# Tag Info

## New answers tagged algebraic-geometry

0

Given any vector space $V$ and a subspace $W^* \subset V^*$, $W^*$ is always the annihilator of $\{v|\lambda(v)=0 \forall \lambda \in W^* \}$(one inclusion is clear, the other follows from dimension reasons). Now apply this statement to $V=Sym V$ and $W^*=I$.

1

I feel like I'm basically reiterating what Dtseng said, but maybe this will clear some things up. Fix $X$ a scheme, and a set of line bundles $L_1,\ldots,L_n$. Denote $E = \bigoplus_{i=1}^n L_i$. We start by giving your graded ring a name. Definition. The section ring or ring of sections of $L_1,\ldots,L_n$ is defined to be $$R(X;L_1,\ldots,L_n) := ... 0 It may be that, once I develop a better intuition, this part of the proof will seem natural, but for now, I find it implausible and incomprehensible. I want to propose an alternative for verifying that the relative conormal sequence is exact in the case of a nonsingular subvariety of a nonsingular variety (per the attendant definitions in H) using an ... 0 It may be that, once I develop a better intuition, this part of the proof will seem natural, but for now, I find it implausible and incomprehensible. I want to propose an alternative for verifying that the relative conormal sequence is exact in the case of a nonsingular subvariety of a nonsingular variety (per the attendant definitions in H) using an ... 0 Let f \in S=k[x_1,x_2,x_3,x_4] be a quadratic form. Then we have an isomorphism (f) \cong S(-2) of graded S-modules, and so the regularity of (f) is 2 (irrespectively of whether the corresponding conic lies in a plane or not). Now let \ell_1,\ell_2 be two disjoint lines in \mathbb{P}^3, and so the ideal of their union is I_{\ell_1} \cap ... 1 Try A =\mathbb C, B = \mathbb C[x], and \mathfrak q = (x). 0 The dimension at x can be greater than n: for instance take X to be line and a plane intersecting at a point, and let U be the part of the line that does not intersect the plane. Then the dimension increases from one to two at the singular point. When x is a closed point, the dimension at x can only increase: here I'm going to use a ... 2 The question you are asking is ill posed. In order to 'differentiate' a section of a vector bundle, you need a connection of the vector bundle \mathcal{E}. There are a few perspectives on this, but roughly a connection should give you something like \mathcal{E}\rightarrow \mathcal{E} \otimes_{\mathcal{O}_X} \Omega_X^1. The de Rham differential is a ... 2 As for your example with polynomials: Let me just introduce some notation: Let's call Functions(\mathbb{R}^2)=\mathbb{R}[x,y], Functions(\mathbb{R})=\mathbb{R}[x]. And the projections, in coordinates, are p_1: \mathbb{R}^2 \rightarrow \mathbb{R}, (x,y) \mapsto x, p_2: \mathbb{R}^2 \rightarrow \mathbb{R}, (x,y) \mapsto y. The conormal sequence ... 0 Here is a possible way of computing the degree. First note that the map Y^d\to X is invariant under the action of S_d, the symmetric group. Also, it is easy to see that the generic degree of this map is precisely the order of S_d, namely d!. So, to calculate the degree of X, suffices to identify the pull back of the hyperplane section in Y^d. If ... 1 Recall that$$\mathrm{reg}(M) := \sup\{\mathrm{end}(H^i_{R_+}(M)) + i : i ∈\mathbb N\}=\inf\{m \in\mathbb Z: H^{i}_{R_+}(M)_{n-i}=0\ \forall i \in \mathbb N\ \forall n > m\},$$where \mathrm{end}(N) := \sup\{n ∈ \mathbb Z: N_n\ne0\}. Set d=\mathrm{reg}(M_1). If d>\mathrm{reg}(M_2) and d>\mathrm{reg}(M_3)+1, then H^i_{R_+}(M_2)_{d-i}=0\ ... 0 Suppose that T acts on G. Then we have an action T \times G \to G. This gives a coation \varphi: \mathbb{C}[G] \to \mathbb{C}[T] \otimes \mathbb{C}[G]. An element f \in \mathbb{C}[G] is called homogeneous of degree \lambda \in \mathbb{C}[T] if \varphi(f) = \lambda \otimes f. Therefore there is a multigrading on \mathbb{C}[G] corresponding to ... 1 Let R = \mathcal{O}_X(U). Take x\in U, with \mathfrak{p} \subset R the corresponding prime ideal. If x\notin V(b), this says that b\notin \mathfrak{p}, so b_\mathfrak{p} is a unit in R_\mathfrak{p}. Likewise, if x\notin V(a), then a_\mathfrak{p} is a unit in R_\mathfrak{p}. So a_\mathfrak{p} / b_\mathfrak{p} is a well-defined unit ... 1 Let us denote by (\;\,.\;) the intersection pairing on your cubic surface X and by \alpha the divisor class e_0-e_1-e_2-e_3, then the quadratic transformation corresponds to$$s_{\alpha} : \left\{\begin{array}{ccc} \mathrm{Pic\;} X & \longrightarrow & \mathrm{Pic\;} X \\ v & \longmapsto & v+(v.\alpha)\,\alpha ...

