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0

Your argument is right, but I don't know a way to transfer it to the other eigenvalues. The analysis of the covering $C\rightarrow C/\varphi$ gives another interpretation of the eigenvalue 1: four points on a $P^1$ give a one-dimensional moduli space (it's $M_{1,4}$) and (most probably - I didn't spell it out) give via the covering of the mentioned type a ...


1

Posting this as an answer because it is too long for comments. Below is the SAGE sessions I used to compute. M = matrix([[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1],[1,-1,0,0],[-1,1,0,0],[-1,0,0,0],[0,-1,0,0],[1,-1,-1,0],[-1,1,0,-1]]) P = Polyhedron(M); P L = LatticePolytope(M.rows()); L print L.poly_x("") print P.f_vector() ​ A 4-dimensional polyhedron in ...


2

Here's a result for $n=2$. As discussed at Bounds for the size of a circle with a fixed number of integer points, a circle of radius $R$ centred at the origin passes through at most $4(2\log_5R+1)$ integer lattice points. Our circles need not be centred at the origin, but we can get an upper bound for the denominator of the rational coordinates of their ...


0

I wrote about this counterexample in another answer, but I suppose it'd be useful to list the relevant articles here. In addition to the sources below, a good friend of mine wrote a nice exposition of this topic, and there is a very nice overview by Kraft: Hanspeter Kraft, Challenging problems on affine $n$-space, Astérisque (1996), no. 237, Exp. No. ...


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

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

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) ...


0

In a projective geometry, the points are the one-dimensional subspaces (a point is associated with a line through the origin in the vector space), as you have said. The LINES are the TWO-dimensional vector subspaces (so a line is associated with a plane through the origin). Three points are on a common line if the one-dimensional subspaces they determine ...


0

Let $\mathcal{D}^p(X)$ denote the space of compactly-supported smooth $p$-forms on a smooth manifold $X$ (topologized in the same way as the space of distributions), then a $p$-current is defined to be a continuous linear functional $\mathcal{D}^p(X) \to \mathbb{R}$. An important class of currents arise from the oriented $p$-submanifolds $S \subset X$ ...


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 $t$, $g$ ...


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 ...


0

Okay, so $$xy - wz = 0$$ cuts a quadric surface -- that is, a two-dimensional variety of degree 2 -- out of $\mathbb{P}^3$. In order to see the dimension you really have to work scheme-theoretically, so that you see all of the non-$\mathbb{F}_p$-rational points of the variety -- otherwise it just appears to be a finite set of points, which would be ...


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

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 $\operatorname {codim} _X(Y)=1$ ...


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

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.


0

So, I assume you mean "lines through the origin" because otherwise, a dimension count suffices (lines are $ax+by+c=0$ and so have three parameters, but modulo a $\mathbb{C}^\times$ gives a $2$ dimensional family). A quick, but admittedly non-rigorous (though it can be made rigorous) way to see that it is not is that the moduli space of lines through the ...


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 ...


0

I think the answer is yes. Take an affine cover of $\mathbb{P}(2, 1, 1)$. The open affines are given by the rings: when $y \ne 0$, $k[\frac{x}{y^2} = D, \frac{z}{y} = E]$ with the curve given by $D(1 + E) - 1 + E^3=0$. The Jacobian for this is the matrix $(1 + E, D + 3E^2)$, which is less than full rank when $E = -1, D = -3$. This point is not on the ...


0

I think exactly how to answer your question may depend on the definitions you are using. To set definitions: On the side of varieties: $K(V') = frac(k[X]/I(V'))$, where $k[X]$ is the coordinate ring of $X$. (And $V'$ is serving double duty, here as an algebraic set.) The residue field of the scheme $X$ at $V'$ is the residue field of the ring $k[X]$ ...


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.


2

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 ...


0

We see the desired parallelogram in this diagram. (I emphasize the points here rather than the vectors.) The relevant property of a parallelogram here is that opposites sides have the same length. Therefore, the length of segment $AB$ equals the length of segment $CD$. One way to find point $D$ is to draw the circles with the appropriates radii to ...


0

This paper might give you some ideas.


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.


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 ...


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).


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: $l=\sin\theta\cos\phi$, ...


3

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 ...


3

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

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 ...


0

I think I figured out how to do this after following the suggestion given in the comment, but I still have some questions which I will write after explaining what I ended up doing. Let $\phi_i : M \times N \rightarrow M \otimes_{A_i} N$ be the maps taking $(m, n) \mapsto (m \otimes n)$ in $M \otimes_{A_i} N$ and $\psi : M \times N \rightarrow P$ be a ...


2

1) That is correct. 2) A few things. First, there is no intrinsic $\mathcal O_X(1)$, and once the surface is embedded, then clearly not every curve is a hyperplane section of the surface (this isn't even true in $\mathbb P^2$ for instance). More important, though, is that not every curve is even a candidate. For example, a single ruling on a smooth quadric ...


1

Here is a nice conceptual way which makes the continuity much clearer. It may not be in the form that you want, but you can certainly use it to prove the result in your language. So, let's assume that $f:X\to\mathrm{Spec}(\mathbb{Q})$ is smooth proper (as we always do). We then have the following nice fact: Theorem(étale Ehrassman's theorem): Let ...


