Prove $X^2+Y^2-1$ is irreducible using geometrical tools. I'm trying to understand what is meant in this paragraph:
 of "Conics and Cubics. A Concrete Introduction to Algebraic Curves (by Robert Byx)": 

He wants to prove that the polynomial $X^2+Y^2-1$ is irreducible using geometrical  tools. I have the following doubts:


*

*What he means by Since every line has at least three points on it ?

*Why the fact that $X^2+Y^2-1=F(X,Y)\cdot G(X,Y)$, where F, G are lines implies that the circle $X^2+Y^2-1=0$ is made up of two lines?
I would really grateful if anyone could help me.

If the field $k$ is the finite field of 2 elements then this argument cannot work - there are lines in $\mathbb{F}_2[x,y]$ whose locus includes only two points, e.g. $x+y=1$, and indeed $x^2 + y^2 - 1 = (x+y+1)^2$ over $\mathbb{F}_2$.  So, let's assume $\text{char } k$ is at least 3.
In that case, a line of the form $ax+by = c$ has at least 3 solutions $(x_0,y_0)$.  Since our field $k$ has at least 3 different elements, we can substitute each of these in for x or y (a or b could be 0) and solve for the remaining variable.  This gives us three solutions to the equation $ax+by=c$.
If $x^2 + y^2 - 1 = F(x,y) \cdot G(x,y)$, then any solution $(x_0,y_0)$ to the equation $x^2 + y^2 - 1 = 0$ must also then satisfy $F(x,y) = 0$ or $G(x,y) = 0$.  This means that the circle $x^2 + y^2 - 1 = 0$ is actually a union of two lines. 
