have non-trivial intersection. By passing to the cones $C(V_+(f))$ and $C(V_+(g))$ in $\mathbb{A}^3_k$, it suffices to show that two hypersufaces of $\mathbb{A}^3_k$ intersecting at the origin, cannot have intersection precisely the origin. 
To see this, we know that our hypersufaces are of the form $V(f)$ and $V(g)$ for some $f,g\in k[x,y,z]$. We may as well assume that $f,g$ are irreducible (why?). Since both curves contain the origin $(f),(g)\subseteq (x,y,z)$. We want then to see why $V(f)\cap V(g)=V(f,g)$ is not precisely the origin or, said differently, why $I(V(f,g))\ne (x,y,z)$. But, any minimal prime containing $(f,g)$ (or equivalently any minimal prime containing $I(V(f,g))=\sqrt{(f,g)}$) must have height at most $2$ by Krull's Height theorem. Thus, in particular, if $(x,y,z)=I(V(f,g))$ then $\text{ht}((x,y,z))\leqslant 2$, which is ridiculous. 
Of course, geometrically, this is describing a bound on the codimension of the intersection of hypersurfaces.
This can be generalized to show that if $X,Y$ are closed, pure dimensional, subvarieties of $\mathbb{P}^n_k$ with $\text{codim}(X)+\text{codim}(Y)\leqslant n$, then $X$ and $Y$ intersect.