Tag Info

New answers tagged

1

Here is an explicit definition of the pairing in the case of the projective line. For $\mathcal{M}$ a coherent sheaf, an element of $H^1(\mathcal{M})$ represents an extension$$0 \to \mathcal{M} \to \text{?} \to \mathcal{O} \to 0.$$If we are given a homomorphism $\mathcal{M} \to \mathcal{O}(-2)$, this induces an extension$$0 \to \mathcal{O}(-2) \to \text{?} ...


3

In response to user264059's comment, the decomposition you presented in your original post is generally called the Hodge decomposition, so it is not really clear what you are looking for here. (A more direct proof of the Hodge decomposition than what?) That said, there are several approaches. The theory of elliptic PDE's gives a proof of the Hodge ...


3

(i) If $a\neq0$ the prime ideals you want are the maximal ideals $(x-a,y-b)\; (b\in \mathbb C)$ and $(x,y-b)\; (b\in \mathbb C)$ about which you already know PLUS the two prime but not maximal ideals $(x)$ and $(x-a)$. (ii) If $a=0$ the prime ideals you want are the maximal ideals $(x,y-b)\; (b\in \mathbb C)$ PLUS the prime ideal $(x)$ Thinking ...


0

Note that $(x(x-a))$ decomposes as $(x) \cap (x-a)$. Thus $\textrm {Spec}$ of this corresponds to the disjoint union of two lines in $\mathbb C^2$. However, if $a=0$, geometrically, this is only one line, but it is a so-called "fat line", in that the ideal is representend by $x^2$.


1

Let's observe that if $a \neq 0$ the ideals $(x)$ and $(x-a)$ are coprime, by CRT you obtain: $$\frac{\mathbb{C}[x,y]}{x(x-a)} \simeq \frac{\mathbb{C}[x,y]}{(x)} \oplus \frac{\mathbb{C}[x,y]}{(x-a)} \simeq \mathbb{C}[y]\oplus \mathbb{C}[y]$$ Now is easy to look at the prime ideals of $\mathbb{C}[x]\oplus \mathbb{C}[x]$ and I think you can easely ...


2

I'm not sure this answers your first question, but here's one way to prove $k[x,y,z]/(y - x^2, z - x^3) \cong k[x]$. The intuition is that $$ k[x,y,z]/(y - x^2, z - x^3) \cong k[x,x^2,x^3] = k[x] $$ but I'm guessing you want something more rigorous. First, let's consider the following lemma. Lemma. Let $R$ be a unital commutative ring and $R[x]$ be the ...


2

There is a unique $k$-linear ring homomorphism $\phi:k[x,y,z]\to k[t]$ such that $\phi(x)=t$, $\phi(y)=t^2$ and $\phi(z)=t^3$, and the ideal $I=(y-x^2,z-y^3)$ is conttained in its kernel, as its generators are. If $f=\sum_{i=0}^na_ix^i$ is an elemntt of of $k[x]$ which is in $I$, then $\phi(f)$ is zero. But $\phi(f)$ is just $\sum_{i=0}^na_it^i$, whose ...


0

One way to explain "very small" is that if you think about $\mathbb A^n$ over $\mathbb C$ as $\mathbb C^n$ with the Euclidean topology (which strictly contains the Zariski topology in the sense that Zariski-closed implies Euclidean-closed), all proper Zariski-closed subsets have Lebesgue measure $0$.


0

If we consider the Zariski topology on the spectrum of a ring (and not only on its maximal ideals), a point is not necessarily closed. Actually the closed points correspond to maximal ideals. For instance if $A$ is an integral domain,the $0$ prime ideal is dense in $\operatorname{Spec}A$. Zariski topology is not Hausdorff, but it is Kolmogorov, i.e., given ...


0

1) Closed sets are defined by some zero set of an ideal, say $I$, so dimension means the dimension of the ring $k[x_1,...,x_n]/I$. 2) For example, closed sets in $\mathbb{A}^1$ are finite set of points as the zero set of a polynomial is bounded by the degree. 3) Zariski topology is not Hausdorff.


0

The vanishing of $H^1(X,\mathcal O_X)$ implies that $Pic^0$ is trivial (in your situation, as the base field is of characteristic zero).


0

A line containing the north pole, not tangent to sphere, will intersect the sphere once in, let's say M point and the plane $z=0$ in one point P(u, v, 0). The association P -> M will give you the mapping. The only point that cannot be mapped this way is the north pole.


0

Sit the sphere on the plane. Center at $(0,0,1)$, North Pole at $(0,0,2)$. Draw a line from $(u,v,0)$ to the North Pole; it will cut the sphere at one other point.


