New answers tagged

1

In the affine plane $\mathbb A^2_k=\operatorname { Spec}k[x,y]$ over a field $k$ the closed affine subset ("the cross") $$X:=V(x\cdot y)=\{x\cdot y=0\}\subset \mathbb A^2_k$$ is a connected noetherian topological space (in the Zariski topology), but its open subset $$X_0:=X\setminus V(x,y)=X\setminus \{(0,0)\}$$ is no longer connected. Indeed it is the ...


1

Suppose we have some equation with $n$ variables (assuming this is all over a field $\mathbb{F} $): $$x_1 + x_2 +... + x_n = c $$ For some constant $c \in \mathbb{F}$. Then, we can take this equation and subtract $x_n$, and $c$ from both sides: $$-x_n = -c + x_1 + ... + x_{n-1} $$ $$\Leftrightarrow x_n = c - x_1 - ... - x_{n-1} $$ From here, any ...


23

Sheaves and sheaf cohomology were invented not by Serre, but by Jean Leray while he was a World War II prisoner in Oflag XVII (Offizierlager=Officer Camp) in Austria. After the war he published his results in 1945 in the Journal de Liouville. His remarkable but rather obscure results were clarified by Borel, Henri Cartan, Koszul, Serre and Weil in the late ...


17

Sheaf cohomology is just the elaboration of the following problem: you have a space and a covering, and you can do something you want on each set of the covering: can you do it on the whole space? This occurs in many places, from the Cousin problem in complex analysis to the construction of sections of fiber bundles to the pasting together of solutions ...


6

Hartshorne says that cohomology was first introduced to abstract algebraic geometry by Serre in his Faisceaux Algebriques Coherents paper (translated to English by a friend of mine). The FAC says We know that the cohomological methods, in particular sheaf theory, play an increasing role not only in the theory of several complex variables ([5]), but ...


2

Let $X=\{0,1,2\}$ with open sets $\varnothing,\{0\},\{1\},\{0,1\}$, and $X$. $X$ is Noetherian and connected, but the open set $\{0,1\}$ is not connected.


0

Here, I just cover the key part: Expressing $$a+b+c+d=R$$ as a polynomial equation in $R$ and $(A,B,C,D)=(a^2,b^2,c^2,d^2)$. Then you can plug in the distance squares $A=x^2+y^2$ etc and do all the messy expansions. Let $S_k = a^k+b^k+c^k+d^k$. We note $$\begin{align} S_1 &= R & S_{2k} &= A^k+B^k+C^k+D^k \end{align}$$ And for every ...


0

I came up with an alternative answer. Let $\mathcal{M}_{g,n}$ be the moduli space of Riemann surfaces of genus $g$ with $n$ marked points. Then our sequence $$0 \rightarrow {p_1}_* p_2^* K_C^{\otimes 2} \rightarrow {p_1}_* (p_2^* K_C^{\otimes 2} \otimes \mathcal{O}_{C \times C}(\Delta)) \rightarrow K_C \rightarrow 0$$ is in fact the same as $$0 ...


2

As a start, let us first derive the formula when the unit square is centered at origin. Let $a = x^2+y^2+\frac12$, the distance to vertex $v$ is: $$\begin{cases} r_{++} &= \sqrt{a - x - y}, & v = (+\frac12,+\frac12)\\ r_{+-} &= \sqrt{a - x + y}, & v = (+\frac12,-\frac12)\\ r_{-+} &= \sqrt{a + x - y}, & v = (-\frac12,+\frac12)\\ ...


0

This is a $2$-dimension cyclic quotient singularity. Here is a very nice PDF file which explains the classification of two dimensional rational double points. Here is a sketch on how to blow up. You start by blowing up $\mathbb A^3$ in the origin. This is given by the set of equations $xs=ry,xt=rz$ and $yt=sz$, where $(r:s:t)$ are coordinates in $\mathbb ...


