About a theorem in Beauville's book.

Question

Look at the following theorem (due to Castelnuovo) taken from Beauville's book on complex surfaces:

I have a question about the behavior of $f$ on the exceptional curves generated by $\eta$.

Suppose that $E$ is an exceptional curve coming from a blow up of $\eta$ at a point $p\in S$, then what is $f(E)$? I think that $f(E)$ can't be a point, otherwise I could define $\phi(p):=f(E)$. This is a contradiction with the definition of rational map. Hence I conclude that $f(E)$ is a curve on $X$. Is my reasoning correct?

Edit: Maybe the above reasoning is wrong, but at least I'd like to know which are the exceptional curves (coming from blow-ups) that are not contracted by $f$ (if there is any). I'm sure that I've read somewhere the following thing:

The exceptional curve of the last blow-up can't be contracted by $f$.

Can you explain this sentence?

Answer 1

There is a little bit of ambiguity in your question: are you asking this for any $\eta$ that satisfies the conclusion of the theorem, or specifically for the one that is constructed in the proof?

The first interpretation is silly: for any $S \dashrightarrow X$, given one $(S',\eta,f)$ satisfying the conclusion, we could just do another 'unnecessary' blow-up. Then clearly the exceptional divisor of the last blow-up maps to a single point.

So let's assume that we chose a 'minimal' such $S'$ (whatever that means). Then your question is answered by Remark II.13 (2):

Remark II.13 (2). The blow-up $\epsilon \colon \hat S \to S$ at a point $p$ also has a universal property 'in the other direction': every morphism $f$ from $\hat S$ to a variety $X$ that contracts $E$ to a point factors through $S$.

Since his explanation is a bit minimalistic, I will expand a little bit on this. In fact it's easier to prove the statement in the bigger generality that $S$ is not necessarily projective (but we do need smoothness!). Write $f(E) = q \in X$.

Remark. Note that such a factorisation, if it exists, is unique (for example, use Hartshorne Exercise II.4.2, but that's overkill).

Reduction 1. Reduce to the case where $X$ is affine.

Let $V = S\setminus\{p\}$; thus, we automatically get a map on $V$. Take an affine open neighbourhood $W$ of $q$, and let $U$ be an open neighbourhood of $p \in S$ such that $U\setminus \{p\}$ maps into $W$. If we can construct a map $U \to W$ factoring $f$, then the maps on $U$ and $V$ automatically agree on $U \cap V$ (by the uniqueness remark above), so they glue to $S$.

(The above step is why I dropped the projectivity assumption.)

Reduction 2. Reduce to $X = \mathbb A^n$.

Any affine $X$ is a closed subvariety of $\mathbb A^n$. Moreover, if the image of $S\setminus\{p\}$ lands in $X$, then so does $p$, by (Zariski) continuity.

Reduction 3. Reduce to $X = \mathbb A^1$.

Giving a morphism $Y \to \mathbb A^n$ is the same thing as giving $n$ morphisms $Y \to \mathbb A^1$ (by the universal property of the product).

And finally:

Proof for $X = \mathbb A^1$. As above, let $V = S \setminus \{p\}$. For any affine open $U \subseteq S$, the restriction map $$\mathcal O_S(U) \to \mathcal O_S(U \cap V)$$ is an isomorphism, by "Hartog's theorem" [Stacks project, Tag 031T, (2)] (this is where we use smoothness). Hence it is true for all opens, so taking global sections gives $$\Gamma(S, \mathcal O_S) = \Gamma(V, \mathcal O_V).$$ The result now follows for $\mathbb A^1$ since elements of $\Gamma(Y,\mathcal O_Y)$ correspond to maps $Y \to \mathbb A^1$: the map $V \to \mathbb A^1$ that we assume given extends (uniquely) to a map $S \to \mathbb A^1$.

Conclusion. If $S' \to S$ is obtained by $n$ blow-ups $S_i \to S_{i-1}$ such that there exists no morphism $S_{n-1} \to X$ extending $\phi$, then the exceptional divisor of the last blow-up does not map to a point in $X$. Thus, it has to map onto a (possibly singular) (complete) curve.

It's probably a good exercise to think about what the image of the exceptional divisor is in each of the examples that Beauville gives in II.14.

Another interesting exercise: it's not so clear where we used the assumption that $f(E)$ is a point! Try to spot which step goes wrong if the image is a curve.

Answer 2

Let think of a rational map $\phi: S\dashrightarrow X$, and $\phi$ is not defined at an isolated point $p\in X$. What happened such that $\phi$ can not be extend to p? there must be 2 points near p such that their images are far away to each other. Otherwise by continuity, the map can be extended. So take a neighborhood of p, the image of this ball(removing its centre p) will be quite stretched in X.

Let's seem a concrete example: $\phi : \mathbb{C}^2 \longrightarrow S^3$, where $S^3$ is the unite 3-sphere $\phi(r_1e^{i\theta_1},r_2e^{i\theta_2})=(e^{i\theta_1}, e^{i\theta_2})$ is not defined at (0,0). However any small neigborhood of origin (removing the origin) has image equals the whole $S^3$.

If you blow up the origin, then the image of exceptional curve can not be any point. because any point infinitely near the origin is also infinitely near each point on the exceptional divisor. If you want the rational map has an analytical continuation to the exceptional divisor, then you must let the image of the exceptional curve as stretched as the image of the neigborhood (removing the origin).