5

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 ...


4

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 ...


4

$$w^4 + x^4 = y^4 + z^4 \tag{1}$$ 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 $\...


3

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(...


2

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 ...


2

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 ...


2

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:=(...


2

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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