Let $X$ be a smooth, affine complex surface, and $M$ be a coherent $\mathcal{O}_X$module. Denote by $D:=\mbox{Spec}(\mathbb{C}[t]/(t^2))$ and $X_D:=X \times_{\mathbb{C}} D$, the trivial deformation of $X$. Let $$0 \to M \to M' \to M \to 0$$ be a short exact sequence of $\mathcal{O}_X$modules. Under the natural morphism $$ \mathcal{O}_{X_D} \to \mathcal{O}_X,$$ we can consider any $\mathcal{O}_X$module as an $\mathcal{O}_{X_D}$module. My question is: Is $M'$ going to be $D$flat, considered as an $\mathcal{O}_{X_D}$module?
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$\begingroup$ Your formulation is unclear. Is the action of $t$ on $M'$ intended to be the composition of the epimorphism $M'\to M$ and the monomorphism $M\to M'$? $\endgroup$ – Jason Starr Feb 5 at 13:20

2$\begingroup$ As worded I think Chen is not asking for that, and so the answer is no. $\endgroup$ – Phil Tosteson Feb 5 at 14:41

3$\begingroup$ @PhilTosteson. I agree with you. As formulated, the OP seems to be asking whether $M'$, considered as an $\mathcal{O}_X$module, is then flat when considered as an $\mathcal{O}_{X_D}$module, and the answer to that question is "no". However, if the OP changes the $\mathcal{O}_{X_D}$module structure as in my comment, then $M'$ is flat as an $\mathcal{O}_{X_D}$module. $\endgroup$ – Jason Starr Feb 5 at 14:50

$\begingroup$ @JasonStarr and Phil: Thank you for the comments. I was considering $M'$ as an $\mathcal{O}_X$module. $\endgroup$ – Chen Feb 6 at 20:59