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I want to use Bertini's theorem for a linear system $|D|$ in a smooth projective variety $X$. I know the definitions that are necessary to understand what Bertini's theorem says, for a less general case as I have described above. However, I have read Bertini's theorem in "Principles of Algebraic Geometry-Phillips Griffiths and Joe Harris", where it is written as follows:

Bertini's Theorem: The generic element of a linear system is smooth away from the base locus of the system.

Seeking an alternative, perhaps more detailed, statement I found:

Theorem (Bertini): Let $X$ be a smooth complex variety and let $|D|$ be a positive dimensional linear system on $X$. Then the general element of $|D|$ is smooth away from the base locus $B_{|D|}$. That is, the set $$ \{H \in |D| \,|\, D_{H} \,\text {is smooth away from} \,B_{D} \} $$ is a Zariski dense open subset of |D|.

Now comes my stupid question, which means: $|D|$ be a positive dimensional linear system?

Linear system, ok! But, positive dimensional, would simply be dim$|D|$>$0$???

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    $\begingroup$ Yes, you need at least two generators for $|D|$. $\endgroup$
    – Alan Muniz
    Nov 5, 2018 at 20:18
  • $\begingroup$ Koé, @AlanMuniz! Valeu mlk! $\endgroup$
    – Manoel
    Nov 5, 2018 at 22:17

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You should definitely read Chapter II, Section 7 of Hartshorne's Algebraic Geometry (here). It gives you the definition of the dimension of a linear system (page 157). Briefly: a linear system corresponds to a sub-vector space of the global sections of some invertible sheaf; the non-zero global sections of an invertible sheaf modulo $k^*$ can be viewed as the set of closed points of a projective space, and so a linear system corresponds to a linear projective variety in this projective space, and its dimension is the dimension of this projective variety.

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