# Tag Info

27

Sheaves and sheaf cohomology were invented not by Serre, but by Jean Leray while he was a World War II prisoner in Oflag XVII (Offizierlager=Officer Camp) in Austria. After the war he published his results in 1945 in the Journal de Liouville. His remarkable but rather obscure results were clarified by Borel, Henri Cartan, Koszul, Serre and Weil in the late ...

17

Sheaf cohomology is just the elaboration of the following problem: you have a space and a covering, and you can do something you want on each set of the covering: can you do it on the whole space? This occurs in many places, from the Cousin problem in complex analysis to the construction of sections of fiber bundles to the pasting together of solutions ...

7

Hartshorne says that cohomology was first introduced to abstract algebraic geometry by Serre in his Faisceaux Algebriques Coherents paper (translated to English by a friend of mine). The FAC says We know that the cohomological methods, in particular sheaf theory, play an increasing role not only in the theory of several complex variables ([5]), but ...

3

The answer is still negative, though I feel the previous answer is incorrect. Let $L=p_1^*K_C, M=p_2^*K_C$ and then saying $0\to p_{1*}M^2\to p_{1*}M^2(\Delta)\to K_C\to 0$ (the last is surjective, since the corresponding $R^1$ is zero) splits is same as sayining after twisting by $K_C^{-1}$, the section $1$ can be lifted. This implies in particular ...

3

No, in any characteristic. Let $f:\mathbb{P}^1\times\mathbb{P}^1=X\to Y=\mathbb{P}^2$ be such a morphism. Then, the map can not be finite, since it is of degree one and $Y$ is smooth would imply that $f$ is an isomorphism. This is not true by Picard group considerations or many other considerations. So, one must have an irreducible curve $E\subset X$ such ...

3

a) An algebraic subset $X\subset\mathbb C^n$ has all its irreducible components of dimension $n-1$ if and only if it is the zero set $X=V(P)$ of some polynomial $P\in \mathbb C[T_1,...,T_n]$. b) Beware that $X=\{(1,0)\}\cup V(T_1)\subset \mathbb C^2$ is an algebraic subset of dimension $1$ (=the maximum dimension of its irreducible components) but ...

