# base locus under blowup in a threefold

Let $X$ be a smooth complete (not-necessarily projective) complex algebraic threefold, and $D$ an effective divisor with one dimensional (stable?) base locus $Bs(|D|)=Z$, i.e. $D$ is base point free off a finite union of curves. Let $\pi: X' \rightarrow X$ be the blow up of $X$ along a smooth curve $C_0 \nsubseteq Z$ that is also not contained $Supp(D)$, with exceptional divisor $E$. Write $C_0 \cap Z = \{ x_1, \ldots, x_n \}$.

Then $\pi^{-1}(Z)=Z' \cup \bigcup l_j$, where $Z'$ is the strict transform of $Z$ and $l_j=\pi^{-1}(x_j)$. Is $mD - E$ base point free off of $Z'$ for $m>>0$, i.e. do we have $Bs(|mD-E|) \subset Z'$?

• Does D in the second paragraph now denote the pullback to X' of the original D? – user5117 Jun 20 '11 at 9:48

If you keep all of your assumptions, I don't think this will hold. Let $D'$ denote the pull-back of $D$ to $X'$. (I am assuming this is what you meant).

My first instinct was that I do not see why $mD'-E$ would even have a non-empty linear system. I guess that could actually happen, but generally it would not.

I think the following is a reasonable setup for the situation you're describing. Let $X$ be an arbitrary (smooth projective) threefold and $D\subseteq X$ an irreducible effective divisor such that $\dim |mD|=1$ and $D\neq\mathrm{Bs}(|mD|)\neq\emptyset$ for all $m>0$. Then in particular $\mathrm{Bs}(|mD|)=Z_m$ is a curve for all $m>0$.

Now let $C_0\not\subseteq Z_m$ be a smooth curve. Notice that there is at most $1$ member $D_0\in|mD|$ such that $C_0\subseteq \mathrm{supp}(D_0)$. This is because any two members generate the linear system, so their intersection is equal to $Z_m$.

On the other hand, if $D'_0\sim mD'-E$ is effective, then the corresponding $D_0\in|mD|$ satisfies that $C_0\subseteq \mathrm{supp}(D_0)$. Here is why: Consider the short exact sequence: $$0\to \mathscr O_{X'}(mD'-E)\to \mathscr O_{X'}(mD') \to \mathscr O_{E}(mD') \to 0$$ and observe that if $D'_0\sim mD'-E$ is effective, then there exists a section in $$H^0(X',\mathscr O_{X'}(mD')) = H^0(X',\pi^*\mathscr O_{X}(mD))= H^0(X,\mathscr O_{X}(mD))$$ whose restriction to $E$ is zero, in other words, a divisor in the linear system $|mD|$ that contains $C_0$.

So, this means that $|mD'-E|\neq\emptyset$ implies the existence of such a divisor, but we had observed that there can be at most one such so it follows that $\dim |mD'-E|\leq 0$. But then its base locus is at least $2$-dimensional.

And now I've just realized that this could have been done much simpler.

(Proof II)

The original condition on the linear systems is the same as to say that $\dim H^0(X,\mathscr O_{X}(mD))=2$ and the observation that $C_0\not\subseteq Z_m$ implies that not all of these sections can restrict to $0$ on $E$ means that the map $H^0(X',\mathscr O_{X'}(mD'-E))\to H^0(X',\mathscr O_{X'}(mD'))$ cannot be surjective, so $\dim H^0(X',\mathscr O_{X'}(mD'-E))\leq 1$ and hence the base locus of $|mD'-E|$ contains a divisor.