New answers tagged

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 \begin{equation} \forall x\in X,\,\mathcal{F}^{\prime}_x\to\mathcal{F}_x\to\mathcal{F}^{\prime\prime}_x\to0\,\,(*) \end{equation} are right-exact sequence of $\mathcal{O}_{X,x}$-modules; then: ...


1

Show $$I(V):I(W) \supset I(V-W)$$ 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 ...


5

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 ...


2

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.


2

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: $$C_{1}: z_{0}z_{3} - z_{1}^{2} = 0$$ $$C_{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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