4. Given $f(x)=\sum_{k=0}^{9} x^{k}$. Then the remainder of the polynomial $f\left(x^{1999}\right)$ divided by $f(x)$ is ( ).
(A) 0
(B) 1
(C) $f(x)-x^{9}$
(D) $f(x)-x^{9}-1$
4. (A).
$\because f(x)=\frac{x^{10}-1}{x-1}$,
$\therefore \omega$ is a tenth root of unity of $f(x)=0$.
That is, $\omega^{10}=1$, and $\omega \neq 1$.
Furthermore, $\omega^{n}$ is a root of $f(x)=0$, where $n=1,2, \cdots, 9$.
Also, $f\left(\left(\omega^{n}\right)^{1099}\right)=\frac{\left[\left(\omega^{n}\right)^{1990}\right]^{10}-1}{\left(\omega^{n}\right)^{1990}-1}$
$$
=\frac{\left(\omega^{10}\right)^{1099 n}-1}{\omega^{1999 n}-1}=0 \text {. }
$$

Then $\omega^{n}$ is a root of the equation $f\left(x^{1999}\right)=0$, where $n=1,2$, $\cdots, 9$.
Therefore, $f(x) \mid f\left(x^{1099}\right)$.