# Tag Info

2

Consider the map $f:k[u,v]\rightarrow k[x,y]$ sending $u\mapsto x^2$, $v\mapsto xy$. Note that $f(u)$ and $f(v)$ are homogeneous polynomials of degree $2$. Also $f$ maps the monomials $u^av^b$ to distinct monomials in $k[x,y]$, so $f$ is injective. Thus $I=0$ and every point is a common zero of $I$. However there are no $x,y\in k$ for which $(x^2,xy)=(0,1)$.

0

The statement is that every divisor $Y$ on $Q$ is a complete intersection in $\mathbf P^n$. In other words, if $Y=V(I)\subset\mathbf P^n$ is the structure of closed subscheme, then the homogeneous ideal $k[x_0,\ldots,x_n]\supset I = (f,g)$ where $f=x_0^2+x_1^2+\cdots+x_r^2$ and $g$ is some irreducible homogeneous polynomial. Truly this is equivalent to ...

1

If the number is rational, using multiplication formulas for $\cos$ and $\sin$ you can express the original parametrization in terms of $\cos$ and $\sin$ of multiples of a single argument. Then you can use a rational parametrization of the circle to replace those $\cos$ and $\sin$ by rational functions of the parameters, and then you can eliminate the ...

1

I have this weird thing where I stare at something for two hours, rage quit, then I get on stackexchange and type up the question. Then right after I post the question I realize how to do it. The functions $f, g, h$ are just the compositions of $k^{\ast} \rightarrow k \times k^{\ast} \times k$ with the projection maps.

1

Consider an ideal $I=(f_1,\dotsc, f_s)$ in a noetherian factorial ring and let $f$ be the greatest common divisor of the $f_i$. Then $I$ is principal if and only if $f \in (f_1, \dotsc, f_s)$ holds. In practice, working in the polynomial ring, figuring out whether this holds, you need Groebner bases. Thus, in practice, you can also directly use the ...

0

The idea of the proof is this: Let $P$ be a rational point on $C$. Then look at all the lines through $P$. Each such line intersect $C$ in one more point. This lets us define a rational map $\mathbb P^1 \to C$ since lines are parametrized by $\mathbb P^1$ (it is rational because the tangent line only meets $C$ at one point). The map is well defined and ...

0

Setup Following page 32 of Vakil's notes,let S be a multiplicative subring of a ring A; i.e., $1 ∈ S ∧ x,y ∈ S ⇒ x · y ∈ S$. Then we consider “formal fractions”, S⁻¹A ≔ { a / s ∣ a ∈ A , s ∈ S } The property we're interested in is 𝒫 : A-algebra → Bool 𝒫 f ≔ for every e in S, f e ∈ B is invertible Want to show: S⁻¹A is initial among A-algebras B ...

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Since $\mathbb{P}^n$ is a rational homogeneous variety this follows from the fact that an exceptional sheaf is already a homogeneous bundle, see for example Proposition 2.1.4 here.

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'Diophantine Geometry: An Introduction', by Silverman and Hindry, contains a proof in section A.4.3 (Page 74). A few minor steps are skipped, but it should be fairly easy to follow.

2

A morphism $\phi: \mathbb{P}^m \rightarrow \mathbb{P}^n$ can be given by $\phi =(F_0: \dotsc : F_n)$, where the $F_i$ are homogeneous polynomials of same degree in $m$ variables (See Remark 3.2 on the same chapter of Silverman). The only problem is that the $F_i$ might have common zeros in $\mathbb{P}^m$. Say $P$ is one such point, then $\phi(P) = (0: \... 3 Ok Slup, here goes. Let$R$be any commutative ring and let$A$be a polynomial ring over$R$. Let$P$be any projective module over$R$. Then Quillen (and Suslin a bit later in this generality) proved that if for every maximal ideal$\mathfrak{m}$of$R$,$P_{\mathfrak{m}}$is of the form$Q\otimes_{R_{\mathfrak{m}}} A_{\mathfrak{m}}$for some projective$...

0

It's a really vague question: what you want can be achieved in a lot of different ways. This one has just come to mind: $$SPARSITY([0=a_0, a_1, \ldots, a_{n-1}, a_n=1])= {1\over1+\sum_{k=1}^{n-1}(a_k-k/n)^2}$$ Could that work for you?

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This is true. Some local maneuvering seems unavoidable, and there are ways to say essentially what you did under the guise of algebraic geometry (etale-locally, etc.). Here's an attempt to do it as you request. We'll use the standard exact sequence for normal bundles: $$0\to N_{E_s/E}\to N_{E_s/M_S}\to N_{E/M_S}\otimes \mathcal{O}_{E_s}\to 0$$ From the ...

