Let $X$ be a normal integral variety over $\mathbb{C}$ and $D \subset X$ be a Cartier divisor in $X$. Is the associated reduced scheme $D\_{\mathrm{red}}$ also necessarily a Cartier divisor in $X$?



Is the reduced scheme associated to a Cartier divisor always Cartier?
No. Consider, for instance, the quadratic cone
$$
X = \{xz - y^2 = 0\} \subset \mathbb{A}^3
$$
and the double line
$$
D = X \cap \{x = 0\} = \{x = y^2 = 0\}
$$
on $X$. Then $D$ is a Cartier divisor, but
$$
D\_{\mathrm{red}} = \{x = y = 0\}
$$
is not Cartier (this is the simplest example of a Weil divisor which is not Cartier).


