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If $f : X \to Y$ is a continuous map and $\mathscr{F}$ is a sheaf on $X$, then $f_* \mathscr{F}$ is the sheaf on $Y$ defined by the following formula: $$f_* \mathscr{F} (V) = \mathscr{F} (f^{-1} V)$$ This is called the direct image sheaf. You should verify that the formula indeed defines a sheaf.

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For real $x>1+\frac{1}{n-1}$ then $$x^n = (1+(x-1))^n> 1+n(x-1) = x+(n-1)(x-1)> x+1$$ Now if $z^n+z+1=0$ then $\|z\|^n = \|z+1\| \leq \|z\|+1$. So, with $x=\|z\|$, we have $x^n\leq x+1$, so $\|z\|=x\leq 1+\frac 1{n-1}$.

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This is indeed quite an interesting and subtle question! The point is that when $\mathcal{L}(-P)\subset \mathcal{L}$, every section $s$ of $\mathcal{L}(-P)$ is a section of $\mathcal{L}$ but its order of vanishing at $P$ seen in $\mathcal{L}(-P)$ is one less than its order of vanishing seen in $\mathcal{L}$. Proof: Since the problem is local, we may ...

3

No. Assume that $R$ is a field and $(A,\mathfrak{m})$ is a local ring such that $\mathrm{Spec}(R) \cong \mathrm{Spec}(A) \setminus \{\mathfrak{m}\}$. Then $A$ has exactly two prime ideals $\mathfrak{p} \subset \mathfrak{m}$ and $R \cong A_{\mathfrak{p}}$. The general fact $Q(A/\mathfrak{p}) \cong A_{\mathfrak{p}} / \mathfrak{p} A_\mathfrak{p}$ implies here ...

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A morphism of affine varieties is invertible if and only if it the associated map of affine coordinate rings is an isomorphism. In your case, this says that $F: k^n \rightarrow k^n$ is invertible if and only if the map $F^*: k[x_1,\dots,x_n] \rightarrow k[x_1,\dots,x_n]$ given by $g(x_1,\dots,x_n) \mapsto g(F_1,\dots,F_n)$ is a ring isomorphism. In ...

3

For a smooth affine scheme $X$ of finite type over $\mathbb C$ the de Rham cohomology $H^*_{dR}(X)$ is just the cohomology of the complex of global differential forms: $$H^*_{dR}(X)=H^*(\Gamma(X,\Omega^*_X))$$ In your case you have to compute the cohomology of the complex $$0 \longrightarrow \mathbb{C}[x,x^{-1}] \stackrel {d}{\longrightarrow} ... 3 In the case of line bundles, your question is related to the behaviour of the Picard variety in families, and counterexamples can be found by looking for examples in which the Neron--Severi rank jumps. E.g. let E_t be a one-parameter family of elliptic curves, chosen so that E_0 is CM, but E_t is not CM for t \neq 0. More precisely, choose the ... 2 Let me explain why the statement in the last paragraph of the question is false as written; it is true under the extra assumption that h has connected fibres. A counterexample for the statement in the question is the following: let h:C \rightarrow \mathbf P^1 be a 2:1 map from an elliptic curve. Let L=O(2). Then h^0(L)=3, but h^*(L) has ... 2 The idea underlying these questions is the structure group of the fiber bundle, and the issues raised here are related to the fact that the structure group is somehow specified by other structures and not determined by the topology of the bundle see this question for a general discussion of this. For both the Mobius band and the cylinder, it is possible to ... 2 \DeclareMathOperator{\Spec}{\operatorname{Spec}}$$\DeclareMathOperator{\codim}{codim}$$\Spec(\mathbb{Z}) cannot be realized as \Spec(A) \setminus \{m\} for a local ring (A,m) - the argument given here can essentially be adapted verbatim. Set X := \Spec(A), Y := \{m\}, U := X \setminus Y \ne \emptyset. Notice that \dim U < \infty \iff \dim X ... 2 Presumably, \pi(z,\ell)=z? It's not bijective, at least for n>1.$$\pi^{-1}(0)=\{(0,\ell)\mid\ell\in\mathbb CP^{n-1}\}$$It is a bijection when excluding the 0\in \mathbb C^n and \pi^{-1}(0) from the sets, because when z\in\mathbb C^n\setminus\{0\}, there is exactly one line \ell through zero containing z. 2 I dont think you have defined an action yet. If we take g \in K then we use the map \varphi to define the action where if (u,h) \in U\times K then$$g(u,h)=(u,gh)$$now you need to use the fact that K is the structural group to show this this is well defined. The orbit space is B since K acts transitively. If you have a section, set the ... 2 Yes, you essentially understand the situation : here is the classification of the points of a scheme X of finite type over a (completely arbitrary) field k. 0) The closed points are the points x\in X whose residue field \kappa(x) is a finite extension of k, i.e. [\kappa(x):k]\lt \infty 1) The other points y\in Y correspond bijectively ... 2 Torsion means torsion over \mathcal O_X. For a coherent sheaf on a curve, it is equivalent to having support a finite number of closed points. 1 I'd like to make some complement to Mohamed Hashi's answer. Let X be the blow-up of \mathbb{A}^2 at O=(0,0),that is$$X=\{((x,y),(u:t))\in \mathbb{A}^2 \times \mathbb{P}^1\mid xu=yt\}.$$Let \mathbb{A}^1_u,\mathbb{A}^1_t be the affine cover of \mathbb{P}^1,then$$\mathbb{A}^2 \times \mathbb{P}^1=(\mathbb{A}^2 \times\mathbb{A}^1_u) ...

