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The book says homogeneous polynomial. I think he wants to emphasize on the fact that $P(t_1,\dots,t_n)$ is a homogeneous element in $\bar A$. Let $f\in A[T]$, $f=a_0+a_1T+\cdots+a_nT^n$, such that $\phi(f)=0$. Suppose $(a)=I$. Then $f(a)=0$, that is, $a_0+a_1a+\cdots+a_na^n$. In $\bar A$ this leads to $a_ia^i=0$ for all $i$, so $a_i=0$ for all $i$ and thus ...

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Let me give a two-part answer: I will give a counterexample when $X$ is not quasi-compact, and I will prove that for $X$ integral and quasi-compact the result holds (without quasi-separatedness). To ease notation, since the only sheaf we use is $\mathcal O_X$, I will just write $\Gamma(X)_f \to \Gamma(X_f)$. (You'll thank me later.) Example. Let $X = ... 1 Just use the group law, which says if$L$, a line intersects$C$at 3 points$A,B,C$, then$A+B+C=O$. If$P$is of order 6, the tangent line at$P$meets it at least twice and if$B$is the third point, then$2\cdot P+B=O$and since$6P=O$, we get$B=4P$. Identical argument for$5P$, since$2\cdot 5P+2P=O$. 0 This question has been asked and answered on MathOverflow. I have replicated the accepted answer by user2035 below. If$f\in\mathbf Z[X]$is any monic polynomial, the solutions of$x(1-x)\cdot f(x)=1$are solutions of the unit equation. Take some$y\in U\setminus\mathbf R$. Since the substitution$z\mapsto1/(1-z)$leaves$S$invariant, we may assume ... 4 No, incidence geometry will not in the least help you with algebraic geometry. If you want to immerse yourself in classsical algebraic geometry consider Semple and Kneebone's Algebraic Projective Geometry (first published in 1952) but beware that you will probably find it difficult or at least disconcerting. Here is a review by Du Val of that book. In ... 0 I see the flaw in my argument in part 2). If have$d.C_1'\sim d.C_2'$. This means that$\mathcal{O}(C_1')^d=\mathcal{O}(C_2')^d$. But this does not mean that$\mathcal{O}(C_1')=\mathcal{O}(C_2')$. For example we for a degree$d $morphism, we have non trivial line bundle$L$on$Y$such that$L^d=\mathcal {O}_Y$1 To compute the$\operatorname{Ext}$in this case, you can indeed use the spectral sequence $$H^p(X, \mathscr Ext^q(\mathscr F, \mathscr G)) \Rightarrow \operatorname{Ext}^{p+q}(\mathscr F, \mathscr G),$$ arising as the Grothendieck spectral sequence of the composition of functors $$\operatorname{Mod}_{\mathcal O_X} \to \operatorname{Mod}_{\mathcal O_X} \to ... 3 This is not true. Let X be any topological space, equipped with a nonzero sheaf \mathcal{F} that has no global sections except 0 (such spaces and sheaves exist). Consider the trivial 1-element cover of X, and let \phi be the 0-morphism \mathcal{F}\to\mathcal{F}. Clearly, the morphism \phi induces an isormorphism ... 0 The group law is equivalent to the following. If L is any line and if it meets C at A,B,C (counted with multiplicity), then A+B+C=O. So, one easily sees that 2P, P, 3P are collinear and similarly for the other 3-tuple. Only thing remaining is to check that 2P, 4P are flexes. For 2P, let L be the tangent line at 2P. Then L meets C at ... 1 If I understand you correctly, you want to know why the derived category is additive, and what represents the sum f+g of two morphisms. (Note that the derived category of any abelian category is additive, not just the derived category of coherent sheaves). In terms of zig-zag, you have the following description of f+g : f and g are represented by ... 2 Low-tech way, following your hint. The family of conics passing through those points is 1-dimensional. This can be seen by counting dimensions (the space of conics is 5-dimensional, and each constraint subtracts one) or more explicitly by noting that the family is given by$$a(X^2-Z^2-YZ)+bZY=0$$for [a:b]\in \mathbb P^1. Conics are degree 2, and your ... 3 For a counterexample, consider X = \mathbb P^1_k, and let \mathscr F be any sheaf of \mathcal O_X-modules. Note that \Gamma(X,\mathcal O_X) = k, and every nonzero element in k is already invertible on all of X. Thus, X_s = X for all s \in \Gamma(X,\mathcal O_X)\setminus\{0\}, so the condition on \mathscr F is vacuous. (To find a sheaf of ... 