# Tag Info

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 ...

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 ...

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 ...

5

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 ...

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, $$\sum_i ... 4 Instead of following blindly Wikipedia's formulas, it is best to understand how to calculate P+Q, or P+P, given an elliptic curve. Let us assume for simplicity that the curve is given by E:y^2=x^3+Ax+B, and P,Q\in E. In order to find P+Q, first find the equation of the line L through P and Q, find the third point R of intersection of L ... 4 If \operatorname{char}(k)=3 and d is divisible by 3, then the curve$$ C=V(X^d+ZX^{d-1}+XY^{d-1}+YZ^{d-1}) \subset \Bbb P^2_{k} $$is non-singular of degree d. Indeed, if [x:y:z] \in C is a singular point, then, by considering the partial derivatives of the equation defining C, we obtain zx^{d-2}=y^{d-1}, xy^{d-2}=z^{d-1} and ... 4 Notice that (y^2+x^3-17)\subset (y^2,x^3-17). Therefore, the quotient of the ring \mathbb{C}[x,y]/I by the ideal, generated by (y^2,x^3-17) is actually isomorphic to \mathbb{C}[x,y]/(y^2,x^3-17). All I am using here is that for any ring A and ideals I\subset J, we have an isomorphism (A/I)/(J/I)\simeq A/J. So the question boils down to ... 4 The theory of curve counting in CY manifolds is well established in dimension 3, and it goes under the name of Donaldson-Thomas theory. Example. The general quintic threefold (which is a CY) is expected to contain a finite number N_d of degree d rational curves, for any d. The history of this sequence \{N_d\} is fascinating. The number of lines, ... 3 (Though it's arguably poor taste to make such extensive edits, particularly after an answer has been up-voted, in this case it seemed justified.) First, here's an intuitive geometric argument: The space of lines through the point x is a copy of \mathbf{P}^2. If l and l' are distinct lines through x, their strict transforms in the blow up are ... 3 Here is an elementary proof using only the part of Hartshorne preceding the exercise on page 21. Suppose two curves X,Y\subset \mathbb P^2 have empty intersection. Then Y\subset U:=\mathbb P^2\setminus X. However U is affine as can be seen through the d-uple embedding of Exercise 2.12, page 13: see the answer here. But this is absurd because the ... 3 Yes, it is. Assume the transcendence degree of L is equal to n, and choose n algebraically independent elements \xi_1,...,\xi_n \in \Bbb C \setminus k. We have an isomorphism$$ k(X_1,...,X_n) \cong k(\xi_1,...,\xi_n) \subset \Bbb C. $$Now by definition, L is a finite algebraic extension of k(X_1,...,X_n), hence isomorphic to a finite ... 2 For the compatibility note that D_+(f_i) \cap D_+(f_j)=D_+(f_if_j) and also \mathcal{O}_X(X_{f_i}) \cap \mathcal{O}_X(X_{f_j})=\mathcal{O}_X(X_{f_if_j}). Now you need to check that if we localize g_i : A[T_i^{-1}T_k] \to \mathcal{O}_X(X_{f_i}) at \frac{T_j}{T_i} and g_j : A[T_j^{-1}T_l] \to \mathcal{O}_X(X_{f_j}) at \frac{T_i}{T_j} we obtain the ... 2 Yes everything you write is correct. In particular your very last inequality follows from the implication for divisors on S (or on any smooth variety for that matter):$$ 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 ... 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 ... 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 ... 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 ... 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 ... 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 ... 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 Yes, since your map f is given by a rational function of the form f(z) = \frac{p(z)}{q(z)}, with p,q coprime polynomials of degree at most d, with the larger of the two having degree d. Hence f is determined by (for example) its zeros, its poles, and one other point (for a scalar multiplier), or 2d+1 points in all (counting with multiplicity, ... 2 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 - ... 2 The mere fact that f and g are irreducible in k(x)[y] does not necessarily mean that they are relatively prime. The could still be associates in k(x)[y], say f = y and g = xy. You'll need some machinery, using the assumptions that they're irreducible in k[x,y] and not associates in k[x,y], to argue that this can't happen. It looks like that ... 1 Since no positive power z^k of z divides the polynomial F(x,y,z)=y^2z-x^3+xz^2, it is irreducible, if and only if its dehomogenization with respect to z$$ F_*(x,y)=y^2-(x^3-x)  is irreducible. Now consider $F_*$ as a quadratic polynomial in $y$. It is reducible, if and only if $x^3-x$ is a square in $k[x]$, which it obviously isn't, as $x^3-x$ ...

1

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 ...

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 ...

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 ...

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, $Q_1\cap Q_2 ... 1 As my reputation is still too low, I could not add a comment to your answer Cantlog. What I wanted to say is the following: shouldn't it be true that the divisor of the rational function$f_{i}$you define equals$2(w_{1} - w_{i})$, as the order of vanishing of the rational function$x$at the point$(0,0)$is$2$? Then indeed the class of the divisor$w_{1} ...

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, ...

Only top voted, non community-wiki answers of a minimum length are eligible