1

I don't know if there is a nice formula for stalks and I doubt of it because it is not mentionned in reference books. Moreover the exact sequences you mentioned are generally not proved with stalks. Here is a way to do it. The first thing to notice is that if $F$ is flabby, then $\Gamma_Z(F)$ is flabby. Once you have proved that, you can for example show ...

0

You can construct a "universal (or tautological)" family of extensions $\mathcal{E}$ on $X\times Ext^1(F'',F')$: $0\rightarrow p_1^{*}F'\rightarrow \mathcal{E} \rightarrow p_1^{*}F''\rightarrow 0$, with $p_1: X\times Ext^1(F'',F')\rightarrow X$, such that for every $e\in Ext^1(F'',F')$ the restriction $\mathcal{E}_e$ is just $0\rightarrow F'\rightarrow ... 4 If$g$acts by scalar multiplication, then its image in$\text{GL}(W_{\lambda})$is central. This need not imply that$g$itself is central; for example,$G$could be a product of two simple groups,$W_{\lambda}$could be an irrep of one of them, and$g$could live in the other. But it's true if$W_{\lambda}$is a faithful representation. 1 The following is an idea based on Cassel's Lectures on Elliptic Curves (see chapter 20, exercise 3). Consider the real third-root of$a$, say$a^{1/3}$(the other roots are$\rho{a^{1/3}}$and$\rho^{2}{a^{1/3}}$where$\rho$is a primitive third-root of unity). Define$\phi:C\rightarrow{E}$by sending$(x,y,z)$to ... 0 I'm beginning to doubt about my claim. Firstly, consider the prime polynomial$Q=Y^2+(X^2-1)^2\in\mathbb{R}[X,Y]$. It generates a prime ideal in$\mathcal{O}(\mathbb{R}^2)$that has 2 disconnected points :$(1,0)$and$(-1,0)$. So its variety is reducible, even though$Q$is prime. Secondly, Bochnak, Coste and Roy, in their Real Algebraic Geometry, don't ... 3 Using the Segre embedding$\mathbb P^1 \times \mathbb P^1 \hookrightarrow \mathbb P^3$given by$(x_0:x_1)x(y_0:y_1) \mapsto (x_0y_0,x_0y_1,x_1y_0,x_1y_1)$, we can rewrite the curve as an intersection of two hypersurfaces in$\mathbb P^3$. Explicitly, these hypersurfaces will be$xw=yz$and$x^2+y^2+z^2+w^2=xw$. These are both quadrics, so that we have a ... 1 It seems to me you're asking what the so-called "Mapping class group" of these manifolds are. This is the group of diffeomorphisms, modulo the subgroups of diffeomorphisms isotopic to the identity (those that can be "continuously deformed" into the identity diffeomorphism). The mapping class group of$S^1$is$\Bbb Z/2$(the nontrivial element is ... 0 I was so sure that this result was going to be false, and wrote something below saying to look at using relative proj instead. If that's what you were looking for, the original answer is at the bottom. However, I think the claim is true if$X$is projective, so, if you're interested, please check my work on this. If$X$is quasi-projective, you have way too ... 1 This answer is a bit verbose, since I don't know where your confusion lies. Hopefully it helps! Let us first record the version of Bertini that Hartshorne wants to use: Bertini's Theorem [Hartshorne II, 8.18, 8.18.1 and III, 7.9.1]. Let$Y$be a subvariety of$\mathbf{P}^n_k$where$k$is an algebraically closed field, and where$Y$has at most finitely ... 0 If the intersection is empty, then it is finite, since the empty set is a finite set. So you don't have to consider that possibility separately. And in fact, it can be empty: for instance, if$f=x$and$g=x+1$. 2 The answer depends on what your ambient space is, either projective or affine. In projective space the intersection will be non-empty. In affine space the intersection may be empty to take a simple example. The curves$x=0$and$x=1$have no point of intersection on affine plane, they are parallel lines. But the projective curves$x=0$and$x=z$have and ... 3 The result is true, and it is a particular case of Proposition 2.2 from this paper. 