I think that Grothendieck had already observed that the map from rational points to sections is injective (for curves of genus at least 2 over a number field) and I believe that his proof works even ...

Yes. Suslin has proved that every stably trivial vector bundle of rank $n$ on an affine variety of dimension $n$ over an algebraically closed field is trivial. See: Suslin, A. A. Stably free modules. ...

Let $E$ be an elliptic curve. The moduli space $M_E$ of line bundles with a connection on $E$ is an $\mathbb{A}^1$ bundle over $Pic^0(E) \cong E$. In particular, $E$ can be recoved from $M_E$ as the ...

No, one cannot find the Hodge numbers this way. For an example, consider $X_0$ the Kummer surface associated to a product of supersingular elliptic curves $E_1$ and $E_2$. Recall that this is the ...

Here is an example showing the answer is no: Start with $Z =\mathbb{P}^2_R$, $R$ an arbitrary dvr. Let $P$ be a section of $Z \to Spec(R)$ and let $W$ be the blowup of $Z$ along the image of the ...
I don't think the assumption of characteristic zero simplifies the proof a great deal. However, it does allow us to avoid having to give a more involved proof for the $p$-power torsion (where $p = ... View answer Accepted answer 15 votes See Proposition 1 on p.231 of Cassels and Frohlich for a proof of the claim in the textbook: The point is that if such an$L$exists then$K_1L$is abelian and unramified over$L$so it is contained ... View answer 14 votes No, toric surfaces can have only finitely many$(-1)$-curves. Since$(-1)$-curves are rigid, i.e., cannot form a non-trivial family, it follows that any$(-1)$-curve must be contained in the ... View answer Accepted answer 14 votes Your definition of normal crossings divisor is, as you say, often called a strict normal crossings divisor. People who use this terminology allow normal crossings divisors to have components which are ... View answer 14 votes It follows from Theorem 6.12 of Borel and Harsh-Chandra, "Arithmetic subgroups of algebraic groups", that$G(\mathbb{Z})$is finitely generated if$G$is affine. Perhaps one can combine this with ... View answer 14 votes For any$g >0$there exists an abelian surface$A$containing a smooth curve$C$of genus$g$; the surface can be assumed to be simple if$g > 1$: For$g=1$, one can just consider$C \times C$.... View answer Accepted answer 14 votes You can probably just take Y to be the spectrum of a valuation ring A which is not a DVR, for example the integral closure of C[[t]] in an algebraic closure of C((t)). In this case any homomorphism ... View answer Accepted answer 13 votes I think the notation might be explained by the explicit construction of the twisted affine Lie algebras as fixed points of automorphisms of the untwisted ones: the$r$indicates the order of the ... View answer Accepted answer 13 votes The induced morphism of Hodge structures for any map of varieties will be an isomorphism if and only if the induced map on cohomology (forgetting the Hodge structure) is an isomorphism. This follows ... View answer Accepted answer 12 votes Yes, this is true. One implication is immediate: if$X$has a$k((t))$point then by the valuative criterion of properness there is a map$Spec(k[[t]])$to any compactification of$X$, so the image ... View answer Accepted answer 12 votes As you suspect, the answer to your first question is yes. A. Roĭtman in "Rational equivalence of zero-dimensional cycles". Math. USSR-Sb. 18 (1974), 571--588, generalised Mumford's theorem to show ... View answer Accepted answer 12 votes Yes. By fixing the three points$\{0,1,\infty\}$one sees that$M_{0,n}$is isomorphic to an open subscheme of$\mathbb{A}^{n-3}$which has trivial Picard group. Since it is smooth, the Picard group ... View answer Accepted answer 12 votes Yes. One may also replace$\mathbb{F}_p$by any perfect field$F$. The reason is that any non-perfect algebraic extension$L$of$k$has a unique inseparable extension of degree$p$, i.e.$L^{1/p}$, ... View answer Accepted answer 12 votes There is an example due to Hirokado of a Calabi-Yau threefold in characteristic 3 with third Betti number zero which implies that it cannot be lifted to characteristic zero. See: Hirokado, Masayuki -... View answer 12 votes Jouanolou's trick has been extended to schemes with an "ample family of line bundles" by Thomason; see Weibel: Homotopy Algebraic K-theory, Proposition 4.4. This includes all smooth varieties and more ... View answer Accepted answer 11 votes It follows from results of Ribet in "On l-adic representations attached to modular forms" (Invent. Math. 28 (1975), 245–275) that the Mumford-Tate conjecture holds for the motives attached to modular ... View answer 11 votes It seems to me that in the proof they only use the fact for abelian varieties: For this you can use the fact that$[n]^*$acts on invariant$1$-forms by multiplication by$n$-- this follows from the ... View answer Accepted answer 10 votes I combine user37314's answer and my comments; the claim is that any smooth projective complex algebraic surface with$\pi_1$abelian and$\pi_2$finite has a finite cover which must be an abelian ... View answer Accepted answer 10 votes It can happen that$Pic(X) = Pic(Y) = Pic(Z) = 0$but$Pic(W) \neq 0$! For example, let$f: \mathcal{E} \to Z$be a non-isotrivial family of elliptic curves, where$Z$is a smooth rational curve. ... View answer Accepted answer 10 votes What you need is the Hilbert irreducibility theorem. This implies that the Galois group for "most" rational points (i.e. outside a thin subset) is the full symmetric group. More generally one can ... View answer 10 votes For varieties of dimension 1, 2 and 3 it is known that derived equivalent varieties have the same Hodge numbers. As far as I know, this is open (though believed to be true) for higher dimensional ... View answer Accepted answer 10 votes No, this is not even true for line bundles. For an example, let$X = \mathbb{P}^2$and$Y$a smooth curve in$X$of genus$>0$(over an algebraically closed field). Since$X$is smooth, any line ... View answer Accepted answer 9 votes Yes, this is true. One can argue as follows: Since$J(C)$has good reduction, your other hypothsism implies that the special fibre of its Neron model at$p$is an absolutely simple abelian variety. ... View answer 9 votes No. If$L$is a local system on a smooth variety$X$of dimension$d$then$L[d]$is perverse. As suggested by BBD, if we take$L$to be a rank$n$local system of$\overline{\mathbb{Q}}_l$vector ... View answer Accepted answer 9 votes In general, for a finite group$G$acting faithfully on a smooth variety$X$, whether or not the quotient is smooth is determined by the Chevalley-Shephard-Todd theorem: For$x \in X$, let$G_x\...