Flatness of direct image sheaf over local artinian ring 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 $\mbox{Spec}(\mathbb{C}[t]/(t^2))$?
 A: $\newcommand{\C}{\mathbb{C}}\newcommand{\D}{\C[t]/t^2}\newcommand{\im}{\mathrm{im}\,}$A module $M$ over $\D$ is flat if and only if the inclusion $(t)\subset \D$ remains injective after tensoring with $M$ over $\D$. In other words, the map $M/tM\xrightarrow{t} M$ is injective which is equivalent to saying that the inclusion $\im (M\xrightarrow{t} M)\subset \ker\, (M\xrightarrow{t} M)$ is equality. If $M$ is finite-dimensional as a $\C$-vector space, then this condition is equivalent to $\dim\im t=\dim\ker t=\dim M-\dim \im t$ which can be rephrased as $\dim\im t=\frac{1}{2}\dim M$(for a non-flat module the LHS is strictly smaller).
(All of the above is more or less tautological since any finitely generated module over $\D$ is a direct sum of some number of copies of $\C$ and a few copies of $\D$ and a module is flat iff it is free)
Denote by $i:X_0\to X$ the immersion of the closed fiber. we have an exact sequence of sheaves on $X$ $$0\to L\otimes_{\D} \C\xrightarrow{t} L\to L\otimes_{\D}\C\to 0$$ which can be rewritten as $$0\to i_*L_0\to L\to i_*L_0\to 0$$ Applying $\pi_*$ we get a sequence of $\D$-modules which is in general not necessarily exact on the right, but you're assuming it is
$$0\to\pi_*i_*L_0\to\pi_*L\to\pi_*i_*L_0\to 0$$ Since $\pi$ is projective, $\pi_*L$ is a finitely generated $\D$-module and from the exact sequence we see that $\dim_{\C} \pi_*L=2\dim H^0(L_0)$. A priori, this exact sequence doesn't imply that $\pi_*L$ is flat, but let's recall the meaning of the arrow $\pi_*i_*L_0\to\pi_*L$. Consider multiplication by $t$ as an endomorphism of $L$. Since $L$ is flat over $\D$, it factors as $$L\twoheadrightarrow i_*L_0\hookrightarrow L$$
Taking global sections and using your assumption again, we get $H^0(L_0)\subset\im (\pi_*L\xrightarrow{t}\pi_*L)$ an this implies the desired equality $\dim\im t=\frac{1}{2}\dim \pi_*L$
A: 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 $N = \operatorname{im}(fg) = tM$. $\square$