0 To avoid any misunderstanding, the way I read your definition, if$k=\mathbb{R}$and$V=\mathbb{R}$, you are considering the ring of rational functions that have no poles on real numbers, so it includes things like$\frac{1}{x^2+1}$but not$\frac{1}{x}$. And you are asking (in that particular case) whether the maximal ideals of that ring are the real ... 0 Now that OP has precised is definition of regular functions, I believe that this result is not true since for$k^n$,$(0)$is a prime ideal for the set of regular functions, 0 Take$x,y$as "parameters"and$z$as your variable. Then by the argument principle the number of zeros in circle about$z$is $$n(x,y)=\frac{1}{2\pi i}\int_C\frac{f^{\prime}}{f} dz$$ Now this is a continuous function of$x,y$so an isolated point is impossible. 0 The existence of local lifts is open in general, even for$F = \mathbf{Q}_p$and general$n$. One expects the stronger result that there exist de Rham lifts. There is recent work of Gee-Hezig-Liu-Savitt on this problem (https://cms.math.ca/Events/winter15/abs/pdf/ant-fh.pdf). However, I do know some people who have an idea to answer the general case. I can ... 1 I know of two more references, but I'm not an expert in the topic, so I don't know how good they will be for you: Section II.7 in Borel–Ji's Compactifications of Symmetric and Locally Symmetric Spaces discusses the construction, but it is really just an outline that refers to the de Concini–Procesi paper. Chapter 6 in Brion–Kumar's Frobenius Splitting ... 1 Just so this has an answer, let me expand upon what I said in the comments slightly. So, let us begin with the following trivial observation: Observation: If$f:E\to E'$is an isogeny of elliptic curves over$k$, with$\mathrm{char}(k)=p>0$, then$f$and$\widehat{f}$are separable if and only if$p\nmid\deg(f)$. Indeed,this follows immediately ... 1 Since$\pi$is locally trivial, it is an open map. Thus the result follows by http://stacks.math.columbia.edu/tag/004Z which states: Let$f : X \to Y$be a continuous map of topological spaces. Assume that (a)$Y$is irreducible, (b)$f$is open, and (c) there exists a dense collection of points$y \in Y$such that$f^{-1}(y)$is irreducible. Then$X$is ... 2 This follows easily from representation theory of group$G = Spin(10)$(the simply connected group of type$D_5$). The spinor variety is the homogeneous space$G/P$, where$P$is the maximal parabolic group corresponding to the simple root$i_5$(for the standard enumeration of vertices). The 16-dimensional spinor representation --- the space of global ... 2 Let$A=$focus of parabola$=(-1,-1)P=(7,13)G=$foot of perpendicular from$A$to tangent at$PV=$vertex of parabola$a=AV\ell=AP=\sqrt{260}\theta=$angle between$AP$and$PG$. The tangent at$P$is$y=3x-8$(given) which is the equation for$PG$. Slope of$AP$,$m_1$=$\dfrac74$. Slope of$PG$,$m_2$=$3$. ... 0 You can consult the following article: W. Lichtenstein "A system of quadrics describing the orbit of the highest weight vector", Proc. AMS 84 (1982), no. 4, 605–608, where it is proved that homogeneous varieties in their minimal embedding have ideal generated by quadratic forms. You can deduce that they are ten going to lines section and finding a canonical ... 1 Let$\mathbb{G}_m=\operatorname{Spec}k[t,t^{-1}]$be the multiplicative group. Consider the map$\pi:\mathbb{G}_m\rightarrow\mathbb{G}_m, t\mapsto t^2$. The finite group$\mathbb{Z}/2\mathbb{Z}$acts on the fiber by$t\mapsto -t$and the action is sharply transitive. However the map is not locally trivial for the Zariski topology. In fact the preimage of ... 2 Let$P(7,13),F(-1,-1)$. Also, let$T$be the intersection point of the line$y=3x-8$with the axis of symmetry. Let$V$be the vertex, and let$K$be the point on the axis of symmetry such that$PK$is perpendicular to the axis. We use the followings (for the proof, see the end of this answer) : (1)$PF=TF$(2)$VT=VK$(3)$\text{(the length of the ...

