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$\newcommand{\sheaf}[1]{{\mathcal #1}}$ $\newcommand{\Ohol}{\sheaf{O}}$ If $\sheaf{K}$ is the constant sheaf of rational functions on $X$, one can assume, that $\sheaf{L} \subseteq \sheaf{K}$ for every line bundle on $X$. The embedding goes as follows: Take an open affine $U \subseteq X$ with $\sheaf{L}|_U \cong \Ohol_U$. Then for every $V$ open in $X$ and ...


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Let $P_1,\ldots,P_r$ be the minimal primes of $A$. Next let $U=X - Z$ with $I \subseteq A$ an ideal of $A$ and $Z=V(I)$. Now as $U$ is dense in $X$ we have $I \subsetneq P_i$ for all $i$. Therefore $I \subsetneq P_1 \cup \cdots \cup P_r$ by the prime avoidance lemma. Now $P_1 \cup \cdots \cup P_r$ is the set of zero-divisors of $A$. (More generally the set ...


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There is no hope of making a justifiably good estimation! Lacking any details about the path inbetween, an adequate guess would be to approximate the curve with a cubic Bezier curve, where $P_0$ is the starting point, $P_3$ the end point, and $\vec{P_0P_1}$, $\vec{P_2P_3}$ are proportional to the measured velocities. This still leaves one degree of freedom, ...


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It seems to me that you'd need more information. I drew this in GeoGebra based on what you wrote: The particle starts at point $A$, goes through some unknown path, and ends at point $B$. The larger angle between the $y$-axes and the segment $AB$ is $245^{\circ}$. I had initially thought that we could get an approximate answer by assuming that the ...


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It is not claimed that the category of $A$-schemes has a zero object. But the category of group $A$-schemes has a zero object. This has nothing to do with schemes. If $C$ is any category with finite products, then $\mathsf{Grp}(C)$ has a zero object, the "trivial group" $T$. The underlying object is $1$, the final object, and the multiplication is the unique ...


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I would say that your proof is not particularly common, certainly I've never seen it before (admittedly this doesn't mean much). The proof in Hartshorne is brief because it is one of those scenarios where doing the "obvious" thing works. Here is the proof that I think Hartshorne had in mind (which is entirely constructive): Let $T$ be the tensor pre-sheaf ...


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Does the tautological line bundle have a nowhere-zero section $\sigma$? (Consider $\sigma([x]) = f(x)x$ for some continuous function $f\colon S^n\to\Bbb R$.) The same argument will work in the case of $\Bbb CP^n$ if you think a bit.


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For two different choices $(x_i)$ and $(y_i)$ of your $n$ points the varieties $X=\mathbb A^1 \setminus \{x_1, \ldots, x_n\}$ and $Y=\mathbb A^1 \setminus \{x_1, \ldots, x_n\}$ will not be isomorphic as soon as $n\geq 3$. Here is why: Any isomorphism $f:X\to Y$ is a birational map and thus extends to a biregular automorphism $F:\mathbb P^1\to \mathbb ...


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The top of page 2 in the linked paper on the arXiv seems to suggest this is the definition: Let $\mathbb{G}$ be a connected simply connected, semisimple algebraic group defined over $\mathbb{Q}$ with a fixed embedding $\mathbb{G} \hookrightarrow GL_n$. Then $$\mathbb{G}(\mathbb{Z}) = \mathbb{G}(\mathbb{Q}) \cap GL_n(\mathbb{Z}).$$ This allows one to ...


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The class group of $U$ is zero, dear Fellow Mathematician. Indeed if you remove the line $z=0$ from $\mathbb P^2$, you are left with $\mathbb A^2$. And if you then remove the points where $x=0$ or $y=0$ you are left with the product $U=(\mathbb A^1\setminus\{0\})\times (A^1\setminus\{0\})$. This is an affine variety with coordinate ring $$A=\mathcal ...


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Hope the graph and given spiral equation would clarify. Else, meet in chat. $$ r = a e^{ \cot m . \theta }$$


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Theorem : A topological space $X$ has a simply connected covering space if and only if $X$ is path connected, locally path connected and semi-locally simply connected. We are interested in the "only if" part of the story. Let $(E, e_0) \stackrel{p}{\to} (X, x_0)$ be a based covering map. Pick a neighborhood $U$ of $x_0$ that is evenly covered by $p$. $U_i$ ...


