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7

First, it is not quite true that there is "only one cohomology theory" for topological spaces. One should rather say that there are several such theories, but that for reasonable spaces, they give the same results. The reason why this is true in a category of reasonable topological spaces is that nice topological spaces can be "built up" from smaller ones ...


6

If $X =\mathbb C^n$ with coordinates $z_1,\dots,z_n$ and $p=(0,\dots,0)$ write $$0\to z_1\mathcal O_X+\dots+z_n\mathcal O_X\hookrightarrow \mathcal O_X\to \mathbb{C}_p\to 0\quad (\bigstar)$$ If $X$ is an arbitrary manifold of dimension $n$ write $$0\to \mathcal I_p\hookrightarrow \mathcal O_X\to \mathbb{C}_p\to 0 \quad (\bigstar \bigstar)$$ where ...


4

The space of conics in $\mathbf P^2$ can be thought of as the space of symmetric $3 \times 3$ matrices $M$, modulo scalars, and the rank of a conic $C$ is exactly the rank of a corresponding matrix $M$. (There is a 1-parameter family of such $M$ for a given $C$, but they are all multiples of each other, so the rank is well-defined.) So the hypersurface of ...


3

$\def\HH{{\mathcal Hom}}\def\Hom{{\operatorname{Hom}}}\def\N{{\mathcal N}}\def\P{{\mathcal P}}$This comes from checking the natural isomorphism from adjunction on each open set and using naturality to see it induces a morphism of sheaves not just sets. Explicitly, take an open set $U \subset Y$. Then $$ f_*\HH_X(f^*\N,\P)(U) = ...


3

Well, first you need a Hermitian form $H:\mathbb{C}\times\mathbb{C}\to\mathbb{C}$ such that $\Im H(\Lambda\times\Lambda)\subseteq\mathbb{Z}$. Using a little bit of linear algebra, it is easy to see that there is only one $\mbox{mod }\mathbb{Z}$. Explicitly, if $\Lambda=\langle 1,\tau\rangle$ for $\tau\in\mathbb{H}$, then every Hermitian form that satisfies ...


3

Consider the ideal $I = \langle x_0x_2 - x_1^2, x_0x_3 - x_1x_2, x_1x_3 - x_2^2 \rangle$. Denote $W = \mathbb{V}(I)$. Clearly $V \subset W$; the claim is $W \subset V$. Let $p = (a_0, \dots, a_3)$ be an element of $W$. If $a_0 = a_3 = 0$, then $a_1^2 = a_0a_2 = 0$ and $a_2^2 = a_1a_3 = 0$ so that $p = (0, 0, 0, 0)$, which is in $V$. Therefore assume $a_0 ...


3

Good observation! Indeed it happens often that filtered colimits preserve finite limits, and as for a morphism $f: X\to Y$ being mono is equivalent to $id, id: X\rightrightarrows X$ being a pullback of $(f,f)$, in these situations monomorphisms are preserved. You find this imposed as a condition in topos theory, where a geometric morphism $f: {\mathcal ...


3

This is only correct for elliptic normal curves, i.e for genus 1 curves in $\mathbb{P}^r$ with degree $r+1$. To see this, observe by Riemann-Roch that any degree $r+1$ line bundle $L$ on a genus 1 curve $E$, is very ample (for $r\geq 2$) and has $r+1$ sections. Moreover, any line bundle of degree $r+1$ on $E$ is isomorphic to $\mathcal{O}_E((r+1)p)$ for some ...


3

$\def\H{{\mathcal Hom}}\def\HH{{\operatorname{Hom}}}$Everything you wrote seems correct for the locally free case. Though its easy to overlook subtle things with these type of arguments, it looks good to me. For the second I believe there is a natural map $f^*\H_Y(E,F) \to \H_X(f^*E,f^*F)$. I will actually define a map $\H_Y(E,F) \to f_*\H_X(f^*E, f^*F)$ ...


