Understanding and using Bertini's theorem

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$$|D|$$>$$0$$???

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

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.