Exercise 3.7 Hartshorne

Question

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

Answer 1

Here is an elementary proof using only the part of Hartshorne preceding the exercise on page 21.

Suppose two curves $X,Y\subset \mathbb P^2$ have empty intersection.
Then $Y\subset U:=\mathbb P^2\setminus X$.
However $U$ is affine as can be seen through the $d$-uple embedding of Exercise 2.12, page 13: see the answer here.
But this is absurd because the global regular functions on an affine variety separate its points [restrict the coordinate functions of the $\mathbb A^N$ in which it is embedded], whereas for our projective curve $\mathcal O(Y)=k$: Theorem 3.4(a), page 18.

Answer 2

Not a complete answer.

Let's take the case of two lines $\ell_1,\ell_2\subset\mathbb P^2$. Assume the line $\ell_1$ is given by the linear equation $a_1x+b_1y+c_1z=0$, while $\ell_2$ is given by $a_2x+b_2y+c_2z=0$.

(This is usually the point where one says "we may assume that $b_1=c_1=0$" to simplify the setting. We could, but let's not do that).

Suppose $\ell_1\neq\ell_2$, i.e. the vector $(a_2,b_2,c_2)$ is not a multiple of $(a_1,b_1,c_1)$. Then, if we were in the world of linear algebra (vector spaces), we would know there is a one dimensional vector space $V=(\alpha,\beta,\gamma)\cdot\mathbb k\subset \mathbb k^3$ of common solutions. We are not so far away from that world, as we are in its "projectivized" version! and still we can use what we know from linear algebra: indeed, the projectivization $\mathbb P(V)$ of that one-dimensional vector space is a point $(\alpha:\beta:\gamma)$ in the plane, and that point is exactly the intersection $\ell_1\cap\ell_2$.

For the case of a curve $C$ and a line $\ell$, one could argue as follows: If $C$ were disjoint from $\ell$, then $C$ would be entirely contained in the open affine subset $\mathbb P^2\setminus \ell=\mathbb A^2$. Then $C$ is affine, but since it was projective, it has to be discrete, in particular it cannot have dimension one - contradiction.

I let you handle the general case.

Answer 3

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.