8 votes
Accepted

Branched cover in algebraic geometry

Let's tackle ramified/unramified first, since that's something that's pretty uniform across the literature. Definition (ref): A morphism of schemes $f:X\to S$ is unramified at $x\in X$ if there ...
  • 52.5k
7 votes
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Integral cohomology. follows from GAGA?

These are not at all the same. If $X$ is irreducible, then $H^q_{Zar}(X, \mathbb{Z})$ will be $0$ for $q>0$. Proof: If $U_1$, $U_2$, ..., $U_n$ are nonempty Zariski open subsets of $X$, then $\...
6 votes
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Problem in proving that $\mathbb{A}^2$ is not homeomorphic to $\mathbb{P}^2$

There exist two disjoint irreducible closed subsets both containing more than one point in $\mathbb A^2_k$ but not in $\mathbb P^2_k$ (Bézout).
5 votes
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projective space minus a closed point

No. As any map $f:X\to Y$ with $X$ a projective variety is closed, if $\Bbb P^n_k\setminus\{p\}$ were projective, then it's image under the natural embedding in $\Bbb P^n_k$ would be closed. But it's ...
  • 52.5k
5 votes

Function field of $\mathbb{P}^1$

The only regular functions on $\mathbb P^1$ are constants, so you have to pass to nontrivial open subsets to get anything interesting. But once you have deleted a single point, you are now looking at $...
5 votes
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Analytification of a smooth projective variety is a compact Kähler manifold.

Yes, this boils down to two facts, which you should be able to find in e.g. Huybrecht's Complex Geometry or Voisin's Hodge Theory and Complex Algebraic Geometry: I. $\mathbb{P}^n(\mathbb{C})$ is a ...
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5 votes
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Why are meromorphic functions on a smooth projective curve rational?

That’s a good question! Note that it is a (very) special case of the GAGA theorems/philosophy – in ”complete enough” situations (usually properness), analytic objects and morphisms should come from ...
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4 votes
Accepted

Dimension of projection of projective variety on hyperplane

For two varieties $X,Y$ in projective space, we define $J(X,Y)$ the join of $X,Y$, to be the union of all lines in $\Bbb P^n$ connecting distinct points in $X$ and $Y$. Now I claim that $\pi(X) = J(X,...
  • 52.5k
4 votes

Hartshorne's Remark 7.8.2

(7.3 $\implies$ 7.8.2 ) Suppose we are given a closed point $P \in X$ and a tangent vector $t \in T_P(X)$. Then, there exist $s \in V = \Gamma(X, \mathcal{L})$ by Proposition 7.7 such that $\phi(s_P) \...
4 votes
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Clarification about Ideal and zero sets of empty set in Varieties

There are many reasons, but I guess what I would think as the most important one is that it is the only definition for which $S\subseteq T \implies Z(T)\subseteq Z(S)$ (and the obvious similar ...
4 votes
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The projective closure of the twisted cubic curve

First we can verify that the image of $\overline{v}$ is closed: it's the vanishing locus of $xw=yz$, $xz=y^2$, and $z^2=xw$. Thus the image of $\overline{v}$ is a closed set containing $Y$, and it's ...
  • 52.5k
4 votes
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Relation between tautological line bundle and blow up at the origin

The way these definitions "talk to each other" has to do with the two projections onto the factors. If you map to $\mathbb C^{n+1}$, you get the blowup. If you map to $\mathbb P^n$, the same ...
4 votes
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universal property of Albanese variety

This is the definition of the Albanese variety. Presumably you mean something like why the dual of $\mathrm{Pic}^0$ is the Albanese variety in good situations? I've always liked the appendix to this ...
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4 votes

All varieties are quasi-projective?

That assertion is not a theorem, it is just informing you that those are the only varieties consideredin Chapter I and that's what you should understand when you read the word variety within that ...
4 votes
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Algebraic, Projective, and Riemannian Geometry: How do they interact?

Let me try to answer your question to some extent; given the vague nature of the question, there will be no canonical answer. You should think in terms of three different areas of mathematics: ...
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3 votes

Rational points on projective varieties: what is density?

