First of all I want to say that algebraic geometry is not "my field of research" so I apologize if the notation is not standard.
$S$ is a smooth complex projective surface with a fibration $f$ over $\mathbb P^1(\mathbb C)$. Moreover suppose that the following properties hold for $f$:
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$f$ has singular fibers (a finite number).
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Each singular fiber is a curve with a single knot. (In this case I think that $f$ is called a stable fibration.)
Under the above hypotheses I have to prove (or disprove) that the fibration $f$ can't be isotrivial. Here isotrivial means that any two smooth fibers are isomorphic as algebraic varieties.
Many thanks.
Best Answer
There are three important invariants for any relatively minimal fibration $f:\,S \to B$ from a smooth complex projective surface $S$ to a smooth curve $B$: the self-intersection $\omega_{S/B}^2$, the degree of $f_*\omega_{S/B}$, and the singular index $e_f$ of $f$, where $\omega_{S/B}:=\omega_S\otimes\omega_B^{\vee}$ is the relative canonical sheaf of $f$. Here "relatively minimal" means that there is no exceptional curves (i.e., a curve isomorphic to $\mathbb P^1$ and have self-intersection equals $-1$) contained in fibers of $f$. If $f$ is stable (or semi-stable), $e_f$ is actually equal to the number of nodes contained in the singular fibers of $f$. These three invariants are all non-negative and satisfy the Noether formula: $$12\deg f_*\omega_{S/B}=\omega_{S/B}^2+e_f,$$ and the slope inequality $$\omega_{S/B}^2 \geq \frac{4(g-1)}{g} \deg f_*\omega_{S/B},~\text{equivalently,}~ (8g+4)\deg f_*\omega_{S/B}\geq g e_f,$$ where $g$ is genus of a general fiber of $f$.
An important description of these invariants for semi-stable (or stable) fibrations is as follows. Let $j:B \to \overline M_g$ be the induced morphism from $B$ to compactification of the moduli space $M_g$ of smooth curves of genus $g$. Then there exist divisors $\lambda, \kappa, \delta$ on $\overline M_g$ such that $$\deg f_*\omega_{S/B}=\deg j^*\lambda, \quad \omega_{S/B}^2=\deg j^*\kappa,\quad e_f=\deg j^*\delta.$$
Now by your assumption, $f$ is stable and we have $e_f>0$. On the other hand, if $f$ is isotrivial, then the image of $j$ is a point, hence the pulling-back of any divisor is trivial and so $e_f=\deg j^*\delta=0$, which is a contradiction.
We remark that the condition "$e_f >0$" implies in particular that $g>0$, since the moduli space of genus zero is a single point.
PS: from the arguments, the conclusion is also true for any base $B$, not only for $B=\mathbb P^1$.