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The answer is yes (in char. 0). Indeed, it suffices to show that for an infinitesimal deformation $\mathcal{X}$ of $X$ over an artinian algebra $A$, the cohomology $H^0(\mathcal{X}, \omega_{\mathcal{X …

The answer is no, even for locally constant families. Let $F$ be a smooth projective variety with an automorphism $\sigma$, and let $L$ be a line bundle on $F$ such that $\sigma^* L$ is not isomorphic …

In fact, there is a natural such structure, namely $$ R \to k \to k[\varepsilon], $$ corresponding to the constant deformation $X\otimes k[\varepsilon]$, and indeed the automorphisms of $X\otimes k[\v …

The answer is no. Easiest nontrivial case: $n=1$, $I_X=(X_1)$, so $X=Spec(k[X_1]/(X_1))=Spec(k)$, take $I_{X'} = (X_1 - t)$, so that $X' = Spec(k[X_1, t]/(X_1 - t, t^2))$. Now you are asking for an is …

EDIT. New version, addressing questions in the comments. (1) Affine and smooth implies what you want. Indeed, suppose $\mathcal{X}$ is smooth and that $H^1(X_0, T_{X_0/\mathbb{C}}) = 0$ where $X_0/ …

The universal property of the total space $$\mathbf{V}(E^\vee) = \operatorname{Spec}_M \operatorname{Sym} E^\vee $$ of a vector bundle (locally free sheaf) $E$ on some scheme $M$ is: giving a map $T\t …

Yes if $X$ is proper. Note that $\det(V)$ is itself a line bundle on $X\times S$, so the question is: given a line bundle $L$ on $X\times S$, with $X$ proper, is the locus of $s\in S$ such that $L_s$ …

Suppose that $X$ has smaller dimension at $x$ than $Y$. Embed $Y\subseteq\mathbb{P}^n$ using a square of a very ample line bundle. For a general linear subspace $L$ in $\mathbb{P}^n$ through $x$ of th …

Posting my comment as an answer: Since $T_{A/S}$ is trivial locally on $S$, we have $T_{A/S} = p^* p_* T_{A/S} = p^* {\rm Lie}_S A$. By the projection formula, we get $$ R^1 p_* T_{A/S} = R^1 p_* p^* …

This is basically a tautology. A tangent vector $v$ to $Def(X)$ at $0$ is a map $D:=Spec(k[t]/(t^2))\to Def(X)$ mapping the closed point to 0, i.e. a family $X_v\to D$, i.e. a first order deformation …

Imagine that $S$ is an open subset of some fine moduli space which does not contain any rational curves. This does happen although I don't know any examples. Then any family over an affine space will …

To say that a deformation is flat, you should assume that $A$ is flat over $R$, e.g. each $A_n$ is a free $R$-module of finite rank. The answer to the main question is no, even if $A$ is commutative. …

I don't think this is always true. Consider the following situation: Let $B=\mathbb{A}^1$ with coordinate $t$ and let $\mathcal{X}$ be the blowup of $\mathbb{A}^2$ at $(0, 0)$, $\pi:\mathcal{X}\to\m …

On $X'$, there is an exact sequence $$ 0 \to \mathcal{O}_X \to \mathcal{O}_{X'} \to \mathcal{O}_X \to 0. $$ Note that this sequence does not carry a lot of information about the deformation, for examp …

My guess: a "deformation preserving dual graph" $\tilde C$ of $C$ over some Artinian local $A$ is a flat lifting $\tilde C/A$ which formally locally looks like ${\rm Spec}(A[x,y]/(xy))$. Some would ma …