# Tag Info

## New answers tagged algebraic-geometry

0

Given $a+b+c=3$ and $a^2+b^2+c^2 =5$ and $a^3+b^3+c^3=7$ Using $$ab+bc+ca = \frac{1}{2}\left[(a+b+c)^2-(a^2+b^2+c^2)\right] = 2$$ and $$a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]=9$$ So $$7-3abc=9\Rightarrow abc=-\frac{2}{3}$$ Now Let $(t-a)\;,(t-b)\;,(t-c)$ be the root of cubic equation in terms of $t\;,$ Then $$(t-a)(t-b)(t-c) ... 0 Using just Macaulay2, you can do the following Macaulay2, version 1.6.0.1 with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition, ReesAlgebra, TangentCone i1 : R=QQ[a,b,c] o1 = R o1 : PolynomialRing i2 : i1=ideal(a+b+c-3,a^2+b^2+c^2-5,a^3+b^3+c^3-7) 2 2 2 3 ... 3 You have to use Newton Identities. See https://en.wikipedia.org/wiki/Newton%27s_identities In general if you have n variables x_1\ldots.x_n, define the polynomials$$p_k(x_1,\ldots,x_n)=\sum_{i=1}^nx_i^k = x_1^k+\cdots+x_n^k,and \begin{align} e_0(x_1, \ldots, x_n) &= 1,\\ e_1(x_1, \ldots, x_n) &= x_1 + x_2 + \cdots + x_n,\\ e_2(x_1, ... 1 If P=(a_1,\dots,a_n), then M_P=(x_1-a_1,\dots,x_n-a_n)/I(V); see Proving that kernels of evaluation maps are generated by the x_i - a_i (the proof given there works for any field). Then \overline K[V]/M_P\simeq\overline K[x_1,\dots,x_n]/(x_1-a_1,\dots,x_n-a_n)\simeq\overline K; for the last isomorphism see Maximal ideals in K[X_1,\dots,X_n]. 1 Yes, we need the field to be algebraically closed. Then it is clear. B(x,y,0,0) is a homogenous polynomial in x,y. Such a polynomial certainly admits a root other than (x=0,y=0). 2 A is quasi-compact, since it is a closed subset of the quasi-compact set K^n. Any quasi-compact discrete set is finite, this is a very easy exercise in basic topology. 0 By definition, a sequence \mathcal{F}^{\prime}\to\mathcal{F}\to\mathcal{F}^{\prime\prime}\to0 of \mathcal{O}_X-modules is right-exact if and only if $$\forall x\in X,\,\mathcal{F}^{\prime}_x\to\mathcal{F}_x\to\mathcal{F}^{\prime\prime}_x\to0\,\,(*)$$ are right-exact sequence of \mathcal{O}_{X,x}-modules; then: ... 1 ShowI(V):I(W) \supset I(V-W)$$This is the easy part. By the definition of the ideal quotient, we have to show I(V-W)I(W) \subset I(V) and this immediate: If f vanishes on V-W and g vanishes on W, then fg vanishes on all of V. Show$$I(V):I(W) \subset I(V-W)$$For this we need a crucial ingredient: Lemma. If W \subset \mathbb ... 2 Have a look at the formulation of the exercise again. He explicitly states that a regular curve is implicitly assumed to be locally noetherian and in part b) (your exercise) he assumes the curve to be quasi-compact. We have quasi-compact + locally noetherian = noetherian. So you are done, aren't you? 0 See distance is fixed so it will trace a circle whose radius is 5 thus the general solution when a circle isnt at origin is (x-h)^2+(y-k)^2=r^2 thus here its (x-4)^2+(y+3)^2=25 on simplify we get x^2-8x+16+y^2+6y+9=25 thus equation is x^2-8x+y^2+6y=0 3 The story is rather complicated in characteristic p, but I think the answer of Drike takes care of it. On the other hand, Gregory Grant makes a claim that seems to me to be not right. We’re working with the algebraic group (K^3,+), which I think would usually be called \mathbf G_{\mathrm a}^3. Consider now the subgroup of (K^3,+) given ... 1 Well, as Gregory Grant noticed, if K has characteristic \bf 0, then G must be a K-vector space, hence a line, so the answer is obvious. If K has characteristic \bf p, I'd consider the morphism of varieties \varphi:K\times G\rightarrow K^3,\ (k,g)\mapsto kg and say that if Im\varphi is a variety, it is of dimension no more than dim(K\times ... 0 I don't understand you!, but, for simplicity let k=\mathbb{C} be the field of complex numbers: let H={xy-1=0}\subset\mathbb{A}^2_{\mathbb{C}} the hyperbola, then p_1(H)=\{x\in\mathbb{C}\simeq\mathbb{A}^1_{\mathbb{C}}\mid x\neq0\} is an open subset of \mathbb{A}^1_{\mathbb{C}} and p_1 is a polynomial function, in other words p_1(H) is not an ... 0 The fact that a point is singular can be expressed in terms of its local ring: a point is non singular if its local ring is regular. A point which is not singular is singular. The local ring does not depend of the generators of the ideal which defines the variety. 1 The right space is the second! For completeness, let \mathbb{V} be a complex vector space of dimension n; I recall that a flag of \mathbb{V} is a strictly increasing sequence of vector subspaces of \mathbb{V}$$ \{\underline{0}\}=\mathbb{V}_0<\mathbb{V}_1<\dots<\mathbb{V}_r\leq\mathbb{V}\,\text{where:}\,\forall ...

