4. If $a$, $b$, and $c$ are all non-zero real numbers, and $a+b+c = abc = a^3$, then the minimum value of $ab + bc + ca$ is ( ).
(A) 6
(B) 8
(C) 9
(D) 13
4.C.

From the given conditions, we have
$$
\begin{array}{l}
b+c=a b c-a=a(b c-1)=a\left(a^{2}-1\right), \\
b c=a^{2} .
\end{array}
$$

Therefore, $b$ and $c$ are the two roots of the equation $x^{2}-\left(a^{3}-a\right) x+a^{2}=0$.
Since $a$, $b$, and $c$ are real numbers, we have
$$
\Delta=\left(a^{3}-a\right)^{2}-4 a^{2} \geqslant 0,
$$

which simplifies to $a^{2}\left(a^{2}+1\right)\left(a^{2}-3\right) \geqslant 0$.
Given $a \neq 0$, then $a^{2}>0$ and $a^{2}+1>0$.
Thus, $a^{2}-3 \geqslant 0$, which means $a^{2} \geqslant 3$.
Therefore, $a b+b c+c a=a(b+c)+b c$
$$
=a\left(a^{3}-a\right)+a^{2}=a^{4}=\left(a^{2}\right)^{2} \geqslant 9 .
$$