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For example, the curve $C: X^3 + pY^3 + p^2Z^3 = 0$ is smooth and doesn't have a $\mathbb{Q}_p^\text{ur}$-point. If $[x, y, z]$ is a $\mathbb{Q}_p^\text{ur}$-point on $C$, then we may choose the projective coordinates in such a way that $\min(v(x), v(y), v(z)) = 0$, where $v$ stands for the $p$-adic valuation. From the equality $x^3 + py^3 + p^2z^3 = 0$, we ...

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Question: "I'm trying to understand Grassmannian and Plücker relationship and I'm having trouble grasping the basic idea." Answer: If $k$ is any field and $W \subseteq V$ is an $m$-dimensional $k$-vector subspace of an $n$-dimensional $k$-vector space it follows grassmannian $\mathbb{G}(m,V)$ is a parameter space, parametrizing $m$-dimensional sub ...

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$$w^4 + x^4 = y^4 + z^4 \tag{1}$$ Let $w=mt+a, x=t-b, y=t+a, z=mt+b,$ then we get $(-4a-4b-4bm^3+4am^3)t^3+(-6b^2m^2-6a^2+6b^2+6a^2m^2)t^2+(-4b^3m+4a^3m-4a^3-4b^3)t=0$ Hence let $$m = \frac{a^3+b^3}{-b^3+a^3}$$ then we get $$t = \frac{3(-b^3+a^3)b^2a^2}{a^6-2b^2a^4-2b^4a^2+b^6}$$ Thus we get a parametric solution as follows. $w = a(a^6+b^2a^4-2b^4a^2+3b^5a+b^... 3 What you write is not quite correct - the category of quasi-coherent$\mathcal{O}_R$-algebras is equivalent to the category of all$R$-algebras (with no finite-generation hypothesis). This can be broken down in to two parts: first, the fact that categories of$R$-modules and quasi-coherent$\mathcal{O}_R$-modules are equivalent; second, the fact that an ... 3 The point here is to note that the equation $$x^4 + w^4 - y^4 - z^4 = 0$$ cuts out a surface$X \subset \mathbb{P}^3$. Indeed, all (primitive, i.e.,$\operatorname{gcd}(x,y,z,w) = 1$) integral solutions are in bijective correspondence with rational points on the surface$X$. Now, one might hope for a solution with$2$parameters (i.e., a birational map$\...

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From L. E. Dickson, History of the Theory of Numbers, Volume II, Chapter XXII, page $644$ $$A^4+B^4=C^4+D^4.$$ L. Euler$^{165}$ took $\,A=p+q,\;\;D=p-q,\;\;C=r+s,\;\;B=r-s\,$ and derived $$(1)\qquad\qquad\qquad pq(p^2+q^2)=rs(r^2+s^2).$$ Set $\,p=ax,\,q=by,\,r=kx,\,s=y.\,$ Then $$y^2/x^2=(k^3-a^3b)/(ab^3-k).$$ If $\,k=ab,\,x=1,\,$ then $\,y=\pm a,\,C=\... 3 The notation$E(-mp)$means$E\otimes_{\mathcal{O}_X} \mathcal{O}_X(-mp)$, where the latter sheaf is the line bundle associated to the divisor$-mp$. Context clues suggest that$X$is a curve,$p$is a point, and$m$is an integer - then$\mathcal{O}_X(-mp)$can be considered as the sheaf of rational functions which are regular outside of$p$and have poles ... 3 You seem to be missing a small but important detail in the definition of the fiber category. Given a functor$F : \mathcal{C}\to\mathcal{D}$and an object$X\in\mathcal{D},$the fiber of$F$over$X$is defined to be the category of$\widetilde{X}\in\mathcal{C}$such that$F(\widetilde{X}) = X$with morphisms$u : \widetilde{X}\to\widetilde{X}'$such that$F(...

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Just for clarity, the exercise is in section 2.5 of the book. The inclusion $\langle LT(g_1),\ldots,LT(g_t)\rangle \subset \langle LT(I) \rangle$ always holds, as $\{g_1,\ldots,g_t\} \subset I$ and $\langle LT(I) \rangle$ is the ideal generated by elements of $LT(I) = \{LT(f) \mid f \in I, f \neq 0\}$. You have correctly identified the leading terms in the ...

