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I am a bit confused and wondering where a contradiction is in the following argument. Let's do everything over $\mathbb{C}$.

Let $X\to C$ be an elliptic surface over $\mathbb{P}^1$ with no multiple fibers.

The following facts are well known (c.f. Friedman Algebraic Surfaces and Algebraic Vector Bundles)

  1. The canonical bundle is given by the formula $$K_X = \pi^*(K_C \otimes L)$$ where $L$ is a line bundle of degree $d \ge 0$ in particular it satisfies $K_X^2 = 0$.
  2. Noether's formula implies $$e(X) = 12 \chi(\mathcal O_X) = 12d$$ where $e(X) = c_2(X)$ is the topologial Euler characteristic of $X$.

The Euler characteristic can be computed by understanding the singular fibers using cut and paste since the smooth fibers $F$ have $e(F) = 0$.

On the other hand, it is known (e.g. here) that there are isotrivial ellitpic surfaces over $\mathbb{A}^1 $ with a single singular fiber $F_0$ of type $I_0^*, II, II^*, III, III^*, IV, IV^*$ with $e(F_0) \in \{ 2,3,4,6,8,9,10\}$.

Now construct $X$ by gluing one of these surfaces over $\mathbb{A}^1$ to a trivial smooth elliptic fibration over $\mathbb{A}^1$ along $\mathbb{A}^1 - \{ 0\}$. We get a isotrivial surface over $\mathbb{P}^1$ with $e(X) \in \{2,3,4,6,8,9,10\}$ which is not a multiple of 12.

Where is the error?

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    $\begingroup$ Such a surface you tried to construct over $\mathbb{P}^1$ doesn't exist. When you extend your surface over $\mathbb{A}^1$ to one over $\mathbb{P}^1$, it will necessarily have a singular fiber over $\infty$ and that singular fiber will exactly "balance" the Euler characteristic to get a multiple of $12$. $\endgroup$
    – Dori Bejleri
    Commented Jan 19, 2022 at 20:42
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    $\begingroup$ Ahh, I see now that there is monodromy along the generator of the fundamental group of $\mathbb A^1 - \{ 0 \}$ and so we can't glue to a trivial surface over $\infty$. The fact that the monodromies exactly pair off to allow this is remarkable! $\endgroup$
    – Enclitic Sarcool
    Commented Jan 19, 2022 at 21:00
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    $\begingroup$ Yes that's right. In fact one way to construct such surfaces is to take $\mathbb{P}^1 \times E$ and take a subgroup of $\mathrm{Aut}(E)$ (necessarily cyclic) and consider the action on $\mathbb{P}^1 \times E$ where the cyclic group acts on $\mathbb{P}^1$ with two fixed points $0$ and $\infty$. Then take the quotient, resolve singularities and contract $(-1)$-curves in the fibers. The subgroup of $\mathrm{Aut}(E)$ is exactly the monodromy over $\mathbb{A}^1 \setminus \{0\}$ and it acts by dual weights on the fixed points over $0$ and $\infty$ which leads to the singular fibers coming in pairs. $\endgroup$
    – Dori Bejleri
    Commented Jan 19, 2022 at 21:07

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