1 Off the top of my head, the only examples I know where this holds are abelian varieties; in this case X is topologically a torus, and we in fact have H^k(X, \mathbb{Z}) \cong \bigwedge^k H^1(X, \mathbb{Z}). As Roland says in the comments, \mathbb{CP}^n for n \ge 1 is a counterexample. 1 Hartshorne does prove Serre duality for \mathbb P^n over any Noetherian ring (Theorem III.5.1). Locally, \mathbb P(\mathscr E) is of this form (e.g. on an affine cover that trivialises \mathscr E). To check that the duality commutes with restriction (on \operatorname{Spec} A), observe that the pairing$$H^0(\mathbb P^n, \mathcal O(d)) \times ... 1 0) First of all you probably want to assume that the base field is algebraically closed: if the base field is$\mathbb R$and$n=2$the image of$\phi_2$is not even algebraic! So let us suppose that$k=\bar k$. 1)$\textbf n=2k$even It is clear that the image of$\phi_n$is the curve$A_n\subset \mathbb A^2$with equation$y=x^k$, obviously isomorphic ... 1 Okay, assume the characteristic is$0$(over complex numbers). Let's stare at your last exact sequence: $$0 \rightarrow H^0(S,O_S(H-C))\rightarrow H^0(S,O_S(H)) \rightarrow H^0(C,O_C(H)) \rightarrow H^1(S,O_S(H-C)) \rightarrow ....$$ In the surface$S$,$C=H|_S$as divisor classes. So we know your first term has dimension 1 ($S$is smooth and simply ... 1 Since$c^Tx=0$supports a facet of your simplex, it is going to be generated by$n-1$of the vectors$b_i$, say they are the first$n-1$, so that$c^Tb_i=0$for all$i=1,\dots, n-1$. The fact that$H\cap C$is not empty tells you that$c^Tb_n>0$(take any point$r\in H\cap C$, so$r=\sum_{i=1}^n e_ib_i$for real numbers$e_i\geq 0$for all$i$, and ... 2 The "naive" coproduct of functors where you define$(\coprod_i F_i)(T)=\coprod_i F_i(T)$in not a sheaf in general. If you sheafify, (at least in the case you're interested in) you should get a description like this: an element of$(\overline{\coprod_i F_i})(T)$(the sheafification) is a decomposition$T_i$of$T$as a disjoint union$T=\coprod_i T_i$(same ... 1 For the even case, you are showing that the image is isomorphic to$\mathbb{A}^1$. It is not necessary (and not true) that$\varphi_n$is the isomorphism. In fact, your answer almost contains the map from$\mathbb{A}^1$to$\varphi_n(\mathbb{A}^1)$and the map in the other direction. (You will need to check that they are mutual inverses of course). In the ... 2 Let me expand a bit on your idea that$\operatorname{Sh}(X)$is the localisation of$\operatorname{Psh}(X)$at the multiplicative set$S$of morphisms that induce isomorphisms on all stalks (even though this may not be your main question). A better way to put it is: Lemma. Let$\mathscr C$be the Serre subcategory of$\operatorname{Psh}(X)$of presheaves ... 1 This is common notation in algebraic geometry. I have two schemes$X,T$with structure morphisms to another scheme$S$, and now$X(T)$is the set of$S$-morphisms$T \to X$. When$T$is the spectrum of some ring$R$we usually just write$X(R)$. In this situation,$S = \mathbb Z$(we really could have left this bit out here, since ring homomorphisms always ... 2 Your first isomorphism seems believable: just plug in any open set$U$on both sides. You need to check things like $$(f^*\mathscr G)\big|_{f^{-1}U} = \left(f\big|_{f^{-1}U}\right)^* \left(\mathscr G\big|_U\right),$$ etc, because you want to use the adjunction$\left(f\big|_{f^{-1}U}\right)^* \dashv \left(f\big|_{f^{-1}U}\right)_*$. But pulling it through ... 1 There is probably a more direct way to prove this, since it is apparently an exercise in Silverman, but one way to see it is: if$C$has genus$1$, then its Jacobian$E$is an elliptic curve, and$C$and$E$become isomorphic over any field$L$over which$C$obtains a rational point. 0 You are right, there was a tiny mistake in my answer, although it did not change anything. The degree of$V$is$4$, not$2$, but as you point out the calculation still leads to getting$\dfrac 12 E$. Furthermore, since you calculated that$K_X$generates the class group$\mathbb Z/2\mathbb Z$, it shows the original claim that$K_X$is not Cartier, but ... 0 You have slightly misunderstood the situation. There is an additional assumption, namely that the image is non-solvable. Now there is a classification of the subgroups of$\mathrm{PGL}_2(\mathbb{F}_p)$. (Probably the paper of Serre that they cite has it --- did you look at that? Otherwise, one place it is discussed is in Swinnerton-Dyer's article in LNM ... 