1

First of all you must assume more. Namely that a section corresponds to a monomorphism: $$\mathcal{O}_X\rightarrow \mathcal{O}_X(D)$$ You may clearly get rid of this by assuming that $X$ is integral. Then there is a bijection between set of such sections up to a scaling by $\Gamma(X,\mathcal{O}^*_X)$ and effective divisors linearly equivalent with $D$. My ...

1

The category of affine schemes is the opposite of the category of commutative rings, and taking opposites switches limits and colimits, so in general the categorical quotient $\left( \text{Spec } R \right) / G$ is the spectrum $\text{Spec } R^G$ of the fixed point subring. Hence you want to compute the subring of $R = k[x, y]$ ($k$ a field containing $\omega$...

1

First of all, in order that the subset $M\subset \mathbb R^n$ defined by the system \begin{align*} f_1 = 0 \\ \cdots \\ f_k = 0 \end{align*}be a submanifold you have to assume that the Jacobian matrix $J(f)(x)=(\frac{\partial f_i}{\partial x_j}(x))$ has rank $k$ at every $x\in M$. If this is the case the tangent space to $M$ at $x$ consists of the $n-... 0 According to Stacks Project, it is wiser to consider the normalization only for a scheme$X$such that every quasi-compact open subset of$X$has finitely many irreducible components. Then by lemma 28.50.2, the normalization$\widetilde{X}\to X$factors through$X_{red} \to X$and$\widetilde{X} \to X_{red}$is the normalization of$X_{red}$. 1 It should be true for any PID. See the book by Lam, Serre's Conjecture 2 Let us say that a ring$A$satisfies condition$(S)$if every finite type projective$A$-module is free. Clearly a necessary condition for$(S)$to hold is$K_0(A)=\mathbf{Z}$. Example. (i) If$A$is local Noetherian then$(S)$holds. (ii) If$A$is a Dedekind domain then$K_0(A)=\mathbf{Z}\oplus\mathrm{Pic}(A)$, and thus a necessary condition for$(S)$is$...

1

Theorem: If Spec$(R)$ is Noetherian, then so is Spec$(R[X])$. [Theorem 2.5 in Rings with Noetherian spectrum'' by Ohm and Pendleton] So for the example, take $R[X]$, where $R$ is any non-Noetherian ring with Noetherian spectrum.

1

Your proof uses the following claim (paraphrasing slightly): If $r$ is a linear combination of monomials with coef in $\mathbb C$, where no monomial is divisible by $uv$, then $r$ must be a polynomial in $w$ alone. This is not true. For a counterexample, consider $r(u,v,w)=2u+v^3+vw$. This is not a polynomial in $w$ alone, but none of its monomials are ...

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edit: I write sequence other way around, confused myself and gave wrong answer. Now it should be correct. Let $M\subseteq \mathbb{R}^n$ be the manifold. We have the following exact sequence of vector bundles: $$0\rightarrow TM\rightarrow T\mathbb{R}^n_{\mid M}\rightarrow\mathcal{N}_{M/\mathbb{R}^n} \rightarrow 0$$ and now vector fields on $M$ correspond ...

0

In a previous answer in How do I reflect a function about a specific line? I showed how to derive a formula for $y=f(x)$, if its rotated by $y=mx+b$ Using the formula $$\left(\frac{2m}{2m^2+1}x'-\frac{1-m^2}{m^2+1}(y'-b)+b\right)=f\left(\frac{2m}{m^2+1}(y'-b)-\frac{1-m^2}{m^2+1}x'\right)$$ If we substitute $f(x)=80x^5-225x^4+350x^3-300x^2+150x-20$, $m=55$...

1

Assume there are polynomials $f$ with $\phi(\overline f)=0$ and $f\notin (y-x^2,z-x^3)$. Among these. let $f_0$ have minimal degree in $z$. If $f_0$ has positive degree $k$ in $z$, say $f(x,y,z)=p_0(x,y)+p_1(x,y)z+\ldots +p_k(x,y)z^k$, then $f_0(x,y,z)-p(x,y)z^{k-1}x^3$ is equivalent to $f_0$ but of lower degree in $z$. We conclude that $f_0$ has degree $0$ ...

3

If we take the definition of a (global) lefschetz number as in $7.1$ of these notes, then the proof goes as follows: If $f$ is homotopic to $g$, the inclusion maps $id×f$ and $id×g$ of graph $f$ and graph $g$ are homotopic. Thus the global Lefschetz number of a map is homotopy invariant.