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Both directions can be proven if one observes that the map $m \mapsto m \otimes 1$ (which, by abuse of notation, was also denoted as "$\phi$") factors as $M \to M \otimes_R R/\mathrm{ker}(\phi) \to M \otimes_R S$ where the first map is injective (in fact, an isomorphism) if and only if $\mathrm{ker}(\phi) \subseteq \mathrm{Ann}(M)$, and the second ...

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Note that \begin{align*} \mathrm{Res}_{V_i, W}(f_i) &= \mathrm{Res}_{V_i \cap V_j, W}(\mathrm{Res}_{V_i, V_i \cap V_j}(f_i)) \\ \mathrm{Res}_{V_j, W}(f_j) &= \mathrm{Res}_{V_i \cap V_j, W}(\mathrm{Res}_{V_j, V_i \cap V_j}(f_j)) \end{align*} so if $W$ is as you mentioned in the question, we can write $$\mathrm{Res}_{V_i \cap V_j, W}(\mathrm{Res}_{V_i, ... 2 Answer: You may have seen the "flatness criteria" in Milnes "Etale cohomology" (2.7d): Let f:A\rightarrow B be be a flat map of rings with A \neq 0. It follows f is faithfully flat iff for any maximal ideal \mathfrak{m}\subseteq A it follows f(\mathfrak{m})B \subsetneq B is a strict ideal. In your case let$$0 \rightarrow I:=(...