3

I think the following works. In the noetherian case, you can just take $V_i = U_i$ below. Let $S$ be qcqs, let $X \to S$ be separated and of finite type, and let $X = X_1 \cup \cdots \cup X_r$ be a decomposition of $X$ into finitely many irreducible components. Let $U_i = X \setminus \bigcup_{i \ne j} X_j$; note this is a non-empty open subset of $X$ ...


1

To prove the indicated statement from scratch, we first sketch a proof of another result that's called Dickson's lemma: For any infinite sequence of distinct $\alpha_0, \alpha_1\in \mathbb{N}^n$, there exist $i < j$ such that $\alpha_i \leq \alpha_j$ (that is, each component $\alpha_i^k \leq \alpha_j^k$.) Assume the result holds for $(n-1)$-tuples. ...


1

The ideals $J_0 \subset J_1 \subset J_2 \subset \cdots$, and $k[x_1, \dots, x_n]$ is Noetherian.


2

We’re looking in the local, complete situation above $\ell$ at the $p^m$-torsion points of $E$ for all $m$. What does it mean to say that $T_p(E)^{G_v}\ne0$, where $G_v=G_{K_v}$, the Galois group of an algebraic closure of $K_v$ over $K_v$? It would mean that there was a consistent sequence of $p^m$-torsion points of $E$, in particularly infinitely many of ...


2

Yes, the canonical morphism $p: C^n\to C^{(n)}$ is flat by miracle flatness. More precisely that aptly named theorem states that a morphism $p:X\to Y$ of varieties over a field is flat whenever $X$ and $Y$ are regular and all the fibers of $p$ have dimension $\dim X-\dim Y$ ($=0$ in our case). For a proof of miracle flatness see Matsumura, 23.1, page 179, ...


2

We show that$$\dim_\mathbb{C} \mathfrak{m}^k/\mathfrak{m}^{k+1} = k+1$$if $k < m$. Let $p = [a, b, c]$ (we can assume $c \neq 0$). Then we can identify $\mathcal{O}_p$ with $\mathbb{C}[x, y]/(f)$, we can identify $p$ with $(a/c, b/c) \in \mathbb{C}^2$, $f(a/c, b/c) = 0$. In this case, we have$$\mathfrak{m} = \left\langle x - {a\over{c}}, y - ...


1

We may restrict to appropriate affine coordinate patches to check zeros and poles. Indeed, away from $z = 0$, taking the affinization $z=1$, the curve is defined by $$y^2 = x(x-1)(x-\lambda).$$Then it is clear $y/z$ indeed has zeros of degree $1$ at$$p = (0, 0),\text{ }q = (1, 0), \text{ }r = (\lambda, 0),$$and that these are the only zeros of the function ...


3

This does follow from the Hodge decomposition, which lets you view the first de Rham cohomology as the set of harmonic 1-forms. The map you are discussing can be viewed as $\Omega^{1,0} \oplus \overline{\Omega^{1,0}} \rightarrow \mathscr{H}^{(1)}$. This is then a canonical map, and you can see it is an isomorphism pretty easily.


6

(Let me answer the case of coefficients in $\mathbf Z$ (or $\mathbf Q$ or $\mathbf R$ or $\mathbf C$.) I think it should only take a bit of universal coefficient wizardry to deduce the answer in general, but it is too late here for that.) First, by the Lefschetz hyperplane theorem and Poincaré duality, we have an isomorphism $$ H_i(\mathbf P^n, \mathbf Z) ...


5

By the Lefschetz Hyperplane theorem, the inclusion $i : X \hookrightarrow \mathbb{CP}^n$ induces a map $i^* : H^q(\mathbb{CP}^n, \mathbb{Z}) \to H^q(X, \mathbb{Z})$ which is an isomorphism for $q \leq n - 2$ and injective for $q = n - 1$. Recall that $$b_q(\mathbb{CP}^n) = \dim H^q(\mathbb{CP}^n, \mathbb{Z}) = \begin{cases} 1 & q\ \text{even}, 0 \leq q ...


4

There is no uniform answer to your question: the numbers $l(3P)$ and $l(4P)$ you are asking about depend on the curve $C$ and on the point $P$. a) We always have $l(2P)=1$ Indeed, $1\leq l(2P)\leq 2$ for a curve of positive genus while for genus $g\geq 2$ the existence of a point $P$ with $l(2P)=2$ characterizes hyperelliptic curves. However a ...


0

I don't know the contents of Cox et al.'s book; but a good basic introduction to algebraic geometry is Klaus Hulek - Elementary Algebraic Geometry, and if it can be useful for you: David S. Dummit & Richard M. Foote - Abstract Algebra, here you can found a little introduction to Groebner basis and other interesting material. ;) P.S.: "Et" is a Latin ...


1

$\newcommand{\Proj}{\mathbf{P}}$Let $k$ be a positive integer. A monomial $f$ of degree $k$ in two variables satisfies the homogeneity conditon $f(ax, ay) = a^{k} f(x, y)$ (for all non-zero $a$) by elementary algebra. It's not difficult to see that a sum of monomials of different degrees is not homogeneous at all. Consequently if a polynomial $f$ in two ...


