# Tag Info

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 ...

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 ...

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 ...

2

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.

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 ...

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 ...

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 ...

2

Set $I=I_0$, and $J=\cap_{i=1}^nI_i$. We have $\mathrm{reg}(I)\le1$, $\mathrm{reg}(J)\le n$, and $\mathrm{reg}(I\cap J)\le n+1$. (See Theorem 2.1 from this paper.) Now use the following exact sequences: $$0\to I\to I\oplus J\to J\to0.$$ $$0\to I\cap J\to I\oplus J\to I+J\to0.$$ From the first we get $\mathrm{reg}(I\oplus J)\le n$, while form the second ...

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 ...

1

Yes, $O_X(U) \cong k[x,y]/(y^2 - x^3)$. To give this full justification depends on your foundations, I guess, but you can think of this in the following way: The equation $y^2z - x^3$ cuts out some points in projective space, and convenient covers are given by the planes $z = 1$, $x = 1$, $y = 1$, which in projective coordinates means the set of points ...

1

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

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 ...

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 ...

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 ...

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 ...

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 ...

1

The zero locus of a monic (in $x$) polynomial $P(x,y)=0$ in $C^2$ is always path-connected. The multiset of roots varies continuously in $y$ for any suitable distance between multisets, such as edit distance based on a metric in $C$. If you want to trace the path of a particular $x$ as a function of $y$ this is not hard to extract from a given path of all ...

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 ...

1

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

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

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 ...

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

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 ...

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).

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