1

I don't know how to answer for algebraic groups in general, only affine algebraic groups over an algebraically closed field $k$. Remember that an affine algebraic group is up to isomorphism the same thing as a closed subgroup of $\textrm{GL}_n$ for some $n$. However, an isomorphism of such a group into $\textrm{GL}_n$ is not canonical. Question 1: For any ...

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The product in a category which has a terminal object is the fiber product over the terminal object.

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A. Weil, The field of definition of a variety, Amer. J. Math. 78 (1956), 509-524. Theorem 3. J-P. Serre, Groupes algébriques et corps de classes, Publications de l'institut de mathématique de l'université de Nancago, VII. Hermann, Paris 1959. Ch. V-20, Prop. 12, Cor. 2. There is an English translation of Serre's book: Algebraic Groups and Class Fields ...

3

When you write $k$ you should write $Spec \ k$. These schemes are of finite type over $Spec \ k$, because the $k$-algebras $k[x,y,t]/(ty-x^2)$ and $k[t]$ are obviously of finite type (that is, a quotient of a polynomial ring over $k$ in a finite number of variables).

3

In general this is false. For example, take the projective plane, $F=\mathcal{O}$ and $G$ a rank two stable bundle of determinant zero. Then $H^0(G)=0$, so there is no inclusion of $F$ in $G$. But, for a general line $H$, $G_{|H}=\mathcal{O}_H^2$.

1

Are $F,G$ coherent? If not, this is clearly false: Consider $R=K[X], f=X, F=K(X), G=K[X] \oplus K[X]$ and let $F \to G$ be the zero map, which is cleary not injective. We have $F/fF=0$, i.e. the map is injective after tensoring with $R/f$. If $F$ is coherent and $G$ is quasi-coherent, it is true: The assertion local, so we can reduce to the affine case ...

5

Let $I=\mathbb R[x,y](1-x^2-y^2)$, and suppose $f\in\ker(\phi)$. Proof 1 Let $p:\mathbb C\rightarrow\mathbb C^2$ be the map $\theta\mapsto(\cos(\theta),\sin(\theta))$. By assumption, $f\circ p$ vanishes on $\mathbb R$. It's analytic so it vanishes on $\mathbb C$ by the identity theorem. That is, $f$ vanishes on $\mathrm{im}\;p=Z(I)$. By Hilbert's ...

0

Let $(X,\mathcal O_X)$ is a locally ringed space, let $\mathcal F$ be a torsion-free sheaf of $\mathcal O_X$- Modules, let $0\neq s\in \mathcal F(X)$ be a non-zero section and let $\phi=\phi_s:\mathcal O_X\to \mathcal F$ be the associated morphism. a) If all $\mathcal O_{X,x}$ are domains, then $\phi:\mathcal O_X\to \mathcal F$ is injective, since all $... 1 I am not quite sure what you are asking for, at any rate the comment I wanted to make is a bit too long, so I thought will write an `answer'. Choosing$2n$(or for that matter any$k$) points in projective line corresponds to a point in$S^{2n}(\mathbb{P}^1)=\mathbb{P}^{2n}$. Thus, we have the incidence variety$\Gamma\subset \mathbb{P}^{2n}\times\mathbb{P}^...

1

One may assume that $A$ is simple over $K$. Then $\mathrm{End}_K(A)$ is a division ring. Since the morphism into $\mathrm{End}_k(A\otimes k)$ is not zero, the kernel is $0$.

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Upgrading my comment to an answer to get this off the unanswered list: The category of rational modules for a torus $T$ is semisimple (regardless of the characteristic), so all modules are projective and injective and thus all functors of the form $\operatorname{Hom}_T(-,M)$ are exact, whatever module $M$ one picks.

1

Actually, for the special case of plane curves, the connection is immediate: We can assume $p=(0,0)$ of course, so the curve $X= \operatorname{Spec} k[x,y]/(f)$. is given by $f \in k[x,y]$ with $f \in \mathfrak m = (x,y)$. We have $\dim \mathcal O_{X,p} = 1$ and the maximal ideal corresponding to $p$ in $k[x,y]/(f)$ is $\mathfrak m/(f)$, so we can compute ...

0

Let $\alpha \in \bar{\mathbb{Q}}$ and $f(T)$ be its minimal polynomial, let $X = \mathbb{A}^1_\mathbb{Q}$ and $D = Spec( \mathbb{Q}[T]/ (T \cdot f(T)))$. Then, if we considering the singular 1-chain \begin{array}{rrl} \gamma: [0,1] & \rightarrow & X^{an} = \mathbb{C} \\ s & \mapsto & s \alpha \end{array} consisting on a path from 0 to $\... 0 Found it ! Thanks to Slup for making me realize I had overlooked the hypothesis that$K$was a subset of the considered valuation rings. Actually, since$K \subset \cal{O}_{F,P}$,$(\overline{\cal{O}_{F,P}})_{\cal{M}} \subset R(F)=Q(\cal{O}_{F,P})$and the residue field of$\cal{O}_{F,P}$is$K$, an element of$(\overline{\cal{O}_{F,P}})_{\cal{M}}$writes$\...

