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New answers tagged algebraic-geometry

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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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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 $j \leq i \leq ... 4 Given a holomorphic line bundle$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)$is the residue field. So we ... 0 Be aware that the sheaf axioms have to hold for any open covering of any open subset of$X$. I have not checked but, for me, that strongly suggests that any open subset of$X$must be quasi-compact - which is equivalent to the space being Noetherian. (Comment posted as an answer on @sdf's suggestion.) 2 Notice that$f(T+1) = f(T+x) = f(T+x+1) = f(T)$. This gives you explicitly the Galois group and proves that the extension is Galois. 2 The form$\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 ... 1 You want Cardano's formula, which gives you an exact solution. 1 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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