# Tag Info

0

Clearly $\mathcal{F}^g$ is a presheaf. In order to show that it is a sheaf we need to show that for ervery open set $U$ and every covering $\{U_i\}$ then is an equalizer. The two maps $$\prod_i \mathcal{F} ^g (U_i) \longrightarrow \prod_{i,j} \mathcal{F} ^g (U_{i,j})$$ are induced by the two different restrictions into $\mathcal{F} ^g (U_{i,j})$ i.e. ...

1

Hint: Use the inner/dot product interpretation of these angles. Then the statement becomes immediate.

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You can use Magma. The code E:=EllipticCurve("35a2"); E2:=IsogenousCurves(E)[1]; A,B:=IsIsogenous(E,E2); A; B; returns true Elliptic curve isogeny from: CrvEll: E to CrvEll: E2 taking (x : y : 1) to ((1/81*x^9 + 2/9*x^8 + 13*x^7 + 3392/27*x^6 - 30325/9*x^5 - 85999*x^4 - 24206654/27*x^3 - 50989888/9*x^2 - ...

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The theory of curve counting in CY manifolds is well established in dimension $3$, and it goes under the name of Donaldson-Thomas theory. Example. The general quintic threefold (which is a CY) is expected to contain a finite number $N_d$ of degree $d$ rational curves, for any $d$. The history of this sequence $\{N_d\}$ is fascinating. The number of lines, ...

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A friend has pointed out that Deligne has a theorem stating that the Hodge to de Rham spectral sequence vanishes on the first page in characteristic zero for any smooth projective variety $X$. In particular this tells us that $\dim_k (H_{\text{dR}}^n(X)) = \sum_{p+q=n}\dim_k(H^p(X,\Omega^q))$. Deligne and Illusie have then extended this result to certain ...

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In greater generality, suppose $X\subset Y$ is a subvariety. Then there is a restriction exact sequence $$0\to I_{X\subset Y} \to \mathcal O_Y \to \mathcal O_X\to 0$$ where $I_{X\subset Y}$ is the ideal sheaf of $X$ in $Y$. Your question is the case where $Y=D$ and $X=C_1$.

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You haven't introduced a base-ring for the various algebras in play; let me denote it by $A_0$. (In the case of varieties, we would take $A_0$ to be a field, but that doesn't affect anything.) For a finitely presented $A_0$-algebra $A$, formal smoothness is equivalent to smoothness. (And if $A_0$ is Noetherian, e.g. a field, then f.p. is equivalent to ...

4

If $\operatorname{char}(k)=3$ and $d$ is divisible by $3$, then the curve $$C=V(X^d+ZX^{d-1}+XY^{d-1}+YZ^{d-1}) \subset \Bbb P^2_{k}$$ is non-singular of degree $d$. Indeed, if $[x:y:z] \in C$ is a singular point, then, by considering the partial derivatives of the equation defining $C$, we obtain $zx^{d-2}=y^{d-1}$, $xy^{d-2}=z^{d-1}$ and ...

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This is only a comment, not a full answer. I hope it will help you a bit on your way. Let $Y$ be an affine noetherian scheme, and $y_0$ a regular point of $Y$ such that codim $y_0 =1$. Let $Z\to Y$ be a flat morphism of schemes such that $Z_{y_0}\to$ Spec $k(y_0)$ is a complete intersection. Then, we can use Proposition 2.1.12 in Olivier Benoist's ...

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Notice that $(y^2+x^3-17)\subset (y^2,x^3-17)$. Therefore, the quotient of the ring $\mathbb{C}[x,y]/I$ by the ideal, generated by $(y^2,x^3-17)$ is actually isomorphic to $\mathbb{C}[x,y]/(y^2,x^3-17)$. All I am using here is that for any ring $A$ and ideals $I\subset J$, we have an isomorphism $(A/I)/(J/I)\simeq A/J$. So the question boils down to ...

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You could try reading (the relevant parts of) Qing Liu's book on Algebraic geometry or the book on Neron models by Bosch-Lutkebohmert-Raynaud to get a feeling for elliptic curves over one-dimensional schemes. You could also try reading some papers where abelian schemes are used, e.g., Szpiro's asterisque (1985) on the Mordell conjecture, or Jinbi Jin's ...