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Mathematica doesn't just applies the Buchberger algorithm and returns a Groebner basis. It returns a reduced Groebner basis. So, if your polynomial set contains any non-zero constant, the result is {1}. For example GroebnerBasis[{ 1/y, z}, {x}] returns {1}. However, if your set consists of constants that can be eventually zero, Mathematica returns a reduced ...


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After getting the point of intersection of the given family of lines we need to simply calculate the distance between the given point $(2,3)$ and the point of intersection of the family of lines i.e., $(1,2)$ which is equal to $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ so $$d= \sqrt{(2-1)^2+(3-2)^2}=\sqrt2$$ Now by the definition of locus of a point (locus of a ...


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Using the criterion you mention in the comments, if you want to know that the Godeaux surface $S$ is of general type, it's enough to know it's irrational. But that is true for a very easy topological reason. Every smooth rational projective variety is simply connected, but $S$ is covered by the Fermat quintic $F$ with deck group $G=\mathbf Z/5 \mathbf Z$, so ...


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The vanishing ideal of $V$ is the radical of $\langle x-y\rangle$ in $k[x,y]$. Since $k[x,y]$ is a factorial ring or even for reasons of polynomial total degree, you should have no problem showing that this ideal is already radical. We are left to show that it is prime. In other words, we have to show that $f\cdot g = h\cdot (x-y)$ implies $f\in\langle ...


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Corollary I.6.10 states that every nonsingular quasi-projective curve is isomorphic to an open subset of a nonsingular projective curve. Thus we have $X - Z \subset W$ where $W$ is a projective nonsingular curve. Now the morphism $X - Z \rightarrow \mathbb{P}^1$ extends uniquely to $W$, and we can apply Theorem 6.10.


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I can try here but I might be wrong. I just write down some ideas.....maybe someone can come up with a better one. I think it is easier to look at the following commutative diagram... $$\require{AMScd} \begin{CD} \text{Spec}(k)\times \mathbb{P}^{1} @>{j}>> \mathbb{P}^{1}\times\mathbb{P}^{1}@>{p}>> \mathbb{P}^{1}\\ @V{f}VV @V{g}VV ...


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Here are some things which make sense on complex manifolds, but not almost complex manifolds: Complex coordinates $(z_1, \ldots, z_n)$. In particular, vectors like $\frac{\partial}{\partial z_k}$ only make sense on a complex manifold. On complex manifolds, we have $\overline{\partial}^2 = 0$. This gives rise to the Dolbeault complex $$\Omega^{p,0}(M) ...


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The Godeaux surface is of general type, hence has kodaira dimension $2$. A Ruled surface has kodaira dimension $-\infty$. Sorry for linking Wikipedia - can also be found in Hartshorne. (http://en.wikipedia.org/wiki/Kodaira_dimension#Dimension_2)


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You can calculate E.F by taking curves representing the classes [E] and [F] which intersect transversely and counting their points of intersection. In particular, pick points $e \in E$ and $f \in F$ and take $E \times \{f\}$ and $\{e\} \times F$. Clearly these only meet at the single point $(e, f)$, and moreover it's easy to calculate the tangent spaces ...


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Regarding your first question, there are certainly theorems ruling out the existence of such $f$ assuming $f$ is continuous or smooth or algebraic or whatever. (I'm thinking of proofs using (co)homological methods.) But I would expect that these proofs ultimately make use of linear mappings as a base case somewhere (especially when $f$ is assumed ...


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If you believe in computers, you can do this in Macaulay2 with the following commands. i1 : R = QQ[x,y,z,w] o1 = R o1 : PolynomialRing i2 : I = ideal(x^2-z*w, z^2-y*w, y^3-x*w, w^3-x*y^2*z) 2 2 3 2 3 o2 = ideal (x - z*w, z - y*w, y - x*w, - x*y z + w ) o2 : Ideal of R i3 : isPrime I o3 = true ...


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Sure. You write a matrix $M$ for the transformation (which arguably uses the notion of basis**, but...) and do row reduction to produce a matrix $M'$. Row-reduction doesn't change the kernel. But in the row-reduced matrix, pick column that does not contain the leading "1" in a row, say column $i$. (There has to be such a column because $m < n$). Then ...


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Contacting the author J.P. Demailly himself about this question has lead to the following conclusion: In the algebraic setting, $X$ is quasi-projective, so by the Noetherian property, we have Zariski-compactness, so we can choose $\alpha$ globally. In the analytic setting, (the maximal choice of) $\alpha$ can get arbitrarily small when approaching ...