3

Here is an alternative to the above answers with a more algebra-y flavor. (For those interested, this is exercise 1.2.8 of Cox, Little, and O'Shea.) Given $f(x,y) \in \mathbb{V}(X)$, let $g(t) = f(t,t)$. Then $g \in \mathbb{R}[t]$ vanishes on $\mathbb{R} \setminus \{1\}$. But a nonzero polynomial of one variable over a field has only finitely many roots, ...


3

For these kinds of questions, if you can see how to establish an isomorphism stalkwise, the key thing you need to do to promote this to a complete proof is to construct a morphism between the sheaves that you are trying to prove are isomorphic. Once you have a morphism, in order to check that it is an isomorphism, you can work on stalks; and then hopefully ...


2

Two lines in the projective plane intersect. This is a quadric surface in $\Bbb P^3$. (Think of the saddle surface $z=xy$ in $\Bbb R^3$. Note that if you fix $x=c$, you get a line $(c,y,cy)\subset\Bbb R^3$. For different values of $c$, these lines are disjoint.)


2

$T_1$ is equivalent to single point sets being closed. Let $(a_1,\dots,a_n)\in\mathbb{A}^n$. Consider the zero locus of the polynomials $x_i-a_i$ where $i$ ranges from $1$ to $n$. For your proof of non-Hausdorffness it looks like you are assuming that $\mathbb{A}^n$ has the cofinite topology, which is not true in general. $\mathbb{A}^1$ does have the ...


2

Perhaps I've misunderstood what you wrote, but it seems to me that your statement about the presheaf cokernel being a sheaf is false. E.g. take $X := \mathbb C^{\times}$ with its usual topology, let $\mathscr G := \mathscr O_X$ denote the sheaf of holomorphic functions on $X$, and let $\mathscr F$ denote the sheaf of locally constant functions on $X$ which ...


2

The ring you describe is the ring of global sections of the total space of the vector bundle $\mathcal{L}$, i.e. $\mathbf{Spec}(\mathrm{Sym}^\bullet(\mathcal{L}))$. (This notation is the "global Spec" construction for a sheaf of $\mathcal{O}_X$-algebras.) In general, this does not need to be a finitely-generated $k$-algebra (see this shocking ...


2

This actually depends on your viewpoint. For example $Y=\mathbb{R}\setminus \{0\}$ is not an affine subset (i.e. a Zariski-closed subset) of $\mathbb{R}$, but it is an affine variety, when viewed in a different way. We can define a map $V(xy-1)\rightarrow Y$ which is an isomorphism. $V(xy-1)$ of course is a Zariski-closed subset in $\mathbb{R}^2$. The ...


2

No. For any compact Riemann surface $X$ of genus $g\geq2$ the anticanonical bundle is a line bundle of degree $2-2g\lt 0$ and has thus no non-zero holomorphic global section: $\Gamma (X,K_X^{-1})=0 $ Edit This is an answer to the original question, which did not suppose $\Gamma (X,K_X)=0 $


2

A preliminary comment: you ask Does such a global section correspond to a meromorphic top form (giving a divisor in the divisor class of $K_X$? No, a section of $K_X^{-1}$ would give a divisor in the class $-|K_X|$. In general, for any line bundle $L$ with associated divisor class $|L|$, a nonzero global section of $L$ gives a divisor in $|L|$. More ...


2

It is essentially the preimage of $E$ but with multiplicities. If $p\in C$ is a point, then $$\pi^*(p):=\sum_{x\in\pi^{-1}(p)}(\mbox{mult}_x\pi)x$$ where $\mbox{mult}_x\pi$ is the local multiplicity of $\pi$ at $x$ (i.e., $\pi$ looks like $z\mapsto z^{\mbox{mult}_x\pi}$ at $x$). You can then extend this definition linearly to all divisors.


2

This is direct by Minimal Model Program on smooth surfaces. Since $-K_X$ is big, after running MMP, $X$ ends up with a Mori fiber space, which is $\mathbb{P}^2$ or $\mathbb{F}_n$ with $n\neq 1$. On the other hand, if $X$ is not a Mori fiber space, then by the Picard number assumption, $X$ can only be $\mathbb{P}^2$ blowing up one point, which is ...


