# Tag Info

0

It's been a while since this was posted, but I'd like to note that the reason it is hard to show is that it is false. If $I$ is the ideal sheaf of a codimension $1$ subvariety (say $X$ is projective), then the determinant is not trivial: $I$ is reflexive, and $\det I = I$. If $I$ is the ideal sheaf of a subvariety of codimension $\geq 2$, then the result ...

0

I'll give an answer considering them as proper varieties over $\mathbb{C}$. I'm pretty sure this should then be true in the holomorphic case as well (maybe by GAGA though I can't say I understand that). Suppose $Y$ is connected, then I claim that $i_*E$ is locally free (that is, a vector bundle) to begin with. $i$ is proper since each both $X$ and $Y$ are ...

8

I suppose what you are looking for is an introduction to classical topos theory. A topos is a category which can be thought of as a category of sheaves. One thing to keep in mind when reading about topos is that there are three equivalent definitions. The first is a left exact localization of a presheaf category. The second constructs sheaves from a ...

1

Dominant still means the same thing. If we have a rational map $f: X \dashrightarrow Y$ we can look at the maximal open subvariety $U \subset X$ where $f$ is regular and call $f$ dominant if $f(U)$ is dense in $Y$. So for this problem, try finding where the projection is regular (I think this was discussed in your previous question) and then show that the ...

10

I have three suggestions: Mac Lane, S., and Moerdijk, I., "Sheaves in Geometry and Logic: A First Introduction to Topos Theory" Kashiwara, M., and Schapira, P., "Categories and Sheaves" The first is my favorite. The latter is more advanced, and doesn't really start talking about sheaves until late in the book. It's a quality text nonetheless. Finally, ...

2

The second formula you quote is true in the following case: $Z \overset{\pi}\to X$ is a closed immersion with $Z\,$ equal to the support of a coherent sheaf $\mathcal{F}$ on $X$. Here is a proof of this fact. We may assume that $X$ is affine (equal to $\operatorname{Spec} A$) with $\mathcal{F} = \widetilde{M}$ for some finite type $A$-module $M$. ...

0

I thought more about it, and I'm now sure this is almost correct. The only mistake is in the final part, when I say notice that $u:X_d \to W_d$ is a surjective map of schemes In fact, $u$ is not always surjective, and anyway we don't need its surjectivity to conclude that pulling back the constant sheaf $\underline{H^1(\mathcal{O}_X)}$ on $W_d$ we get ...