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(1) The "old" $\operatorname{Proj}$ of a graded ring $S$ came with a canonical morphism $\operatorname{Proj} S \to \operatorname{Spec} S_0$. It seems to me that one has to define this on standard opens via $S_0 \to S_{(f)}$ and then glue. Maybe that's what your notation means. (2) Having $\mathscr{F}(\operatorname{Spec} A_f) = ... 1 Let$D = C_1\cup C_2$be any projective reducible curve with two components meeting transversally in a point$C_1\cap C_2 = \{p\}$. A regular function$f$on$D$is the datum of a regular functions$f_1$on$C_1$and$f_2$on$C_2$such that$f_1(p) = f_2(p)$. The restriction map $$\mathcal{O}_{D}\rightarrow\mathcal{O}_{C_1},\: f\mapsto f_{|C_1}$$ is clearly ... 2 You are completely on the right track.$U \cap V$contains an affine$W$, and$W$will have a closed point$x$(closed in$W$). But since$X$is of finite type over a field,$x$is in fact closed in$X$(and hence in particular closed in$U \cap V$) as well. This is an immediate consequence of the following result. Let$X$be a scheme, which is locally ... 4 Instead of following blindly Wikipedia's formulas, it is best to understand how to calculate$P+Q$, or$P+P$, given an elliptic curve. Let us assume for simplicity that the curve is given by$E:y^2=x^3+Ax+B$, and$P,Q\in E$. In order to find$P+Q$, first find the equation of the line$L$through$P$and$Q$, find the third point$R$of intersection of$L$... 1 Let$C$be the twisted cubic. The space$V$of quadric surfaces containing$p_1,...,p_7$has projective dimension$9-7 = 2$. Now, any quadric$Q$containing$p_1,...,p_7$intersects$C$in$7 > 2deg(C) = 6$points. Therefore$C\subset Q$. We have three independet quadrics$Q_1,Q_2,Q_3\in V$such that$C\subset Q_i$for any$i = 1,2,3$. Now,$Q_1\cap Q_2 ...

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It just a transverse Mercator projection. The library GeographicLib (written by me) has a class OSGB which converts eastings + northings to latitude + longitude (and also deals with OS grid references). If you want to know how to calculate the transverse Mercator projection accurately, see Transverse Mercator with an accuracy of a few nanometers (a ...

1

Let's define vectors $\mathbf{x} = (x,y,z)$ and $\mathbf{n} = (a,b,c)$, and let $\mathbf{o} = (0,0,0)$ be the origin. Then we have $$ax+by+cz = (\mathbf{x} - \mathbf{o}) \cdot \mathbf{n}$$ where the dot on the right denotes a vector "dot" product. So, the equation of the plane is $(\mathbf{x} - \mathbf{o}) \cdot \mathbf{n}=0$. The dot product of two ...

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By the definition of a plane, $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$, which is equivalent to the equation $Ax+By+Cz=0$ , the normal vector to the plane is $(a,b,c)$

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If $(x,y,z)\in P$, $(x,y,z)\cdot(A,B,C) = Ax+By+Cz = 0$: $$P\subset \{(x,y,z) | Ax+By+Cz = 0\}$$But as both are planes, they have the same dimension and are equal.

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For the compatibility note that $D_+(f_i) \cap D_+(f_j)=D_+(f_if_j)$ and also $\mathcal{O}_X(X_{f_i}) \cap \mathcal{O}_X(X_{f_j})=\mathcal{O}_X(X_{f_if_j})$. Now you need to check that if we localize $g_i : A[T_i^{-1}T_k] \to \mathcal{O}_X(X_{f_i})$ at $\frac{T_j}{T_i}$ and $g_j : A[T_j^{-1}T_l] \to \mathcal{O}_X(X_{f_j})$ at $\frac{T_i}{T_j}$ we obtain the ...

1

It's not the topological approach you ask about, but one standard algebraic technique is to write your curve as the locus of a homogeneous polynomial. For your first curve, you have $2zw - w^2 - 1 = 0$; treating $z = Z/T$ and $w = W/T$ as ratios of homogeneous coordinates in $\mathbf{P}^2$, the preceding equation becomes $2ZW - W^2 - T^2 = 0$. This is a ...