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If by "curve of degree $n$" you mean the zero-set of an $n$-th degree polynomial in two variables, such polynomials have $\binom{n+2}2$ coefficients, one of which can always be set to $1$ by scaling. Each point specified to be on the curve provides an equation. With $\binom{n+2}2 - 1$ points we have $\binom{n+2}2 - 1$ equations in the $\binom{n+2}2 ... 2 This map is not the identity. It should have a graded kernel $$I = \bigoplus_{m \ge 0} I_m \subset \bigoplus_{m \ge 0} S^m H^0(X, \mathcal{O}_X(1)),$$ which will tell you what subvariety of$\mathbb{P}^n$is the image of the embedding$X \hookrightarrow \mathbb{P}^n$: we have$\mathop{im}{X} = V(I)$. Note also that it is certainly not necessarily the case ... 2 Let's assume$n\ge 3$, so the target is not$\mathbb P^1$and$\pi_p|_C:C\to \pi_p(C)$is birational. Choose a general hyperplane$H\subseteq \mathbb P^{n-1}$that intersects$\pi_p(C)$transversely at$m$points. So $$\deg(\pi_p(C))=\#(H\cap \pi_p(C))=m.$$ Now, the cone of$H$and$p$defines a hyperplane$\tilde{H}$of$\mathbb P^{n}$, which intersects$C$... 2 You don't need proper here, just Krull's height theorem. First, replace$0$or$\infty$in$\Bbb P^1$by$0$in$\Bbb A^1$by taking the fiber product along either standard open immersion$\Bbb A^1\to\Bbb P^1$so that we're considering the map$f':X\times_{\Bbb P^1}\Bbb A^1\to \Bbb A^1$and looking at$f'^{-1}(0)$. Next, cover$X\times_{\Bbb P^1}\Bbb A^1$by ... 2 There is a corresponding statement for varieties, and it's exactly what you intuit. To see why this is the case, recall we have an (inclusion reversing) correspondence between radical ideals and subvarieties of$V$. So asking if $$\mathcal{V}(\mathfrak{a}) \supseteq \{ x \}$$ is asking if the variety of$\mathfrak{a}$contains a point$x$(that is, a minimal ... 2 The hypothesis here is not merely that$I_\mathfrak{p}$is free of rank$0$or$1$, but that it is equal as an ideal to$(0)$or$(1)$. That is, in the case where it is rank$1$, it is required to be equal to the entire ring. That fails for$I=(2)$if you localize at$\mathfrak{p}=(2)$. 2 It is more convenient to think here of$w\mathbb{P}$as of a quotient stack; anyway, if$Y$i smooth it does not pass through the stacky points $$(0,0,0,1,0), (0,0,0,0,1) \in w\mathbb{P}$$ and therefore the stacky structure of$w\mathbb{P}$plays no role for$Y$. The advantage of$w\mathbb{P} = \mathbb{P}(w_0,w_1,\dots,w_n)$as of a stack is that it comes ... 2 In an integral scheme, you can check that all restriction maps are injective. Thus, it follows that if$U \subset X$is an open subset covered by the affine open subsets$V_i$,$\mathcal{O}(U) = \bigcap_i\mathcal{O}(V_i)$(as subrings of$K$) by the sheaf property. So it is enough to show the statement for affine open subsets$U$. In other words, we just ... 1 Question: "My question is if we can express the kernel of this morphism in general, that is when f is not smooth." Answer: If$Y:=Spec(A/I), X:= Spec(A)$and$f:Y \rightarrow X$is the canonical map and$I \neq (0)$it follows$f$is not smooth. There is an exact sequence ($B:=A/I$) $$I/I^2 \rightarrow^{\delta} B \otimes \Omega^1_{A} \rightarrow^{\... 1 You're right, \mathscr{N}^{i-1} \scr{F}/\mathscr{N}^i \mathscr{F} is exactly i^*\mathscr{N}^{i-1} \scr{F}. It suffices to verify this affine-locally, so take X=\operatorname{Spec} A, \mathscr{F}=\widetilde{F}, and \mathscr{N}=\widetilde{N}. Then i^*\mathscr{N}^{i-1}\mathscr{F}\cong (N^iF\otimes_A A/N)^\sim, and as R/I\otimes_R M\cong M/IM, we ... 1 I believe he wants to apply the gluability axiom of the structure sheaf, so he is looking for f_1\in A_x and f_2\in A_y such that Res_{D(x),D(xy)}f_1=Res_{D(y),D(xy)} f_2, so writes: So we are looking for functions on D(x) and D(y) that agree on D(x)\cap D(y)=D(xy) Yes, by the sheaf axioms, since D(x) and D(y) cover D(x)\cup D(y) with ... 1 Siu's extension theorem tells us that the restriction map \Gamma(X, m K_{X})\to\Gamma(X_{0}, m K_{X_{0}}) is surjective. This implies that the complex vector space \Gamma(X_{0}, m K_{X_{0}}) is isomorphic to \pi_*\mathcal{O}(mK_X)/\mathfrak{m}_0\cdot\pi_*\mathcal{O}(mK_X), where \mathfrak{m}_0 is the defining ideal of 0\in\Delta. Then (by, e.g., ... 1 First consider the easier task of obtaining \Gamma as the zero-locus of a family of polynomials of degree d or less: If q \not \in \Gamma and p \in \Gamma there exists a hyperplane H_p such that p \in H_p and q \not \in H_p. Let L_p be a homogenous polynomial of degree one that cuts out H_p. Choose such an L_p for a every p \in \Gamma ... 1 You have to show that for any U=SpecB, open affine subescheme of X, the restriction of your closed immersion f^{-1}(U) \rightarrow U, is a closed immersion too, then you have that f^{-1}(U) is SpecB/I, for some I ideal of B. 1 Question: "Closed points should correspond to prime ideals of some ring, and some of them should be identified. Is the ring the polynomial ring in n variables? Or one in n+1 variables?" Answer: You define projective space using the "Proj"-construction: \mathbb{P}^n_k:=Proj(k[x_0,..,x_n]) and you may find it proved (in Hartshorne or any ... 1 Question: "Is it always true that dim(V)≤n− rank (J_p(I))?" Answer: I use the notation of Hartshorne, Chapter I. If (A, \mathfrak{m}) is a noetherian local ring with residue field k it follows (HH.I.Prop.5.2A)$$krdim(A) \leq dim_k(\mathfrak{m}/\mathfrak{m}^2).$$There is a formula (see the proof of HH.Thm.I.5.1) saying$$rk(J_p)=n-dim_k(\... 1 Question: "Are the irreducible components of a codimension one subvariety also codimension one?" Answer: If$A:=k[x,y,z]$and$I:=(x),J:=(y,z)$, it follows the ideal$IJ$define a codimension one subscheme$X:=V(IJ)⊆\mathbb{A}^3$(the dimension of the component of largest dimension is one).$X$has two irreducible components: One of dimension$2$... 1 With the help of @AlexWertheim I managed to understand where my confusion was coming from. An irreducible component$V'$of$V:=\mathcal{V}(X^2−YZ,XZ−X)\subseteq \mathbb{A}_k^3$is an irreducible closed subsets of the affine space$\mathbb{A}_k^3$. Throughout, let${\frak a}:= (X^2−YZ,XZ−X)$. Since$V'\subseteq \mathbb{A}_k^3\$ is irreducible if and only if ...

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