0

$\mathcal O_{X,\bar x}$ is an $\mathcal O_{X,x}$-algebra means that we get a ring map $\mathcal O_{X,x} \to \mathcal O_{X,\bar x}$, not the other way around. Given an open $U \ni x$, we have $\bar x \in U$, hence we get map $\mathcal O_{X}(U) \to \mathcal O_{X,\bar x}$. Taking the direct limit over all such $U$, we get the desired map $\mathcal O_{X,x} \to \... 1 Recall that a section$s$of$L$is a map$s : Y \to L$such that$\pi\circ s = \operatorname{id}_Y$. Given$f : X \to Y$, we can form the composition$s\circ f : X \to L$. Now$s\circ f$is not a section of$L$as$L$is a bundle on$Y$, not$X$. However, we can form the pullback bundle$f^*L$, then$f^*s := s\circ f$is a section of$f^*L$. 0 A cubic surface (at least over$\mathbb{C}$) has a full exceptional collection. In fact many full exceptional collections. One of them consists of sheaves $$(O_{e_1}(-1),\dots,O_{e_6}(-1),O(-2\ell),O(-\ell),O)$$ in the standard notation. Consequently, the objects of the derived category whose hypercohomology vanish are just objects of the subcategory ... 2 These type of questions should be dealt with using standard universal properties. Let$X$be any variety,$L$a line bundle and assume$s_0,\ldots, s_g\in H^0(L)$. Then we get a homomorphism$\mathcal{O}_X^{g+1}\to L$, using these sections. These sections have no common zeros is equivalent to saying the above map is onto. On the other hand, giving such a ... 2 First, there is a minor error in your sequence; the middle term should be$\mathcal{O}_{\mathbb{P}^r}^{r+1}$. Can you see that$\sum\deg\mathcal{L}_i=\deg \mathcal{M}_{|X}$? Then$\deg\mathcal{M}_{|X}=\deg \mathcal{O}_{\mathbb{P}^r}(-1)_{|X}$? The latter of course is just$-\deg X$. 1 First off: what is the affine cone here?$v_d(\mathbb P^n)$is the image of all monomials of degree$d$in$n+1$variables. So the affine cone is$Y = \mathrm{Spec}k[x^I]$, where$x^I$ranges through all monomials of degree$d$. According to Wikipedia, we have to check two things. First, that$Y$is invariant under$G$, and secondly that any invariant ... 2 This is not true. Intuitively: When thinking of sheaves of functions on spaces, the surjectivity on fields is a statement about the value types of the functions only, while being a sheaf epimorphism would incorporate the local nature of the functions. Hence, if we seek to formalize something like the inclusion of continuous/regular/well-behaved functions ... 1 For your statement to be true you should assume that these discrete valuation rings contain your field$K$. You may assume that$F$is a normal curve. You know that for a closed point$p\in F$there exists an affine open neighbourhood$U\subseteq F$of$p$. Write$U=\mathrm{Spec}(A)$for some finitely generated$K$-algebra. Now$p$corresponds to some ... 0 Thanks to Hoot's remark, I tried working it out in more specific case. Let$k$be a field, and define$A=k[x,y]$,$f=x^2+y^2-1$,$B=A_f$,$C=A/(f)$,$X=\mathbb{A}_k^2$,$Y=X_f\cong Spec(A_f)$, and$Z=V(f)\cong Spec(A/(f))$. Now I'll let$\phi:Y\rightarrow X$and$\psi:Z\rightarrow X$be the maps induced from the natural maps$A\rightarrow B$and$A\...

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I think I'm missing something as it seems the answer is yes to me: Since $f$ is defined over $k$ (and $\sigma$ leaves both $k \subseteq K$ invariant), $f(\sigma{\overline{x}})=\sigma (f(\overline{x}))$. And since $f(\overline{x})$ is a $K$-point, $\sigma$ leaves it invariant, i.e. $\sigma(f(\overline{x}))=f(\overline{x})=y$. I didn't use $f$ is etale ...

0

Consider (1)...The boundary operator commutes with affine maps and (2)...The projection operator of k-space onto the (x,y) plane is affine. The assertion can be proved by contradiction once you have proved that (3)...the boundary of the boundary of any 2-D chain is zero. Indeed, if the boundary of the boundary of a k-D polyhedron was not zero then the ...

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