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For an effective divisor $D = \Sigma n_x \overline{\{x\}}$, we consider it as a subscheme $\operatorname{supp}D = \cup_{n_x \neq 0}\overline{\{x\}} $ with the structure sheaf $\mathscr{O}_X / \mathscr{O}_X(-D)$.

This is irreducible iff $n_x \neq 0 $ for only one $x$. So for example $D = nY$ is irreducible curve. ($Y$ is a prime Weil divisor.) For this, $D$ (as a scheme) is not reduced unless $n = 1$, since for $ x \in D$, $\mathscr{O}_{D,x} = \mathscr{O}_{X,x}/\mathfrak{m}_{x}^n$.

So I wonder that V 1.2. of Hartshorne's Algebraic Geometry does not work, since Bertini's theorem requires, in particular, reducedness.

Is this wrong?

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    $\begingroup$ Some people take curves to be reduced, which makes sense if you want to say that a curve is a variety of dimension one, e.g., but there is no general agreement. One always has to check the precise assumptions. $\endgroup$ – Ben Apr 10 '18 at 7:47
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As I wrote in the comment, there is no general agreement whether a curve is reduced or not. Furthermore, there is not even an agreement whether a curve is assumed to be irreducible or not. In fact, in Hartshorne's Algebraic geometry, the definition of curve on p. 105 says it is an abstract variety of dimension one, implying that it is integral, hence reduced and irreducible. However, in later chapters, most notably chapter V, curves (on surfaces) might be reducible (by default, it seems). I cannot tell if this is mentioned somewhere explicitly; neither can I say for sure whether all curves in this book are really reduced. But from the statement of V 1.2, we can infer that the curves in question (explicitly assumed to be irreducible) are assumed to be reduced, for otherwise no curve will intersect them smoothly.

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  • $\begingroup$ Thank you so much! $\endgroup$ – agababibu Apr 10 '18 at 17:11

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