A scheme $X$ has an underlying topological space $|X|$. When one asks whether the $k$-points are dense, one usually refers to density as a subset of $|X|$.
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3 votes
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Intersection between projective curves

Bezout's Theorem depends on the field you are working over. Roughly, the Theorem states that if you have curves of degree $d$ and $e$, then the curves will intersect in exactly $de$ points in the ...
  • 1,536
3 votes
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Isomorphism between projective varieties $\mathbf{P}^{1}$ and a conic in $\mathbf{P}^{2}$

1) The isomorphism $\phi$ is between $\mathbb P^1$ and a conic $Y\subset\mathbb P^2$, not a cone. (This conic is related to the cone in $\mathbb A^3$ with the same equation, but you should not ...
3 votes
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Existence of a hyper surface containing the singular locus of projective variety.

The following more general fact is true. Proposition: For any closed subset $S \subset \mathbf P^n$ and any point $p \notin S$ there is a hypersurface $W$ which contains $S$ but not $p$. (Your ...
3 votes
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Projecting a projective variety away from a linear subspace

Any linear subspace $L$ of dimension, after change of variables in $\mathbb{P}^{2n+1}$ can be assumed to be given by $x_0=x_1=\ldots=x_n=0$. Then the projection from $L$, $\mathbb{P}^{2n+1}\to \mathbb{...
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3 votes
Accepted

Is the Zariski topology on a variety $V$ a maximal Noetherian topology?

Given any Noetherian topological space $(X,\tau)$, if you adjoin one more closed set $C$ the topology always remains Noetherian. That is, let $\tau'$ be the topology generated by $\tau\cup\{X\...
3 votes
Accepted

Showing the join of two disjoint projective varieties is a projective variety.

From my reading, it appears that the assumption of disjointness is used in defining the "join" construction, not the proof of showing that it is projective. Indeed, if $X\cap Y$ is nonempty, then the ...
  • 52.5k
3 votes
Accepted

Why can a projective variety of dimension $n$ be covered by $n+1$ affine open subsets?

Let $X_d\subseteq \mathbb{P}^n$ be a projective variety of dimension $d$. Pick a hyperplane $H_d$ which does not contain any irreducible component of $X_d$. Then $X_{d-1}=X_d\cap H_d$ has dimension $...
3 votes
Accepted

Problem in proving a statement regarding projective closure of an affine variety.

Hint: we want to use a combination of $\alpha$ and $\beta$ (defined in the proof of proposition 2.2) to make this conclusion. Are there elements $f\in k[x_0,\cdots,x_n]$ so that $\beta(\alpha(f))=f$? ...
  • 52.5k
3 votes
Accepted

Degree of a finite morphism $f: X\to \mathbb P^1$ induced from a non-constant rational function in the Riemann-Roch space of a divisor

The divisor of $f$ is of the form $E_1-E_2$ where $E_1$, $E_2$ are two effective divisors with disjoint support such that $\deg(f) = \deg(E_1) = \deg(E_2)$. If $div(f)+D = E_1-E_2+D$ is effective then ...
  • 3,626
3 votes
Accepted

How to understand group action especially Galois action on a scheme?

Your attempt at writing things down more explicitly is good. The important thing to realize is that because the equivalence between affine schemes and rings is contravariant, the order of composition ...
  • 52.5k
3 votes
Accepted

There are no sections of blowup $Bl_0(\mathbb{A}^2)$ of affine plane at the origin

Here's one simple geometric reason: Consider any two distinct lines $L_1, L_2$ through the origin in $\newcommand{\AA}{\mathbb{A}}\AA^2$. The proper transforms of these two lines under the blowup at ...
  • 4,116
3 votes
Accepted

Isomorphism map keep smoothness?

Yes. An isomorphism $\varphi$ of varieties induces an isomorphism of local rings $\mathcal O_x\cong \mathcal O_{\varphi(x)}$ at every point, hence an isomorphism of the maximal ideals $\mathfrak m_x\...
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3 votes
Accepted

There is a line in $\mathbb{P^2}$ not passing through any of a finite collection of points

I’m assuming the base field is infinite, otherwise it can be a little trickier to state. This is equivalent to the statement: for any $p_1,\ldots,p_s \in \mathbb{A}^3 \backslash \{0\}$, there are $a,b,...
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