0

Here is a sketch without many details, just intuition (but it can be made rigourous). Since $X$ locally looks like $\mathbb R$, the stalk $\mathscr G_x$ can be thought of as functions $\mathbb R \to X$. These are periodic functions. But the stalk $\mathscr F_x$ can be thought of as arbitrary continous functions $\mathbb R \to \mathbb R$. Hence we have an ...


0

You don't say what is $t$. May be you mean the following. Consider the map (which is an inclusion) $R=k[x,y]/(y^2-x^2-x^3)\to k[t]$ given by $x\mapsto t^2-1, y\mapsto t(t^2-1)$. Then $k[t]$ is a generated by $1,t$ as a module over $R$. The rest should be clear. You can see that $k[t]$ is just the integral closure of $R$.


0

If you assume the curve smooth, then yes. This is because for a smooth curve, the local rings $\mathcal O_{X,x}$ are DVRs or fields, e.g. PIDs. In particular, a submodule of a free module over a PID is again free. Therefore, the exact sequence $$ 0 \to \ker \varphi \to \mathcal F \to \mathcal G $$ can be seen at the stalks as the sequence $$ 0 \to \ker ...


1

Hint. An isomorphism of local rings maps the maximal ideal to the maximal ideal.


1

Here is a partial answer to (a): We know $$z_0z_2=z_1^2 $$ and $$z_0z_3 = z_1z_2.$$ Suppose first that $z_0$ and $z_2$ are non-zero. Multiply $F_2$ by $z_0z_2$ to get $$ z_0z_2F_2 = z_0z_1z_2z_3 - z_2^2z_0z_2$$ which becomes $$(z_0z_3)^2 - z_2^2z_1^2 $$ or after factoring $$z_0z_2F_2 = (z_0z_3-z_1z_2)(z_0z_3 + z_1z_2) = F_1(z_0z_3 + z_1z_2) = 0$$ since ...


1

I'll look at problem (a). On $U_0 = \{Z_0 \ne 0\}$ you are intersecting the affine curves $$z_2 = z_1^2, \; \; z_3 = z_1 z_2.$$ From these equations already follows $$z_2^2 = z_1^2 z_2 = z_1 z_3,$$ so it looks like the twisted cubic. When $Z_0 = 0$, $V(F_1,F_2)$ reduces to $Z_1 = 0;$ i.e. the projective line $$\{[0 : 0 : z_2 : z_3]\}.$$ So $V(F_1,F_2)$ is ...


3

No, in any characteristic. Let $f:\mathbb{P}^1\times\mathbb{P}^1=X\to Y=\mathbb{P}^2$ be such a morphism. Then, the map can not be finite, since it is of degree one and $Y$ is smooth would imply that $f$ is an isomorphism. This is not true by Picard group considerations or many other considerations. So, one must have an irreducible curve $E\subset X$ such ...


-1

Let $f\in A$ and $f\neq 0$, where $A$ is a noetherian domain. Let $Y$ be a prime divisor, i.e. an integral closed subset of codimension 1 of $Spec A$. It is $v_Y(f)>0$ if and only if $Y\subset V(f)$. Write $Y=V(\mathfrak{p})$ where $\mathfrak{p}$ is a height 1 prime ideal. So $Y\subset V(f)$ means $f\in \mathfrak{p}$. The primes containing $f$ are the ...


2

I will give an outline of the proof of this fact if all the sheaves involved are quasi-coherent. I claim that the original exact sequence does split. First recall the Proj construction. Let $A=\oplus_{n\geq 0}A_n$ be a graded commutative ring, then $\operatorname{Proj} A$ is as a set the set of homogeneous prime ideas of $A$ not containing the trivial ideal ...


0

Well, I think you have to make a choice for $t_{P}$. But apparently you do not need to make a choice for $t_X$. This happens because the role of $t_X$ in the duality isomorphism gets cancelled when you evaluate the class using $t_X(. )$. At first, for a closed immersion $j: Y \to X$, let us look at the definition of the map $H^{n-p}(X, \Omega_X^{n-p}) \to ...