4

What have you tried with this problem? I will help you out with (1) but you must show effort with (2). Here's how (1) goes. For every $x \in \ker \varphi(U)$ we can consider it as an element of $\mathcal{F}(U)$ since $\ker \varphi(U)$ is a subgroup of $\mathcal{F}(U)$. Since $\mathcal{F}$ is a sheaf the fact that $x|_{V_i} = 0$ implies immediately that $x = ... 0 There was an excellent answer to your previous question, where "John" showed you how to decompose your transformation into rotate, shear, and scale pieces. To get the "half-way" transformation, just take the values in the rotation, shear, and scale matrices and divide them by two. By the way, what you're doing has a name -- it's called "tweening". If you ... 0 We have reduced the question to showing$(S/\mathfrak{p})(n)_d \cong \Gamma(X,\widetilde{S/\mathfrak{p}}(n+d))$for large enough$d$. We can assume$n=0$since$(S/\mathfrak{p})(n)_d = (S/\mathfrak{p})_{n+d}$. We claim$\Gamma(X,\widetilde{S/\mathfrak{p}}(d)) \cong \Gamma(V(\mathfrak{p}),\widetilde{S/\mathfrak{p}}(d))$. This is a bijection by looking at the ... 1 The point is that in a factorial domain, the height one prime ideas are principal. By definition a Weil divisor gives a height one prime ideal in the local ring a each point (this is the ideal that cuts out the Weil divisor), and if this local ring is factorial, it is principal, so we get an equation that cuts out the Weil divisor in a n.h. of this point. ... 1 We expect this to be true since proper is the algebraic geometry version of compact and we know continuous images of compact sets are compact. In fact over$\mathbb{C}$, we could view this as the proof since$X/\mathbb{C}$is proper if and only if it is compact in the analytic topology. In general, suppose the base is$S$(you could pick your favorite ... 1 Have you shown that$f(X)$is closed in$X'$? If not, do this first. Now you know that$f(X)$is a variety (being closed in a variety). (Without this, the question doesn't quite make sense; we need to knowthat$f(X)$is a variety to talk about it being complete.) Next, show that if$X \to Z$is a surjection of varieties and$X$is complete, so is$Z$. ... 0 Yes: As you suggested,$V(x_0^3-x_2) = V(x_0^3 - x_2x_1^2) \cap \{x_1\neq 0\}$is an open set of$V(x_0^3-x_2)$. Now use that two varieties are birrational if you can find isomorphic open sets in them. 3 First of all you must suppose$p_1\neq0$, else the projection clearly does not make sense (you would always project onto$p$!). A parametric description of your line is$L_x(t)=p+t(x-p)$. It will hit the hyperplane$H=V(x_1)$for$t_0=-\frac{p_1}{x_1-p_1}$and the hitting point is thus the required projection$\pi(x)=L_x(t_0)=p-\frac{p_1}{x_1-p_1}(x-p)$. ... 1 Let$Z$be the support of$F$. Then$\mathcal F(n) := \mathcal F \otimes \mathcal O(n) = \mathcal F \otimes (\mathcal O(n)_{| Z})$(i.e. we can restrict$\mathcal O(n)$to$Z$before computing the tensor product). Because$Z$is zero-dimensional, it is a union of Specs of Artinian local rings, and so (exercise) any invertible sheaf on$Z$is trivial. Thus ... 1 One should read the comments by Matt E before reading the following. I am keeping this just as a record. Let$X = \hbox{Proj} \, (K [x,y])$and$Y = \hbox{Proj} \, (K[x,y]/(x^2))$. Write$i: Y \hookrightarrow X$and$F = i_* \mathcal{O}_Y$. Then$\dim \hbox{Supp} \, F = 0$. One has$\Gamma(X, F) = 1$, but$\Gamma(X, F(i)) = 2$for$i \ge 1$. I belive ... 0 If you scale, shear, and rotate (in that order) then you'll have: $$\begin{bmatrix} A_{11} & A_{12} \ \\ A_{21} & A_{22} \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} 1 & m \\ 0 & 1 \end{bmatrix} \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix}.$$ Then ... 1 From my cursory reading of what slerp interpolation is, it sounds like it only deals with rotations, so it may or may not be helpful for general affine transformations. It sounds like you wish to decompose the transformation$ax+b$(where$a\in GL(\Bbb R,2)$is the skewing and$ b\in \Bbb R^2$is the translation, into a transformation$cx+d$such composing ... 6 Neither is true. In the first your mistake is in this step: $$(\pi^{-1}F)(\pi^{-1}V) = \lim_{W \supset V}F(W)$$ It should be $$(\pi^{-1}F)(\pi^{-1}V) = \lim_{W \supset \pi\pi^{-1}V}F(W)$$ because in general we don't have$\pi(\pi^{-1}(V)) = V$. Take, for example,$\pi$to be the inclusion map of a point. For the second assume$X$and$Y$are affine ... 2 I don't think this is true. In fact if we denote a rigidified line bundle as$(L,\sigma)$where we make explicit the isomorphism$\sigma:\varepsilon^*L \to \mathcal{O}_S$, then the set of rigidified line bundles forms a group under $$(L,\sigma_L) \otimes (M, \sigma_M) = (L\otimes M, \sigma_L \otimes \sigma_M).$$ We can verify$\sigma_L \otimes \sigma_M$... 3 It's looking like the answer is "yes." That said, in case it helps anyone who hits this via Google, the relevant definitions are actually written down in: EGA O, 3.5.1 - 3.5.2, 4.1.1 (composition of morphisms of ringed spaces) EGA O, 4.1.2 (the inclusion morphism and restriction of morphisms) 10 I agree. On p. 72 locally ringed spaces and their morphisms are defined. Then on p. 73 we have the definition of an isomorphism of locally ringed spaces as a morphism with a two-sided inverse. In principle we need a composition of morphisms for this definition, but Hartshorne doesn't define it there. Instead, he characterizes isomorphisms$(f,f^\#)$by the ... 2 This is true for a separable field extension. In fact, this works more generally than just curves. If$X$is any normal scheme over$k$and$K/k$is a separable field extension, then$X_K$is normal. I'll sketch this proof with appropriate references to technical results in the Stacks Project. Normal is a local condition so we can assume$X$is an affine ... 2 Let$X=\{a,b\}\subset \mathbb P^n\; (n\geq 1)$be a two-point subset of$n$-dimensional projective space and$\mathcal F=\mathcal I_X$the ideal sheaf of functions vanishing at$a$and$b$. This means that for an open subset$U\subset \mathbb P^n$a section$s\in F(U)=\mathcal I_X(U)\subset \mathcal O(U)$is a section in$\mathcal O(U)$vanishing on$U\cap ...