2

First let me correct a mistake in your question: is it WRONG to think of this as an astroid. It is just 4 arcs of circles, NOT an astroid, though it looks somewhat similar. Second. The conformal map is certainly NOT fractional-linear (which you call Mobius). This conformal map is written explicitly in the paper I already referred to on MO: arXiv:1110.2696. ...


2

KDong has already given the answer. There are some points that might help understand how the equations describing $V$ were arrived at. Note that your set $V$ is defined by points whose co-ordinates are all monomials of degree 4 in $s$ and $t$. They are ordered in ascending powers of $t$. In general one can do the same with sets defined by all monomials ...


2

Try a change of variables $X = x +\alpha$ and $Y = y + \beta$ for suitable $\alpha$ and $\beta$, to put it in the required form. This will give equations on $\alpha,\beta$, that you will be able to solve because the characteristic of $k$ is not $2$ not $3$.


2

If the morphism of varieties $f:X\to Y$ is ├ętale it is automatically quasi-finite. One of the versions of Zariski's main theorem (see our friend Akhil's notes, Theorem 8.5) then implies that $f$ can be factored as $f=f'\circ j: X \stackrel {j} {\hookrightarrow} Y' \stackrel {g} {\rightarrow} Y$ where $j$ is an open immersion and $g$ is finite. Since you ...


2

Since $V$ is smooth, the interpretation is the same. The intersection $V\cdot V \in A^*(V)$ is $c_d(N_{V/X}) \cap [V]$ where $d$ is the codimension of $V$ in $X$ (i.e. the rank of $N_{V/X}$) and $c_d(N_{V/X})$ is the top chern class of the normal bundle. In the case that $V$ is a divisor, $d = 1$ so $N_{V/X}$ is a line bundle on $V$ and $c_1(N_{V/X}) \in ...


2

The answer is affirmative if $X$ is compact and compactness is also necessary. A reference for the first statement can be found in the discussion after Corollary 5.1.4, the item (ii) (page 235) in the book Complex Geometry by Huybrechts. This is also known as the Gauss-Bonnet-Formula and this Wikipedia Link contains a counterexample for the noncompact case. ...


2

Rewrite everything as $$3(x+y)^2+2x^2=(y+z)^2$$ looking at this modulo $3$, you can show that $3$ divides both $x$ and $(y+z)$ (an idea to show this: squares modulo $3$ are just $0,1$, and $3$ is a prime). In particular $9$ divides both $x^2$ and $(y+z)^2$. But now, looking this modulo $9$, you have $$(x+y)^2 \equiv 0 \mod{9}$$ hence $3$ divides $(x+y)$ as ...


1

As you correctly observe, there is no such isomorphism in general, just a morphism from the left hand side to the right hand side. There is a correct formula involving sheaf-hom, which is as follows: $$f_* \mathscr{H}{om}_X(f^* A, B) \cong \mathscr{H}{om}_Y(A, f_* B),$$ for a sheaf $A$ on $Y$ and a sheaf $B$ on $X$. This does follow from the definition ...


1

(I will take $n=m=1$, you will adapt in higher dimensions.) Let $P = X-Y$ and $Z = V(P)$, so that $Z$ is closed. Now $Z = \{(x,x)\;|\; x\in \mathbf{A}_k^1\}$. Now (easy) $Z$ is closed in the product topology if and only if $\mathbf{A}_k^1$ is Hausdorff, and this is not the case as $k$ is infinite, being algebraically closed, and as (independently of $k$ ...


1

The answer to the three questions is yes. If one wants to make it a little more precise, one could add the functor in play. This however isn't done in one line, which is why many people suppress it. The main ingredient of the proof is in fact Chow's theorem. Serre's GAGA is mainly concerned with the additional equivalence of coherent sheaves and cohomology ...



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