# Tag Info

17

A set of points that generate $E(\mathbb{Q})$ modulo torsion is given by (1955516573881233507049678279 : -86467145649172260650105545143411861089140 : 1), (49225691888888099223656060329/10201 : 67749663895993353685065159554645568700902610/1030301 : 1), (61339810590192565389735634 : -440289331793622522908840423931186017125 : 1), (...

7

There is an obvious relation: the pushforward map $f_*:|mD|\rightarrow |f_*(mD)|=|mf_*D|$ is injective, hence $\kappa (X,D)\leq \kappa (X,f_*D)$. It is easy to see that you cannot get more: for instance, take for $f$ a general projection from a cubic surface $X\subset \mathbb{P}^3$ to $\mathbb{P}^2$, and for $D$ a line in $S$. Then $\kappa (S,D)=0$, but $f_*... 6 First of all, it's unclear what you mean by$\Omega^1_{X_0}(\log D)$since$X_0$is singular. Second, if you make up such a definition then most probably such a vector bundle will not exist. Note that the Euler characteristics$\chi(X_\eta, \Omega^1_\eta)$and$\chi(X_0, \Omega^1_{X_0}(\log D)$have to agree. I suggest checking this for an elliptic curve ... 5 Let$\tilde{X}_k$be the normalization of the closed$k$-codimension stratum, so$\tilde{X}_0$is the normalization of$X$. Then there is a diagram of pullback functors between the categories$\text{Perf}(\tilde{X}_k),$such that$\text{Perf}(X)$is the pullback of this diagram in the$\infty$-category of derived categories (for example perfect derived ... 5 I see this is one of your first question on MO -- welcome! The topic of the question is certainly interesting but I think you need to put a little more effort in future questions into being clear (first and foremost with yourself) about the nature of the objects you're asking about. You don't need to understand the details of all the definitions in order to ... 4 Another nice example is by Oguiso and Truong, "Explicit Examples of rational and Calabi-Yau threefolds with primitive automorphisms of positive entropy". Briefly, take$E$to be an elliptic curve with an order$3$automorphism$\tau$and form the abelian variety$A = E \times E \times E$. The quotient of$A$by the diagonal action of$\tau$is mildly ... 4 The universal map is not a map of formal groups without some extra condition. An easy class of counterexamples comes from completions of an affine group along a closed subscheme that does not contain the identity element. In general, when you want to complete an affine group to get a formal group, you set$I_n = I^n$for an ideal$I$that defines a closed ... 2 This is an expanded version of my comment: The main claim is that the question can be answered from basic facts in (local) commutative algebra. In particular, the same statement holds for a complete intersection in affine space, which I explain below. We work over any dvr$R$with residue field$k$. Let$X$be of codimension$d$in$\mathbb{A}^n_k$... 2 Heuristically, we can compute the genus. This follows essentially the srategy of damiano and David Lampert in the comments. First note that it is possible to compute from this function the number$\#'C(L)$of points over$L$with field of definition exactly$L$. (Here the field of definition is the smallest field containing$K$that the point is defined ... 2 The projection$\pi \colon C \to C'$is induced by a$2$-dimensional sub-linear system of the complete linear system$|L|$, so$C'$is not linearly normal (unless$H^0(C, \, L)=3$) and$H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'})$has the same dimension as$H^0(C, \, L)$. It follows that every global section$\sigma' \in H^0(C', \, \mathcal{O}_{\mathbb{...

2

As requested in the comments, here is Proposition 6.1.6.1 of Lurie's "Spectral Algebraic Geometry", specialized to the case of ordinary commutative rings: Let $n\ge 0$ be an integer, let $A_0$ be a commutative ring and let $B_0$ be an $A_0$-algebra of finite presentation. Suppose we are given a diagram $\{A_\alpha\}_{\alpha\in I}$ of $A_0$-algebras ...

2

By the Bonnet Myers theorem, bounded positive Ricci curvature and complete Riemannian metric forces compact. David Wraith once explained to me that if the Ricci decays more slowly than quadratically in distance from a given point, on a complete Riemannian manifold, then the manifold is compact, a stronger result than Bonnet-Myers. Wraith's result does not ...

1

Here is an answer for the naive de Rham cohomology $\mathbb{H}(X, \Omega_X^{\bullet})$ (not the more sophisticated one of the linked article by Hartshorne (which involves chooseing an embedding $X \subset Y$ in a smooth variety $Y$, completion along $X$ etc.)). We can use the hypercohomology spectral sequence  E_1^{pq} = H^q(X, \Omega_X^p) \implies \...

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