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6

The natural transformation $\alpha$ is a monomorphism precisely when all of its components are injections. That way we may see $X$ as just a subpresheaf of $Y$. That is, for each object $C$ in $\mathbb{C}$ we have $X(C) \subseteq Y(C)$ and $\alpha_C$ is just this inclusion. I do not know what definition of separated you precisely use, but since you are ...

4

Given $\sigma \in Gal(K^s/K)$, let $\sigma_{K'} \in Gal(K'/K)$ be restriction of $\sigma$ to $K'$. The map $\sigma_{K'}:K' \to K'$ gives rise to $F(\sigma_{K'}):F(K') \to F(K')$ (by definition of a presheaf as a covariant functor on category of $K$-algebras) and so for $a_{K'} \in F(K')$, define $$\sigma_{K'} \cdot a_{K'} = F(\sigma)(a_{K'}) \in F(K')$$ ...

3

The standard argument for this question is as follows. Since $A$ is Noetherian and $M$ is finitely generated, it is finitely presented, so we have an exact sequence $A^p\to A^q\to M\to 0$ for some $p,q\in\mathbb{N}$. The map $A^p\to A^q$ is given by a $p\times q$ matrix over $A$. Then $U_n$ is given by the open set of points where at least one of the $q-n\... 3 Another way of defining sheaves that I find elegant is the following:$Y$is a sheaf (with respect to a given covering) if and only if for every covering sieve$S$on$U$represented as a subfunctor of$\mathsf{Hom}(-,U)$we have$y\mapsto(f\mapsto Y(f)(y)):Y(U)\to\mathsf{Nat}(S,Y)$is an isomorphism. Here,$\mathsf{Nat}(S,Y)$is the set of natural ... 3 Q1: On an affine scheme$Z$, we know that for any quasicoherent sheaf$\mathcal{A}$we have that$\mathcal{A}\cong \widetilde{\mathcal{A}(Z)}$. Applying that to the case at hand, we know that$f_*\mathcal{F}$is a quasicoherent sheaf on$Y$, so$f_*\mathcal{F}\cong \widetilde{f_*\mathcal{F}(Y)}$. But by the definition of the pushforward, we have that$f_*\...

3

One sentence answer would be look up Grothendieck complex and semi-continuity. Under your hypothesis there is a finite complex of $A$-projective modules, $P_.: 0\to P_0\to P_1\to\cdots\to P_n\to 0$ such that for any $Y\to \mathrm{Spec}\, A$, if $g:X\times_A Y\to Y$ is the induced morphism and $G$ is the pull back of $F$ to $X\times_A Y$, $R^kg_*G$ is the $k^{... 2 My question is: is the moral of the story here that$(\text{Spec}(R), \mathcal{O}_{\text{Spec}(R)})$as a pair remembers more information (e.g. about nilpotency) than the pair$(V(I), \mathcal{O}_{V(I)})$? Exactly. Let's take a look at the examples you consider$R_n=\mathbb{C}[x]/(x^n)$. Prime ideals$\mathfrak{p}\subset R_n$are in one-to-one ... 2 A very simple example is to take$X$to be the Riemann sphere$\Bbb C_\infty$. A differential is$\omega=dz$which has a double pole at$\infty$. So its divisor is$-2(\infty)$which is not principal, since its degree is nonzero. In general, a canonical divisor has degree$2g-2$where$g$is the genus, so it can only be principal if$g=1$, that is on an ... 2 Let$F$be an$\mathcal{O}_X$module. Given a global section$s\in F(X)$, one can define$\alpha: \mathcal{O}_X\to F$as follows: \begin{equation} \alpha(U): g\mapsto g.s|_U \end{equation} Here I mean by$g.$the action of$\mathcal{O}_X$on$F$. Conversely given a homomorphism$\alpha: \mathcal{O}_X\to F$, define \begin{equation} s=\alpha(X)(1)\in F(X) \... 2 The higher right derived functors generalize cohomology. If you take the map$X$to a point, then the right derived functors are exactly the cohomology of the the global sections functor. In particular, cohomology of the constant sheaf$\mathbb Z$gives singular/cw/de-rham cohomology. You should think of the higher derived functors as a way to patch ... 1 As you said, the assignment$(T\to S)\mapsto\operatorname{Mod}(T)$is a pseudo-functor, but this implies that$(h:T\to S)\mapsto \Gamma(T,h^*F)$is a functor. Indeed, if$u$is a map from$h:T\to S$to$h':T'\to S$(in other words that there is a commutative triangle) then there is a well-defined morphism :$$u^*:\Gamma(T',h'^*F)\to\Gamma(T',u_*u^*h'^*F)=\... 1 You will have exactness at the stalks, but in general you do not have exactness at sections over$U$. 1 Nicolas Hemelsoet has already written an answer, which is correct, but this answer will contain a different perspective and address the OPs questions in a different manner. Yes. What you've written correctly describes the action of$k$.$k(x)\simeq k$, so since$\newcommand\calF{\mathcal{F}}\calF(x)=M\otimes_A k(x)$is always a$k(x)$module, it is also ... 1 1 : I'm not sure I understand your notation, but since$\mathcal F$is a$\mathcal O_X-$module there is a map$R \times M \to M$and tensoring everything by$k(x)$gives the multiplication map$R \times M(x) \to M(x)$(it factors through$k(x) \times M(x) \to M(x)$. 2 : "Fiber" is fine I think. 3 : The intuition is as follows : imagine for simplicity ... 1 Denote$i: Z \to X$the closed immersion, assume$f$factors as$i \circ g$with$g: Y \to Z$, then$f^* = g^* \circ i^* $. So it suffice to show this with$f = i$, that is, the map$i^*\mathcal{I} \to i^*\mathcal{O}_X = \mathcal{O}_Z$is zero. It then suffices to show that$(i^*\mathcal{I})_x = 0$for every stalk. But$(i^*\mathcal{I})_x = \mathcal{I}_{i(...

1

The answer is no. Let our ring be $A=\Bbb{Z}/90\Bbb{Z}$, and let $M=A$. Let $\mathfrak{a} = (6,15) = (3)$. Then $(6)(15)=(0)$, so regardless of $g_1$ and $g_2$, they will agree on $(f_if_j)$. Defining $g_1(6)=12$, $g_2(15)=15$, we see that $g_1(30) = 60$, but $g_2(30)=30$. Thus $(g_1,g_2)$ can't be extended to a morphism on $(3)$. We'd need $g_i$ and $g_j$...

1

(To start, note that $U$ should be an affine open.) No, this is nowhere close to being true. Consider $X=\Bbb P^1_k$, $U=\Bbb A^1_k\subset \Bbb P^1_k$, and $I$ the sheaf of ideals of a closed point in $\Bbb A^1_k$, which WLOG we can take to be $0$. $I(U)=(x)$, which is infinite-dimensional as a $k$-vector space. On the other hand $0\to I\to \mathcal{O}_{\... 1 Use the universal property of the sheafification and the fact that a sheaf morphism is an isomorphism if and only if it is an isomorphism on all stalks. Let$\eta: \mathcal{F}|_U \to \mathcal{F}|_U^+$and$\theta: \mathcal{F} \to \mathcal{F}^+$be the canonical sheafification maps, which is are isomorphisms on the stalks. Restricting$\theta$to$U\$ we get ...

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