3

a) An algebraic subset $X\subset\mathbb C^n$ has all its irreducible components of dimension $n-1$ if and only if it is the zero set $X=V(P)$ of some polynomial $P\in \mathbb C[T_1,...,T_n]$. b) Beware that $X=\{(1,0)\}\cup V(T_1)\subset \mathbb C^2$ is an algebraic subset of dimension $1$ (=the maximum dimension of its irreducible components) but ...


3

The answer is still negative, though I feel the previous answer is incorrect. Let $L=p_1^*K_C, M=p_2^*K_C$ and then saying $0\to p_{1*}M^2\to p_{1*}M^2(\Delta)\to K_C\to 0$ (the last is surjective, since the corresponding $R^1$ is zero) splits is same as sayining after twisting by $K_C^{-1}$, the section $1$ can be lifted. This implies in particular ...


0

This is an application of 5.3.2(ii). If $\phi: X \rightarrow Y$ is an equivariant morphism of homogeneous $G$-spaces, $r = \textrm{Dim } Y - \textrm{Dim } X$, $Y'$ is an irreducible component of $Y$, $X'$ is an irreducible component of $\phi^{-1}Y'$, then $\textrm{Dim } X' = \textrm{Dim } Y' +r$. In particular, if $y \in Y$, then $\textrm{Dim } X' = r$ ...


0

Rotate the ellipse by $ \tan^{-1}\frac{ B} { A-C } $ to bring them to form where axes are parallel to axes. To get a feel, plotting them parametrically will help.The ellipse plots with $$ x = H + A \cos t ,\,y= K + B \sin t ;$$ Next plot circles by varying $ h,k,a$ $$ x = h + a \cos u, \, y=k + a \sin u $$ along with ellipse to see if and where they ...


1

The linear forms, quadratics, cubics, and quartics on the plane are vector spaces of dimensions 3,6,10,15 respectively. If we choose each point P generally, imposing the extra condition of vanishing at P will decrease each of the dimensions above by 1 (unless the dimension is already 0). If this helps, we are just saying that, if we have chosen $P_1$ ...


1

Try the zero set of $X$ and $Y$ in $\mathbb A^2$.


2

Note that the $d-$tule embedding is an isomorphic copy of $\mathbb P^n$ where the linear forms $O_X(1)$ corresponds to the $d$-forms of $\mathbb P^n$.($O_{\mathbb P^n}(d)$). (look up [Hartshorne II, Ex. 5.13]). This should help you to see your second question. To answer your original question, I will just compute the Hilbert polynomial of $v_d: \mathbb P^n ...


2

Let $L_i=\mathcal{O}_{\mathbb{P}^3}(1)|_{X_i}$. If the isomorphism $f:X_0\to X_1$ preserves $L_i$ (that is, $f^*(L_1)=L_0$), rest will easily follow. Now, since $K_{X_i}=L_i^{d-4}$, we have $L_0^{d-4}=f^*L_1^{d-4}$. But (may be one should assume char 0, though may not be necessary), Picard group of hypersurfaces in 3-space is torsion free and since $d>4$, ...


1

This is not a complete answer, but it is too long to write a single comment. Let $V$ be a vector space of dimension $n+1$. Let $S^d V$ denote the space of homogeneous polynomials of degree $d$ on $V^*$ (you can think $S^d V$ as polynomials whose variables are the elements of a basis of $V$). Then we have $$ \mathbb{P} V \to \mathbb{P} S^d V \to \mathbb{P} ...


1

If the homogeneous coordinate ring of your projectively embedded $X$ was $k[x_0, \ldots, x_n]/(f_1, \ldots, f_m)$, then your "trivially embedded $X$" has homogeneous coordinate ring $k[x_0, \ldots, x_n, x_{n+1}]/(f_1, \ldots, f_m, x_{n+1})$, which is obviously isomorphic to the ring you started with. This will be true whenever you take a linear embedding of ...