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Vakil's question is about the fibres at various points, not the stalk. The fibre at points away from the cusp is one-dimensional, while at the cusp, the fibre is two-dimensional. This shows that the module is not free (it has torsion supported at the cusp, as Martin Brandenburg remarks). You are suggesting that you should localize at $(X,Y)$, but to ...

1

The case $ax^4-by^2\pm\ldots=0$ with $a,b>0$ is an open curve pretty much for the same reason that $ax^2-by^2\pm\ldots=0$ is an open curve as well, called hyperbola. Then all you have to do is compute $y_{_{1,2}}(x)=\dfrac{-(s+qx)\pm\sqrt{(s+qx)^2-4r(t+px)^4}}{2r}$ , the area being $\displaystyle\int_v^w|y_{_1}(x)-y_{_2}(x)|~dx=$ ...

1

A $2 \times 2$ symmetric matrix $M = \begin{bmatrix} a& b\\ b& c \end{bmatrix}$ defines a quadratic form by $f(x,y) = \begin{bmatrix} x & y \end{bmatrix} M \begin{bmatrix} x \\ y \end{bmatrix} = ax^2 + 2bxy + cy^2$. Thus a conic section (centered at the origin) can be regarded as a $2 \times 2$ symmetric matrix and vice-versa. An ellipse ...

1

Because $X$ and $Y$ are non-sigular, the conormal bundle to $Y$ in $X$ is a locally free sheaf on $Y$ of rank $n$. Choose an affine n.h. $U =$ Spec $A$ of $x$ in $X$ such that over the intersection $U \cap Y$ the conormal bundle becomes free. If we let $I$ be the ideal in $A$ cutting out $Y$ (so $U \cap Y =$ Spec $A/I$), then the conormal bundle is the ...

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Since any abelian variety is isogenous to its dual, the answer is yes.

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By definition of fiber bundle, there exists $\theta_{U,V}:U\cap V\to K$ such that $\phi_V(b,k)=\phi_U(b,k\cdot\theta_{U,V}(b))$ on $p^{-1}(U\cap V)$. Evaluating this equality on $(p(x),h_1)$, we get: $$\phi_U(p(x),h)=x=\phi_V(p(x),h_1)=\phi_U(p(x),h_1\cdot\theta(p(x)))$$ This means that $\theta_{U,V}(p(x))=h_1^{-1}\cdot h$. In particular, ...

1

Assume that a root $w$ of your polynomial lies outside the disk $\|z\|\leq \frac{n}{n-1}$. Then: $$\|w^n\| = \|w+1\|, \tag{1}$$ but, due to the following stronger version of the Bernoulli inequality, the LHS is greater than: $$\left(1+\frac{1}{n-1}\right)^n \geq e\left(1+\frac{1}{2n-1}\right),$$ while the RHS is at most $2+\frac{1}{n-1}$, so the equality in ...

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