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To develop what Jason says: if your curve deforms in a family of rational curves, it means that you can find a dominant rational map from a ruled surface onto your K3. This is forbidden (over $\mathbb …

Put $d:=\deg(X)$. From the exact sequence $$0\rightarrow \mathcal{O}_X(-d)\rightarrow \Omega ^1_{\mathbb{P}^n|X}\rightarrow \Omega ^1_X\rightarrow 0$$you get an exact sequence $\ 0\rightarrow T_X\righ …

This is actually true for all Chern classes, but you must first say how you identify $H^*(X,\mathbb{Z})$ and $H^*(X_t,\mathbb{Z})$. There is no problem for small deformations, that is if $B$ is a ball …

No. Take $X=\mathbb{P}^1\times \mathbb{P}^1$, $Y=\mathbb{P}^1$, $f$ the first projection, $\mathcal{L}=\mathcal{O}_{\mathbb{P}^1}(1)\boxtimes \mathcal{O}_{\mathbb{P}^1}(1)$, $s=X\otimes X'$, where $(X …

This map is injective. There is a Hochschild-Serre spectral sequence with $E^{pq}_2=H^p(\mathrm{Gal}(\bar{K}/K), H^q(X_{\bar{K}},\mathbb{G}_m))$ converging towards $H^{p+q}(X_{K},\mathbb{G}_m)$. This …

The answer to the first question is no. In the moduli space $\mathcal{M}_{10}$ of curves of genus 10, the complete intersections $(3,3)$ in $\Bbb{P}^3$ form a strict subvariety $\mathcal{CI}$. Pick fo …

Not in general. Perhaps the simplest example is given by the Hilbert scheme of curves $C$ of degree 3 in $\mathbb{P}^3$ with $\chi (\mathscr{O}_C)=1$. This has 2 components, one (of dimension 12) corr …

These are two different questions. 1) No. The arithmetic genus of a degree $d$ surface $Y\subset\mathbb{P}^3$ is $\chi (\mathcal{O}_Y)-1=\binom{d-1}{3}$, regardless of the singularities of $Y$. If $ …

First of all, the image of your homomorphism is invariant under the Galois group $G:=\mathrm{Gal}(\bar{K}/K)$. So the right question is to ask whether the induced homomorphism $\mathrm{Pic}(X_{K})\rig …

No. First of all note that your line bundle $\mathcal{N}$ on $Y$ is trivial, so your assertion is $\mathcal{L}\cong\mathcal{M}$. Take a smooth projective curve $C$ of genus $\geq 1$ (say over $\math …

It is an inclusion of analytic spaces — $\Delta $ is not a scheme. If $\mathfrak{A}=(f_1,\ldots ,f_p)$, $\Delta _{\mathfrak{A}}$ is the subspace of $\Delta $ defined by $f_1=\ldots =f_p=0$. I think $ …

Let $f:X\rightarrow \Delta $ be your family. $\ (*)$ implies that $f_*K_{X/\Delta }^{N}$ is a vector bundle on $\Delta $, with fiber $H^0(X_t, K_{X_t}^N)$ at $t\in\Delta$. The canonical homomorphism $ …

The space $H^{0}(X,f_{0}^{*}T_{Y})$ is the tangent space at $f_0$ to the variety $\mathrm{Hom}(X,Y)$, see J. Kollár, Rational curves on algebraic varieties. Thus if it is zero, $f_0$ is an isolated po …

This is true if you assume moreover that $(\mathcal{X}_K)_{red}$ is geometrically reduced -- in particular, in characteristic $0$. First of all, note that set-theoretically $\mathcal{X}'_b=\mathcal{X} …

I think this is not true, at least if $k\geq 6$. The Euler exact sequence pulled back to $\mathbb{P}^1$ is $$0\rightarrow \mathscr{O}_{\mathbb{P}^1}\rightarrow \mathscr{O}_{\mathbb{P}^1}(d)^3\rightarr …