0

Here's the idea: (a) Suppose that $\lambda \in \text{Aut}(X/S)$ fixes a geometric point $\bar{z}$ of $X$. Then $\lambda\circ\bar{z} = \bar{z} = \text{id}_X\circ\bar{z}$. Corollary 5.3.3. then implies that $\lambda = \text{id}_X$. So nontrivial automorphisms can't fix any geometric points. (b) Assuming that I'm interpreting your notation correctly (I don't ...


1

I can't totally parse the post, but A Gröbner basis is a special generating set for an ideal in a ring $R$ considered as an $R$-submodule of $R$. and A basis for a vector space is a special generating set for the vector space as an $\Bbb F$-module for some field (or even division ring) $\Bbb F$. Yes, if $1$ were part of a Gröbner basis for an ...


2

Your question covers a vast amount of material and without specifying further it is hard to say exactly what you're looking for. Also, you did not state the level of material you were looking for. Almost any text on Algebraic Geometry (or more specifically Arithmetic Algebraic Geometry), Elliptic Curves, or Number Theory related to these areas would contain ...


1

Here is an argument for $n=2$, the general case is not different. I will just write $\mathcal{O}$ instead of $\mathcal{O}_{\mathbb{P}^2}$. Let $0\to\mathcal{O}(m)\to E\to\mathcal{O}\to 0$ be an extension. Let $S=\oplus_k\mathrm{H}^0(\mathcal{O}(k))$ be the homogeneous coordinate ring. Applying this to our exact sequence, we get, $0\to S\to M\to S$, an exact ...


0

The model of the affine curves $E: y^{2} = g(x)$ is reminiscent of the fact that they are ramified coverings of the projective line $\mathbb{P}^{1}$ ramified at four points (this is well known: look at the image of the rational function $y/x$) and the ramification points are the roots of $g(x)$ and it is unramified at infinity. So if we had an isomorphism ...


2

No. This is basically the point of sheaf cohomology -- global Ext is a global computation that relies on how the different pieces of the sheaf patch together. For example, suppose $F$ and $G$ are coherent and locally free. Then $\mathrm{Ext}^i(F,G) = H^i(F^* \otimes G)$, whereas over any local ring $\mathcal{O}_{X,x}$, $Ext^i(F_x,G_x)$ vanishes for $i>0$ ...


2

No. Here is an explicit example. Let $X=\mathbf A^3$ and $Z=\{0\}$. The blow-up $Y$ of $X$ at $Z$ is isomorphic to the total space of the line bundle $\mathcal O(-1)$ over $\mathbf P^2$. Take a sheaf on $\mathbf P^2$ having a nonzero $H^1$, for example, the ideal sheaf $\mathcal I$ of the union of two distinct closed points $P$ and $Q$ on $\mathbf P^2$. The ...


2

Here is an outline. I'll just assume $Y$ is affine to make the problem easier. We'll have to use the second part of the lemma here. This says: $$ A(Y)_{\mathfrak{m}_P} = \bigcap_{\mathfrak q \text{ height 1 prime of $A(Y)_{\mathfrak m_P}$}} (A(Y)_{\mathfrak{m}_P})_{\mathfrak{q}} = \bigcap_{\mathfrak q \text{ height 1 prime of $A(Y)$ contained in $\mathfrak ...


1

A good way to visualize this is to look the set $Q$ of all possible exponents. In this case, it's the set of all $\Bbb N$-linear combinations of $(4,0), (3,1),(1,3),(0,4)$. For our ring to have even a chance to be normal, $Q$ certainly (prove it!) must satisfy the following criterion (in which case we say $Q$ is saturated): If $\mathbf a \in \Bbb Z Q$ ...