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The operation is called projection. Tbe first-order theory of the complex field (which is the same as the first-order theory of algebraically closed fields of characteristic $0$) admits quantifier elimination. This means that $\exists x_n (x_1, \ldots, x_n) \in Z$ is equivalent to a propositional combination of formulas of the form $p_j(x_1, \ldots, x_n) = ... 7 Here's an example showing that$S$is not always a finite union of algebraic sets. Let$Z$be the zero locus of the single polynomial$x_1x_2 - 1$. Then$S = \mathbb{A}^1\setminus \{0\}$. What is true is that$S$is always a finite union of sets defined by finitely many polynomial equations (basic Zariski closed sets) and negated equations (basic Zariski ... 1 Let$A$be a commutative ring and$S$a multiplicative set. Then the family of rings$\left\{A_s \right\}_{s \in S}$forms a directed family. To see this, first we define a partial order on$S$by$s \le t$if$t = u s$for some$u \in S$. Next for$s \le t$with$t = u s$, there exists a ring homomorphism$f_{s,t}: A_s \rightarrow A_t$, which is defined by ... 1 If you have a subspace of$K^3$that is also a subgroup of the additive group$\langle K,+\rangle$, then that subspace must be a plane or a line. In this case choose any line that does not lie in the plane and the intersection will be just the origin. 0 If$U$is a linear algebraic group, then you can embed it at a subgroup of$\operatorname{GL}_n$. Furthermore, if$U$is unipotent, then$U$can be embedded as a subgroup of$U_n\subseteq\operatorname{GL}_n$, where$U_n$denotes the group of all upper triangular unipotent matrices, i.e. upper triangular matrices with$1$on the diagonal. Since$U_n$is ... 3 You can also use the Jacobian criterion directly for the homogenous polynomial and notice that the only point, where all partials vanish, is$(0,0,0)$, which is not a point of the quadric (it is not even a point of the projective space). You can check Hartshorne, Exercise$I.5.8$for details. A key ingredient is Euler's Lemma. 0 I figured it out. Let me prove a more general result in a very long winded way: Theorem: Let$Z, X$be algebraic$k$-schemes. An immersion of$Z$into$X$is an open subscheme of a closed subscheme of$X$(or the other way around, same thing). If$\iota: Z \rightarrow X$is an immersion, then$\iota_{\ast}:Z(k') \rightarrow X(k'), \phi \mapsto \iota ...

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Let $\DeclareMathOperator{Spec}{\operatorname{Spec}}\mathfrak p \in \Spec A$. We first show that the map $f^\#_\mathfrak p : \mathcal O_{\Spec A, \mathfrak p} \to f_* \mathcal O_{\Spec B, \mathfrak p}$ is the homomorphism $\varphi_\mathfrak p : A_\mathfrak p \to B_\mathfrak p$. Since the basic open subsets $D(g)$ for $g \in A$ form a basis for $\Spec A$, we ...

0

Is the locus a circle ? In general, no. It is a multifocal ellipse. More information on this topic can be found here, along with a video showing how to actually draw one.

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Unless I missed something, $R[x]/\mathscr{P}$ is generated as an algebra over $R/\mathfrak{m}$ by the class $\bar x$ of $x$ (because any element of $R[x]$ is a polynomial in $x$ with coefficients in $R$, and its class mod $\mathscr{P}$ is therefore a polynomial in $\bar x$ with coefficients in $R/\mathfrak{m}$). But a field extension which is finitely ...