0 Point one.$K_X$is indeed$\mathbb{Q}$-Cartier but not Cartier. I calculated class group (see my question). It is$ \mathbb{Z} / 2 \mathbb{Z} $.$K_X$represents the nontrivial element of class group. To see this we will construct an explicit section$\omega$, which equals zero precisely on divisor$D_1$. Pull back of this section on$\mathbb{A}^3$equals ... 2 Are you asking for integer pairs? If so,$(x+3)^2+y^2=13=2^2+3^2$, so the integer pairs on this circle can only be$(-3 \pm 2, 0 \pm 3)$, and$(-3 \pm 3, 0\pm 2)$. 0 I assume that by "the variety associated with$f_1,\dots,f_n$" you mean the variety$Y\subseteq k^n$cut out by the ideal$I=\{g\in k[x_1,\dots,x_n]:g(f_1,\dots,f_n)=0\}$. The image of$p$is always contained in$Y$, since for any$a\in X$and any$g\in I$,$g(p(a))=g(f_1(a),\dots,f_n(a))=0$by definition of$I$. So$p$can always be considered as a ... 1 Usually Schubert cells are constructed using the structure theory of semisimple Lie groups (in this case$SL(3,\mathbb C)$), but for this case, one can give an explicit description as follows. Start by fixing one flag, say the standard one$\mathbb C\subset\mathbb C^2\subset\mathbb C^3$. This will be the unique Schubert-cell of dimension$0$. The further ... 4 As it is pointed out in Matsumura, CRT, page 31, for a noetherian local domain$A$the equality$\operatorname{ht}\mathfrak p+\dim A/\mathfrak p=\dim A$holds for any prime ideal$\mathfrak p$iff$A$is catenary. So, we are looking for a noetherian local domain which is not catenary. There are no trivial examples, but you can find one at ... 2 There are prime ideals$\mathfrak p$in noetherian local rings$A$such that$\operatorname{ht}\mathfrak p+\dim A/\mathfrak p<\dim A$. For instance, let$A=K[X,Y,Z]_{(X,Y,Z)}/(XY,XZ)$, and$\mathfrak p=(y,z)$. Then$\operatorname{ht}\mathfrak p=0$(since$\mathfrak p$is minimal), and$\dim A/\mathfrak p=1$(since$A/\mathfrak p\simeq K[X]_{(X)}$). On ... 1 By your definition of regular functions ("A regular function on$U$is a rational function that is well-defined at all points of$U$"), they indeed do not form a sheaf, as$O_X(\emptyset)$is the entire field of rational functions on$X$, rather than$0$. An easy (if inelegant) way to fix this is to say that your definition is only the definition of a ... 1 Great question! Here is a counterexample: Let$(A,\mathfrak m)$be a DVR, and let$A \subseteq B$be a finite extension of domains such that$B$has exactly two primes above$\mathfrak m$. For example,$A = \mathbb Z_{(5)}$, and$B = \mathbb Z_{(5)}[i]$, with the primes$(1+2i), (1-2i)$lying above$(5)$. Then$Y = \operatorname{Spec} A$is the space ... 2 Yes, that sounds about right. Yes. The superscript means taking the invariant subspace under the action of$\tau$. In general, if$V$is a vector space and$G$is a group acting on it, then$V^G = \{ v \in V \mid gv = v \forall g \in G \}$. 2 Careful. You have shown that a the sections of a presheaf$\mathscr G$over a base for the topology on some space$X$are isomorphic to those of a sheaf$\mathscr F$($\mathscr G$,$\mathscr F$are assumed to be presheaves on$X$). This is not enough to conclude abstractly that$\mathscr G$is a sheaf isomorphic to$\mathscr F$; such a thing is clearly not ... 1 There is a problem in the conjecture. Over$\mathbb C$it is really easy to prove. I believe you mean$\mathbb{A}^n(\mathbb{C})$. Here is a way of proving it that I like and that can work for proving many things over rings whence you know it over$\mathbb{C}$First notice that for countability reasons,$\mathbb{C}$has an infinite transcendance basis. This ... 3 Here is a (translated) quote from G. Castelnuovo, "Sur les intègrales de différentielles totales appartenant à une surface irrégulaière", Comptes rendus hebdomadaires des séances de l'Acadeémie des sciences, Paris. 140, 23 Jan 1905. pp. 220-222: out of respect for Picard's profound research on surfaces admitting a group of birational automorphisms, ... 