2

Note that the $d-$tule embedding is an isomorphic copy of $\mathbb P^n$ where the linear forms $O_X(1)$ corresponds to the $d$-forms of $\mathbb P^n$.($O_{\mathbb P^n}(d)$). (look up [Hartshorne II, Ex. 5.13]). This should help you to see your second question. To answer your original question, I will just compute the Hilbert polynomial of $v_d: \mathbb P^n ... 2 Let$L_i=\mathcal{O}_{\mathbb{P}^3}(1)|_{X_i}$. If the isomorphism$f:X_0\to X_1$preserves$L_i$(that is,$f^*(L_1)=L_0$), rest will easily follow. Now, since$K_{X_i}=L_i^{d-4}$, we have$L_0^{d-4}=f^*L_1^{d-4}$. But (may be one should assume char 0, though may not be necessary), Picard group of hypersurfaces in 3-space is torsion free and since$d>4$, ... 2 I will give an outline of the proof of this fact if all the sheaves involved are quasi-coherent. I claim that the original exact sequence does split. First recall the Proj construction. Let$A=\oplus_{n\geq 0}A_n$be a graded commutative ring, then$\operatorname{Proj} A$is as a set the set of homogeneous prime ideas of$A$not containing the trivial ideal ... 2 In the affine plane$\mathbb A^2_k=\operatorname { Spec}k[x,y]$over a field$k$the closed affine subset ("the cross") $$X:=V(x\cdot y)=\{x\cdot y=0\}\subset \mathbb A^2_k$$ is a connected noetherian topological space (in the Zariski topology), but its open subset $$X_0:=X\setminus V(x,y)=X\setminus \{(0,0)\}$$ is no longer connected. Indeed it is the ... 2 Let$X=\{0,1,2\}$with open sets$\varnothing,\{0\},\{1\},\{0,1\}$, and$X$.$X$is Noetherian and connected, but the open set$\{0,1\}$is not connected. 2 As a start, let us first derive the formula when the unit square is centered at origin. Let$a = x^2+y^2+\frac12$, the distance to vertex$v$is: $$\begin{cases} r_{++} &= \sqrt{a - x - y}, & v = (+\frac12,+\frac12)\\ r_{+-} &= \sqrt{a - x + y}, & v = (+\frac12,-\frac12)\\ r_{-+} &= \sqrt{a + x - y}, & v = (-\frac12,+\frac12)\\ ... 2 The answer is yes; Claborn showed every abelian group is a class group of a Dedekind domain, and C. R. Leedham-Green, "The class group of Dedekind domains," Trans. Amer. Math. Soc. 163 (1972), 493–500. doi:10.2307/1995734 gives a geometric construction using affine curves, if I remember correctly. 1 It's in the lemma 2.1. V(\mathfrak{a}) \cap D(f) = \varnothing is equivalent to V(\mathfrak{a}) \subset V(f), which according to lemma 2.1 means that \sqrt{(f)} \subset \sqrt{\mathfrak{a}}, so f \in \sqrt{\mathfrak{a}}. 1 This is carried out in a lot of detail in Voisin – Hodge theory and complex algebraic geometry I, section 11.1.2. The particular result you're interested in seems to be Corollary 11.15. 1 Let me try to say something as promised. First, an obvious point: slogans like this can never give the full picture. If you take this slogan completely literally, it would imply that if you knew the cone of curves, then you would know everything about the birational geometry of X. That would include, say, the whole birational automorphism group of X. ... 1 The group SO(n) \subset SO(n+1) by an n-1-connected map. Consequently for k < n-1 \pi_k(O(n+1)) = \pi_k(O(n)). I am euclideanizing SO(4,2) \rightarrow SO(6), and not considering for the time the hyperbolic aspects. So all we have to consider is the fundamental group \pi_1(SO(4,2)) The Serre fibration$$ SO(n) \rightarrow SO(n+1) \rightarrow ... 1 Suppose we have some equation with$n$variables (assuming this is all over a field$\mathbb{F} $): $$x_1 + x_2 +... + x_n = c$$ For some constant$c \in \mathbb{F}$. Then, we can take this equation and subtract$x_n$, and$c$from both sides: $$-x_n = -c + x_1 + ... + x_{n-1}$$ $$\Leftrightarrow x_n = c - x_1 - ... - x_{n-1}$$ From here, any ... 1 Hint. An isomorphism of local rings maps the maximal ideal to the maximal ideal. 1 Here is a partial answer to (a): We know $$z_0z_2=z_1^2$$ and $$z_0z_3 = z_1z_2.$$ Suppose first that$z_0$and$z_2$are non-zero. Multiply$F_2$by$z_0z_2$to get $$z_0z_2F_2 = z_0z_1z_2z_3 - z_2^2z_0z_2$$ which becomes $$(z_0z_3)^2 - z_2^2z_1^2$$ or after factoring $$z_0z_2F_2 = (z_0z_3-z_1z_2)(z_0z_3 + z_1z_2) = F_1(z_0z_3 + z_1z_2) = 0$$ since ... 1 I'll look at problem (a). On$U_0 = \{Z_0 \ne 0\}$you are intersecting the affine curves $$z_2 = z_1^2, \; \; z_3 = z_1 z_2.$$ From these equations already follows $$z_2^2 = z_1^2 z_2 = z_1 z_3,$$ so it looks like the twisted cubic. When$Z_0 = 0$,$V(F_1,F_2)$reduces to$Z_1 = 0;$i.e. the projective line $$\{[0 : 0 : z_2 : z_3]\}.$$ So$V(F_1,F_2)$is ... 1 The linear forms, quadratics, cubics, and quartics on the plane are vector spaces of dimensions 3,6,10,15 respectively. If we choose each point P generally, imposing the extra condition of vanishing at P will decrease each of the dimensions above by 1 (unless the dimension is already 0). If this helps, we are just saying that, if we have chosen$P_1$... 1 If the homogeneous coordinate ring of your projectively embedded$X$was$k[x_0, \ldots, x_n]/(f_1, \ldots, f_m)$, then your "trivially embedded$X$" has homogeneous coordinate ring$k[x_0, \ldots, x_n, x_{n+1}]/(f_1, \ldots, f_m, x_{n+1})$, which is obviously isomorphic to the ring you started with. This will be true whenever you take a linear embedding of ... 1 This is not a complete answer, but it is too long to write a single comment. Let$V$be a vector space of dimension$n+1$. Let$S^d V$denote the space of homogeneous polynomials of degree$d$on$V^*$(you can think$S^d V$as polynomials whose variables are the elements of a basis of$V$). Then we have$$\mathbb{P} V \to \mathbb{P} S^d V \to \mathbb{P} ... 1 I can't totally parse the post, but A Gröbner basis is a special generating set for an ideal in a ring$R$considered as an$R$-submodule of$R$. and A basis for a vector space is a special generating set for the vector space as an$\Bbb F$-module for some field (or even division ring)$\Bbb F$. Yes, if$1\$ were part of a Gröbner basis for an ...

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