1

The cycle map $CH^*\rightarrow H^{2*}$ is almost never injective, in fact the group $CH^*$ is really huge. Most of the time, it is not finitely generated. Take the example of a smooth projective curve $C$. Then $CH^1(C)$ is the Picard group and $H^2(C)=\mathbb{Z}$. The kernel of the cycle map is an abelian variety of dimension $g$, where $g$ is the genus of ...


0

In your situation $\overline{\rho(U)}$ is a proper closed subset of $W$, so the corresponding ideal in $k[W]$ is not just $\{0\}$.


1

$W$ is affine which means that its closed subsets are in one-to-one correspondence with radical ideals of $k[W]$ (this being the correspondence coming from the Nullstellensatz). Since $W$ is a variety, $k[W]$ has no nilpotent elements, so $(0)$ is radical, and corresponds to the entire closed set $W$ itself. Hence if $\overline{\rho(U)} \not= W$ then, since ...


0

There are points of the variety, and there are lattice points in the characters $M$, and there are lattice points in the dual lattice $N$. What you have in the picture you link are the points in $N$. What the homomorphism is defined on is the points of $S$, which is a subset of $M$. Finally, the special point is in $V$. Your special point is $[1:0:0]$ in ...


0

For $\mathbb{R}^3$, you have $$ \frac{1}{p_x}x+\frac{1}{p_y}y+\frac{1}{p_z}z=1. $$ Generalize for higher dimensions.


3

This appears to be false. See the paper Misconceptions about $K_X$ by Steven Kleiman.


2

Yes, it's not quite correct as written, but the argument is straightforward—or, at least, as straightforward as other things omitted in the book. Given a Weil divisor $D$ on $X$, if $D$ maps to zero in $\operatorname{Cl} U$, then there is a rational function $f$ on $U$ with $(f) = D\cap U$ on $U$. Then, on $X$, $D-(f)$ is supported on $Z$, so $D$ is ...


2

It need not be true in general. For example, take $X = \mathbb{P}^1$, $n=1$, $i$ the identity map, $N = 2$ and $j$ the degree 2 Veronese map into $\mathbb{P}^2$. In this situation $\mathcal{L} \cong \mathcal{O}_{\mathbb{P}^1}(1)$ but $\mathcal{L}' \cong \mathcal{O}_{\mathbb{P}^1}(2)$.


1

Now solved. If $\mathscr I_X$ is the ideal sheaf of $X\subset A$, then the ideal sheaf of $T_a(X)\subset X$ is $T_{-a}^\ast\mathscr I_X$. Indeed, for any morphism $f:Z\to Y$ and a subvariety $Y^\prime\subset Y$ with ideal $I$, the ideal of $Z\times_YY^\prime\subset Z$ is $f^\ast I\cdot \mathscr O_Z$. Apply this with $f=T_{-a}:A\to A$ and $Y^\prime = X$. ...


0

For convenience of notation let $A:=\{(x,y):x^2+y^2=1,x\neq 0\}$, $B:=\{(x,y): x^2+y^2=1, x=0\}$ so that $A\cup B = V(x^2+y^2-1)$ and $A\cap B=\emptyset$. This implies $A=B^c$ when considered as subsets in the induced topology on $V(x^2+y^2-1)$. Notice that, for any $p\in B$ we have $x^2+y^2=1$ and $x=0$ so that $y^2=1$. But this means $y=\pm 1$ and $B$ is ...


0

$\{(x,y):x^2+y^2=1,x \neq 0\}=V(x^2+y^2-1) \cap \{(x,y): x \neq 0\}$ now take closure of the both sides, and observe that $\{(x,y) : x \neq 0\}$ is dense.


1

Locally $D$ is given by the vanishing of $f \in \mathbb{Z}[x_0, \ldots, x_n]$. Let $Spec\ A$ be an affine open of $X$, so $A$ in a Noetherian integral domain, then $X \times Z$ is locally given by $f \in A[x_0,\ldots, x_n]$, this latter ring which is a Noetherian domain, and hence $f$ is a non-zero divisor and we can apply Krull's Principal Ideal theorem, ...


3

Write $f(x,y)=(x^2+y^2-1)g(x,y)+a(x)y+b(x)$ with $a(x), b(x)\in\mathbb R[x]$. Since $f(x_0,y_0)=0$ for all pairs $(x_0,y_0)$ such that $x_0^2+y_0^2=1$, $x_0\ne0$ it follows $a(x_0)y_0+b(x_0)=0$ for all pairs $(x_0,y_0)$ such that $x_0^2+y_0^2=1$, $x_0\ne0$. In particular, $a(\sin t)\cos t+b(\sin t)=0$ for $t\ne k\pi$, so $a^2(\sin t)(1-\sin^2 t)=b^2(\sin t)$ ...


3

A polynomial $f\in \mathbb{R}[x,y]$, considered as a function $f:\mathbb{R}^2\to\mathbb{R}$, is continuous (where $\mathbb{R}^2$ and $\mathbb{R}^2$ are given their usual topologies). Therefore if $f\in I(\{(x,y): x^2+y^2=1, x \neq 0 \})$, we have $$f(0,1)=\lim_{t\to \pi/2}f(\cos(t),\sin(t))=\lim_{t\to 0}0=0\\ f(0,-1)=\lim_{t\to ...



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