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Let $X$ be a smooth scheme over $k[t]/(t^2),$ where $k$ is a field of characteristic 0 (the case when $X$ is a projective curve is already interesting). Let $X_{0} \to X$ denote $X$ with the reduced induced structure, i.e. just the fiber product of $X$ with $k$ over $k[t]/(t^2).$ Let $\beta \in H^{0}(X_{0},\Omega^{1} _{X_{0}/k}).$

Is there a smooth $Y_0/k$ and a map $f:X\to Y_{0}$ as schemes over $k$ and an element $\alpha \in H^{0}(Y_{0},\Omega^{1} _{X_{0}/k}) $ such that $f^* \alpha =\beta. $

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    $\begingroup$ There is no $k[t]/(t^2)$-morphism from $X$ to a $k$-scheme $Y_0$, unless $X$ is empty. Indeed, $t$ vanishes in $\mathscr{O}_{Y_0}$ but not in $\mathscr{O}_X$. $\endgroup$
    – Laurent Moret-Bailly
    Commented Dec 9, 2018 at 8:19
  • $\begingroup$ Thanks very much for the remark, I am editing the original post. I mean map of $k$-schemes. Sorry for the confusion. $\endgroup$
    – dererumnatura
    Commented Dec 9, 2018 at 8:26
  • $\begingroup$ You mean f is defined from X_0, not X, right? $\endgroup$
    – user148212
    Commented Dec 9, 2018 at 9:02
  • $\begingroup$ No $f$ is defined from $X.$ To be precise let $f_0$ be the map from $X_0 \to Y_0$ which is the composition of the inclusion and $f$ then I require $f_0 ^* \alpha=\beta$ $\endgroup$
    – dererumnatura
    Commented Dec 9, 2018 at 9:11

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Welcome new contributor. Your post still has typos, so it is not completely clear what you mean. However, my best guess of your meaning has a negative answer. The simplest examples have $X_0$ a smooth projective curve of genus $g=2$ or a non-hyperelliptic curve of genus $g>2$. This follows from the proof (rather than the statement) of the infinitesimal Torelli theorem for nonhyperelliptic, smooth, projective curves $X_0$ of genus $g> 2$.

Denote by $D$ the $k$-algebra $k[t]/\langle t^2\rangle$. Thus, the $k$-vector space $\Omega_{D/k}$ is $1$-dimensional. The sheaf of relative differentials $\Omega_{X/k}$ is a coherent $\mathcal{O}_X$-module. By functoriality / transitivity for differentials, for every $k$-morphism $f:X\to Y_0$, there are induced morphisms of $k$-vector spaces, $$H^0(Y_0,\Omega_{Y_0/k}) \to H^0(X,\Omega_{X/k}) \to H^0(X_0,\Omega_{X_0/k}).$$ Thus, if there exists $f$ and $\beta$ as you state, then there exists a section $f^*\beta\in H^0(X,\Omega_{X/k})$ such that the image in $H^0(X_0,\Omega_{X_0/k})$ equals $\alpha$. To understand this better, consider the short exact sequence of locally free $\mathcal{O}_{X_0}$-modules, $$0 \to \Omega_{D/k}\otimes_k \mathcal{O}_{X_0} \to \Omega_{X/k}\otimes_D (D/tD) \to \Omega_{X_0/k} \to 0.$$ The long exact sequence of cohomology gives a connecting map $$ \alpha(\xi):H^0(X_0,\Omega_{X_0/k}) \to \Omega_{D/k}\otimes_k H^1(X_0,\mathcal{O}_{X_0}). $$

Now assume that the Kuranishi map of $X/S$ is nonzero, $$\kappa:k \to \text{Ext}^1_{\mathcal{O}_{X_0}}(\Omega_{X_0/k},\mathcal{O}_{X_0}).$$ Also assume that the Torelli map is an immersion at $X_0$. This holds if $X_0$ has genus $g=2$ or if $X_0$ is a non-hyperelliptic curve of genus $g>2$. Then it follows by the proof of the infinitesimal Torelli theorem that there exists $\alpha\in H^0(X_0,\Omega_{X_0/k})$ that does not lift to a global section in $H^0(X,\Omega_{X/S})$. For instance, confer Corollary 1.9, p. 8 of the following.

Chris Peters.
Lectures on Torelli Theorems. Spring School Rennes 2014.
https://www.lebesgue.fr/sites/default/files/attach/hodge-peters.pdf

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  • $\begingroup$ This is great, thanks very much for the answer! $\endgroup$
    – dererumnatura
    Commented Dec 9, 2018 at 17:59
  • $\begingroup$ You are welcome. $\endgroup$
    – Jason Starr
    Commented Dec 9, 2018 at 18:10

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