2

This problem is called implicitization and can be solved systematically but not easily by hand. In Mathematica (and Wolfram Alpha), you can use Eliminate. For $n=2$, Eliminate[{x==z(1-z), y==(z-1)-(z-1)^2/2}, {z}] gives $$ (6-4 x) y+4 y^2 = (-x-6) x $$ For $n=3$, Eliminate[{x==z(1-z), y==(z-1)-(z-1)^2/2+(z-1)^3/3}, {z}] gives $$ (-36 y-66) y = 4 x^3-9 ...


3

One can show that the ring of coordinates of $(y^2-x^3)$ is $\Bbb R[t^2,t^3]$, and that of the line is, of course, $\Bbb R[t]$. You're reduced to show these are not isomorphic algebras. One way to show these are not isomorphic is to note the former is not a unique factorization domain, for example.


1

Okay, you've established an isomorphism $A(C\cup D) \cong A(C)\times A(D)$, so you know how to produce a polynomial $h:C\cup D\to\mathbb{C}$ given polynomials $f:C\to\mathbb{C}$ and $g:D\to\mathbb{C}$. Presumably the thing you're wondering is: is this $h$ actually an extension of $f$ and $g$? To see that it is, you need to actually look at how the ...


1

It is certainly not true in general that a $\Bbb R$ subgroup is dense. What you want, for the general story, is the following. 1) The closure of an abelian Lie subgroup of $G'$ is an abelian Lie subgroup of $G'$. 2) Every compact abelian Lie group is a torus. From this, we see that the closure if your $\Bbb R$ is a torus. It can't be a circle, so in the ...


1

Define the codimension of a subvariety of $\mathbf P^n$ or $\mathbf A^n$ à la Krull, and use the elementary fact from commutative algebra that the sum of the dimension and the codimension of a subvariety is the dimension of the ambient variety. Then, a similar argument as yours shows that $$ \mathrm{codim}(C(X))\geq\mathrm{codim}(X)=n-\dim X. $$ Since $$ ...


2

Let $X=\operatorname{Spec} A$, and pick some nonzero $f\in A$ with $\operatorname{Spec} A_f = D(f) \subset U$. Then we can lift chains of closed sets to see that $\dim D(f) \leq \dim U \leq \dim X$. But as shown here, we have $\dim D(f) = \dim A_f = \dim A = \dim X$. We can also see this quite quickly by noting that $A$ and $A_f$ have the same fraction ...


0

You should note that any nonempty open subset of an irreducible variety (or topological space) is dense : https://en.wikipedia.org/wiki/Hyperconnected_space. Then you can use the definition of the dimension to conclude.


2

Let $\mathcal{A}$ be a locally small category with a terminal object. Then $\mathbf{Fam} (\mathcal{A})$ is also locally small with a terminal object $1$, so we can define $\Gamma = \mathrm{Hom}(1, -) : \mathbf{Fam} (\mathcal{A}) \to \mathbf{Set}$. This functor has a left adjoint $\Delta : \mathbf{Set} \to \mathbf{Fam} (\mathcal{A})$, defined on objects by $X ...


1

The intersection of sets translates into a logical conjunction, thus: $$ z\in Z(S) \Leftrightarrow \forall i: z\in Z(S_i) \Leftrightarrow z\in \bigcap_i Z(S_i) \; .$$ And this means that the sets $Z(S)$ and $\bigcap_i Z(S_i)$ are equal.


1

Yes, except that for functions on a compact manifold with boundary, harmonic is no longer equivalent to being annihilated by $d$. Essentially, the boundary interferes with integrating by parts, as I'll explain. On functions $f$ with domain (a subset of) $\mathbb{R}^2$, the classical Laplacian $\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial ...


0

I hope the following is correct. Sorry, but I'll be using slightly different notations than what you have above. I think I'm using the same notation of Shafarevich. I think the main confusing point is that in his Theorem 4.8, I believe it should really say "the composite rational map $\psi = \dots: X_m \to \mathbb P^n$ can be extended to a regular map". Or ...



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