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We need the hypothesis that $f$ is (either zero or) nonconstant. Assume WLOG that $f(x, y)$ is nonconstant in $x$. Write $f(x, y) = x^n g(y) + \text{lower terms}$ where $x^n$ is the highest order term in $x$ and $g(y)$ is a nonzero polynomial in $y$. For all but finitely many values $y_0$ of $y$, $g(y_0)$ is nonzero, so for all but finitely many values $y_0$ ...


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Hint : for all $y \in k$, consider the polynomial $P_y(X) = f(X,y) \in k[X]$.


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$O(X)$ is a finitely generated algebra over $k$, so we can write $O(X) = k[\xi_1,\dots,\xi_n]$. Since $k$ is algebraically closed, every maximal ideal $m$ will be of the form $m=(\xi_1 - a_1,\dots,\xi_n - a_n)$ with $a_i \in k$. From this it is seen that $O(X)/m = k[\xi_1,\dots,\xi_n] / (\xi_1 - a_1,\dots,\xi_n - a_n) \cong k$.


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There is no such field. As $k[x,y]/(y) \cong k[x]$ is an integral domain for any field $k$, the ideal $(y) \subset k[x,y]$ is a prime ideal. Thus, the variety $V(y)$ is irreducible.


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If $(u_1, u_2)$ is an orthogonal basis than the kernel of the projector to the subspace generated by $u_1$ is the span of $u_2$ i.e. all the vectors of the form $a u_1$ with $a \in\mathbb{R}$ ( if the vector space is over $\mathbb{R}$), not only $u_2$. The same is true for the projector to the subspace generated by $u_2$ and, yes, the whole space is the ...


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It is. This is the Zariski Lemma.


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Up to order of coordinates, and perhaps some signs (depending on the convention you follow), this looks like Plücker coordinates. I think about this as the line joining the point $\mathbf a$ with the point at infinity in direction $\mathbf b$. In homogeneous coordinates you could write both as the columns of a matrix: $$\begin{pmatrix} a_1 & b_1 \\ a_2 ...


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An alternative description is that (x,y,z) belongs to the tangent surface if and only if the polynomial (in the variable $s$) $1+3sx+3s^2y+s^3z$ has a double root. The corresponding M2 commands to get the equation of the tangent surface are R=QQ[s,x,y,z] factor (discriminant(1+3*s*x+3*s^2*y+s^3*z,t)) The reason is that the twisted cubic corresponds to ...


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The morphism you describe is known as the "Segre embedding" or "Segre map". By doing a google search you will find a lot of relevant information, and even proofs of the relation you are interested in. Note that the image of a point under this map is a tuple (coordinate vector); the fact that it is organized in a matrix is not of any significance. The ...


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$\newcommand{\CC}{{\mathbb C}}$ $\newcommand{\im}{\text{im}}$ First $\CC[x,y]/(\ker f) \subseteq \CC[t]$. So as $\CC[t]$ is a domain $Q=\ker f$ must be a prime of $\CC[x,y]$. So we have a chain of strict inclusions $(0) \subsetneq P \subsetneq Q$. As $\dim \CC[x,y] = 2$ the ideal $Q$ must be maximal in $\CC[x,y]$. Therefore $\CC[x,y]/Q = \CC$ and $\im f = ...


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It's true that an algebraic map of Zariski-closed subsets of $K^n$ (this is not what "affine variety over $K$" means unless $K$ is algebraically closed) is an isomorphism iff the induced map on rings of functions is an isomorphism. The point is that an inverse map on rings of functions provides the components of an algebraic map which inverts the original ...


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The Plücker embedding is an isometry of $G(k,\Bbb C^n)$ to its image in $\Bbb P(\Lambda^k\Bbb C^n)$ with the standard Fubini-Study metric. In the moving frames notation, for example, the Kähler form on $\Bbb P^N$ is given by $$\frac i2\sum_{j=1}^N \omega_{0\bar j}\wedge\overline\omega_{0\bar j},$$ where $\{f_0;f_1,\dots,f_N\}$ is a unitary frame at the ...


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It is true and the proof is fine. Here is a very similar one, based on the observation that, since $(f^{-1}\mathcal{O}_Y)_x = \mathcal{O}_{Y,f(x)}$ for all $x\in X$, $ F $ is flat over $Y$ if and only if it is flat as $f^{-1}\mathcal{O}_Y$-module. We don't work with the sheaf of rings $f^{-1}\mathcal{O}_Y$ very often, because it's not an ...


