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Take $X$ to be $\mathbb P^2$ blown up at three $R$-points which are colinear on the special fiber and not on the generic, and take $L$ to be $\mathcal O(2)$ minus the three exceptional divisors. The l …

Because the symplectic form on $H_1(C_t, \mathbb Z)$ is a perfect pairing, it suffices to check that there is a group homomorphism $H_1(C_t,\mathbb Z) \to \mathbb Z$ that sends $ \gamma$ to $1$, which …

You need a characteristic zero assumption, and you need some additional axioms on the homorphism to make it a bijection. The map from derivations $D$ to homomorphisms sends a function $y \in \mathcal …

Claim: Given a representation from the Galois group to $GL_2(\mathbb F_q)$ (maybe $p>2$ to be safe) whose image contains $SL_2(\mathbb F_q)$, if $R$ is any quotient of the deformation ring of that rep …

The Hilbert scheme is universal for flat families of closed subschemes of $\mathbb P^{n-1}$. So if $\mathcal X_1(U) \cap H$ is flat over $U$, then we obtain a map $U \to Hilb( \mathbb P^{n-1})$ by the …

I think the answer is "basically never". Let $B$ be the simplest kind of blowup, that is, let $N$ be a single smooth point. Take a map from $M$ to some Hilbert scheme and a map from $E$ to some Hilber …

Since every degree $d$ hypersurface in $\mathbb P^3$ has the same Hilbert polynomial, $Q(n)=\left(\begin{array}{c} n+3 \\ 3 \end{array}\right) - \left(\begin{array}{c} n+3-d \\ 3 \end{array}\right)$, …

We can choose $d_1,g_1,d_2,g_2,d$ such that every pair of curves in $H$ contained in a single degree $d$ surface must intersect. The most well-behaved hypersurface that is not a plane is of course a q …

The normal definition of a lattice can be stated as: A lattice in a $\mathbb R$-module $M$ is a finitely generated $\mathbb Z$-submodule $L \subset M$ such that $\mathbb R \cdot L = M$, Now, $\mathb …

Anything with the same Hilbert polynomial as a globally complete intersection. This is true because two fibers of the same flat projective family have the same Hilbert polynomial, and because the Hilb …