1

For $T$ noetherian, you can prove it by showing that $f_*\mathcal{O}_{X_T}\otimes k(t)\rightarrow H^0(X_t, \mathcal{O}_{X_t})$ is surjective: since $H^0(X_t, \mathcal{O}_{X_t})\cong H^0(X, \mathcal{O}_{X})\otimes k(t)\cong k(t)$, hence $\mathcal{O}_T\otimes k(t)\cong H^0(X_t, \mathcal{O}_{X_t})$, so $f_*\mathcal{O}_{X_T}\otimes k(t)\rightarrow H^0(X_t, ... 5 As Robert Auffarth said in a comment, a$\mathbb{Q}$-isogeny of degree$p$is a non-constant morphism of elliptic curves$E\to E'$defined over$\mathbb{Q}$, that sends zero to zero, i.e.,$\mathcal{O}_E\mapsto \mathcal{O}_{E'}$, and has degree$p$. Since the degree of the map equals the size of the kernel, the degree$p$condition means that the kernel has ... 2 Yes, it is. Assume the transcendence degree of$L$is equal to$n$, and choose$n$algebraically independent elements$\xi_1,...,\xi_n \in \Bbb C \setminus k$. We have an isomorphism $$k(X_1,...,X_n) \cong k(\xi_1,...,\xi_n) \subset \Bbb C.$$ Now by definition,$L$is a finite algebraic extension of$k(X_1,...,X_n)$, hence isomorphic to a finite ... 1 Since rationality can be tested on any non-empty open subset let's just study the affine part$C_0\subset \mathbb A^2_k=\{Z\neq0\}$of your curve$C$. The equation of$C_0$is then$x^p=y^{2p-1}$. But in general the equation$x^a=y^b$cries out to be parametrized by$x=s^b, y=s^a$so that in your case the parametrization is$x=s^{2p-1}, y=s^{p}$. If you add ... 2 Yes everything you write is correct. In particular your very last inequality follows from the implication for divisors on$S$(or on any smooth variety for that matter): $$D\leq E\implies H^0(S, \mathcal O(D))\subset H^0(S, \mathcal O(E))$$ This implication is evident by interpreting$H^0(S, \mathcal O(D))$as the vector space of rational functions ... 0 The set$\mathrm{Hom}_{R-{\mathrm{Alg}}}(A,B)$is never an algebra if$B\neq 0$because the zero map is not a morphism. Indeed, the zero map sends$1_A$to$0_B$instead of sending$1_A$to$1_B$like any honest morphism should. 2 No. For example,$A$be the trivial algebra$\{ 0 \}$and let$B$be any non-trivial algebra. Then$\mathrm{Hom}(A, B) = \emptyset$. 7 The zero set of a polynomial$p\in \Bbb{C}[x,y]$is unbounded (FTA), but the torus is compact, so that doesn't work. OTOH you can get many toruses as algebraic varieties in the projective space$\Bbb{C}P^2$. Look up Elliptic curves. These can also be described as sets of solutions of an equation of the form $$y^2=x^3+Ax+B$$ together with a point at ... 1 OK, the discussion in the comments is getting a little confusing, so let me write a full answer instead. I'll try to spell everything out in detail. We are assuming statement (a) and trying to deduce (b). Of course, we can assume that$C_1 \cap C_2$contains at least 4 points. Claim 1: No 3 points of$C_1 \cap C_2$lie on a line. Proof of claim: A ... 2 Suppose$A$is a finitely generated algebra (not necessarily reduced) over a field$k$(not necessarily algebraically closed). Then Noether's normalization theorem says that there exist$n\geq 0$elements$y_1,\cdots, y_n\in A$, algebraically independent over$k$, such that$A$is module-finite over its sub-algebra$k[y_1,\cdots,y_n]$. Since$\text {Spec} ...