0

Probably no longer useful, but I find that Murre's Introduction to Grothendieck's theory of the etale fundamental group is quite an excellent source. http://www.math.tifr.res.in/~publ/ln/tifr40.pdf


4

The most natural family of rings which comes to mind as having every commutative ring as a quotient is $\mathbb{Z}[\{t_i \mid i \in I\}]$, i.e., a polynomial ring in an arbitrary set of indeterminates over $\mathbb{Z}$. These rings are all UFDs -- see e.g. Corollary 15.27 of these notes -- hence integrally closed ("normal").


4

The intersection of two varieties is given by the sum of their ideals. So you're looking at the ideal generated by the union of the generators of $V_V$ and $V_S$. Immediately we see that both $x_1^2-x_0x_3$ and $x_0x_3-x_1x_2$ are generators of $I=I_{S \cap V}$. But this implies that $x_1^2-x_1x_2=x_1(x_1-x_2)$ is a generator of $I$. Thus $I$ is reducible. ...


0

In other way: you remember that $X$ has finitely many irreducible components, and these are connected subsets of $X$.


0

I think this is right... maybe somebody can confirm the translation is good and possibly fix it up a little bit. http://attach3.bdwm.net/attach/0Announce/groups/GROUP_3/Mathematics/D4335A993//M.1104367920.A/RS.pdf


0

Let's use the following form for the parabola: $$y=a(x-h)^2+k$$ Distance from vertex $(h,k)$ to your directrix and focus is calculated using the following formula where $p$ is the distance. $$a=\frac{1}{4p}$$ If $a$ is positive, the equation for your directrix will be as follows: $$y=k-p$$ Also, the coordinates of the focus will be the following with ...


0

The standard equation of the parabola with its axis (symmetric axis=OX) coincident with the x-axis & vertex at the point $(k, 0)$ on the x-axis is $$\color{blue}{y^2=4a(x-x_1)}$$ Where, $a$ & $k$ are arbitrary constants. Now, satisfying the above equation of parabola by the coordinates of two given point $P_1(x_1, y_1)$ & $P_2(x_2, y_2)$. We ...


2

The question reduces to a question about sheaves of sets, so I will just work with those (instead of modules or whatever). The point is that we have the following commutative diagram of right adjoint functors, $$\require{AMScd} \begin{CD} \mathbf{Sh}(X) @>>> \mathbf{Sh}(Y) \\ @VVV @VVV \\ \mathbf{Psh}(X) @>>> \mathbf{Psh}(Y) \end{CD}$$ ...


1

Not sure if I am allowed to use the Nullstellensatz here (i.e. if it is instructive in your current context) but if I can, then we just get that $V(b) = \emptyset \subseteq \mathbb C^{n+1}$ so in particular for any $p = (x,t)\in \mathbb C^{n+1}$ we get the following: if $p\in V(f_1, \dots, f_r)\subset \mathbb C^{n+1}$, i.e. $f_1(x) = 0, \dots, f_r(x) = 0$ ...


1

Suppose $1\in b$. This means we can write $1=\sum c_i f_i+ d(1-yg)$, for some $c_1,\dots,c_r,d\in \mathbb{C}[x_1,\dots,x_n,y]$. Write $c_i=\sum c_{ij} y^j$ and $d=\sum d_j y^j$, for $c_{ij},d_j\in \mathbb{C}[x_1,\dots,x_n]$. Consider the right-hand side of this equation as a polynomial in $y$ with coefficients in $\mathbb{C}[x_1,\dots,x_n]$, and consider ...


2

No, this is not true. There is a famous example of Nagata of a Noetherian local domain $A$ of dimension $2$ which has a finite overring $A \subset B$ with two maximal ideals $\mathfrak m, \mathfrak n \subset B$ with $\dim(B_{\mathfrak m}) = 2$ and $\dim(B_{\mathfrak n}) = 1$. Let $X$ be the spectrum of $B$ and $Y$ be the spectrum of $A$ and let $x$ be the ...


0

Maybe a bit late, but how about this construction? By definition, $\Omega_{A/R} = I/I^2$, where $I= \ker[A\otimes_RA\rightarrow A, a\otimes b\mapsto ab]$. We have a direct sum decomposition $(A\otimes_RA)/I^2\cong A \oplus \Omega_{A/R}$ induced by $a\otimes b\mapsto (ab, a\cdot db)$ (where $d\colon A\rightarrow \Omega_{A/R}$, $a\mapsto (1\otimes a - a\otimes ...


