# Tag Info

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A very late answer, my apologies if I am not contributing anything that the very wonderful answers above have already contributed. Generic points $\Leftrightarrow$ irreducible subsets... as long as we are doing classical geometry, there really is no major difference between using $\text{Spec}$ and using $\text{MaxSpec}$ and talking about things "being true ...

3

1) Your varieties $V_{f_1,\dots,f_t}$ are almost never affine, except in very degenerate cases (like for instance $f_1=\dots=f_t$ !). The basic tool is the following very general and powerful theorem: Given an arbitrary subvariety $Y\subset X$ of the arbitrary variety $X$, if the complement $X\setminus Y$ is affine, then $\operatorname {codim} _X(Y)=1$ ...

3

I doubt it. Presumably that's a typo, and it should read "geometric reasons". The usual proof identifies the coefficients of the product of two Schubert polynomials on the basis of Schubert polynomials as intersection numbers of transverse subvarieties, which are therefore non-negative integers. It has been generalised in various directions by increasingly ...

3

When $\mathscr{F}$ and $\mathscr{G}$ are both invertible sheaves of $\mathcal{O}_X$-modules, the failure of the presheaf $U \mapsto \mathscr{F}(U) \otimes_{\mathcal{O}(U)} \mathscr{G}(U)$ to satisfy the sheaf axiom can be viewed as a failure of the $\mathscr{F}$ and $\mathscr{G}$ to be trivialized on a common open cover of $X$ "in a compatible manner". ...

3

The answer is yes. Here is the basic idea. Take a very large degree line bundle $L$ and consider the sections $s$ of $L$ such that $s(x)=s(y)$ for the two given points $x,y$. These sections define a morphism from the curve to a nodal curve identifying just the two points $x,y$ if the degree of $L$ is sufficiently large.

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Try the following online documents: "The Malcev Completion for Groups" http://math.univ-lille1.fr/~fresse/OperadHomotopyBook/MalcevCompletion.pdf "Relative Malcev Completion" by P. Dalakov http://math.mit.edu/conferences/talbot/2011/notes/talbot_2011_16_peter_notes.pdf "Fillings in Nilpotent Groups" by A. Lukyanenko ...

3

The surface described by the following parametric equations should be close to what you are after: \eqalign{ &x=\alpha\ l^3+\beta\ l\ (1-|l^2-1/3|-|m^2-1/3|-|n^2-1/3|)^5\cr &y=\alpha\ m^3+\beta\ m\ (1-|l^2-1/3|-|m^2-1/3|-|n^2-1/3|)^5\cr &z=\alpha\ n^3+\beta\ n\ (1-|l^2-1/3|-|m^2-1/3|-|n^2-1/3|)^5\cr } where: $l=\sin\theta\cos\phi$, ...

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Hint. Let $p, q \in C$ be two points of inflection, and let $L$ be the line between them. Let $L \cdot C = p + q + r$. Let $M$ the tangent line to $C$ at $r$, and $M \cdot C = 2r + s$. Show that $r = s$ by showing that $s$ lies on the line through $p$ and $q$. In the interests of completeness, we provide a complete solution. Let $M_1$ be the line tangent ...

2

A less direct explanation than that of Kevin Dong: your curve is elliptic, and any point of inflection may be taken to be the identity $\Bbb O$, after which choice you can use chord-and-tangent to describe the addition. Then, any point of inflection is a $3$-torsion point, and the set of all these is a group.

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Let $Y\subset X$ be closed and $E$ a vector bundle on $X$. If $E$ restricted to $Y$ is trivial and $H^0(X,E)\to H^0(Y,E_{|Y})$ is onto, then $E$ is trivial in a neighbourhood of $Y$. In particular this is true if $X$ is affine. On the other hand, in general, this is not true. For example, take $C$ to be an elliptic curve and let $V$ be the non-split ...

2

Yes, the projective closure depends on the choice of embedding. For example, the closure of $\{ x^3 - y = 0 \} \subset \mathbb{A}^2$ in $\mathbb{P}^2$ is singular, but it can also be embedded in a non-singular projective curve. Functoriality also fails in general: the morphism $\{ x^3 - y = 0 \} \subset \mathbb{A}^2 \to \mathbb{A}^1$ given by $(x, y) ... 2 First of all, in the case when$X : f(x,y) = 0$is hyperelliptic there are$2g + 2$Weierstrass points on$X$. I believe that in general the number of points is bounded above by$g^3 - g$. Keep this in mind in case you venture into non-hyperelliptic territory. Second, you may need to check more than just the places$P \in X$with$x$-projections equal to ... 2 This is just my comment, with some renaming of variables. The point is that$g$is not a degree$d$map. Letting$\omega$be a primitive$d$th root of unity, you have that if for$[x_0:x_1:x_2] \in \mathbf{P}^2$, and a choice$r_0,r_1,r_2$of$d$th roots of$x_0,x_1,x_2$, respectively, then$[\omega^{a_0}r_0 : \omega^{a_1}r_1 : \omega^{a_2}r_2]$is a point ... 2 The short answer is that taking a variety$X$to its$k$-algebra of regular functions$\mathscr{O}(X)$defines a contravariant functor from the category of varieties to the category of$k$-algebras, and that this functor becomes a duality when restricted to the category of affine varieties. More precisely: for any variety$X$, you associate its ring of ... 1 Yes. Twist the inclusion by$\mathcal{L}^{-1}$, so that$\mathcal{L}^{-1} \otimes \mathcal{L}' \subset \mathcal{O}_X$, in particular it is an ideal sheaf (hence has non-positive degree). 1 Every non-zero holomorphic function has a holomorphic square root on any simply connected open set. Pick any open$\Omega$with a function$f:\Omega\to\mathbb C$which does not has a square root and consider the covering of$\Omega$by the discs it contains. 1 Try thinking about a simple case. Let$V=\mathbb{C}^2$, an affine variety with co-ordinates$x,y$. Then$\mathbb{C}^2-\{0\}=V_{x,y}$in your notation. Show that$\mathbb{C}[V_{x,y}]=\mathbb{C}[x,y]$, thus showing that$V_{x,y}$is not affine and answering your 3 negatively. For 2, no quasi affine varieties are projective except in the trivial case of ... 1 Here is a sketch (and there are other methods too) for jorst. For any point$p\in\mathbb{P}^2$, one has an exact sequence,$0\to\mathcal{O}_{\mathbb{P}^2}(-2)\to\mathcal{O}_{\mathbb{P}^2}(-1)\oplus \mathcal{O}_{\mathbb{P}^2}(-1)\to \mathfrak{m}_p\to 0$, where$\mathfrak{m}_p$is the ideal sheaf defining$p$. Take a quadric$Q$not passing through$p$and ... 1 I'm not entirely clear on what's meant by the phrase a clear geometric meaning that is coordinate free Whether or not a curve is a line is not intrinsic to that curve; it depends on the embedding, and hence on the coordinate system. In the context of algebraic geometry, lines and circles are isomorphic to one another, and you can't talk about a circle ... 1 Since the first part has been dealt with in the comments: Assume$f(X,Y)=\sum_{i,j\ge 0}a_{i,j}X^iY^j\in\mathbb R[X,Y]$is a polynomial with$f(x,y)=0$whenever$x^2+y^2<1$. Let$\alpha\in\mathbb R$and consider the polynomial$g(T)=f(T,\alpha T)\in\mathbb R[T]$. Then$g(t)=0$whenever$t^2(1+\alpha^2)<1$. As there are infinitely many such$t$,$g\$ ...

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