5

I imagine one could write books on the subject, but probably the first moment one encounters homological algebra in algebraic geometry is the following. If $f: X\to Y$ is a proper map of varieties, then given a coherent sheaf $\mathcal{F}$ on $Y$ one can form the pull-back $f^{*} \mathcal{F}$. Now, the pull-back functor $f^{*}: \mathcal{C}oh(Y) \rightarrow ... 3 It's not hard to find a field extension$k\subset K$such that$K\otimes_kK$is not reduced. 2 In the affine case, this is the well-known classification of prime ideals of a localization of a commutative ring at a prime ideal. In the general case, glue everything. 2 I think the tangent bundle you should consider is spanned by the Lie algebra generated by$L_i$, not just$L_{ij}$. In particular, if the Lie algebra is commutative then$L_{ij}=0$, which is not what you are looking for. My favourite reference on these things is "Mathematical Methods of Classical Mechanics" by V.I. Arnold. In the chapter on Hamiltonian ... 0 Here are some preliminary computations and a first attempt. Please let me know if you see mistakes. To simplify, let$\mathbf a=(1:0)$and$\mathbf b=(0:1)$in$\mathbb P^1$over an algebraically closed field$k$and$\mathscr F$be the ideal sheaf of$\mathcal O_{\mathbb P^1}$vanishing on$\mathbf a$and$\mathbf b$. I wish to compute the cohomology ... 0 We have to be maximally careful with the way function fields embed, so let me translate your map to a map on rings. The corresponding map is given by: $$\begin{eqnarray*} \varphi : &k[y]& \to k[x] \\ &y& \mapsto x^2. \end{eqnarray*}$$ The first$k[y]$is the coordinate ring of$Y = \Bbb{A}^1 $, and the second of$X = \Bbb{A}^1$. Taking into ... 0 At each vertex of the polygon, the interior angle and exterior angle must sum to$180$. You can use this to find the value of each exterior angle. 1 The notation$G(-)$is the functor of points of the algebraic group$G$. The general yoga of Grothendieck style algebraic geometry is that a space$X$should be studied by looking at all the maps into from all spaces at once. More precisely, to each space$X$(scheme or variety in this case but really this works in any category) we can associate a functor ... 1 I think I've figured it out: The map$\alpha$that Abramo defines, which for clarification purposes I'll call$\beta$, is not the map taking an open subset$U$of$X$to the closure of its points (i.e. the map$U\mapsto \left\{\overline{\{P\}}:P\in U\right\}$). What it is is the inverse to the pre-image map$\alpha^{-1}$of the original map$\alpha$. To see ... 0 Following Wikipedia, the affine Grassmannian is defined for an algebraic group$G$defined over a field$k$. Since$\mathcal{K}$and$\mathcal{O}$are$k$-algebras, you can look at the points of$G$in$\mathcal{K}$or$\mathcal{O}$and these points are denoted by$G(\mathcal{K})$and$G(\mathcal{O})$. Maybe, aren't you familiar with the fact that an ... 1 According to SAGE, the points$(0,\pm 1)$are the only integral points on this curve. 3 Don't work with elements, work with morphisms. This general lesson of category theory is also very useful for algebraic geometry. Let$g := p_1 h = p_2 h : T \to X$. Then$h$is equal to$T \xrightarrow{g} X \xrightarrow{\Delta} X \times_Y X$, since$p_i \Delta g = g = p_i h$. Hence,$h$factors through$\Delta$. By the way, you are right with your guess ... 5 Regarding #2, here is a related question. The idea is rather simple: the existence of nontrivial automorphisms implies existence of nontrivial isotrivial families (i.e. with all fibres isomorphic/equivalent). To see why this messes up representability, suppose we are given a nontrivial isotrivial family$\mathcal X\to B$. There exists a unique map$B\to P$, ... 