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| visits | member for | 2 years, 5 months |
| seen | 27 mins ago | |
| stats | profile views | 4,739 |
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May 23 |
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uniformization theorem - squares and circles The closed square is not a Riemann surface – it isn't even a manifold without boundary! |
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May 23 |
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Grothendieck topology on pre/sheaves That defines, at best, a pretopology. However the sieve generated by any such family is a covering sieve in the canonical topology. |
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May 23 |
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What are some examples of subtle logical pitfalls? Actually, the construction of JDH indicated here shows that Fact I is enough to construct, within any given model of ZFC, a transitive set that externally is a model of ZFC! |
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May 22 |
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a group is not the union of two proper subgroups - how to internalize this into other categories? Actually, there's an easy counterexample in $\mathbf{Grp}$ as well, since internal groups there are just abelian groups. |
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May 22 |
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When is the pullback functor on sheaves faithful? It's true at the level of toposes and at the level of sheaves of abelian groups as soon as the map is surjective, but I think it is also true at level of quasicoherent sheaves if the morphism is faithfully flat. So what kind of sheaves are you asking about? |
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May 21 |
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A question regarding etale morphisms of affine varieties Well, the closed subvarieties are either the whole variety or finite sets of points, so those won't do either. And I don't think you really want to consider subvarieties that are neither open nor closed... |
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May 21 |
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A question regarding etale morphisms of affine varieties I suppose you mean to either have $p$ surjective, or "locally isomorphic" instead of "isomorphic". |
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May 20 |
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Does every category have a functor? Your answer is false if taken literally: $1$ does not embed into the empty category, and the category $1$ only has the identity endofunctor. |
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May 20 |
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Does every category have a functor? Of course, if $D$ is empty... |
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May 20 |
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Question about comultiplication This is formally dual to the following problem: Suppose $A$ is an algebra over the field $k$. Show that the multiplication map $\mu : A \otimes A \to A$ is an algebra map if and only if $A$ is commutative. |
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May 19 |
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ZF Extensionality axiom The converse is a logical tautology, however. |
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May 19 |
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Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme $f^*$ is an isomorphism, but whether or not it is the identity depends on the explicit construction of the structure sheaf etc. |
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May 19 |
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Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme No, you start with $f$ and then calculate $f^*$. |
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May 19 |
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Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme You can do that if you like, but that is not necessary. |
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May 19 |
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Does $\,x>0\,$ hint that $\,x\in\mathbb R\,$? It looks like 1.5cm to me... |
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May 18 |
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Degree of effective Cartier divisor Perhaps the degree is supposed to be a function, rather than a number? After all, the LHS of the equation appearing in Theorem 8.3 is also a function... |
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May 18 |
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Uniformly solvable families of polynomials It's basically a translation of your problem into algebraic geometry. Rather than thinking of $f(a_1, \ldots, a_n; x)$ as a family of one-variable polynomials, think of it as just one; then it defines a hypersurface $V$ in $U \times \mathbb{A}^1$ where $U$ is an open subset of $\mathbb{A}^n$, and what you want to do is to study the properties of the projection $V \to U$ in terms of the generic fibre. |
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May 18 |
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Topology of the Segre product vs. the product topology I think you've missed the point, though: it's not important that the product of varieties very rarely has the product topology; rather, it's more important to know that it doesn't happen in all cases! |
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May 18 |
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Uniformly solvable families of polynomials Very interesting! Perhaps it would be better to consider the question in the setting of commutative algebra: assume $f$ is monic, irreducible, and separable; let $A$ be the subring of the coefficient field generated by $\mathbb{Q}[a_0, \ldots, a_k]$ and the coefficients of $f$, and consider the $A$-algebra $A [x] / (f)$. Then the automorphism group of $A [x] / (f)$ (as an $A$-algebra) will also act on each of the specialisations $A [x] / (f) \otimes_A A / \mathfrak{m}$, for each maximal ideal $\mathfrak{m}$ of $A$, and the "generic" situation is obtained by considering $\mathfrak{p} = (0)$. |
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May 17 |
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Graph of a morphism of affine schemes In what sense is the image of a scheme morphism a sheaf? |
