Let $X$ be a smooth projective surface and $f:X\to\mathbb{P}^1$ be a $\mathbb{P}^1$-fibration with a singular fiber consisting of a tree with three irreducible rational ($-2$)-curves $D\_1$, $D\_2$, $D\_3$ and a ($-1$)-curve $L$ intersecting one of $D\_i$'s. I want to know is it possible to calculate multiplicity of $L$ in the scheme theoretic fiber of this singular fiber? Is it possible to calculate multiplicity of the singular fiber from above information?



Multiplicity of irreducible component of a singular fiber of a $\mathbb{P}^1$-fibration
If we assume that the rational curves intersect in nodes, then we can compute it from the data of the dual graph. Since $L$ intersects only one of the $D\_i$, then the $D\_i$ necessarily form a chain, say $D\_1 \cup D\_2 \cup D\_3$ and without loss of generality, there are two possibilities: 1) $L$ intersects $D\_1$, or 2) $L$ intersects $D\_2$. With that in mind we can compute the multiplicities of the curves in the fiber using the following basic lemma.


**Lemma:** If $F = f^\*p$ is a scheme theoretic fiber, then $F.E = 0$ for all curves $E$ contained in a fiber of $f$.


Then writing our fiber $F = a\_1D\_1 + a\_2D\_2 + a\_3D\_3 + bL$, we have a system of equations $F.D\_i = F.L = 0$. On the other hand, we know the self intersections $D\_i^2 = -2$ and $L^2 = -1$, we know that $D\_1.D\_2 = D\_2.D\_3 = 1$ and $D\_1.D\_3 = 0$, and finally we know $L.D\_i$ in each of the two cases. Using this we can solve completely for the $a\_i$ and $b$ in each of the two cases.


**Edit:** I think the curves actually have to intersect in nodes under the assumption that $X$ is smooth. This is just a question about the reduced fiber, call it $C$, which is a tree of rational curves on a smooth surface $X$. The idea is to compute the contribution of the $\delta$-invariant of the singularity to the arithmetic genus.


Let's suppose that $p \in C$ is a singular point and let $C' \to C$ be the partial normalization at $p$. If $p$ lies on a unique irreducible component of $C$ then $p\_a(C) - p\_a(C') = \delta > 0$ which is impossible as $p\_a(C) = 0$. Otherwise, $p$ is at the intersection of exactly two components of $C$ (by the assumption that $C$ is a tree). Then $C'$ is a disjoint union of two components $A$ and $B$ so $p\_a(C') = p\_a(A) + p\_a(B) - 1 = -1$ and $p\_a(C) - p\_a(C') = \delta$ so we conclude that $\delta = 1$. Then its not hard to see that the node is the only plane curve with $2$ branches and $\delta = 1$.


**Edit 2:** It turns out that in the first case there are no solutions which means that such a configuration cannot appear as the fiber of a genus $0$ fibration on a smooth surface. We can see this geometrically as follows. Since $L$ is a $(-1)$-curve we can contract it to obtain a smooth surface that the fibration factors through, but then $D\_1$ becomes a $(-1)$-curve which can contract to another smooth surface through which the fibration factors. Now $D\_2$ becomes a $(-1)$ curve so after one more contraction we have a smooth surface with a fibration and a fiber that is set theoretically just $D\_3$, but we also have that $D\_3^2 = -1$ (after these three contractions) and this is impossible.


In the second case solving the linear system just yields that $a\_2 = b = 2a\_3$ and $a\_1 = a\_3$ so we still need one more input. This comes from Tsen's theorem which guarantees (at least over an algebraically closed field) that the fibration $f$ has a section. This means that the fiber $F$ contains at least one reduced component. Since the $a\_i$ and $b$ are all positive integers, this gives us the unique solution $a\_1 = a\_3 = 1$ and $a\_2 = b = 2$. We can achieve such a fiber by starting with a fiber that is just $\mathbb{P}^1$, blowing up once, then blowing up again at the new node we produced which gives us a fiber that looks like $D\_1 + 2D\_2 + D\_3$, and finally blowing up a generic point on $D\_2$ to obtain the $D\_1 + 2D\_2 + D\_3 + 2L$ fiber.


