Let $\pi:X \to \mbox{Spec}(\mathbb{C}[t]/(t^2))$ be a smooth, projective morphism and $L$ be an invertible sheaf on $X$. Denote by $L\_0$ the restriction of $L$ to the closed fiber, say $X\_0$ of $\pi$. Suppose that the natural morphism $H^0(L) \to H^0(L\_0)$ is surjective. Can we conclude that $\pi\_\*L$ is flat over $\mbox{Spec}(\mathbb{C}[t]/(t^2))$?



Flatness of direct image sheaf over local artinian ring
This follows easily from the theory of modules over $R = \mathbb C[t]/(t^2)$. Indeed, we have a short exact sequence
$$0 \to H^0(\mathscr L\_0) \to H^0(\mathscr L) \to H^0(\mathscr L\_0) \to 0,$$
induced by the observation that $(t) \cong R/(t)$ and your assumption that the map $H^0(\mathscr L) \to H^0(\mathscr L\_0)$ is surjective. Now we have the following lemma.


**Lemma.** *Let $0 \to N \stackrel f\to M \stackrel g\to N \to 0$ be a short exact sequence of finitely generated $R$-modules, and assume that $fg \colon M \to M$ is multiplication by $t$. Then $M$ is free and $N = tM \cong M/tM$.*


*Proof.* Because $N \stackrel f\to M \stackrel g\to N$ is exact, $M \stackrel g\to N$ is surjective, and $N \stackrel f\to M$ is injective, the sequence
$$M \stackrel{fg} \longrightarrow M \stackrel{fg}\longrightarrow M\tag{1}\label{1}$$
is exact. But we also have $fg = t$. From the structure theory of $R$-modules, we may write
$$M \cong R^m \oplus (R/(t))^n$$
for some $m,n \in \mathbb Z\_{\geq 0}$. If $n > 0$, then any nonzero element in $(R/(t))^n$ is killed by $t$ but not in the image of $t$, contradicting exactness of (\ref{1}). We conclude that $n = 0$, so $M$ is free. The final statement follows since $N = \operatorname{im}(fg) = tM$. $\square$