0 According to wikipedia "The name is in honor of Émile Picard's theories, in particular of divisors on algebraic surfaces." I would assume one of the references on the page probably contains what you want. 2 For a quick and dirty explanantion that offers no insight (as requested), take a$3\times 3$permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any$1$. Put$0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix. For example, for the ... 4 I think the following works, maybe there is an easier answer... First note that$\det E_{|L}=\bigotimes_{i=1}^{\dim V-1}\mathcal{O}_L(a_i(L))=\mathcal{O}_L(\sum a_i(L))$. Note also that$\det E_{|L}\otimes \mathcal{O}_L(2)=\det (V\otimes\mathcal{O}_{\mathbb{P}^n})_{|L}=0$. So$\det E_{|L}=\mathcal{O}_L(-2)$and$\sum a_i(L)=-2$. Then, note that, because ... 0 I found two mistakes in my proof. First, like Hoot said, I tried to shove the$D(f_{i})$into the same ambient affine space as closed subsets. Second,$(1−f_{i}T,1−f_{j}T)=1$does not imply$(1−f_{i}T)+(1−f_{j}T)=(1)$when$f_{i}$and$f_{j}$is in an integral domain. 4 They agree. In fact, you only need to assume quasicoherence; the key point is that the étale cohomology of a quasicoherent sheaf on an affine scheme vanishes in degrees$> 0$, just as for the Zariski topology. For more details, see tag 03DW in the Stacks project. 4 Sorry for resurrecting such an old post, but Section §1.6 of Hartshorne annoys me in the sense that the concept of "abstract nonsingular curve" is not used very much (if at all) in algebraic geometry. We provide an alternative version of that section (specifically, Lemma 1.6.5 through Theorem 1.6.9) that avoids use of said object. We start with a ... 1 Let$R$be the quotient of a polynomial ring$k[x_1,x_2,\dots]$in infinitely many variables over a field by the ideal generated by all products$x_ix_j$for$i\neq j$. Note that if$P\subset R$is a prime ideal, then there can be at most one$i$such that$x_i\not\in P$. It follows that if$P_i$is the ideal generated by all the$x_j$for$j\neq i$, ... 3 If one of the quadrics has maximal rank$4$(or better one of the quadrics in the pencil, which is in fact true), then it is projectively equivalent to the Segre variety. In this case, the other quadric cuts out a curve of bidegree$(2, 2)$on$\mathbb{P}^1 \times \mathbb{P}^1$. As the twisted cubic has bidegree$(1, 2)$, it follows that we get not only the ... 1 Lemma. Let$f \colon X \to Y$be a finite morphism, and let$\mathcal F$be any sheaf on$X_{\operatorname{ét}}$. Then$R^if_* \mathcal F = 0$for all$i > 0$. Proof. It suffices to show that$(R^if_* \mathcal F)_{\bar y} = 0$for all geometric points$\bar y \to Y$. But the stalk is computed by$H^i_{\operatorname{ét}}(X', \mathcal \pi^* F)$, where$X' ...

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First, your even more basic question. Indeed $\pi_*\mathbb{Z}_{\mathbb{A}^1}$ and $i_*\mathbb{Z}_x$ are not coherent sheaves. In fact, they are not even sheaves of $\mathcal{O}_X$ modules. And as Remy points out, $H^1_{ét}(X,\pi_*\mathbb{Z}_{\mathbb{A}^1})$ and $R^1\pi_*\mathbb{Z}_{\mathbb{A}^1}$ are not the same kind of objects, the first being a group, the ...

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The isomorphism comes about by looking at lines through the singular point $P$. Indeed, for any point $Q \neq P$ on $C$, the unique line in $\mathbb P^2$ through $P$ and $Q$ does not intersect $C$ anywhere else, because it already intersects with multiplicity $3$ (namely, once at $Q$ and twice at $P$). Thus, we get a map f \colon C\setminus\{P\} \to ...

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Since $T\subseteq\mathfrak{a}$, clearly $Z(\mathfrak{a})\subseteq Z(T)$. On the other hand, if $P\in Z(T)$, that means $f(P)=0$ for all $f\in T$. For any $P\in Z(T)$, the set $I=\{f\in A:f(P)=0\}$ is an ideal, and so since it contains $T$, it must contain $\mathfrak{a}$. Thus $P\in Z(\mathfrak{a})$. Since $P\in Z(T)$ was arbitrary, this shows ...

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Lemma. Let $X$ be a variety over $k$. Then $X$ is proper if and only if for every smooth proper curve $C$ and every $U \subseteq C$ open, any morphism $f \colon U \to X$ can be extended to a map $C \to X$. Proof. We extend one point at a time. Suppose $P \in C\setminus U$, and let $V = \operatorname{Spec} B$ be an affine open neighbourhood of $P$. Let \$\eta ...

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