Reducible cubic surface are always singular.

Question

I want to prove that

Any reducible cubic surface are always singular.

A possible way may be to take a look at the intersection of the irreducible components. But I don't know how. Thanks for any help.

Answer 1

More generally, any reducible variety is either disconnected or singular (or both!). This is because if there are two components that intersect nontrivially, a point in the intersection would not have its local ring be a domain, much less a regular local ring.

Now you have a reducible cubic surface which is necessarily not disconnected because two hypersurfaces in $\mathbb{P}^3$ necessarily intersect in at least a point. Hence a reducible cubic surface is singular.