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As @MooS says, your hypotheses imply surjective. Then since $f$ is birational and $Y$ is normal, we have $f_* O_X = O_Y$ by ZMT. Then $H^0(X,f^*D) = H^0(Y,f_*f^*D) = H^0(Y,D)$ by the projection formula. In particular, one of these spaces of sections is nonzero iff the other is.


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First, assume that a polynomial $p(t) \in \Bbb C[t]$ satisfies $p(0) = p(1) = 0$. Then there is a polynomial $q(t)\in \Bbb C[t]$ such that $p(t) = t(t-1)q(t)$. Expanding $q(t) = c_1 + c_2t^1 + \cdots c_nt^{n-1}$, we get that $$ p(t) = \sum_{i = 1}^n c_n t^n(t-1) $$ In general, the term $t^i(t-1)^j$ is the image of a monomial in $x$ and $y$ iff $j \leq i \leq ...


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Given a holomorphic line bundle $L$ with trivialising open cover $\mathcal{U} = \{U_{\alpha} \mid \alpha \in A\}$ and trivialisations $\{\phi_{\alpha} \mid \alpha \in A\}$, Griffiths & Harris construct an element $t_L \in H^1(\mathcal{U}, \mathcal{O}^*)$. They then show that $t_L$ does not depend on the choice of trivialisations. In particular, if $L$ ...


2

Let $\mathcal{O}_{T,t}$ be the local ring at $t\in T$. Then we want to show that $\mathcal{F_y}/\mathcal{O}_{C\times T,y}$ is flat over this ring, where $y\in C\times t$. It is well known that to check flatness of a coherent module $M$ over a Noetherian local ring $R$, it suffices to show that $Tor_1(M,k(R))=0$, where $k(R)$ is the residue field. So we ...


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Be aware that the sheaf axioms have to hold for any open covering of any open subset of $X$. I have not checked but, for me, that strongly suggests that any open subset of $X$ must be quasi-compact - which is equivalent to the space being Noetherian. (Comment posted as an answer on @sdf's suggestion.)


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Notice that $f(T+1) = f(T+x) = f(T+x+1) = f(T)$. This gives you explicitly the Galois group and proves that the extension is Galois.


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The form $\omega$ is clearly holomorphic except maybe at the three points at a finite distance $P_i=(x_i,0)$ where $y=0, x_i=0,1,-1$ or at the point $Q_\infty=(0:0:1)$ . 1) At a finite distance we write $y^2=x^3-x$ so that $2ydy=(3x^2-1)dx$ and thus $\omega= \frac {dx}{y}=\frac{2dy}{3x^2-1}$ which has neither zero nor pole at $P_i$. 2) At infinity ...


1

I agree with the OP and Bruno Joyal that the statement of this exercise is faulty. As you say, condition (ii) had better hold for any rational point $O$ given that we've defined the group law in such a way to make $O$ the origin. Unfortunately I could not remember what I had in mind when I wrote this, so I uploaded a new copy in which condition (ii) is ...


2

By orbit-stabilizer, the dimension of the orbit should be the dimension of the group minus the dimension of the stabilizer, so we should compute the dimension of the stabilizer. Suppose $$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & ...


2

The four points at infinity of $C$ are $(\xi_0:\xi_1:\xi_2)=P_i=(0:a_i:1)$ where the $a_i$'s are the four complex roots of $z^4+1=0$. In the affine coordinates $(v_0,v_1)$ we have $$v_1^4-v_0^4+1=0,\quad v_1^3dv_1-v_0^3dv_0=0\quad (\bigstar)$$ and our points at infinity have coordinates $P_i=(v_0,v_1)=(0,a_i)$. At $P_i$ the implicit function theorem ...


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You want Cardano's formula, which gives you an exact solution.


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The meromorphic form $\omega$ has no poles and its only zeros are at the four points $P _i$ whose homogeneous coordinates $(\xi_0:\xi_1:\xi_2)$ are $(0:a_i:1)$, where the $a_i$ 's are the complex solutions of the equation $z^4=-1$. In the coordinates $v_0,v_1$ the differential form $\omega$ becomes $\omega=v_0^2dv_1-v_0v_1dv_0$ and at the points $P_i$ ...



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