0 More can be said: if$A$is any subset of high-dimensional sphere$S^n$such that$\operatorname{vol}_nA=\frac12\operatorname{vol}_n S^n$, then the volume is concentrated near the boundary of$A$; in the sense that an$\epsilon$-neighborhood of the boundary contains most of the measure. When$A$is a hemisphere, you get the statement about equator. Even ... 1 Recasting Matt E's answer in terms I feel more comfortable with: Let$L=(Q_1,Q_2)$. By the unmixedness theorem, the associated primes of$L$are precisely the minimal primes over$L$and all such have height 2. Since$I$is such a prime, the minimal primary decomposition of$L$looks like$q \cap q_1 \cap ... \cap q_l$, where$q$is$I$-primary. Since each ... 4 I'm not sure exactly how Eisenbud intends to argue, because I don't know what he's assuming known at this point. But here is some kind of explanation, which perhaps you can adapt to what you know: The inclusion$(Q_1,Q_2) \subset I$shows that$V(Q_1,Q_2) \supset V(I) = C$. Now$V(Q_1,Q_2)$is a curve of degree$4$(by Bezout), as is$C$(by assumption) ... 5 It is just the disjoint union of the underlying sets. The elements of$\mathcal{O}_{\text{Spec}A}(U)$are functions on$U$whose value at the point$\mathfrak{p}$is an element of the local ring$A_{\mathfrak{p}}$2 An indirect answer is that Hartshorne's sheaf constructions are often a little lacking in the elegance department. This is one example; the definition of the structure sheaf is another. If you are looking for a better way to understand the reduced scheme associated with an arbitrary scheme, think of it this way: it's a closed subscheme and thus given by a ... 6 Take$X=\coprod_n \mathrm{Spec}(\mathbb{Z}/p^n)$for a prime$p$. Then$s:=(p)_n \in \prod_n \mathbb{Z}/p^n = \mathcal{O}_X(X)$is not nilpotent, but the restriction to each$\mathrm{Spec}(\mathbb{Z}/p^n)$is nilpotent. Hence, we have$[s] \neq [0]$in$\mathcal{O}_X(X)_{red}$, although we have$[s]=[0]$in each ... 1 Mumford's Lectures on curves on an algebraic surface is a wonderful book, which develops ideas related to deformation theory (especially embedded deformations, as part of the infinitesimal theory of Hilbert schemes), and many other fundamental topics in algebraic geometry. 3 What does "canonical" mean? It's not actually a technical term. If you mean something like I mean in this paper, then under the hypotheses given there, the answer is yes just because the main theorem implies that the stalk maps are in fact induced by the global map. Note that your hypotheses on the sheaves are lacking: the induced maps on fibers being ... 1 Of course not, take two non-isomorphic line bundles, say, on$X$and take the trivial homomorphism between them (sending any to$0$). 1 Why is$\phi(W)=X_f$Take and element in W say$(X_1,X_2,...X_n,X_{n+1})$, i.e all the polynomials$G_1,G_2,...G_r,FT_{n+1}-1$vanishes on this point. In particular$FT_{n+1}-1$vanishes on$(X_1,X_2,...X_n)$i.e.$F(X_1,X_2,...X_n).X_{n+1}-1=0$. Therefore,$F(X_1,X_2,...X_n)$cannot be equal to zero (Note: it also says$F$does not belong to$I(X)$. ... 3 This is indeed true. As Keenan's answer stated, the geometric version of the notion you're talking about is geometric irreducibility and reducedness. Consider the scheme$X$cut out by$I$. Then$X$is irreducible if and only if$\sqrt{I}$is prime. Then$X$is geometrically irreducible if it stays irreducible after any base extension. Similarly,$X$is ... 5 Instead of giving a vague a-priori reason or the usual proof, let me give a quite unknown direct proof, which even works in the larger category of locally ringed spaces: If$X \to S$and$Y \to S$are morphisms of locally ringed spaces, then you can construct their fiber product$X \times_S Y\$ explicitly, see here for a summary. The set consists of all ...

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