1

Considering the map $\pi$ is the right idea. Recall (or try prove) the following lemma. $\textbf{Lemma:}$ Let $f\colon A\to B$ be a ring homomorphism. Then the continuous map on the spectra $F=\operatorname{Spec}f$ can be extended to a map of schemes $(F,F^\sharp)\colon \operatorname{Spec}B\to \operatorname{Spec}A$ such that on the principal open sets ...

1

Let $C$ be an algebraic curve. A collection of points $P_1,…,P_n$ in $C$ with assigned integer multiplicities $k_1,…,k_n$ is called a divisor on $C$ and it is denoted $$D=k_1P_1+...+k_nP_n.$$ And that is it, it is a formal sum, so as it is defined it does not have any immediate meaning. For instance, on your example we can define the divisors $D_1=(i,1), ... 1$U$is contained in the image of$\rho^\ast$, which we can see by covering$U$with principal opens$D(f)\subset U \subset \operatorname{Spec}R$, giving us restriction maps$R\to \Gamma(U) \to R_f$that allow us to pull back primes of$R_f$. But$\rho^\ast$need not have image$U$. Let$R=k[X,Y]$,$U=\operatorname{Spec}R \setminus \{(X,Y)\}$, so that$U$... 0 Hint: $$\sin a = 2\sin(a/2)\cos(a/2) = 2\tan(a/2)\cos^2(a/2) = \frac{2\tan(a/2)}{1+\tan^2(a/2)}$$ $$\cos a =\cdots$$ 1 This is kind of useless at this point, but since I was able to get a copy of one of Olivier's original articles on weakly étale/absolutely flat morphisms, I thought I'd share what I found. The article Ferrand, Daniel. "Epimorphismes d'anneaux et algèbres séparables." C. R. Acad. Sci. Paris Sér. A-B 265 1967 A411–A414. MR0244313 (39 #5628) is cited as ... 2 In general,$k'$-points of$X$need not be determined by their image (and I'm not sure where you're seeing Milne treat them as if they were, except in the case$k'=k$). For instance, if you take$X=Y$, then$X(k')$is in bijection with the set of automorphisms of$k'$over$k$(of which there can be many), but all of these maps send the unique point of$Y$... 4 The module of Kahler differentials is spanned by$dx$and$dy$, and differentiating the relation$y^2 = x^3$gives the relation$2y \, dy = 3x^2 \, dx$. Multiplying by$y$gives $$2y^2 \, dy = 2x^3 \, dy = 3 x^2 y \, dx$$ from which it follows that$2x \, dy - 3y \, dx$is torsion, since it is annihilated by multiplication by$x^2$. (Of course then we need ... 2 The definition is stating that the map$f \mapsto \{f(s)\}_{s \in S}$is injective, meaning that a function is completely determined by its values on$S$. This is just the scheme theoretic version of the statement that a standard continuous function in topology (e.g.$\mathbf{R} \to \mathbf{R}$) is determined by its values on a dense subset. 0 Here is an attempt. So if you start with a morphism$f:X\to Y$and a subvariety$W\subset X$, as you said you define the function $$f_\ast \mathbb 1_W:Y\to \mathbb Z, \qquad p\mapsto \chi(f^{-1}(p)\cap W).$$ By definition, in order to check that$f_\ast\mathbb 1_W$is constructible, you want to write it as $$f_\ast\mathbb 1_W=\sum_{n\in\mathbb Z}n\cdot ... 1 m_p is the maximal (irrelevant) ideal of K[X_1,\dots,X_n]/I(V), that is, the ideal generated by the images of X_1,\dots,X_n hence m_p=M_P/I(V). Then m_p^2=(M_P^2+I(V))/I(V). Now it's clear why$$m_p/m_p^2=M_P/(M_P^2+I(V)).$$2 Sketch of the solution: Consider ellipse \cal E with foci B and C which is tangent to the graph of f. Observe that the point of tangency is the point P. Indeed, each point Q(s, f(s)) lies inside the ellipse \cal E which means that QB+QC\le PB+PC. The tangent \ell to the graph of f at point P is tangent to the ellipse \cal E. But ... 1 Thanks Greg Martin for the hint. We observe that \dim(\pi(V))=\dim(\pi(\cup_i V_i))=\dim(\cup_i \pi(V_i)). Now assuming \dim(\cup_i \pi(V_i)) is finite, any chain of irreducible closed sets in \cup_i \pi(V_i) must be contained in some \pi(V_j). Thus, \dim(\cup_i \pi(V_i))=\dim(\pi(V_j))=\dim(\overline{\pi(V_j)})\leq \dim(V_j)\leq \dim(V). 2 If U\subseteq X is an open set and V\subseteq Y is any open set containing f(U), then the restriction of f^*\mathcal{G} to U depends only on the restriction of \mathcal{G} to V (more precisely, (f^*\mathcal{G})|_U=g^*(\mathcal{G}|_V), where g:U\to V is the restriction of the morphism f to a morphism U\to V). To show f^*\mathcal{G} is ... 