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 $X$?
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8$\begingroup$ No. Take for $X$ the quadric cone $xy+z^2=0$ in $\mathbb{C}^3$, and for $D$ the divisor $x=0$. Then $D$ is Cartier, but $D_{\operatorname{red}}$ is not. $\endgroup$– abxCommented May 12, 2022 at 10:13
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2$\begingroup$ This is true if $X$ is locally factorial (because any prime ideal of height one is principal in a UFD, see Th. 1.12A in Chap. I of Hartshorne). $\endgroup$– Damian RösslerCommented May 12, 2022 at 11:47
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$\begingroup$ @DamianRössler Thanks for the answer in the positive direction. abx: Thanks $\endgroup$– user45397Commented May 12, 2022 at 18:50
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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).
