Homogenous polynomials with disjoint zero sets

Question

Let $P, Q:\Bbb C^n \to \Bbb C$ be two homogenous polynomials of the same degree $d$. Denote their zero-sets in $\Bbb{CP}^{n-1}$ by $N(P), N(Q)$. My question is:

If $n≥3$ is the intersection $N(P)\cap N(Q)$ always non-empty?

For $n=2$ one can find $P(z_1,z_2)=z_1$, $Q(z_1,z_2)=z_2$ as simple counter-examples. Every other example I can find (like $P=z_1^2+z_1z_2$, $Q=(z_1-z_2)^2$) appears to be a variation of this example. This strategy does not seem extend to $n≥3$.

Answer 1

Here’s a somewhat more elementary argument.

Assume the intersection is empty, for a given $n \geq 1$. Then the zero locus $V(P,Q) \subset \mathbb{C}^n$ is $\{0\}$. By the Nullstellensatz, it means that $(x_1,\ldots,x_n)=\sqrt{V(P,Q)}$.

Let $R$ be the Noetherian local ring $\mathbb{C}[x_1,\ldots,x_n]_{(x_1,\ldots,x_n)}$, $m$ its maximal ideal, $I$ the ideal generated by $P,Q$.

If you know dimension theory, you can easily conclude – $R$ has an ideal of definition with two generators so $n=\dim{R} \leq 2$. What follows is an elementary version of this argument.

Now, as $I$ has two generators, for any $l \geq 1$, $I^l/I^{l+1}$ is an $R/I$-module with $l+1$ generators, so that the $R$-length of $I^l/I^{l+1}$ is at most $t(l+1)$, where $t$ is the $R$-length of $R/I$.

This means that the $R$-length of $R/I^l$ is $O(l^2)$, and in particular that the $R$-length of $R/m^l$ is $O(l^2)$.

But the $R$-length of $m^l/m^{l+1}$ is the dimension over $\mathbb{C}$ of the space of homogeneous polynomials of degree $l$ in $n$ variables, so that the $R$-length of $R/m^l$ is the dimension over $\mathbb{C}$ of the space of polynomials in $n$ variables of total degree less than $l$, which is $\binom{n+l-1}{l-1}=\binom{n+l-1}{n}$, ie a polynomial in $l$ of degree $n$. It follows therefore that $n\leq 2$.

Answer 2

By Bézout's theorem, the intersection $n-1$ hypersurfaces in $\mathbb C\mathbb P^{n-1}$ (i.e. zero sets of $n-1$ homogeneous polynomials in $n$ variables) is never empty.

Moreover, 2 hypersurfaces in $\mathbb C\mathbb P^{n-1}$ ($n\geq 4$) meet in infinitely many points, since they cut out an algebraic subvariety of codimension at most 2.