1

We have $H^n_{\mathfrak m}(I)=H^n_{\mathfrak m}(S)\ne0$. But $\max\{i:H^n_{\mathfrak m}(S)_i≠0\}=−n$, so for $d≥−n+1$ we have $H^n_{\mathfrak m}(S)_d=0$. This shows that the same conclusion holds for $m≥0$ and $d≥m−n+1$. At this point I guess (but I'm not sure!) they assume $m≥0$ in this lemma (otherwise (c) is automatically satisfied).

1

https://www.researchgate.net/publication/251217821_General_Theory_of_Embeddings But you have to have an account to see it

4

There's not going to be an associated vector bundle in general; you need your input data to have some connection to the topology on $\mathbb{R}$. Whatever construction you make, to have any chance of being "correct", it should definitely be functorial with respect to isomorphisms. But in the case $\mathcal{E}=\mathcal{O}_X$ (where the associated vector ...

1

When you visualize what $\text{Spec } \mathbb{R}[x, y]/(y^2 - x^3 + x)$ looks like to an algebraic geometer, you should not be visualizing its real points: rather, you should be visualizing its complex points, together with the action of complex conjugation on them. The fixed points of complex conjugation are the real points, but there is a much richer set ...

0

No, there is no such formula. If the sides are of length $\ell$, then all you know is that the area is between $0$ (when you have a very very thin, elongated rhombus) and $\ell^2$ (when you have a square). That is, $0 < \text{Area} \leq \ell^2$.

1

For your first question: $X_g\subseteq X_f$ means every prime ideal containing $f$ contains $g$. However, the intersection of the prime ideal containing $f$ is the radical of the ideal generated by $f$. In particular, $g$ is in this radical. So by definition of radical, $g^n\in (f)$ for some $n$. For your second question: The term localisation map refers to ...

2

If $X_g \subset X_f$ then $V(f) \subset V(g)$. So, every prime ideal that contains $f$ also contains $g$, hence $g \in \sqrt{(f)}$. It means there exist some $n$ such that $g^n \in (f)$. (Edit: leibnewtz has pointed out a typo in the definition of the map) Let's say $g^n = fh$, for some $h \in R$. The restriction map is just the map $R_f \to R_g$ that ...

1

Why do you think $\operatorname{depth}_SI=3$? I'd say that it is $1$.

1

The statement is weirdly written. Let me try to say it more conceptually. Assume $X$ is irreducible, for convenience. (And also the ideal cutting out $X$ is radical.) In this setting, it is a (difficult) theorem that $X$ is smooth on an open set, hence "generically" the dimension of $m_p/m_p^2$ is the dimension of the variety. Hence the set described is ...

2

Yes: Since $\mathcal O_{Z,z}\to\mathcal O_{Y,y}$ is integral, the induced morphism $\mathcal O_{Z,z}/\mathfrak m_z \to \mathcal O_{Y,y}/\mathfrak m_y$ is also integral. In other words, $K(y)=\mathcal O_{Y,y}/\mathfrak m_y$ is an algebraic extension of $K(z)=\mathcal O_{Z,z}/\mathfrak m_z$ via $f^\sharp$, in particular $y$ and $z$ are birational. With ...

3

These two schemes are certainly not isomorphic to each other, as the first has one dimensional tangent space and the second two dimensional. Clearly any such scheme must be affine, so it is just a case of classifying local rings $A$ of vector space dimension three over an algebraically closed field $k$. Let $m$ be the maximal ideal. Then $m$ indues a ...

1

There are many ways of proving this. I assume that when you say `dominated' you mean $mA-D$ is effective and $f$ is non-constant. Clearly, suffices to prove this for a single point $P$ as $D$. If $P$ is in the support of $A$, then $A-P$ is effective and we are done. So, assume not. So, $f$ is regular at $P$ and change $f$ to $f-f(P)$ and then $A$ is ...

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