2

a) The simplest example is with $n=r=1$: take $\mathfrak a=\langle X_1(X_1-1)\rangle \subset k[X_1]$. Then $Z(\mathfrak a)=\{0,1\}$ has two irreducible components $\{0\}, \{1\}$ each of dimension $n-r=1-1=0$. b) If that was too degenerate for your taste, the next simplest example is with $n=2,r=1$ and $\mathfrak a=\langle X_1X_2\rangle \subset ... 5 If you've done any number theory, you are probably aware of the following classic formula: Let$L/K$be an extension of number fields,$\mathfrak{p}\in\text{Spec}(\mathcal{O}_K)$and$\mathfrak{p}\mathcal{O}_L=\mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_n^{e_n}$. Define$f_i$to be$\,[\mathcal{O}_L/\mathfrak{P}_i:\mathcal{O}_K/\mathfrak{p}]$. Then, $$\sum_i ... 1 If n=2, take for X the circle x^2+y^2=1, for H the "hypersurface" y=0 ( a good old line!) and then X\cap H consists of the two irreducible components (=points) \{(-1,0)\} and \{(1,0)\}. 0 I'm going to post this as an answer since it received positive reviews in the comments. This is probably the intended solution (also, by the way the OP responded to my comments, not the one he was thinking), but is certainly not as nice as Georges's or Brenin's answer. We want to take two projective curves V_+(f) and V_+(g) (where f,g\in k[x,y,z] are ... 1 I don't know if this is the simplest example, but it's the one that I thought to try first. Let K=\mathbb{Q}(\sqrt{-15}). Then, L=\mathbb{Q}(\sqrt{-3},\sqrt{5}) is the Hilbert class field for K and so, in particular, \mathcal{O}_L/\mathcal{O}_K is unramified. But, \mathcal{O}_K is a Dedekind domain, and thus \mathcal{O}_L is flat over ... 6 Try the first chapters of Arithmetic Moduli of Elliptic Curves, by N. Katz and B. Mazur, Annals of Math Studies 108 2 I don't know your background in algebraic geometry, but one way of seeing this is the following: It is well-known that on a variety (the vanishing set of a set of polynomials) over an algebraically closed field, the singular points form a proper closed set of the original variety, where "closed" means closed in the Zariski topology. In your case, it is ... 0 The claim is true, but I don't know of any easy way of reducing to the affine case. There is some real work to be done here, in my opinion! See e.g. Theorem 2.55 in Vistoli's notes. 2 As A is a finitely generated k-algebra, we have a surjection k[x_1,\ldots x_r] \to A and thus A \cong k[x_1,\ldots ,x_r]/I. Assume that Spec A is finite, then there are only finitely many maximal ideals containing I in k[x_1, \ldots ,x_r], say m_1, \ldots m_n. As these are co-maximal, we have \cap m_i = \prod m_i and thus this product ... 3 Here is an elementary proof using only the part of Hartshorne preceding the exercise on page 21. Suppose two curves X,Y\subset \mathbb P^2 have empty intersection. Then Y\subset U:=\mathbb P^2\setminus X. However U is affine as can be seen through the d-uple embedding of Exercise 2.12, page 13: see the answer here. But this is absurd because the ... 8 This is achieved by the Veronese map. Namely, consider all monomials M_{i_0\cdots i_n}(x)=x_0^{i_0}\cdots x_n^{i_n} (there are N+1=\binom {n+d}{d} of them) and embed \mathbb P ^n into \mathbb P ^N by ver:x\mapsto (\cdots :M_{i_0\cdots i_n}(x)):\cdots) . The image of \mathbb P^n under ver is a subvariety V\subset \mathbb P ^N and the image ... 0 I think I can try to make a partial answer to my question. And the answer is yes, it is indeed true that one has the claimed isomorphism \hat{\mathcal{O}}_{C,P}\cong k[[x_i]]/(x_i-h_i(x_n)), and this can be shown either abstractly or "geometrically". Approach 1: the ring on the right is clearly isomorphic to k[[t]], power series ring in 1-variable. ... 1 Let \infty =(0:1:0)\in \mathbb P^2 and consider the projection from \infty onto the projective line L=\{y=0\}, which is the map$$\Pi:\mathbb P^2\setminus \{\infty\}\to L:(u:v:w)\mapsto (u:0:w)$$Its restriction to X is precisely \pi. So we have to show that \Pi|X extends to \overline X. For that we use the affine coordinates \xi= \frac ... 2 The associated ideal sheaf you you define makes sense only for affine schemes. For a projective scheme X with projective coordinate ring S, given a homogeneous element f \in S_+ the basic open set it defines is D_+(f)=Spec{ S_{(f)}}. So the corresponding ideal should be the zero grading of the ideal I \cdot S_f. Globally you just need to take the ... 1 Use Eulers Lemma for a homogeneous polynomial f, of degree d$$\sum_{i=0}^{n}x_{i}\dfrac{\partial f}{\partial x_{i}}=df$$1 Three numbers is not enough to represent a 3D (infinite) line. You need at least 4 numbers, and using 5 or 6 makes life easier. The obvious representation with 6 numbers is a point (3 numbers) and a vector (3 more numbers). To get down to 5, you use a unit vector, which can be represented by two numbers (e.g. spherical polar angles). To use the technique ... 0 Edit: Yes, you're quite correct that you get only a point. In order to define a line, you need either two points$p_1$and$p_2$, or you need$p_1$and a direction vector$v$. I don't believe there is a standard way for doing this using spherical coordinates. 3 Not a complete answer. Let's take the case of two lines$\ell_1,\ell_2\subset\mathbb P^2$. Assume the line$\ell_1$is given by the linear equation$a_1x+b_1y+c_1z=0$, while$\ell_2$is given by$a_2x+b_2y+c_2z=0$. (This is usually the point where one says "we may assume that$b_1=c_1=0\$" to simplify the setting. We could, but let's not do that). Suppose ...

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