There is a line in $\mathbb{P^2}$ not passing through any of a finite collection of points

Question

I'm working on Fulton's Algebraic Curves and having a little bit of trouble with an exercise from Chapter 4 on projective varieties (4.25). The question is as follows:

Let $P = [x:y:z] \in \mathbb{P}^2$

(a) Show that $\{(a,b,c) \in \mathbb{A}^3| ax+by+cz = 0\}$ is a a hyperplane in $\mathbb{A}^3$

This part seems easy since we can take the hyperplane $\mathcal{Z}_a(xX+yY+zZ)$ (where $\mathcal{Z}_a$ denotes the affine zero set of a polynomial) and this is visibly equal to the set of points in $\mathbb{A}^3$ with the desired property.

(b) Show that for any finite set of points in $\mathbb{P}^2$, there is a line not passing through any of them.

This is the part I'm confused about. I can see how the defining polynomials of the hyperplanes in $\mathbb{A}^3$ from (a) define lines in $\mathbb{P}^2$ but I don't see how to apply that to this problem. Any help or hints would be appreciated.

Edit: Thanks for the helpful answer and comment. I think Fulton probably means to assume an infinite field of characteristic $0$.

Answer 1

I’m assuming the base field is infinite, otherwise it can be a little trickier to state. This is equivalent to the statement: for any $p_1,\ldots,p_s \in \mathbb{A}^3 \backslash \{0\}$, there are $a,b,c \in k$ such that $(a,b,c) \cdot p_i \neq 0$ for each $i$ (a line of $\mathbb{P}^2$ is a plane of $\mathbb{A}^3$).

Write $p_i=(x_i,y_i,z_i)$. If the statement is false, then it means that the polynomial $P(X,Y,Z)=\prod_{i=1}^s{(x_iX+y_iY+z_iZ)}$ vanishes at every point, so is zero (as $k$ is infinite). This implies that one of its factors is zero, ie that some $p_i$ is zero.

Answer 2

It is sometimes conventient to think of the dual $\check{\mathbb P}^2$, which is a projective plane, whose points $l = [a_0, a_1, a_2] \in \check{\mathbb P}^2$ correspond to lines $$ L = \{[x_0, x_1, x_2] \,|\, a_0x_0 + a_1x_1 + a_2x_2 = 0\} \subset \mathbb P^2.$$ Conversely, a point $p = [x_0, x_1, x_2] \subset \mathbb P^2$ defines a line in the dual space, by $$ P = \{[a_0, a_1, a_2] \,|\, a_0 x_0 + a_1 x_1 + a_2 x_2 = 0\} \subset \check{\mathbb P}^2.$$ This is the set of all lines passing through $p$. So in this way, finitely many points $p_1, \dotsc, p_n$ define finitely many lines $P_1, \dotsc, P_n \subset \check{\mathbb P}^2,$ and any point $$ l \in \check{\mathbb P}^2 \setminus (P_1 \cup P_2 \cup \dotsb \cup P_n)$$ will define a line $L \subset \mathbb P^2$ which misses $p_1, \dotsc, p_n$. If $k$ is infinite of characteristic $0$ (e.g. $k = \mathbb C$), then there are many points $l$ not lying on the finitely many lines $P_1, \dotsc, P_n$.