Could anyone explain why singular cubic has only one singularity? I'm reading Klaus Hulek's algebraic geometry and have encountered this. It is the page 124 of the book and C means irreducible singular cubic plane curve.

Here the sentece says if C has more than two singularities, the line through two singular points intersect C in at least 4 points. Why is it so? Could anyone explain?
 A: The explanation is already written in that text. The intersection multiplicity of the curve with a line passing through a singular point is larger than one ($\geq 2$). If you take a line passing through both singularities (assuming more than one) then you have $\geq 4$ multiplicity. But for a cubic we must have $\leq 3$.
The counting of $4$ is including multiplicity. So the intersection of $y=x^3$ and $y=0$ is one point (but $3$ counting multiplicity).
A: This is appealing, I'd imagine, to what is often times called 'Bezout's theorem'. It says that for two projective plane curves $C,C'\subseteq\mathbb{P}^2$ of degrees $d$ and $d'$, then their intersection has 'size' $dd'$. Here 'size' means number of intersection points, counted with multiplicities. 
Now, for a cubic curve $C$ and a line $\ell$, the above says that $C\cap\ell$ has 'size' $3$. Now, suppose that $C$ has two singular points, $p,p'$. This means that at each point, $C$ can't be 'locally generated' by a single function. If the line intersected $C$ with multiplicity $1$ at either $p$ or $p'$, then the equation of the line would generate $C$ locally, contradicting that the point is singular. So, $\ell$ has to intersect each $p$ and $p'$ with multiplicity greater than $1$. But, this means then that $C$ and $\ell$ have intersect with size at least $2+2=4$, this contradicts Bezout's theorem.
More rigorously, if you know the language, here's what's happening. The intersection multiplicity of two curves $C$ and $D$ in $\mathbb{P}^2$ is defined to be 
$$\langle C,D\rangle=\sum_{x\in C\cap D}m_x(C,D)$$
where 
$$m_x(C,D)=\ell_{\mathcal{O}_{\mathbb{P}^2,x}}\mathcal{O}_{\mathbb{P}^2,x}/(f,g)$$
where $f,g$ are equations cutting $C$ and $D$ out respectively at $p$. 
Bezout's theorem then says that if $C$ and $D$ are degrees $d$ and $d'$ respectively (meaning they are cut out by polynomials of that degree) that
$$\langle C,D\rangle=dd'$$ 
Now, we say that $C$ and $D$ intersect transversely at $x$ if $m_x(C,D)=1$. It's clear then that $C$ and $D$ must be non-singular at $x$ since then the maximal ideals of $\mathcal{O}_{C,x}$ and $\mathcal{O}_{D,x}$ are principle. 
So, assume that $C$ is a degree $3$ curve with two singularities, $p$ and $p'$. Let $\ell\subseteq\mathbb{P}^2$ be the line connecting these two points. Then, since $\ell$ is degree $1$, Bezout's lemma implies that $\langle C,\ell\rangle=3\cdot 1=3$. That said, since $p$ and $p'$ were assumed singular points of $C$, $C$ and $\ell$ can't meet transversely at $p$ or $p'$. So
$$3=\langle C,C'\rangle\geqslant m_p(C,\ell)+m_{p'}(C,\ell)\geqslant 2+2=4$$
which is a contradiction.
