I'm trying to learn some algebraic geometry, and I would like to know if the following approach is correct.
I would like to show that a fourth degree curve in $P\mathbb C^2$ with four singular points is reducible. Suppose for the sake of contradiction that the given curve is not reducible.
The intersection multiplicity where a line intersects a singular point is at least 2. So we see by Bezout's theorem that each singular point must be a double point, and no three can be collinear. For example, if three were collinear, the line and the curve would intersect in at least $2\cdot 3$, points, contradicting Bezout's theorem (which says they can intersect in at most $4$ points, counting multiplicity) and showing the line is a component of the curve (and hence the curve is reducible).
Using a projective linear transformation (change of coordinates), we may assume the four points are $[1:0:0]$, $[0:1:0]$, $[0:0:1]$, and $[1:1:1]$.
It is not hard to see that any quadric that passes through these points must have the following form:
$$(-a-b)XY+aXZ+bYZ=0.$$
On the affine piece with $Z\neq 0$, this is $$(-a-b)xy+ax+by=0.$$
By varying $a$ and $b$, we may arrange it so that the tangent line at $0$ is any line we desire. In particular, we may arrange the tangent line at zero so it coincides with a tangent line of the degree $4$ curve at that point. Hence, the intersection multiplicity summed over the four points is at least $$3+2+2+2>4\cdot 2,$$ contradicting Bezout's theorem and showing this quadric must be a component of the curve.
Is this right?