1

Sorry to resurrect my old question, but for the sake of completeness, I will provide a complete solution. Consider the tangent space at the origin of our divisor $D$. We have$$k[D] = k[a, b, c, d]/(a, b, ac - bd) \cong k[c, d].$$This Zariski cotangent space has dimension $2$ here (generated by $c$, $d$). Now, let us consider a tangent space of $X \cap ...


2

Let us work with schemes over $\mathbf{C}$ (and not algebraic spaces). Let $F$ a contravariant functor from the category of schemes over $\mathbf{C}$ to the category of sets. A scheme $M$ and a transformation of functors $\phi : F \to h_M$ is called a coarse moduli scheme $M$ if (a) $\phi$ is a categorical quotient, and (b) the map $F(\text{Spec}(k)) \to ...


3

We have that $X$ is a smooth projective algebraic variety and $X$ is also good reduction. Therefore, we can apply the Weil conjectures and have that$$\text{dim}_\mathbb{Q} H^m(X, \mathbb{Q}) = \deg(P_mn)),$$where$$Z(X/\mathbb{F}_q, n) = {{P_1(n)P_3(n) \dots}\over{P_0(n)P_2(n)\dots}}$$with$$P_m(n) = \prod_{j=1}^{b_m} (1 - \alpha_{mj}n),\text{ }|\alpha_{mj}| = ...


1

With somewhat-altered labels, we have this figure for $\square ABOC$ with incenters $P$ and $Q$ (and inradii $p$ and $q$) and excenters $R$ and $S$ (and exradii $r$ and $s$). We embed the figure in the coordinate plane with $O$ at the origin, and: $$A = a\;(1,0) \qquad B = b\;(\cos 2\beta,\sin 2\beta) \qquad C = c\;(\cos 2\gamma,-\sin 2\gamma)$$ ...


3

The answer is no in general, because if $X$ is any nonempty set with the indiscrete topology (only $\emptyset$ and itself are open) then $\mathcal{O}_X \cong \mathbb{C}$, while two indiscrete spaces of different cardinality are not homeomorphic. When you restrict to compact Hausdorff spaces, this becomes true. (See A theorem due to Gelfand and Kolmogorov ...


2

This group is isomorphic to $\Bbb C^{*2}$ via $(a,b) \mapsto (a+ib,a-ib)$ so it is a torus. However it is not anisotropic over $\Bbb R$ because it has characters defined over $\Bbb R$, such as $(a,b) \mapsto a^2+b^2$.


1

Okay so I'll just dive in here. What is the logic supposed to be? As written the proof makes almost no sense to me, and while it is 4 in the morning here I think there are some serious errors. You have a variety $X = \mathbb{V}(I)$ and you start by assuming $\mathbb{V}(I) \subseteq \mathbb{V}(m)$ which is only true if $m \subseteq I$ which is only true if ...


1

The $A$-algebra $S^{-1}A$ is a member of this category: an element of the multiplicative subset $s \in S$ is a unit as it is inverted by $1/s$. It is furthermore initial among these algebras because the unique morphism$$\overline{\varphi}: S^{-1}A \to B$$is given by$$\overline{\varphi}\left({r\over{s}}\right) = \varphi(r)\varphi(s)^{-1},$$where $\varphi: A ...


4

We expand on Takumi Murayama's comment. The standard example is to take $$X = \mathbb{A}_k^1, \text{ }Y = V(x^3 - y^2) \subset \mathbb{A}_k^2,$$ and define$$F: X \to Y,\text{ }F(t) = (t^2, t^3).$$It is not hard to see this is a bijection. Because the Zariski closed subset of $X$, resp. $Y$, are $X$ itself, resp. $Y$ itself, together with all finite subsets, ...


1

Let $U$ be a nonempty open subset of an irreducible topological space $X$. Denote by $\overline{U}$ the closure of $U$ in $X$. Then $(X - U, \overline{U})$ is a decomposition of $X$. Because $X$ is irreducible, one of these sets equals $X$. Since $U$ is nonempty,$$X \setminus U \neq U \implies \overline{U} = X.$$


2

This is false: let $C = \mathbb{P}^1_{\mathbb{C}}$ and let $L = \mathcal{O}(-1)$ and $L' = \mathcal{O}(1)$. Then, $$2 = \underbrace{h^0(C,L)}_{=0} + \underbrace{h^0(C,L')}_{=2} \neq h^0(C, L \otimes L') = 1.$$ It might be more reasonable to expect the dimensions to multiply i.e. have $$ h^0(C,L) \cdot h^0(C,L') = h^0(C,L\otimes L'), $$ because there is a ...


1

I've made this community wiki for obvious reasons. Arithmetic Moduli of Elliptic Curves by Katz and Mazur. This is the ideal comprehensive reference if you want to work with just elliptic curves. It has pretty much everything known at the time (and much more—a lot of the work in their is original). But, it's super long, and super detailed. So, it can be ...



Top 50 recent answers are included