0 Step 1: Suppose f: V \rightarrow W is a morphism of affine varieties, and let f^{\#}: A(W) \rightarrow A(V) be the induced homomorphism of affine coordinate rings. Let Y be a closed irreducible subset of V corresponding to the prime ideal P of A(V). Then the the zero set of \left(f^{\#}\right)^{-1} (P) in W is precisely the closure of f(V) ... 3 The answer is no. Why did you expect that? For example, consider, say A is a domain and B=A[x], I=(x-a)B for some 0\neq a\in A. Then the natural map A\to B/I is an isomorphism and in particular flat. If J=I+xB, clearly A\to B/J=A/aA is not flat. 3 The forgetful functor \mathsf{LRS} \to \mathsf{RS} has a right adjoint. The right adjoint "\mathrm{Spec}" is a rather direct generalization of the spectrum of a commutative ring. You can find the construction in W. D. Gillam's Localization of ringed spaces, for instance. The underlying set of \mathrm{Spec}(X,\mathcal{O}_X) consists of all pairs ... 1 I prefer work in the framework of (affine) schemes. Let X be a scheme and let Y be a closed (non empty) subet of X with immersion i:Y\hookrightarrow X; we would like define a(n affine) scheme structure over Y as the scheme structure induces from (X,\mathcal{O}_X). A possible solution is check that (Y,i^{-1}\mathcal{O}_X), where ... 1 If F=f^{-1}(p), then the normal bundle is the pull back of the normal bundle of p\in B and this of course is trivial. The second sequence doesn't look right. You should have \omega_{F_i} s on the right. You always have the natural exact sequence, 0\to O_X\to O_X(F_1+\cdots+F_r)\to \oplus O_{F_i}\to 0, since O_{F}(F)=O_F. Tensoring with \omega_X ... 1 First of all, "equivariant version of \mathcal{O}(1)" sounds a bit confusing to me. What we actually do, we endow \mathcal{O}(1) with equivariant structure. U(1) acts as follows. Point of total space is a pair (x, \xi), where x \in \mathbb{P}^1 i.e. x is line in \mathbb{C}^2. \xi is a linear function on this line l. U(1) acts on hole ... 2 Strictly speaking the association of D_f to R_f only defines the sections on a basis; to construct the regular functions on an arbitrary open U, we have to take the inverse limit of the D_f contained in U. The fact that this is a contravariant functor is pretty straightforward from the universal property of inverse limits. Separatedness is also ... 4 Many of these are proved using the universal hypersurface. Let P the projective space of all degree d forms (since the equation and any non-zero constant multiple give the same variety) and consider Z\subset P\times\mathbb{P}^n the universal hypersurface defined in the obvious way - these are pairs (f,p) with f(p)=0. Consider T\subset Z, defined ... 0 Let \pi:Grass(\mathcal{E})\rightarrow X be the projection, and \mathbf{q}:\pi^*\mathcal{E}\twoheadrightarrow \mathcal{Q} the universal quotient rank l quotient. One has an exact sequence 0\rightarrow T_{G/X}\rightarrow T_{G}\rightarrow \pi^* T_{X}\rightarrow 0. Since the tangent space to the ordinary Grassmannian at a point ... 1 I will use the following notation The definition of the Veronese map will be taken from the question above. I will denote the coordinates in \mathbb{P}^{5} by [z_{0}:\ldots:z_{5}] and the coordinates in \mathbb{P}^{2} by [x_{0}:x_{1}: x_{2}]. Claim The Veronese surface is cut out by three quadrics:$$C_{1}: z_{0}z_{3} - z_{1}^{2} = 0C_{2}: ... 2 Section 2.4 of Waterhouse's Introduction to Affine Group Schemes (he doesn't number theorems for some reason). You forgot "commutative." 2 This is actually fairly easy. Indeed, if$f \colon A \to B$is of finite presentation, then so is$B \otimes_A B \to B\$ by Tag 00F4 (4). Moreover, the latter is flat by assumption, and it is always a surjective ring map (i.e., closed immersion). Thus, by Tag 0819, it is an open immersion onto a clopen subset. Thus, the diagonal is an open immersion, which ...

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