I have tried to solve the question below from Igor R. Shafarevich's Basic Algebraic Geometry I
Prove that if a plane curve of degree 3 has three singular points then it breaks up as a union of 3 lines.
I tried to solve this using the dimension of the tangent space of this plane curve since the singular points have a higher dimension than the minimum dimension of the tangent space of the curve. I connected two of three singular points to make lines. However, I do not know how to go further from here.
You have the right idea that you need to connect the singular points by lines. If you can show that the plane curve must contain each of these 3 lines, then because the degree of the curve is 3, the polynomial for the curve must decompose as the product of 3 linear terms. Then the plane curve is just the union of those 3 lines.
Do you think you can show that the line connecting a pair of the singular points is contained in the plane curve?