On a surface $X$, does the word “an irreducible curve on X” implies the reducedness?

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?

• 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. – Ben Apr 10 '18 at 7:47